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https://www.asknumbers.com/lbs-to-oz/3.5-lbs-to-oz.aspx	[ "# How Many Ounces in 3.5 Pounds?\n\n3.5 Pounds to ounces (lbs to oz) converter. How many ounces in 3.5 pounds?\n\n3.5 Lbs equal to 56.0 oz or there are 56.0 ounces in 3.5 pounds.\n\n←→\nstep\nRound:\nEnter Pound\nEnter Ounce\n\n## How to convert 3.5 pounds to ounces?\n\nThe conversion factor from pound to ounce is 16. To convert any value of pounds to ounces, multiply the pound value by the conversion factor.\n\nTo convert 3.5 lbs to oz, multiply 3.5 by 16, that makes 3.5 lbs equal to 56.0 oz.\n\n3.5 lbs to oz formula\n\noz = lbs value * 16\n\noz = 3.5 * 16\n\noz = 56.0\n\nCommon conversions from 3.5x lbs to oz:\n(rounded to 3 decimals)\n\n• 3.5 lbs = 56.0 oz\n• 3.51 lbs = 56.16 oz\n• 3.52 lbs = 56.32 oz\n• 3.53 lbs = 56.48 oz\n• 3.54 lbs = 56.64 oz\n• 3.55 lbs = 56.8 oz\n• 3.56 lbs = 56.96 oz\n• 3.57 lbs = 57.12 oz\n• 3.58 lbs = 57.28 oz\n• 3.59 lbs = 57.44 oz\n• 3.6 lbs = 57.6 oz\n\nWhat is a Pound?\n\nPound (lb) is an Imperial system mass unit. 1 Pound = 16 Ounces. 1 Troy pound = 12 Troy ounces. The symbol is \"lb\".\n\nWhat is a Ounce?\n\nOunce is an Imperial system mass unit. 1 oz = 0.0625 lb. 1 troy oz = 0.08334 troy lb. The symbol is \"oz\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7594062,"math_prob":0.9994017,"size":884,"snap":"2023-40-2023-50","text_gpt3_token_len":359,"char_repetition_ratio":0.20454545,"word_repetition_ratio":0.030456852,"special_character_ratio":0.4638009,"punctuation_ratio":0.2007874,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966448,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T06:59:37Z\",\"WARC-Record-ID\":\"<urn:uuid:b00eecb1-ccec-4f6f-a6e1-5ec14b392616>\",\"Content-Length\":\"48635\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a61d0446-4764-44b3-a11d-04bb98206a78>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d168dc4-1866-4871-a3e6-6993f10a09d6>\",\"WARC-IP-Address\":\"172.67.189.12\",\"WARC-Target-URI\":\"https://www.asknumbers.com/lbs-to-oz/3.5-lbs-to-oz.aspx\",\"WARC-Payload-Digest\":\"sha1:TWMA5CTH4FUXM52VE4BOCGGHW5UAAFUQ\",\"WARC-Block-Digest\":\"sha1:R7HL5UPMH6ETN2VAGT55EPIDD73ZYG2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103558.93_warc_CC-MAIN-20231211045204-20231211075204-00724.warc.gz\"}"}
https://onlinecalculator.guru/algebra/boolean-algebra-calculator/	[ "# Boolean Algebra Calculator\n\nWith the help of our handy Boolean Algebra Calculator tool, you can easily solve any difficult boolean algebraic expression in seconds. Provide your boolean expression as the input and press the calculate button to get the result as early as possible.\n\nBoolean Algebra Calculator: Evaluating the boolean algebraic expressions is not like solving any other mathematical expressions. It is possible by taking the help of various boolean laws and proper knowledge on them. Without all these, you can simply solve your equation by using our free online boolean algebra calculator tool. From this article, you can find the detailed procedure for computing your questions.\n\n## How to Solve Boolean Algebra Expression?\n\nIn the following sections you can get the step by step process to solve a boolean expression. Go through the below segments and follow them. Two simple steps to solve the boolean expression is by doing the truth table for each operation and finding the result. Another easy step is right here.\n\n• Take any boolean expression\n• Know all the Laws of Boolean Algebra\n• Replace the Boolean Algebra Laws at each possible step with proper knowledge\n• Keep on doing the step 3 till you reach a point where you can’t substitute any law\n\nLaws of Boolean Algebra\n\nHere, we are providing the basic laws of boolean algebra that assist you when solving the boolean algebra expression.\n\n• Idempotent Law\n• A * A = A\n\nA + A = A\n\n• Associative Law\n• (A * B) * C = A * (B * C)\n\n(A + B) + C = A + (B + C)\n\n• Commutative Law\n• A * B = B * A\n\nA + B = B + A\n\n• Distributive Law\n• A * (B + C) = A * B + A * C\n\nA + (B * C) = (A + B) * (A + C)\n\n• Identity Law\n• A * 0 = 0 and A * 1 = A\n\nA + 1 = 1 and A + 0 = A\n\n• Complement Law\n• A * ~A = 0\n\nA + ~A = 1\n\n• Involution Law\n• ~(~A) = A\n\n• DeMorgan's Law\n• ~(A * B) = ~A + ~B\n\n~(A + B) = ~A * ~B\n\n• Redundancy Laws\n• Absorption\n\nA + (A * B) = A\n\nA * (A + B) = A\n\n(A * B) + (A * ~B) = A\n\n(A + B) * (A + ~B) = A\n\nA + (~A * B) = A + B\n\nA * (~A + B) = A * B\n\nExample:\n\nQuestion: Solve ~(A * B) * (~A + B) * (~B + B)?\n\nSolution:\n\nGiven expression is ~(A * B) * (~A + B) * (~B + B)\n\nBy applying Complement law i.e ~B + B=1\n\n=~(A * B) * (~A + B) * 1\n\nApply Identity law i.e (~A + B) * 1=~A + B\n\n=~(A * B) * (~A + B)\n\nApply DeMorgan's law i.e ~(A * B)=(~A + ~B)\n\n=(~A + ~B) * (~A + B)\n\nDistributive Law is ~A + B=B\n\n=(~A + ~B) * B\n\nComplement law says ~B * B=0\n\n=~A + 0\n\nApply Identity law\n\n=~A\n\n~(A * B) * (~A + B) * (~B + B)=~A\n\nOnlinecalculator.guru contains a comprehensive array of calculators designed for the people with any level of mathematical knowledge to solve various questions effortlessly.", null, "### Frequently Asked Questions on Boolean Algebra\n\n1. What is meant by Boolean Algebra?\n\nBoolean algebra is a branch of mathematics that deals with the operations on logical values. It returns only two values i.e true or false or represented by 0 and 1.\n\n2. What are the operations used in the boolean algebra?\n\nThe various basic operations used in the boolean algebra are Conjunction (AND), Disjunction(OR), and Negotiation (NOT).\n\n3. How do you calculate the Boolean Algebra Expression using a calculator?\n\nEnter a valid boolean expression and hit on the calculate button to get your answer quickly.\n\n4. What are the 7 logic gates?\n\nThere are seven basic logic gates. They are AND, OR, XOR, NOT, NAND, and XNOR.\n\n5. What is the other name of Boolean Algebra?\n\nBoolean Algebra is used to simplify and analyze the digital (logic) circuits. It has only the binary numbers i.e. 0 (False) and 1(True). It is also called Binary Algebra or logical Algebra." ]	[ null, "https://onlinecalculator.guru/static/images/Algebra/Boolean-Algebra-Calculator.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8794097,"math_prob":0.9966886,"size":3027,"snap":"2021-43-2021-49","text_gpt3_token_len":876,"char_repetition_ratio":0.166391,"word_repetition_ratio":0.054858934,"special_character_ratio":0.32077965,"punctuation_ratio":0.08985025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99976057,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T04:53:43Z\",\"WARC-Record-ID\":\"<urn:uuid:88c78db8-4907-46f5-a841-9df6f9acd6a8>\",\"Content-Length\":\"25010\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2bb3e164-1919-4e58-87cd-3f127b6a4912>\",\"WARC-Concurrent-To\":\"<urn:uuid:61572c80-32e3-47a9-9e60-92357d74a578>\",\"WARC-IP-Address\":\"104.26.4.233\",\"WARC-Target-URI\":\"https://onlinecalculator.guru/algebra/boolean-algebra-calculator/\",\"WARC-Payload-Digest\":\"sha1:AZWQDPPIWHIACEHI66HKR6CSU3LS4KY4\",\"WARC-Block-Digest\":\"sha1:CNGW3OKJSCHMGJLYTK2K3ZXCTTALWWLF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585242.44_warc_CC-MAIN-20211019043325-20211019073325-00439.warc.gz\"}"}
https://codeforces.com/problemset/problem/1025/F	[ "Please subscribe to the official Codeforces channel in Telegram via the link https://t.me/codeforces_official. ×\n\nF. Disjoint Triangles\ntime limit per test\n2 seconds\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nA point belongs to a triangle if it lies inside the triangle or on one of its sides. Two triangles are disjoint if there is no point on the plane that belongs to both triangles.\n\nYou are given $n$ points on the plane. No two points coincide and no three points are collinear.\n\nFind the number of different ways to choose two disjoint triangles with vertices in the given points. Two ways which differ only in order of triangles or in order of vertices inside triangles are considered equal.\n\nInput\n\nThe first line of the input contains an integer $n$ ($6 \\le n \\le 2000$) – the number of points.\n\nEach of the next $n$ lines contains two integers $x_i$ and $y_i$ ($\|x_i\|, \|y_i\| \\le 10^9$) – the coordinates of a point.\n\nNo two points coincide and no three points are collinear.\n\nOutput\n\nPrint one integer – the number of ways to choose two disjoint triangles.\n\nExamples\nInput\n61 12 24 64 57 25 3\nOutput\n6\nInput\n70 -1000000000-5 -55 -5-5 05 0-2 22 2\nOutput\n21\nNote\n\nIn the first example there are six pairs of disjoint triangles, they are shown on the picture below.", null, "All other pairs of triangles are not disjoint, for example the following pair:", null, "" ]	[ null, "https://espresso.codeforces.com/d62ab77bce49d2bdb17494a6a01186d60c1a37a0.png", null, "https://espresso.codeforces.com/47cf59062aed5d31d5c681b14885838ceefada12.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8577868,"math_prob":0.9967522,"size":1127,"snap":"2021-31-2021-39","text_gpt3_token_len":304,"char_repetition_ratio":0.16473731,"word_repetition_ratio":0.06603774,"special_character_ratio":0.30700976,"punctuation_ratio":0.06550218,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99777746,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T17:01:46Z\",\"WARC-Record-ID\":\"<urn:uuid:cd336d31-888d-4d1e-b833-d5c1dbaa41fe>\",\"Content-Length\":\"55514\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0567f7e9-cd75-4a37-9855-8368e963baec>\",\"WARC-Concurrent-To\":\"<urn:uuid:0ebfe189-538e-49b6-a567-eac521ed0968>\",\"WARC-IP-Address\":\"213.248.110.126\",\"WARC-Target-URI\":\"https://codeforces.com/problemset/problem/1025/F\",\"WARC-Payload-Digest\":\"sha1:64V5ZT7JZ3P55IGSBKRQGEEW735COGJ4\",\"WARC-Block-Digest\":\"sha1:4AIZVV5WRLD3GNBMDLJEARYHV4QCGN2A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057083.64_warc_CC-MAIN-20210920161518-20210920191518-00638.warc.gz\"}"}
https://lists.boost.org/Archives/boost/2002/10/38309.php	[ "", null, "# Boost :\n\nFrom: Daryle Walker (dwalker07_at_[hidden])\nDate: 2002-10-24 03:26:45\n\nFor now, let's stick with functions that map one real number to another\n(so no complex numbers or vector mapping(s)). A basic function design\ncould be:\n\ntemplate < typename ResultType, typename DomainType, typename RangeType\n>\nResultType\nintegrate\n(\nstd::map<DomainType, RangeType> m,\nResultType initial_condition = ResultType()\n);\n\nThis is pure numerical integration without assuming any distribution\npattern for the domain or range values (except that all the domain\nvalues are unique).\n\nThat function can't work unless all the points are known in advance.\nIf we build the graph as we go, we need something like:\n\ntemplate < typename ResultType, typename DomainType, typename RangeType\n>\nclass integrator\n{\npublic:\nexplicit integrator( ResultType initial_condition = ResultType() );\n\nvoid update( DomainType x, RangeType y );\n\nResultType current_result() const;\n};\n\nWe should insist that updates keep new \"x\" values strictly increasing\n(or risk undefined behavior), so we don't have to keep an arbitrary\nhistory.\n\nNow for the variants:\n\nA. The domain values can be given be an STL iterator sequence; an\ninitial value, step size, and final value or step count; a generator\n(C++) function/functor.\n\nB. The range values can be given in similar ways, or with a (C++)\nfunction/functor that takes processes the current domain value. The\nlatter option could be combined with hints about the (math) function's\ndiscontinuities or undefined regions.\n\nC. The domain and range values can be given in a combined STL iterator\nsequence of tuples.\n\nNote that specific techniques are not mentioned. If there are several\nways to do an integration (rectangular rule, trapezoidal rule, Simpon's\nRule, etc.), we can provide similar (C++) functions or classes with the\nsame signature, or policies could be used. (I don't know too much\nabout the \"policy\" philosophy with C++ programming techniques, could\nsomeone explain it to me privately?)\n\nDaryle" ]	[ null, "https://lists.boost.org/boost/images/boost.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73332894,"math_prob":0.9038375,"size":2107,"snap":"2021-04-2021-17","text_gpt3_token_len":499,"char_repetition_ratio":0.11887779,"word_repetition_ratio":0.05015674,"special_character_ratio":0.22496441,"punctuation_ratio":0.15844156,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96164274,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T06:15:21Z\",\"WARC-Record-ID\":\"<urn:uuid:ed29f4d5-3f1c-4508-bdf0-f35901a01772>\",\"Content-Length\":\"12097\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e4053f4-61b7-451c-91a5-86a1659ff2a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:734100d9-3b67-45c8-b40e-ed3ad43f740c>\",\"WARC-IP-Address\":\"146.20.110.251\",\"WARC-Target-URI\":\"https://lists.boost.org/Archives/boost/2002/10/38309.php\",\"WARC-Payload-Digest\":\"sha1:ZFZXYLVJ55P5DWC6MK7I5CZMUACOVE6Z\",\"WARC-Block-Digest\":\"sha1:3LBXMDEK556YWZGE5RYMHEH4IY2IHS6O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038468066.58_warc_CC-MAIN-20210418043500-20210418073500-00405.warc.gz\"}"}
https://www.boost.org/doc/libs/1_73_0/libs/multiprecision/doc/html/boost_multiprecision/intro.html	[ "#", null, "Boost C++ Libraries\n\n...one of the most highly regarded and expertly designed C++ library projects in the world.\n\n## Introduction\n\nThe Multiprecision Library provides integer, rational, floating-point, and complex types in C++ that have more range and precision than C++'s ordinary fundamental (built-in) types. The big number types in Multiprecision can be used with a wide selection of basic mathematical operations, elementary transcendental functions as well as the functions in Boost.Math. The Multiprecision types can also interoperate with any fundamental (built-in) type in C++ using clearly defined conversion rules. This allows Boost.Multiprecision to be used for all kinds of mathematical calculations involving integer, rational and floating-point types requiring extended range and precision.\n\nMultiprecision consists of a generic interface to the mathematics of large numbers as well as a selection of big number back-ends, with support for integer, rational, floating-point, and complex types. Boost.Multiprecision provides a selection of back-ends provided off-the-rack in including interfaces to GMP, MPFR, MPIR, MPC, TomMath as well as its own collection of Boost-licensed, header-only back-ends for integers, rationals and floats. In addition, user-defined back-ends can be created and used with the interface of Multiprecision, provided the class implementation adheres to the necessary concepts.\n\nDepending upon the number type, precision may be arbitrarily large (limited only by available memory), fixed at compile time (for example, 50 or 100 decimal digits), or a variable controlled at run-time by member functions. The types are expression templates - enabled for better performance than naive user-defined types.\n\nThe Multiprecision library comes in two distinct parts:\n\n• An expression-template-enabled front-end `number` that handles all the operator overloading, expression evaluation optimization, and code reduction.\n• A selection of back-ends that implement the actual arithmetic operations, and need conform only to the reduced interface requirements of the front-end.\n\nSeparation of front-end and back-end allows use of highly refined, but restricted license libraries where possible, but provides Boost license alternatives for users who must have a portable unconstrained license. Which is to say some back-ends rely on 3rd party libraries, but a header-only Boost license version is always available (if somewhat slower).\n\n###### Getting started with Boost.Multiprecision\n\nShould you just wish to 'cut to the chase' just to get bigger integers and/or bigger and more precise reals as simply and portably as possible, close to 'drop-in' replacements for the fundamental (built-in) type analogs, then use a fully Boost-licensed number type, and skip to one of more of :\n\nThe library is very often used via one of the predefined convenience `typedef`s like `boost::multiprecision::int128_t` or `boost::multiprecision::cpp_bin_float_quad`.\n\nFor example, if you want a signed, 128-bit fixed size integer:\n\n```#include <boost/multiprecision/cpp_int.hpp> // Integer types.\n\nboost::multiprecision::int128_t my_128_bit_int;\n```\n\nAlternatively, and more adventurously, if you wanted an arbitrary precision integer type using GMP as the underlying implementation then you could use:\n\n```#include <boost/multiprecision/gmp.hpp> // Defines the wrappers around the GMP library's types\n\nboost::multiprecision::mpz_int myint; // Arbitrary precision integer type.\n```\n\nOr for a simple, portable 128-bit floating-point close to a drop-in for a fundamental (built-in) type like `double`, usually 64-bit\n\n```#include <boost/multiprecision/cpp_bin_float.hpp>\n\n```\n\nAlternatively, you can compose your own 'custom' multiprecision type, by combining `number` with one of the predefined back-end types. For example, suppose you wanted a 300 decimal digit floating-point type based on the MPFR library. In this case, there's no predefined `typedef` with that level of precision, so instead we compose our own:\n\n```#include <boost/multiprecision/mpfr.hpp> // Defines the Backend type that wraps MPFR.\n\nnamespace mp = boost::multiprecision; // Reduce the typing a bit later...\n\ntypedef mp::number<mp::mpfr_float_backend<300> > my_float;\n\nmy_float a, b, c; // These variables have 300 decimal digits precision.\n```\n\nWe can repeat the above example, but with the expression templates disabled (for faster compile times, but slower runtimes) by passing a second template argument to `number`:\n\n```#include <boost/multiprecision/mpfr.hpp> // Defines the Backend type that wraps MPFR.\n\nnamespace mp = boost::multiprecision; // Reduce the typing a bit later...\n\ntypedef mp::number<mp::mpfr_float_backend<300>, et_off> my_float;\n\nmy_float a, b, c; // These variables have 300 decimal digits precision\n```\n\nWe can also mix arithmetic operations between different types, provided there is an unambiguous implicit conversion from one type to the other:\n\n```#include <boost/multiprecision/cpp_int.hpp>\n\nnamespace mp = boost::multiprecision; // Reduce the typing a bit later...\n\nmp::int128_t a(3), b(4);\nmp::int512_t c(50), d;\n\nd = c * a; // OK, result of mixed arithmetic is an int512_t\n```\n\nConversions are also allowed:\n\n```d = a; // OK, widening conversion.\nd = a * b; // OK, can convert from an expression template too.\n```\n\nHowever conversions that are inherently lossy are either declared explicit or else forbidden altogether:\n\n```d = 3.14; // Error implicit conversion from double not allowed.\nd = static_cast<mp::int512_t>(3.14); // OK explicit construction is allowed\n```\n\nMixed arithmetic will fail if the conversion is either ambiguous or explicit:\n\n```number<cpp_int_backend<>, et_off> a(2);\nnumber<cpp_int_backend<>, et_on> b(3);\n\nb = a * b; // Error, implicit conversion could go either way.\nb = a * 3.14; // Error, no operator overload if the conversion would be explicit.\n```\n##### Move Semantics\n\nOn compilers that support rvalue-references, class `number` is move-enabled if the underlying backend is.\n\nIn addition the non-expression template operator overloads (see below) are move aware and have overloads that look something like:\n\n```template <class B>\nnumber<B, et_off> operator + (number<B, et_off>&& a, const number<B, et_off>& b)\n{\nreturn std::move(a += b);\n}\n```\n\nThese operator overloads ensure that many expressions can be evaluated without actually generating any temporaries. However, there are still many simple expressions such as\n\n```a = b * c;\n```\n\nwhich don't noticeably benefit from move support. Therefore, optimal performance comes from having both move-support, and expression templates enabled.\n\nNote that while \"moved-from\" objects are left in a sane state, they have an unspecified value, and the only permitted operations on them are destruction or the assignment of a new value. Any other operation should be considered a programming error and all of our backends will trigger an assertion if any other operation is attempted. This behavior allows for optimal performance on move-construction (i.e. no allocation required, we just take ownership of the existing object's internal state), while maintaining usability in the standard library containers.\n\n##### Expression Templates\n\nClass `number` is expression-template-enabled: that means that rather than having a multiplication operator that looks like this:\n\n```template <class Backend>\nnumber<Backend> operator * (const number<Backend>& a, const number<Backend>& b)\n{\nnumber<Backend> result(a);\nresult = b;\nreturn result;\n}\n```\n\nInstead the operator looks more like this:\n\n```template <class Backend>\nunmentionable-type operator (const number<Backend>& a, const number<Backend>& b);\n```\n\nWhere the 'unmentionable' return type is an implementation detail that, rather than containing the result of the multiplication, contains instructions on how to compute the result. In effect it's just a pair of references to the arguments of the function, plus some compile-time information that stores what the operation is.\n\nThe great advantage of this method is the elimination of temporaries: for example, the \"naive\" implementation of `operator` above, requires one temporary for computing the result, and at least another one to return it. It's true that sometimes this overhead can be reduced by using move-semantics, but it can't be eliminated completely. For example, lets suppose we're evaluating a polynomial via Horner's method, something like this:\n\n```T a = { / some values / };\n//....\ny = (((((a x + a) * x + a) * x + a) * x + a) * x + a) * x + a;\n```\n\nIf type `T` is a `number`, then this expression is evaluated without creating a single temporary value. In contrast, if we were using the mpfr_class C++ wrapper for MPFR - then this expression would result in no less than 11 temporaries (this is true even though mpfr_class does use expression templates to reduce the number of temporaries somewhat). Had we used an even simpler wrapper around MPFR like mpreal things would have been even worse and no less that 24 temporaries are created for this simple expression (note - we actually measure the number of memory allocations performed rather than the number of temporaries directly, note also that the mpf_class wrapper that will be supplied with GMP-5.1 reduces the number of temporaries to pretty much zero). Note that if we compile with expression templates disabled and rvalue-reference support on, then actually still have no wasted memory allocations as even though temporaries are created, their contents are moved rather than copied. \n\nImportant", null, "Expression templates can radically reorder the operations in an expression, for example: a = (b * c) * a; Will get transformed into: a = c; a = b; If this is likely to be an issue for a particular application, then they should be disabled.\n\nThis library also extends expression template support to standard library functions like `abs` or `sin` with `number` arguments. This means that an expression such as:\n\n```y = abs(x);\n```\n\ncan be evaluated without a single temporary being calculated. Even expressions like:\n\n```y = sin(x);\n```\n\nget this treatment, so that variable 'y' is used as \"working storage\" within the implementation of `sin`, thus reducing the number of temporaries used by one. Of course, should you write:\n\n```x = sin(x);\n```\n\nThen we clearly can't use `x` as working storage during the calculation, so then a temporary variable is created in this case.\n\nGiven the comments above, you might be forgiven for thinking that expression-templates are some kind of universal-panacea: sadly though, all tricks like this have their downsides. For one thing, expression template libraries like this one, tend to be slower to compile than their simpler cousins, they're also harder to debug (should you actually want to step through our code!), and rely on compiler optimizations being turned on to give really good performance. Also, since the return type from expressions involving `number`s is an \"unmentionable implementation detail\", you have to be careful to cast the result of an expression to the actual number type when passing an expression to a template function. For example, given:\n\n```template <class T>\nvoid my_proc(const T&);\n```\n\nThen calling:\n\n```my_proc(a+b);\n```\n\nWill very likely result in obscure error messages inside the body of `my_proc` - since we've passed it an expression template type, and not a number type. Instead we probably need:\n\n```my_proc(my_number_type(a+b));\n```\n\nHaving said that, these situations don't occur that often - or indeed not at all for non-template functions. In addition, all the functions in the Boost.Math library will automatically convert expression-template arguments to the underlying number type without you having to do anything, so:\n\n```mpfr_float_100 a(20), delta(0.125);\nboost::math::gamma_p(a, a + delta);\n```\n\nWill work just fine, with the `a + delta` expression template argument getting converted to an `mpfr_float_100` internally by the Boost.Math library.\n\nCaution", null, "In C++11 you should never store an expression template using: ```auto my_expression = a + b - c;``` unless you're absolutely sure that the lifetimes of `a`, `b` and `c` will outlive that of `my_expression`. In fact, it is particularly easy to create dangling references by mixing expression templates with the `auto` keyword, for example: ```auto val = cpp_dec_float_50(\"23.1\") * 100;``` In this situation, the integer literal is stored directly in the expression template - so its use is OK here - but the `cpp_dec_float_50` temporary is stored by reference and then destructed when the statement completes, leaving a dangling reference. If in doubt, do not ever mix expression templates with the `auto` keyword.\n\nAnd finally... the performance improvements from an expression template library like this are often not as dramatic as the reduction in number of temporaries would suggest. For example, if we compare this library with mpfr_class and mpreal, with all three using the underlying MPFR library at 50 decimal digits precision then we see the following typical results for polynomial execution:\n\nTable 1.1. Evaluation of Order 6 Polynomial.\n\nLibrary\n\nRelative Time\n\nRelative number of memory allocations\n\nnumber\n\n1.0 (0.00957s)\n\n1.0 (2996 total)\n\n1.1 (0.0102s)\n\n4.3 (12976 total)\n\n1.6 (0.0151s)\n\n9.3 (27947 total)\n\nAs you can see, the execution time increases a lot more slowly than the number of memory allocations. There are a number of reasons for this:\n\n• The cost of extended-precision multiplication and division is so great, that the times taken for these tend to swamp everything else.\n• The cost of an in-place multiplication (using `operator=`) tends to be more than an out-of-place `operator` (typically `operator =` has to create a temporary workspace to carry out the multiplication, where as `operator` can use the target variable as workspace). Since the expression templates carry out their magic by converting out-of-place operators to in-place ones, we necessarily take this hit. Even so the transformation is more efficient than creating the extra temporary variable, just not by as much as one would hope.\n\nFinally, note that `number` takes a second template argument, which, when set to `et_off` disables all the expression template machinery. The result is much faster to compile, but slower at runtime.\n\nWe'll conclude this section by providing some more performance comparisons between these three libraries, again, all are using MPFR to carry out the underlying arithmetic, and all are operating at the same precision (50 decimal digits):\n\nTable 1.2. Evaluation of Boost.Math's Bessel function test data\n\nLibrary\n\nRelative Time\n\nRelative Number of Memory Allocations\n\nmpfr_float_50\n\n1.0 (5.78s)\n\n1.0 (1611963)\n\nnumber<mpfr_float_backend<50>, et_off>\n(but with rvalue reference support)\n\n1.1 (6.29s)\n\n2.64 (4260868)\n\n1.1 (6.28s)\n\n2.45 (3948316)\n\n1.65 (9.54s)\n\n8.21 (13226029)\n\nTable 1.3. Evaluation of Boost.Math's Non-Central T distribution test data\n\nLibrary\n\nRelative Time\n\nRelative Number of Memory Allocations\n\nnumber\n\n1.0 (263s)\n\n1.0 (127710873)\n\nnumber<mpfr_float_backend<50>, et_off>\n(but with rvalue reference support)\n\n1.0 (260s)\n\n1.2 (156797871)\n\n1.1 (287s)\n\n2.1 (268336640)\n\n1.5 (389s)\n\n3.6 (466960653)\n\nThe above results were generated on Win32 compiling with Visual C++ 2010, all optimizations on (/Ox), with MPFR 3.0 and MPIR 2.3.0.\n\n The actual number generated will depend on the compiler, how well it optimizes the code, and whether it supports rvalue references. 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https://structural-analyser.com/domains/DE/Fourier_DE/	[ "# Fourier serier\n\n## Solving DE by Fourier series\n\nThe Fourier series decomposes periodic or bounded function into simple sinusoids. It is difficult to work with functions as e.g. square waves, sawtooth are and it is easy to work with sines.\n\n### Modal analysis, natural frequencies, vibrations, dynamic behaviour\n\nFourier series are useful to study resonances of a system. Fourier analysis is able to decompose $f(t)$ into pure oscillations\n\n$$f(t) = b_1\\sin \\omega_1 t + b_1\\sin \\omega_2 t + b_1\\sin \\omega_3 t + \\dots \\\\$$\n\nThat is very useful, because then we can analyze which frequencies are main or important within response. We can work with arbitrary periodic function (sine, cosine), solve individually and then recombine to obtain the solution of the original problem (superposition principle).\n\nNatural frequencies characterize the basic behaviour of the system and indicate how the structure will respond to a dynamic loading. There are many reasons to compute the natural frequencies and mode shapes of the structure. For example a rotating machine is to be installed on the floor and it is necessary to determine if the operating frequency of the engine is close to one of the natural frequencies of the building. If the frequencies are too close the consequence might be structural damage or failure.\n\n#### Example\n\nSolve DE describing free undamped motion of spring mass system:\n\n$$my'' + ky = 0\\hskip2em y(0) = 0,\\ y(T)=0$$\n\nNote: the example is not solved using the Fourier series and serves to review spring-mass system and natural frequency.\n\nWe are solving free motion (no external force) of the mass on the spring. The boundary conditions tell us that at $t=0$ and at $t=T$ the mass has to be found in position of equilibrium. If we use natural angular frequency of the system $\\boldsymbol{\\omega_0=\\sqrt{k/m}}$, the same DE can be written as\n\n$$y'' + \\omega^2y = 0.$$\n\nSuch DE is linear with constant coefficients and can be solved by means of auxiliary equation\n\n$$m^2 + \\omega^2 = 0 \\implies m = \\{ -i \\omega ,\\ i\\omega \\}$$\n\nThen the solution is\n\n$$y = C_1e^{-i\\omega t} + C_2e^{i\\omega t} = c_1 \\sin \\omega t + c_2 \\cos \\omega t.$$\n\nLet us use boundary conditions to find $c_1,\\ c_2$:\n\n\\begin{align} y(t=0) = 0: \\hskip2em 0 &= c_1 \\sin (\\omega \\cdot 0) + c_2 \\cos (\\omega \\cdot 0) \\implies c_2 = 0 \\nonumber \\\\ y(t=T) = 0: \\hskip2em 0 &= c_1 \\sin \\omega T \\nonumber \\end{align}\n\nNow if we accept also $c_1 = 0$, we have trivial solution. Such solution is not much useful, because there is no motion described. The mass stays conserved at equilibrium position (does not move). We are looking for nontrivial solution, which comes from the fact that $\\sin n\\pi$ is zero for each $n = 1,\\ 2,\\ 3,\\ \\dots$\n\n$$\\omega t = n \\pi \\implies \\omega = \\frac{n\\pi}{T}$$\n\nIn other words we can expres other (nontrivial) solution (regardless the value of $c_1$):\n\n$$y = \\sin \\left(\\frac{n \\pi }{T}t\\right).$$\n\nThe above solution can be listed as a sequence of solutions:\n\n$$y_1 = \\sin \\frac{\\pi t}{T} ,\\ y_2 = \\sin \\frac{2\\pi t}{T},\\ y_3 = \\sin \\frac{3\\pi t}{T},\\ \\dots$$\n\nAll these solutions are in chord with the complementary function found and at the same time satisfy given boundary conditions. The only issue is that the natural frequency of the system has to fit into given boundary conditions. The half-period of natural frequency has to be $T,\\ 2T,\\ 3T,\\ \\dots$ and then the mass passes the requested position once, twice, three times, ... Otherwise the DE has only the trivial solution.\n\nThe numbers $\\omega_n^2 = {(n^2\\pi^2)}/{T^2}$, for which the problem posseses nontrivial solutions, are known as eigenvalues and the nontrivial solutions $y_1,\\ y_2\\, \\dots,\\ y_n$ are known as eigenfunctions.\n\nNote: $y''+5y = 0,\\ y(0) = 0,\\ y(T)=0$ has no trivial solution. The mass is not going to appear back into requested zero position at the requested time $T$, because its natural frequency is not tuned with the time period $T$.\n\n#### Example\n\nFind the particular solution of the spring/mass system $y''+\\omega_0^2y = f(t)$, if driving force of period $T=2\\ \\text{s}$ is\n\nf(t) = \\left\\{ \\begin{aligned} 1&\\hskip2em 0\\leq t \\lt 1, \\\\ 0&\\hskip2em 1\\leq t \\lt 2. \\\\ \\end{aligned} \\right.\n\nand natural frequency of the system is $\\omega_0 = 10$.\n\nFirstly, let us discuss why we are finding particular solution $y_p$ and not complementary function $y_c$. The complementary function is a transient solution: self motion with no driving force involved. In general, because of damping, $\\lim_{t \\to \\infty} y_c(t) = 0$. We are rather interested into the particular solution (steady periodic solution), which is the response to the driving force $f(t)$.\n\nGiven periodic wave $f(t)$ expressed as a Fourier series is (not solved here)\n\n$$f(t) = \\frac 1 2 + \\frac 2 {\\pi}\\sum_{n=1}^{\\text{odd}}\\frac{\\sin nt\\pi}{n}.$$\n\nSo it is a series of $\\sin nt\\pi$ where each sine has a coefficient $b_n = 2/(\\pi n)$ if $n=1,\\ 3,\\ 5,\\ \\dots$ If we can solve this DE with/for one sine, then the same applies to the series of sines. The nonhomogeneous function $f(t)$ on the right side involves sine. According to the method of undetermined coefficients we have to expect particular solution in the form $y_p = A\\cos nt\\pi + B\\sin nt\\pi$. But because there is no $y'$ within DE, cosine terms have no origin to come from and can be overlooked:\n\n\\begin{align} y_p &= \\color{blue}{\\sum_{n=1}^{\\text{odd}} B_n \\sin nt\\pi + A} \\label{ref:four_ex1} \\\\ y_p'' &= \\color{red}{-\\sum_{n=1}^{\\text{odd}} B_n n^2\\pi^2\\sin nt\\pi} \\nonumber \\\\ \\end{align}\n\nLet us do what we are used to do: substitute $y_p,\\ y''_p$ into the DE $\\color{red}{y''}+ \\color{blue}{\\omega_0^2y} = f(t)$.\n\n$$\\color{red}{-\\sum_{n=1}^{\\text{odd}} B_n n^2\\pi^2\\sin nt\\pi} + \\color{blue}{\\omega_0^2 \\sum_{n=1}^{\\text{odd}} B_n \\sin nt\\pi + \\omega_0^2 A} = \\frac 1 2 + \\frac 2 {\\pi}\\sum_{n=1}^{\\text{odd}}\\frac{\\sin nt\\pi}{n}\\nonumber \\\\$$\n\nBy method of undetermined coefficients, the coefficient $A$ is easy to solve: $A= 1/2\\omega_0^2$. The coefficient $B_n$ can be solved in the same way as there is only $B$ (with no series):\n\n\\begin{align} \\sum_{n=1}^{\\text{odd}}B_n(\\color{blue}{\\omega_0^2} \\color{red}{- n^2\\pi^2}) \\sin nt\\pi &= \\frac 2 {\\pi} \\sum_{n=1}^{\\text{odd}}\\frac{\\sin nt\\pi}{n} \\nonumber \\\\ B_n &= \\frac 2 {\\pi}\\cdot \\frac{1}{n(\\omega_0^2 - n^2\\pi^2)} \\nonumber \\end{align}\n\nThe sine series of the solution is determined and let us list a few terms to investigate dominant frequencies.\n\n\\begin{align} n=1:&\\hskip1em B_1 = \\frac{2}{\\pi}\\frac{1}{1(10^2-1\\pi^2)} = 0.007063 \\nonumber \\\\ n=3:&\\hskip1em B_3 = \\frac{2}{\\pi}\\frac{1}{3(10^2-3^2\\pi^2)} = 0.018992 \\nonumber \\\\ n=5:&\\hskip1em B_5 = \\frac{2}{\\pi}\\frac{1}{5(10^2-5^2\\pi^2)} = -0.000868 \\nonumber \\\\ n=7:&\\hskip1em B_7 = \\frac{2}{\\pi}\\frac{1}{7(10^2-7^2\\pi^2)} = -0.000237 \\nonumber \\\\ n=9:&\\hskip1em B_9 = \\frac{2}{\\pi}\\frac{1}{9(10^2-9^2\\pi^2)} = -0.000101 \\nonumber \\\\ \\end{align}\n\nNow the found coefficients can be substituted back into particular solution $(\\ref{ref:four_ex1})$.\n\n\\begin{align} y_p(t) = \\frac{1}{2\\omega_0^2} & { } + 0.007063 \\sin \\pi t + 0.018992 \\sin 3\\pi t - 0.000868 \\sin 5\\pi t - \\nonumber \\\\ & { } - 0.000237 \\sin 7 \\pi t - 0.000101 \\sin 9 \\pi t - \\dots \\nonumber \\end{align}\n\nThe amplitude of $\\sin 3\\pi t$ (i.e. for $\\omega = 3\\pi$) is the highest. The system is not going to respond equally to all frequencies but favors frequencies close to its natural frequency.\n\n#### Example\n\nSolve the spring/mass system $\\color{red}{y''}\\color{green}{+0.05y'}\\color{blue}{+10.01y} = f(t)$ if $f(t)$ is periodic function\n\nf(t) = \\left\\{ \\begin{aligned} 1&\\hskip2em 0\\leq t \\lt \\pi, \\\\ -1&\\hskip2em \\pi\\leq t \\lt 2\\pi. \\\\ \\end{aligned} \\right.\n\nFrom the given DE can be observed that it is a system with damping, where the mass is 1 kg and spring constant is 10.01 N/m. The natural frequency of the system is ${\\omega_0=\\sqrt{k/m} = \\sqrt{10.01/1} = 3.164\\ \\text{s}^{-1}}$\n\nGiven periodic wave $f(t)$ expressed as a Fourier series is (not solved here)\n\n$$f(t) = \\frac 4 {\\pi}\\sum_{n=1}^{\\text{odd}}\\frac{1}{n}{\\sin nt}.$$\n\nSo now we have DE with sine on the right side and we are looking for a particular solution $y_p$, which is a response to the driving force $f_t(t)$. We are not going to study complementary function $y_c$, which is only a solution of self motion, which is going—due to the damping—to zero anyway and is only a transient solution.\n\nWe have differential equation of the second order with constant coefficients with sine term on the right side. Such DE is easily solved by method of undetermined coefficients: let us collect expected terms $\\sin nt$ and $\\cos nt$ and then we have to solve constants $A,\\ B$:\n\n\\begin{align} \\color{red}{y_p} &\\color{red}{= \\sum_{n=1}^{\\text{odd}}B_n\\sin nt + \\sum_{n=1}^{\\text{odd}}A_n\\cos nt} \\label{ref:four_partic1}\\\\ \\color{green}{y_p'} &\\color{green}{= \\sum_{n=1}^{\\text{odd}}B_n\\cos nt \\cdot n - \\sum_{n=1}^{\\text{odd}}A_n\\sin nt \\cdot n}\\nonumber \\\\ \\color{blue}{y_p''} &\\color{blue}{= -\\sum_{n=1}^{\\text{odd}}B_n\\sin nt\\cdot n^2 - \\sum_{n=1}^{\\text{odd}}A_n\\cos nt \\cdot n^2}\\nonumber \\\\ \\end{align}\n\nLet us substitute $y_p,\\ y_p',\\ y_p''$ into DE and solve coefficients $A,\\ B$ by method of undetermined coefficients as usually:\n\n\\begin{align} \\color{blue}{- \\sum_{n=1}^{\\text{odd}}B_n\\sin nt\\cdot n^2 - \\sum_{n=1}^{\\text{odd}}A_n\\cos nt \\cdot n^2} + &\\nonumber \\\\ + \\color{green}{0.05 \\left(\\sum_{n=1}^{\\text{odd}}B_n\\cos nt \\cdot n - \\sum_{n=1}^{\\text{odd}}A_n\\sin nt \\cdot n\\right)} + &\\nonumber \\\\ + \\color{red}{10.01 \\left(\\sum_{n=1}^{\\text{odd}}B_n\\sin nt + \\sum_{n=1}^{\\text{odd}}A_n\\cos nt \\right)} &= \\frac 4 {\\pi}\\sum_{n=1}^{\\text{odd}}\\frac{1}{n}{\\sin nt} \\nonumber \\\\ \\sum_{n=1}^{\\text{odd}} \\sin nt(-B_nn^2 - 0.05A_nn+10.01B_n) + & \\nonumber \\\\ \\sum_{n=1}^{\\text{odd}} \\cos nt(-A_nn^2 + 0.05B_nn+10.01A_n) &= \\frac 4 {\\pi}\\sum_{n=1}^{\\text{odd}}\\frac{1}{n}{\\sin nt} \\nonumber \\\\ -B_nn^2 - 0.05A_nn + 10.01B_n &= \\frac{4}{\\pi n} \\label{ref:four_ex11} \\\\ -A_nn^2+0.05 B_nn + 10.01A_n &= 0\\label{ref:four_ex12} \\end{align}\n\nFrom $(\\ref{ref:four_ex12})$ express $A_n$:\n\n\\begin{align} A_n(-n^2+10.01) &= -0.05B_nn \\nonumber \\\\ A_n &= \\frac{-0.05B_nn}{10.01-n^2} = \\frac{0.05B_nn}{n^2-10.01} \\nonumber \\\\ \\end{align}\n\nand substitute into $(\\ref{ref:four_ex11})$:\n\n\\begin{align} -B_nn^2 -0.05\\frac{0.05B_nn}{n^2-10.01} \\cdot n + 10.01 B_n &= \\frac{4}{\\pi n} \\nonumber \\\\ \\vdots & \\nonumber \\\\ B_n &= \\frac{4(10.01-n^2)}{\\pi n\\left(0.05^2n^2+(n^2-10.01)^2\\right)} \\nonumber \\\\ A_n &= \\frac{-0.2}{\\pi \\left(0.05^2n^2+(n^2-10.01)^2\\right)} \\nonumber \\end{align}\n\nWe can form the Fourier series of the response $y_p(t)$ if we use coefficients within $(\\ref{ref:four_partic1})$. It is likely to be a messy motion, but this Fourier analysis served to decompose the motion into pure oscillations. The important information can be obtained by studying amplitudes $C_n$ of particular frequencies:\n\nn=1: \\hskip2em \\left . \\begin{aligned} A_1 &= -0.0008 \\\\ B_1 &= +0.1413 \\end{aligned} \\right \\} \\implies C_1 = 0.1413 \\\\ n=3: \\hskip2em \\left . \\begin{aligned} A_3 &= -0.0611 \\\\ B_3 &= +0.4111 \\end{aligned} \\right \\} \\implies C_3 = 0.4156 \\\\ n=5: \\hskip2em \\left . \\begin{aligned} A_5 &= -0.0003 \\\\ B_5 &= -0.0170 \\end{aligned} \\right \\} \\implies C_5 = 0.0170 \\\\ n=7: \\hskip2em \\left . \\begin{aligned} A_7 &= -0.0000 \\\\ B_7 &= -0.0047 \\end{aligned} \\right \\} \\implies C_7 = 0.0047 \\\\\n\nThe first and the second frequencies are the most important. For $n=3$ is $\\omega = 3$ and such frequency is close to the natural frequency $\\omega_0 = 3.164\\ \\text{s}^{-1}$ of the system.\n\nNote: since $\\cos nt$ is phase delayed $\\pi/2 = 90 ^\\circ$ after $\\sin nt$, there is a right angle between the actual amplitudes of sine and cosine. Therefore, the resultant (maximum amplitude) at particular frequency is $C_n = \\sqrt{A_n^2+B_n^2}$.\n\n#### Example\n\nSolve the spring/mass system $2x'' + 18\\pi^2x = f(t)$ if $f(t)$ is periodic function\n\nf(t) = \\left\\{ \\begin{aligned} -1&\\hskip2em -1\\leq t \\lt 0, \\\\ 1&\\hskip2em 1\\leq t \\lt 2. \\\\ \\end{aligned} \\right.\n\nFrom the given DE can be observed that it is a system with no damping, where the mass is likely 2 kg and the natural frequency of the system is $\\omega_0=\\sqrt{k/m} = \\sqrt{18\\pi^2/2} = 3\\pi$.\n\nFrom auxiliary equation\n\n\\begin{align} m^2+\\omega^2 &= 0 \\implies m = \\pm i\\omega \\nonumber \\\\ y_c &= C_1e^{-i\\omega x} + C_2e^{i\\omega x} \\hskip2em \\text{or} \\nonumber \\\\ y_c &= c_1\\sin\\omega x + c_2\\cos\\omega x \\label{ref:Fourier_DE_1} \\\\ \\end{align}\n\nThe complementary function was obtained from associated homogeneous DE and describes self motion with no external load involved. Since we are solving induced motion, $y_c$ can be considered as a transient solution only.\n\nGiven periodic wave $f(t)$ expressed as a Fourier series is\n\n$$f(t) = \\sum_{n=1}^{\\text{odd}}\\frac 4 {n\\pi}{\\sin nt\\pi}.$$\n\nThe function given has half-period $p=1$. The series is odd, so only $b_n \\cos \\omega_nt$ terms are involved. We do not need to solve $a_n$. If we solve them anyway, they are going to turn to be zero.\n\n\\begin{align} b_n&=\\frac 1 {p}\\int_{-p}^{p}f(t)\\sin nt\\frac{\\pi}{p}\\, dt = \\frac 2 p \\int_0^p \\sin nt\\pi\\ dt = \\frac 2 p \\left[-\\cos nt\\pi \\frac{1}{n\\pi}\\right]_0^{p} = \\nonumber \\\\ &= \\frac{2}{p} \\left(-\\frac{1}{n\\pi}\\right)(\\cos np\\pi - \\cos 0) = -\\frac{2}{p}\\cdot\\frac{1}{n\\pi}\\left( \\cos n\\pi-1\\right) = -\\frac{2}{n\\pi}(-2) \\nonumber \\\\ b_n &= \\frac{4}{n\\pi} \\hskip2em \\text{for }\\ n\\ \\text{odd} \\nonumber\\\\ \\end{align}\n\nIf the right side is $4/n\\pi\\sum\\sin nt\\pi$ and there is no $x'$ within DE, the particular solution is expected in the form $x_p = B_n\\sum\\sin nt\\pi$ for $n$ odd, no cosine terms involved. The trouble here is that sumation includes also $\\boldsymbol{\\sin 3t\\pi}$ and that term is already part of the complementary function $\\boldsymbol{(\\ref{ref:Fourier_DE_1})}$ with arbitrary constant $c_1$. As was discussed in chapter Undetermined coefficients—Superposition approach, we have to solve the particular solution for $\\color{red}{t}\\cdot \\sin 3t\\pi$ and its derivatives.\n\nThe particular solution is then expected as\n\n\\begin{align} x_p(t) &= x_{p1}(t) + x_{p_2}(t) \\label{ref:Fourier_DE_3} \\\\ x_{p1}(t) &= \\sum_{n=1, n \\neq 3}^{\\text{odd}} B_n \\sin nt\\pi \\nonumber \\\\ x_{p2}(t) &= A_3\\cos 3t\\pi \\cdot \\color{red}{t} + B_3 \\sin 3t\\pi \\cdot \\color{red}{t} \\color{grey}{+ C\\cos 3t\\pi + D\\sin 3t\\pi} \\\\ \\end{align}\n\nThe terms $\\cos 3t\\pi,\\ \\sin 3t\\pi$ are already included within $y_c$ so they are voided from $x_p$ again.\n\n\\begin{align} x_{p1}(t) &= \\sum_{n=1, n \\neq 3}^{\\text{odd}} B_n \\sin nt\\pi \\nonumber \\\\ x''_{p1}(t) &= - \\sum_{n=1, n \\neq 3}^{\\text{odd}} B_n\\sin nt\\pi\\cdot n^2\\pi^2 \\nonumber \\\\ \\end{align}\n\nLet us substitute solution $x_{p_1}$ into DE to solve the coefficients.\n\n\\begin{align} -2 \\sum B_n \\sin nt\\pi \\cdot n^2\\pi^2 + 18\\pi^2 \\sum B_n \\sin nt\\pi &= \\sum \\frac 4 {n\\pi}\\sin nt\\pi \\nonumber \\\\ \\sum\\left( -2B_n n^2\\pi^2 + 18\\pi^2 B_n\\right)\\sin nt\\pi &= \\sum \\frac 4 {n\\pi} \\nonumber \\\\ -2B_n n^2\\pi^2 + 18\\pi^2 B_n &= \\frac 4 {n\\pi} \\nonumber \\\\ B_n(-2n^2\\pi^2 + 18\\pi^2) &= \\frac 4 {n\\pi} \\nonumber \\\\ B_n = \\frac{4}{-2n^3\\pi^3 + 18n\\pi^3} &= \\frac{2}{n\\pi^3(9-n^2)} \\nonumber \\end{align}\n\\begin{align} x_{p2}(t) &= A_3\\cos 3\\pi t\\cdot \\color{red}{t} + B_3\\sin 3\\pi t\\cdot \\color{red}{t} \\nonumber \\\\ x'_{p2}(t) &= A_3\\cos 3\\pi t- 3\\pi A_3 \\sin 3\\pi t\\cdot t + B_3\\sin 3\\pi t + 3\\pi B_3 \\cos 3\\pi t\\cdot t \\nonumber \\\\ x''_{p2}(t) &= -6\\pi A_3\\sin 3\\pi t - 9A_3\\pi^2 \\cos 3\\pi t\\cdot t + 6\\pi B_3 \\cos 3\\pi t - 9B_3\\pi^2 \\sin 3\\pi t \\cdot t \\nonumber \\end{align}\n\nLet us substitute $x_{p2}$ into DE $\\color{red}{2x''} \\color{blue}{+ 18\\pi^2x} = f(t)$.\n\n\\begin{align} \\color{red}{-12\\pi A_3\\sin 3\\pi t - 18 A_3\\pi^2 \\cos 3\\pi t \\cdot t + 12\\pi B_3 \\cos 3\\pi t -18 B_3 \\pi^2 \\sin 3\\pi t \\cdot t} &+ \\nonumber \\\\ \\color{blue}{+ 18\\pi^2 A_3 \\cos 3\\pi t \\cdot t + 18 \\pi^2 B_3 \\sin 3\\pi t \\cdot t} &= \\frac 4 {3\\pi}\\sin 3\\pi t \\nonumber \\\\ -12\\pi A_3\\sin 3\\pi t + 12\\pi B_3 \\cos 3\\pi t &= \\frac 4 {3\\pi}\\sin 3\\pi t \\nonumber\\\\ \\end{align}\n\nAnd method of undetermined coefficients gives us\n\n\\begin{align} -12\\pi A_3 &= \\frac 4 {3\\pi} \\implies A_3 = -\\frac{1}{9\\pi^2} \\nonumber \\\\ B_3 &= 0 \\nonumber \\end{align}\n\nNow we can go back to $(\\ref{ref:Fourier_DE_3})$ and complete the particular solution.\n\n$$\\underline{\\underline{x_p(t) = -\\frac{1}{9\\pi^2} \\cos 3\\pi t + \\sum_{n=1, n \\neq 3}^{\\text{odd}} \\frac{2}{n\\pi^3(9-n^2)} \\sin nt\\pi}}$$\n\nTo study resonancy and which frequencies are main when the system is loaded by $f(t)$, we have to evaluate the coefficients:\n\n\\begin{align} \\omega_{ 1} = \\pi:&\\hskip2em B_1 = 8.06\\times 10^{-3} \\nonumber \\\\ \\omega_{ 3} = 3\\pi:&\\hskip2em A_3 = -11.25\\times 10^{-3} \\nonumber \\\\ \\omega_{ 5} = 5\\pi:&\\hskip2em B_5 = -0.81\\times 10^{-3} \\nonumber \\\\ \\omega_{ 7} = 7\\pi:&\\hskip2em B_7 = -0.23\\times 10^{-3} \\nonumber \\\\ \\omega_{ 9} = 9\\pi:&\\hskip2em B_9 = -0.10\\times 10^{-3} \\nonumber \\\\ \\omega_{11} = 11\\pi:&\\hskip2em B_{11} = -0.05\\times 10^{-3} \\nonumber \\\\ \\end{align}\n\nList of chapters" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7210159,"math_prob":0.99997413,"size":16260,"snap":"2020-24-2020-29","text_gpt3_token_len":5971,"char_repetition_ratio":0.15803395,"word_repetition_ratio":0.064129084,"special_character_ratio":0.38739237,"punctuation_ratio":0.079350494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999785,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-01T02:47:28Z\",\"WARC-Record-ID\":\"<urn:uuid:303d38e4-9695-4504-adbd-67576511aae4>\",\"Content-Length\":\"25920\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e31d42de-0c3f-4ce1-9378-4849d2ba732f>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f879efc-5fc8-4881-9377-8e96d5eed12b>\",\"WARC-IP-Address\":\"185.183.8.130\",\"WARC-Target-URI\":\"https://structural-analyser.com/domains/DE/Fourier_DE/\",\"WARC-Payload-Digest\":\"sha1:6EQQ6LEJ6B5NVU3FAK5UTWHVVDLLBXWA\",\"WARC-Block-Digest\":\"sha1:DKALEZYDGPBKCWKNKFA2PYSU6SKICZK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413901.34_warc_CC-MAIN-20200601005011-20200601035011-00006.warc.gz\"}"}
https://ncert-books.com/ncert-solutions-for-class-12-maths-chapter-13-probability-ex-13-2/	[ "# NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2\n\n## NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2\n\nThe topics and sub-topics included in Chapter 13 Probability the following:\n\n Section Name Topic Name 13 Probability 13.1 Introduction 13.2 Conditional Probability 13.3 Multiplication Theorem on Probability 13.4 Independent Events 13.5 Bayes’ Theorem 13.6 Random Variables and its Probability Distributions 13.7 Bernoulli Trials and Binomial Distribution\n\nNCERT Solutions for Class 12 Maths Chapter 13 Probability 13.2 are part of NCERT Solutions for Class 12 Maths. Here we have given Class 12 Maths NCERT Solutions Probability Ex 13.2\n\nQuestion 1\nIf, P(A) =3/5 and P(B) =1/5 find P (A ∩ B) if A and B are independent events.\nSolution:", null, "Question 2.\nTwo cards are drawn at random and without replacement from a pack of 52 playing cards.Find the probability that both the cards are black.\nSolution:", null, "Question 3.\nA box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.\nSolution:", null, "Question 4.\nA fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not\nSolution:", null, "Question 5.\nA die marked 1,2,3 in red and 4,5,6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?\nSolution:", null, "Question 6.\nLet E and F be the events with P(E) =", null, "$\\frac { 3 }{ 5 }$, P (F) =", null, "$\\frac { 3 }{ 10 }$and P (E ∩ F) =", null, "$\\frac { 1 }{ 5 }$. Are E and F independent?\nSolution:", null, "Question 7.\nGiven that the events A and B are such that P(A) =", null, "$\\frac { 1 }{ 2 }$,P(A∪B) =", null, "$\\frac { 3 }{ 5 }$and P(B) = p. Find p if they are\n(i) mutually exclusive\n(ii) independent.\nSol:", null, "Question 8.\nLet A and B independent events P(A) = 0.3 and P(B) = 0.4. Find\n(i) P(A∩B)\n(ii) P(A∪B)\n(iii) P (A \| B)\n(iv) P(B \| A)\nSolution:", null, "Question 9.\nIf A and B are two events, such that P (A) =", null, "$\\frac { 1 }{ 4 }$, P(B) =", null, "$\\frac { 1 }{ 2 }$,and P(A∩B) =", null, "$\\frac { 1 }{ 8 }$.Find P (not A and not B)\nSolution:", null, "Question 10\nEvents A and B are such that\nP(A) =", null, "$\\frac { 1 }{ 2 }$,P(B) =", null, "$\\frac { 7 }{ 12 }$and P (not A or not B) =", null, "$\\frac { 1 }{ 4 }$. State whether A and Bare independent\nSolution:", null, "Question 11\nGiven two independent events A and B such that P (A) = 0.3, P(B) = 0.6. Find\n(i) P(A and B)\n(ii) P(A and not B)\n(iii) P (A or B)\n(iv) P (neither A nor B)\nSolution:", null, "Question 12\nA die is tossed thrice. Find the probability of getting an odd number at least once.\nSolution:", null, "Question 13\nTwo balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that\n(i) both balls are red.\n(ii) the first ball is “black and the second is red.\n(iii) one of them is black and other is red.\nSolution:", null, "Question 14\nProbability of solving specific problem independently by A and B are", null, "$\\frac { 1 }{ 2 }$and", null, "$\\frac { 1 }{ 3 }$respectively. If both try to solve the problem independently, find the probability that\n(i) the problem is solved\n(ii) exactly one of them solves the problem.\nSolution:", null, "Question 15\nOne card is drawn at random from a well-shuffled deck of 52 cards. In which of the following cases are the events E and F independent?\n(i) E: ‘the card drawn is a spade’\nF: ‘the card drawn is an ace ’\n(ii) E: ‘the card drawn is black’\nF: ‘the card drawn is a king’\n(iii) E: ‘the card drawn is a king or queen ’\nF: ‘the card drawn is a queen or jack ’.\nSolution:", null, "", null, "Question 16\nIn a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random\n(a) Find the probability that she reads neither Hindi nor English newspapers.\n(b) If she reads Hindi newspaper, find the probability that she reads english newspapers.\n(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.\nSolution:", null, "Choose the correct answer in the following Question 17 and 18:\n\nQuestion 17\nThe probability of obtaining an even prime number on each die when a pair of dice is rolled is\n(a) 0\n(b)", null, "$\\frac { 1 }{ 3 }$\n(c)", null, "$\\frac { 1 }{ 12 }$\n(d)", null, "$\\frac { 1 }{ 36 }$\nSolution:", null, "Question 18\nTwo events A and B are said to be independent, if\n(a) A and B are mutually exclusive\n(b) P(A’B’) = [1 – P(A)] [1 – P(B)]\n(c) P(A) = P(B)\n(d) P (A) + P (B) = 1\nSolution:", null, "+" ]	[ null, "https://i0.wp.com/www.learncbse.in/wp-content/uploads/2018/12/Class-12-Maths-NCERT-Solutions-Chapter-13-Probability-Ex-13.2-Q-1.png", null, 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https://digitalcommons.lsu.edu/gradschool_theses/2678/	[ "## LSU Master's Theses\n\n#### Identifier\n\netd-11152007-173114\n\n#### Degree\n\nMaster of Science in Mechanical Engineering (MSME)\n\n#### Department\n\nMechanical Engineering\n\nThesis\n\n#### Abstract\n\nThe static and dynamic analysis of structures requires us to obtain solutions of their espective governing differential equations subject to appropriate boundary conditions. The dynamic analysis of non-uniform continuous structures is of primary interest, as most traditional methods take the help of discrete models to analyze them. Well established discrete model methods lead to an algebraic eigenvalue problem, the characteristic equation associated with which is a polynomial. The spectral characteristics of a continuous system nevertheless are represented by transcendental functions and cannot be approximated by polynomials efficiently. Hence finite dimensional discrete models are not capable of predicting the response of continuous systems irrespective of the model order used. In this research, a new low order analytical model is developed which approximates the dynamic behavior of the continuous system accurately. The idea here is to replace a non-uniform continuous system by a set of continuous system with piecewise constant physical properties. Such approximations lead to a transcendental eigenvalue problem, i.e. a problem with transcendental characteristic equation. A numerical method has been developed to solve such eigenvalue problems. The spectrum of non-uniform rods and beams are approximated with fair accuracy by solving the corresponding transcendental eigenvalue problem. This mathematical model is extended to reconstruct non-uniform rods and beams using a linear polynomial approximation of piecewise area. A piecewise tapered approximation of the physical parameters in non-uniform rods and beams leads to better accuracy in the solution. The ability to use higher order area functions as basic building blocks profoundly reduces the model order when using the mathematical model to analyze complex geometries. To further study the impact of this method in various problems of engineering the buckling of thin rectangular plates with stepped thickness has been analyzed and compared with the finite element solution. The transcendental eigenvalue method leads to the reduction in matrix sizes when compared with discrete model methods, thus making the solution computationally viable. Finally the transcendental eigenvalue problem associated with the active control of vibration in discrete mass-spring-damper systems has been developed and the proposed mathematical method has been applied.\n\n2007\n\n#### Document Availability at the Time of Submission\n\nRelease the entire work immediately for access worldwide.\n\nYitshak M. Ram\n\nCOinS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9129485,"math_prob":0.86380327,"size":2946,"snap":"2020-45-2020-50","text_gpt3_token_len":533,"char_repetition_ratio":0.111828685,"word_repetition_ratio":0.0,"special_character_ratio":0.16429056,"punctuation_ratio":0.075723834,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9891867,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T05:29:17Z\",\"WARC-Record-ID\":\"<urn:uuid:b496daf9-1352-47fc-aadd-babe8958facc>\",\"Content-Length\":\"41348\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8325d334-412a-4e2d-ad44-0a343ea65525>\",\"WARC-Concurrent-To\":\"<urn:uuid:8737be11-dda4-46a9-9a12-91e0d9f48e33>\",\"WARC-IP-Address\":\"13.57.92.51\",\"WARC-Target-URI\":\"https://digitalcommons.lsu.edu/gradschool_theses/2678/\",\"WARC-Payload-Digest\":\"sha1:LKDEF3RJYZHMRLSK4DQTL5ZCMHWS455Z\",\"WARC-Block-Digest\":\"sha1:2YP7NKN2TIUPLGZASWGHG7O4DPRFXOI6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107912807.78_warc_CC-MAIN-20201031032847-20201031062847-00113.warc.gz\"}"}
http://worldheritage.org/articles/eng/Solar_mass	[ "# Solar mass\n\n### Solar mass", null, "Size and mass of very large stars: Most massive example, the blue Pistol Star (150 M). Others are Rho Cassiopeiae (40 M), Betelgeuse (20 M), and VY Canis Majoris (17 M). The Sun (1 M) which is not visible in this thumbnail is included to illustrate the scale of example stars. Earth's orbit (grey), Jupiter's orbit (red), and Neptune's orbit (blue) are also given.\n\nThe solar mass (M) is a standard unit of mass in astronomy that is used to indicate the masses of other stars, as well as clusters, nebulae and galaxies. It is equal to the mass of the Sun, about two nonillion kilograms:\n\nM = (1.98855±0.00025)×1030 kg\n\nThe above mass is about 332946 times the mass of the Earth (M), or 1048 times the mass of Jupiter (MJ).\n\nBecause the Earth follows an elliptical orbit around the Sun, its solar mass can be computed from the equation for the orbital period of a small body orbiting a central mass. Based upon the length of the year, the distance from the Earth to the Sun (an astronomical unit or AU), and the gravitational constant (G), the mass of the Sun is given by:\n\nM_\\odot = \\frac{4 \\pi^2 \\times (1\\,\\mathrm{AU})^3}{G \\times (1\\,\\mathrm{yr})^2}\n\nThe value of the gravitational constant was first derived from measurements that were made by Henry Cavendish in 1798 with a torsion balance. The value he obtained differs by only 1% from the modern value. The diurnal parallax of the Sun was accurately measured during the transits of Venus in 1761 and 1769, yielding a value of 9″ (9 arcseconds, compared to the present 1976 value of 8.794148). If we know the value of the diurnal parallax, we can determine the distance to the Sun from the geometry of the Earth.\n\nThe first person to estimate the mass of the Sun was Isaac Newton. In his work Principia, he estimated that the ratio of the mass of the Earth to the Sun was about 1/28 700. Later he determined that his value was based upon a faulty value for the solar parallax, which he had used to estimate the distance to the Sun (1 AU). He corrected his estimated ratio to 1/169 282 in the third edition of the Principia. The current value for the solar parallax is smaller still, yielding an estimated mass ratio of 1/332 946.\n\nAs a unit of measurement, the solar mass came into use before the AU and the gravitational constant were precisely measured. This is because the relative mass of another planet in the Solar System or the combined mass of two binary stars can be calculated in units of Solar mass directly from the orbital radius and orbital period of the planet or stars using Kepler's third law, provided that orbital radius is measured in astronomical units and orbital period is measured in years.\n\nThe mass of the Sun has decreased since the time it formed. This has occurred through two processes in nearly equal amounts. First, in the Sun's core hydrogen is converted into helium by nuclear fusion, in particular the pp chain, and this reaction converts some mass into energy in the form of gamma ray photons. Most of this energy eventually radiates away from the Sun. Second, high-energy protons and electrons in the atmosphere of the Sun are ejected directly into outer space as a solar wind.\n\nThe original mass of the Sun at the time it reached the main sequence remains uncertain. The early Sun had much higher mass-loss rates than at present, so it may have lost anywhere from 1–7% of its natal mass over the course of its main-sequence lifetime. The Sun gains a very small mass through the impact of asteroids and comets; however the Sun already holds 99.86% of the Solar System's total mass, so these impacts cannot offset the mass lost by radiation and ejection.\n\n## Contents\n\n• Related units 1" ]	[ null, "http://images.worldlibrary.net/articles/eng/File:Rho_Cassiopeiae_Sol_VY_Canis_Majoris.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8944285,"math_prob":0.96248084,"size":5033,"snap":"2020-10-2020-16","text_gpt3_token_len":1301,"char_repetition_ratio":0.1264665,"word_repetition_ratio":0.040141676,"special_character_ratio":0.27081263,"punctuation_ratio":0.13293651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9892664,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T07:37:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f1bf7740-da0f-4520-ae9e-44d349a04cb0>\",\"Content-Length\":\"136134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6dbb5263-7573-4e1d-85df-d1f2c8737fb9>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f56ae25-c925-4df4-9d86-c57401979f8c>\",\"WARC-IP-Address\":\"66.27.42.21\",\"WARC-Target-URI\":\"http://worldheritage.org/articles/eng/Solar_mass\",\"WARC-Payload-Digest\":\"sha1:IWNFFG2SYGYFOLZVJS2BOYHB5J4VYSCE\",\"WARC-Block-Digest\":\"sha1:6I7SPRE3K2AQJ5KKAG2EWXEBI2CHV4YI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505550.17_warc_CC-MAIN-20200401065031-20200401095031-00396.warc.gz\"}"}
https://www.maplesoft.com/support/help/maple/view.aspx?path=examples%2FCalculus1Integration	[ "", null, "Calculus 1: Integration - Maple Programming Help\n\nHome : Support : Online Help : Education : Student Packages : Calculus 1 : Example Worksheets : examples/Calculus1Integration\n\nCalculus 1: Integration\n\nThe Student[Calculus1] package contains two routines that can be used to both work with and visualize the concepts of approximating integrals and antiderivatives. This worksheet demonstrates this functionality.\n\nFor further information about any command in the Calculus1 package, see the corresponding help page. For a general overview, see Calculus1.\n\nGetting Started\n\nWhile any command in the package can be referred to using the long form, for example, Student[Calculus1][ApproximateInt], it is easier, and often clearer, to load the package, and then use the short form command names.\n\n > $\\mathrm{restart}$\n > $\\mathrm{with}\\left(\\mathrm{Student}\\left[\\mathrm{Calculus1}\\right]\\right):$\n\nThe following sections show how the routines work.\n\nMain: Visualization\n\nPrevious: Applications of Derivatives" ]	[ null, "https://bat.bing.com/action/0", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73742235,"math_prob":0.92245513,"size":2857,"snap":"2020-10-2020-16","text_gpt3_token_len":730,"char_repetition_ratio":0.14686295,"word_repetition_ratio":0.0,"special_character_ratio":0.18375918,"punctuation_ratio":0.16318786,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98810464,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T02:52:59Z\",\"WARC-Record-ID\":\"<urn:uuid:3b8cf5bf-f825-4fb7-8b11-81f69427b5c0>\",\"Content-Length\":\"196393\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37c76376-f733-4376-8556-232ab3919d70>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f59c103-ddb5-41e7-b48f-18b89f3c81e5>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://www.maplesoft.com/support/help/maple/view.aspx?path=examples%2FCalculus1Integration\",\"WARC-Payload-Digest\":\"sha1:RXQXCKRAFBWTWAOGGV4TDHYDHDXYUUWG\",\"WARC-Block-Digest\":\"sha1:IB37ZYVIS6UT5MPO3L666YW6MWXUWREL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505359.23_warc_CC-MAIN-20200401003422-20200401033422-00533.warc.gz\"}"}
https://coq.gitlab.io/zulip-archive/stream/237977-Coq-users/topic/contra.20statement.20.3F.html	[ "## Stream: Coq users\n\n### Topic: contra statement ?\n\n####", null, "zohaze (Nov 09 2021 at 15:01):\n\nI have these three hypothesis . These are contradictory statements, we can prove it ?\nH1 : a > nth 0 l 0\nH2 : max a (list_max l) <= a\n\n####", null, "Laurent Théry (Nov 12 2021 at 13:14):\n\nIf I that `a = 2` and `l = 1 :: nil`, we have both `a > nth 0 l 0` and `max a (list_max l) <= a`\n\n####", null, "sara lee (Nov 13 2021 at 08:40):\n\nWhen we say list is not empty, then we should write a::l instead of writing l in the hypothesis?(which shows at least one element a is present in the list)\n\n####", null, "zohaze (Nov 13 2021 at 08:42):\n\nI forget to write H3: a <= max a (list_max l). From H2 & H3 it is clear that max a (list_max l) = a. Now it is possible to prove contradiction among two(H1 : a > nth 0 l 0 & H2 : max a (list_max l) <= a)?\n\n####", null, "Christian Doczkal (Nov 13 2021 at 12:09):\n\nEven with `H3`, if you have `l = nil`, then `list_max l = 0` so for any `a > 0`, all three assumptions are true, and there is no contradiction. Do you also have `l <> nil`?\n\n####", null, "zohaze (Nov 14 2021 at 17:36):\n\nYes,l have non empty list and max a (list_max l) <>0 . From S (length l) >0 we can derive length l >0 ? Would you like to answer sara's question also? Is there any problem if we say list is non empty then use l instead of a::l in the code?\n\nLast updated: Jan 29 2023 at 06:02 UTC" ]	[ null, "https://coq.gitlab.io/zulip-archive/icon.svg", null, "https://coq.gitlab.io/zulip-archive/icon.svg", null, "https://coq.gitlab.io/zulip-archive/icon.svg", null, "https://coq.gitlab.io/zulip-archive/icon.svg", null, "https://coq.gitlab.io/zulip-archive/icon.svg", null, "https://coq.gitlab.io/zulip-archive/icon.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8118236,"math_prob":0.899855,"size":1025,"snap":"2022-40-2023-06","text_gpt3_token_len":322,"char_repetition_ratio":0.13516161,"word_repetition_ratio":0.08189655,"special_character_ratio":0.34146342,"punctuation_ratio":0.132,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888987,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T06:20:53Z\",\"WARC-Record-ID\":\"<urn:uuid:ec352045-b7a2-45ea-b8d0-691f301762da>\",\"Content-Length\":\"4743\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce9b7574-d2a4-4a82-8538-8752ba83d52c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8517ac4-7a6f-4fc4-8db8-d16ee415f379>\",\"WARC-IP-Address\":\"35.185.44.232\",\"WARC-Target-URI\":\"https://coq.gitlab.io/zulip-archive/stream/237977-Coq-users/topic/contra.20statement.20.3F.html\",\"WARC-Payload-Digest\":\"sha1:MEOC2K77W34M7VWAU4UTKHSURRAGCWHZ\",\"WARC-Block-Digest\":\"sha1:U3VVVQROHTOH3DX4YS7NRBUYD763NBXE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499700.67_warc_CC-MAIN-20230129044527-20230129074527-00085.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/1512.06787/	[ "# The 750 GeV diphoton excess at the LHC and dark matter constraints\n\nXiao-Jun Bi Qian-Fei Xiang Peng-Fei Yin Zhao-Huan Yu Key Laboratory of Particle Astrophysics, Institute of High Energy Physics, Chinese Academy of Sciences, Beijing 100049, China ARC Centre of Excellence for Particle Physics at the Terascale, School of Physics, The University of Melbourne, Victoria 3010, Australia\n###### Abstract\n\nThe recent reported 750 GeV diphoton excess at the 13 TeV LHC is explained in the framework of effective field theory assuming the diphoton resonance is a scalar (pseudoscalar) particle. It is found that the large production rate and the broad width of this resonance are hard to be simultaneously explained if only visible final states are considered. Therefore an invisible decay channel to dark matter (DM) is strongly favored by the diphoton resonance with a broad width, given a large coupling of the new scalar to DM. We set constraints on the parameter space in this scenario using the results from LHC Run 1, DM relic density, and DM direct and indirect detection experiments. We find that the DM searches can exclude a large portion of the parameter regions accounting for the diphoton excess with a broad width.\n\n###### pacs:\n95.35.+d,12.60.-i\n\\setstretch\n\n1.2\n\n## I Introduction\n\nA great success of LHC Run 1 is the discovery of the 125 GeV Higgs boson in the Standard Model (SM). With a higher collision energy of 13 TeV, LHC Run 2 is ideal for probing heavy exotic particles in new physics beyond the Standard Model (BSM). Recently, the CMS and ATLAS collaborations have reported their results based on Run 2 data with an integrated luminosity of ATLAS:2015diphoton ; CMS:2015dxe ; LHC:201512seminar . As the majority of Run 2 searches have not detected exotic signatures, their results have been used to set limits on BSM models, most of which are stronger than those from Run 1 searches. Some anomalies found in Run 1, such as the diboson excess, have not yet been confirmed by the latest Run 2 searches LHC:201512seminar .\n\nHowever, some surprising results have been provided from diphoton resonance searches. Based on a data set of , the ATLAS collaboration found an excess in the diphoton invariant mass distribution around 750 GeV, with a local (global) significance of (ATLAS:2015diphoton . A fit to data suggested that this excess could be due to a resonance with a width of about 45 GeV. A similar excess at has also been reported by the CMS collaboration based on data, with a lower local (global) significance of 2.6 (1.2CMS:2015dxe . Both collaborations claimed that this excess was not excluded by Run 1 data, because of the large uncertainties in this energy range.\n\nThis excess may be just due to statistical fluctuation, as many other disappeared anomalies in high energy physics experiments. Future LHC searches with more data are required to confirm its existence. Undoubtedly, if it is true, it would become the beginning of an epoch of BSM physics. According to the Landau-Yang theorem Landau:1948kw ; Yang:1950rg , a particle decaying into two photons cannot be of spin-1, implying that this 750 GeV resonance should be either a spin-0 or spin-2 particle. Spin-2 tensor fields can be realized in BSM physics related to gravity. For instance, this particle may be an excitation of the graviton in the Randall-Sundrum model. However, this kind of interpretations suffer stringent constraints CMS:2015dxe .\n\nMany studies on the diphoton excess assuming a scalar or pseudoscalar resonance have appeared soon after the reports D'Eramo:2016mgv ; Csaki:2016raa ; Hamada:2015skp ; Bai:2015nbs ; Falkowski:2015swt ; Chakrabortty:2015hff ; Alves:2015jgx ; Csaki:2015vek ; Bian:2015kjt ; Curtin:2015jcv ; Fichet:2015vvy ; Chao:2015ttq ; Demidov:2015zqn ; No:2015bsn ; Becirevic:2015fmu ; Agrawal:2015dbf ; Ahmed:2015uqt ; Cox:2015ckc ; Kobakhidze:2015ldh ; Cao:2015pto ; Dutta:2015wqh ; Petersson:2015mkr ; Low:2015qep ; McDermott:2015sck ; Molinaro:2015cwg ; Gupta:2015zzs ; Ellis:2015oso ; Buttazzo:2015txu ; Knapen:2015dap ; Franceschini:2015kwy ; DiChiara:2015vdm ; Nakai:2015ptz ; Angelescu:2015uiz ; Backovic:2015fnp ; Mambrini:2015wyu ; Harigaya:2015ezk ; Bellazzini:2015nxw ; Dhuria:2015ufo . This resonance, which is denoted as hereafter, can be realized as an axion or composite Higgs boson in models with new strong dynamics, as well as an exotic Higgs boson with weak coupling. The cross section of the signal is of . In general, Higgs-photon couplings induced by top and loops are suppressed. Therefore, heavy Higgs bosons in ordinary two Higgs doublet models (including Higgs bosons in the MSSM) could not account for this excess with such a large production rate Angelescu:2015uiz ; Buttazzo:2015txu ; DiChiara:2015vdm . Thus the excess may suggest the existence of new charged and colored particles coupled to . Furthermore, the broad width of suggested by the ATLAS analysis is also a challenge for model building. On the other hand, the CMS fit seems to favor a narrower width, but the signal significance is too low. Nevertheless, current data with low statistics could not give a final conclusion.\n\nIn this work, we attempt to provide an interpretation of the diphoton excess in the context of effective field theory. We find that in order to explain the diphoton production rate, large couplings to gluons and photons are required. Considering the constraints from resonance searches of LHC Run 1, it is challenging to achieve a broad decay width, if only the visible decay final states, such as dijet, , , and , are taken into account. This implies that there may be some invisible decay channels. An appealing and interesting possibility is that can decay into dark matter (DM) particles, which compose of the Universe energy but leave no energy deposit in the ATLAS and CMS detectors. Thus it is straightforward to assume a scenario where is a portal between DM and SM particles, i.e., DM particles interact with SM particles through the scalar (pseudoscalar) particle.\n\nConstraints on this scenario are investigated in this work. We consider the limits from the LHC monojet search Aad:2015zva , the DM relic density measurement Ade:2015xua , DM direct detection results from the LUX experiment Akerib:2013tjd , and indirect detection results from the Fermi-LAT Ackermann:2012qk ; Ackermann:2015zua ; Ackermann:2015lka and HESS Abramowski:2013ax -ray observations, and the AMS-02 measurement of the ratio of cosmic-ray antiproton and proton fluxes AMS02pp . The expected exclusion limit of XENON1T Aprile:2015uzo is also used to show the sensitivity of future direct detection experiments. We find that these DM experiments set strong constraints on the model parameter space, and that most of the parameter regions accounting for the diphoton signal have been excluded or can be explored.\n\nThis paper is organized as follows. In Sec. II, we provide an interpretation to the diphoton excess in the context of effective field theory and consider the constraints from LHC Run 1 data. In Sec. III, the constraints from DM relic density and direct/indirect detection experiments on the model parameter space are discussed. Finally we give a conclusion in Sec. IV.\n\n## Ii Interpretation of the diphoton signal and LHC constraints\n\nAs there is no measurement on the CP property of at this moment, can be either a CP-even scalar or CP-odd pseudoscalar. In principle, its production could be initiated by annihilation, fusion, and electroweak vector boson fusion. Since LHC is a machine, fusion has higher luminosity than annihilation after considering parton distribution functions. production via fusion is most probably a loop process, like the SM Higgs case, where the major contribution comes from the top quark loop. decay into diphoton can be mediated by loops of charged fermions and bosons. However, LHC Run 1 data have given constraints on its direct couplings to SM particles Chatrchyan:2013lca ; Aad:2015kna ; Khachatryan:2015cwa ; Aad:2015agg ; Aad:2015mna ; CMS:2014onr ; Aad:2014aqa , such as\n\n σpp→ϕBr(ϕ→t¯t)<550 fb, (1) σpp→ϕBr(ϕ→ZZ)<12 fb, (2) σpp→ϕBr(ϕ→W+W−)<40 fb, (3) σpp→ϕBr(ϕ→Zγ)<4 fb, (4) σpp→ϕBr(ϕ→jj)<2.5 pb, (5)\n\nat 95% CL for . Therefore, the contributions to from the top and loops are stringently constrained and unlikely to account for the diphoton excess with a cross section of . In order to increase the cross section, some extra vector-like charged fermions may be required to induce a large effective coupling Angelescu:2015uiz ; Dutta:2015wqh .\n\nIf these new fermions are heavy enough, they can be integrated out and the loop processes become contact interactions. Consequently, the interactions between and gauge bosons can be described by some dimension-5 effective operators. Assuming CP conservation, the effective operators can be given by\n\n L0+=ϕΛ(k1BμνBμν+k2WaμνWaμν+k3GaμνGaμν) (6)\n\nfor a CP-even , and\n\n L0−=ϕΛ(k1Bμν~Bμν+k2Waμν~Waμν+k3Gaμν~Gaμν) (7)\n\nfor a CP-odd . , , and are parameters describing the effective couplings between and SM gauge fields of the , , and groups, respectively. is a cutoff energy scale set by UV theory, and will be taken as a typical value of for simplicity.\n\nIn order to respect SM gauge symmetry, the Lagrangians (6) and (7) are expressed by gauge eigenstates. In terms of the physical fields (photon) and , the effective interactions become\n\n L0+ ⊃ ϕΛ(kAAAμνAμν+kAZAμνZμν+kZZZμνZμν), (8) L0− ⊃ ϕΛ(kAAAμν~Aμν+kAZAμν~Zμν+kZZZμν~Zμν), (9)\n\nwhere , , and\n\n kAA ≡ k1c2W+k2s2W, (10) kAZ ≡ 2sWcW(k2−k1), (11) kZZ ≡ k1s2W+k2c2W. (12)\n\nwith and . We can see that the and couplings generally accompany the coupling.\n\nUnder the narrow width approximation, the production cross section for can be estimated by Agashe:2014kda\n\n dσdE≃2J+1(2S1+1)(2S2+1)2π2k2ΓinΓoutΓϕδ(E−mϕ), (13)\n\nwhere is the center-of-mass energy of the system and is the momentum of one of the initial particles. and are the polarization states of the initial particles. is the spin of the resonance. is the total decay width of the resonance. and are the partial widths for the resonance decaying into initial states and final states, respectively. For the process , the production cross section depends on a factor of .\n\nThe partial widths of and are independent of the CP property of and given by\n\n Γγγ = k2AAm3ϕ4πΛ2=3.4 MeV(kAA0.01)2(Λ1 TeV)−2(mϕ750 GeV)3, (14) Γgg = 2k23m3ϕπΛ2=27 MeV(k30.01)2(Λ1 TeV)−2(mϕ750 GeV)3. (15)\n\nAs an illuminating case, we may set , i.e., assume that the coupling only comes from the coupling to the gauge field. Then the and partial widths can be estimated as\n\n ΓZZ ∼ tan4θWΓγγ∼0.09Γγγ, (16) ΓZγ ∼ 2tan2θWΓγγ∼0.6Γγγ. (17)\n\nCompared with the 8 TeV upper limits (2) and (4), one can easily see that the diphoton excess signal would not be constrained by Run 1 data in this case. The total decay width can be given by\n\n Γϕ = Γgg+Γγγ+ΓZZ+ΓZγ∼Γgg+1.7Γγγ∝k23+0.13k21. (18)\n\nThus, for fusion production, is predominantly determined by a factor of\n\n ΓggΓγγΓϕ∝k23k21k23+0.13k21. (19)\n\nAs the diphoton production cross section through an -channel is under the narrow width approximation, we can separate the cross section calculation into two parts, i.e., 1-body production cross section and decay branching ratio. Other channels can be dealt with in a similar way. We calculate the production cross section with the simulation code Madgraph 5 Alwall:2014hca , to which the effective Lagrangian is added through FeynRules FeynRules . As fusion production would be important when the coupling is much larger than the coupling Franceschini:2015kwy ; Fichet:2015vvy ; Csaki:2015vek ; Csaki:2016raa , we include both fusion and fusion to compute the production cross section, using the parton distribution function set NNPDF2.3 with QED corrections Ball:2013hta . In the rest of this section, we only consider the case of CP-even . The results for CP-odd are similar.", null, "Figure 1: Contours of σγγ at √s=13 TeV (blue dashed lines) and Γϕ (red dashed lines) for a CP-even ϕ. The left (right) panel corresponds to k2=0 (k1=0) and Λ=1 TeV. The green band denotes the favored range σγγ∼5−20 fb at √s=13 TeV. Also shown are the 8 TeV LHC constraints from the ZZ (red solid line), W+W− (black dashed line), γZ (dark green solid line), and dijet (black solid line) channels. The constraints from 8 TeV LHC diphoton resonance searches are indicated by magenta solid lines, where the lower (upper) one corresponds to the assumption of Γϕ=0.1 (75) GeV. The arrows denote the directions of exclusion.\n\nWe present the contours of at and for in the - plane in the left panel of Fig. 1. For large , the contours tend to be parallel with x-axis. This behavior can be easily understood from Eq. (19). For , fusion dominates, and is also independent of . It can be seen that a wide region in the parameter space can satisfy the desired values, . However, in order to simultaneously obtain and , should be as large as , where the channel dominates in decays.\n\n8 TeV LHC diphoton resonance searches give the constraints Khachatryan:2015qba\n\n σpp→ϕBr(ϕ→γγ)<1.5 fb%forΓϕ=0.1 GeV (20)\n\nand\n\n σpp→ϕBr(ϕ→γγ)<2.4 fb%forΓϕ=75 GeV (21)\n\nat 95% CL. They are also plotted in Fig. 1 with magenta solid lines, where the lower (upper) one corresponds to . It seems that the fusion dominant interpretation is incompatible with LHC Run 1 data.\n\nIn many UV models, the interaction between and gauge bosons are induced through loop diagrams containing new charged and colored vector-like fermions. Hence the effective couplings and can be approximately expressed as\n\n kAA∼α4πNcNfIγγ(m2ϕ/4m2f),k3∼αs4πNfIgg(m2ϕ/4m2f), (22)\n\nwhere the functions denote loop factors, is a typical fermion mass, and are the flavor and color number of the new fermions. Therefore, in order to have a large that is greater than , a very large is needed, which may not be reasonable. Furthermore, a large coupling would induce a significant dijet resonance signal via , which could conflict with LHC Run 1 data. The 8 TeV dijet constraint is also plotted in the left panel of Fig. 1, from which we can see that it excludes the possibility of simultaneously having and for .\n\nOn the other hand, we can fixed and assume that the coupling solely comes from the coupling to the gauge field. The result is shown in the right panel of Fig. 1. We find that the situation becomes worse, as the Run 1 constraints from the , , and channels has already excluded the region for desired .", null, "Figure 2: σγγ at √s=13 TeV as functions of 1−BR(ϕ→χχ) for different values of k1 or k3 assuming k2=0 and Λ=1 TeV in the case of CP-even ϕ. Here Γϕ has been set to be 45 GeV. The green band denotes the favored range σγγ∼5−20 fb.\n\nTherefore, we can conclude that the Run 2 diphoton excess indicates that there should be a large branching ratio into invisible final states, whose most tempting candidate is the DM particle . For , the required branching ratio into DM particles is , with denoting the total width of visible channels. In Fig. 2 we demonstrate as a function of for . We assume and plot two sets of lines with either or fixed. When () is fixed, we can obtain the required by adjusting () and and then derive the value of . It can be seen that the broad width can be accommodated for moderate values of when is dominant.", null, "Figure 3: σ(pp→ϕ→χχ) at √s=13 TeV as functions of Γϕ for different values of k1 assuming k2=0 and Λ=1 TeV in the case of CP-even ϕ. Here σγγ has been set to be 10 fb. The shaded region is excluded by the 8 TeV monojet search.\n\nThe invisible channel is actually constrained by searches in Run 1, where denotes transverse missing energy. The 8 TeV ATLAS monojet analysis Aad:2015zva with an integrated luminosity of is adopted to give this constraint. In order to take into account the acceptance and efficiency of the signal regions in the ATLAS analysis, We simulate the process with MadGraph 5, PYTHIA 6 PYTHIA , and Delphes Delphes and apply the same cut conditions in each signal region. It turns out that the signal region SR6 gives the most stringent constraint. Thus we obtain a 95% CL upper limit on the cross section at :\n\n σpp→ϕBr(ϕ→χχ)<0.39 pb. (23)\n\nThis 8 TeV limit can be translated to an upper limit on at , which is when fusion is dominant.\n\nIn Fig. 3, we show at 13 TeV as functions of the total decay width for different values of assuming . Here has been set to be . We find that is basically proportional to , as is actually the dominant decay channel in these cases. Besides, a smaller requires a larger to give a broad total width, but it is easier to be excluded by the monojet search.\n\n## Iii Dark matter searches\n\nIn the previous section, we have shown that a broad scalar resonance accounting for the diphoton excess should have a large branching ratio into invisible decay modes. An appealing assumption is that the invisible channel is into DM particles. In this scenario, the diphoton excess would have a strong correlation with DM phenomenology.\n\nIn this section, we discuss constraints on the parameter space from DM searches. We consider three types of DM particles, i.e., Majorana fermion, real scalar, and real vector. They couple to , which serves as a DM portal to SM particles. Below we study 4 simplified models with respect to CP conservation. In Model M1, we assume is CP-even and is a Majorana fermion, and the Lagrangian with interaction and mass terms is\n\n LM1=L0++12gχϕ¯χχ−12m2ϕϕ2−12mχ¯χχ. (24)\n\nModel M2 also contains a Majorana fermion , but a CP-odd :\n\n LM2=L0−+12gχϕ¯χiγ5χ−12m2ϕϕ2−12mχ¯χχ. (25)\n\nIn Model S, we consider a real scalar with a CP-even :\n\n LS=L0++12gχϕχ2−12m2ϕϕ2−12m2χχ2. (26)\n\nA real vector and a CP-even is assumed in Model V:\n\n LV=L0++12gχϕχμχμ−12m2ϕϕ2+12m2χχμχμ. (27)\n\nNote that the coupling is dimensionless in Models M1 and M2, but has a mass dimension in Models S and V. Each model has 5 free parameters, , , , , and , as we adopt and without loss of generality. A symmetry is imposed to guarantee the stability of the DM particle. Other possible interactions between and described by higher dimensional operators have been neglected.\n\nIn the following analysis, we randomly scan the parameter space to investigate the regions where the diphoton excess can be well explained and study its implication for DM experiments. In the scan, we allow the free parameters varying in the ranges as\n\n 0\n\nWe impose the requirement of . LHC Run 1 bounds from , , , dijet, and monojet searches are taken into account. Since the current statistics of the diphoton excess is quite low and could not allow a very precise measurement of , we adopt a wide range of , and would select some distinct points satisfying a broad resonance condition .\n\nTo begin with, we attempt to find some parameter points that could provide a correct DM relic density. In these -portal simplified models, DM particles can annihilate into a pair of gauge bosons (see e.g. Refs. Chu:2012qy ; Godbole:2015gma ), , , , , and , through the exchange of an -channel and stay in the thermal equilibrium in the early Universe. Thus we assume DM particles are produced via the standard thermal freeze-out mechanism. The DM relic density measured by the Planck experiment Ade:2015xua , , would set strong constraints on the parameter space of the models. In this analysis, we obtain the predicted DM relic density by numerically solving the Boltzmann equation, as described in Appendix B, and require the relic density satisfies a loose criterion of .\n\nAfter imposing these conditions, we project the parameter points in the - plane in Fig. 4. The notations of the parameter points in Fig. 4, as well as in Figs. 5, 6, and 7 below, have the following meaning: red circles represent the parameter points satisfying both a correct relic density and the broad resonance condition; blue circles corresponds to a correct relic density but ; both purple and green crosses cannot give a correct relic density; purple crosses can satisfy the broad resonance condition, while green crosses lead to .\n\nFrom Fig. 4 we can see that red circles favor large values of . This is because the broad resonance condition always requires a large invisible decay width, as discussed in Sec. II. For instance, if the invisible decay width is 45 GeV in Model M1, should be larger than . A correct relic density can be achieved by a canonical value of annihilation cross section , which requires moderate values of (). Since would be constrained by the LHC bounds, a large is helpful for achieving a canonical annihilation cross section. However, if the resonance width condition is relaxed, a smaller may also provide a correct relic density as a result of the resonant annihilation effect. This is the case for many blue circle points corresponding to .\n\nFor heavy with , the contribution from the annihilation process becomes important. In this region, a correct relic density requires a large to enhance this channel. Since the decay is forbidden when , is quite narrow. This is the case for blue circles and green crosses in the heavy DM region. The distributions of parameter points in Models M1 and M2 are similar, due to their similar kinematics at the LHC. However, the distributions in Models S and V are very different. This is because Model V is not a UV-complete model. The decay width would be significantly increased by the longitudinal polarization of for light DM. Thus, the condition excludes a large portion of the parameter space for .\n\nFor DM indirect searches, we consider the limits from the Fermi-LAT gamma-ray line spectrum observation and dwarf galaxy continuous spectrum observation, as well as the AMS-02 ratio measurement. Since DM annihilation in Model M1 is velocity suppressed, it is irrelevant to indirect detection. Therefore, we only consider the constraints for the rest models.\n\nFirstly, we show the constraints from the gamma-ray line spectrum searches on the annihilation cross section in Fig. 5. The corresponding gamma-ray signal is monochromatic at , usually considered as a “smoking-gun” signature for the discovery of the DM particle. Since no significant line-spectrum photon signal was found, the Fermi-LAT collaboration set upper limits on up to based on 5.8 years of data Ackermann:2015lka . The Fermi-LAT limits at 95% CL from the regions R41 and R3 optimized for NFW profiles with and , respectively, are shown in Fig. 5. The R41 limit is at the order of for , and becomes stricter for lighter DM. As the R3 limit is obtained using a denser DM profile around the Galactic Center, it is stricter than the R41 limit. However, it is not very reliable due to the indeterminate DM distribution in the Galactic Center region. For heavy DM with , 95% CL upper limits come from the HESS observation of the central Galactic halo region based on data with 112 hours effective time Abramowski:2013ax . From Fig. 5, we can see that these searches have set stringent constraints for the parameter points satisfying the correct relic density and LHC bounds.\n\nThe annihilation process could give rise to another kind of line spectrum signal at . Published limits for this channel given by Fermi-LAT are based on their 2 years of data Ackermann:2012qk , and weaker than those from searches.\n\nDM annihilation channels into , , , and would produce photons with continuous energy distribution via final state radiation, hadronization, and decay processes. Here we define an effective total cross section for the channels with continuous gamma-ray spectra as\n\n ⟨σannv⟩cont=⟨σannv⟩ZZ+⟨σannv⟩W+W−+12⟨σannv⟩Zγ+⟨σannv⟩gg+2⟨σannv⟩ϕϕ. (29)\n\nDwarf spheroidal satellite galaxies around the Galaxy are ideal targets for probing such gamma-ray signals, since the corresponding astrophysical gamma-ray backgrounds are quite clean. As no signal has been detected, the Fermi-LAT collaboration set stringent constraints on the DM annihilation cross section from a combined analysis of 15 dwarf galaxies Ackermann:2015zua . This analysis is based on 6 years of data. For many benchmark points, the dominant contributions to the continuous gamma-ray spectrum are given by the final states. As the initial gamma-ray spectra induced by the , and light quark final states are similar, the corresponding upper limits from Fermi-LAT observations would also be similar. We show the 95% CL upper limit on in Fig. 6, as the typical continuous spectrum induced by would be analogous to the spectrum here. For pure and final states, the Fermi-LAT limits are weaker than that in the channel by a factor . However, these final states have small fractions in most cases, so we would not treat them separately here.\n\nSince the , , , and channels would also produce antiprotons via hadronic decay and hadronization processes, an important constraint comes from the cosmic-ray antiproton measurement. The latest result of the ratio has been reported by the AMS-02 collaboration AMS02pp . Here we show the AMS-02 antiproton limits on at 95% CL derived from Ref. Lin:2015taa in Fig. 6. In that analysis, two cosmic-ray propagation models, namely the diffusion-convection (DC) and diffusion-reacceleration (DR-2), was adopted. The uncertainties from propagation processes was considered with great care. As can be seen, the constraints from the Fermi-LAT dwarf galaxy observations and the AMS-02 ratio are comparable. Compared with line-spectrum gamma-ray searches, lesser parameter points are excluded by these two searches. For Models M2 and V, the constraints for are not very strong. For Model S, the antiproton constraints can exclude some points with a correct relic density in the low region.\n\nFinally, we discuss the constraints from DM direct detection. DM particles can interact with nuclei through the coupling induced by the coupling. Since the scattering cross section in Model M2 is spin-dependent and momentum suppressed, this model would not predict testable signals in direct detection experiments. In the rest models, DM-nucleus scattering is spin-independent (SI). The content of gluons in a nucleon is given by Shifman:1978zn\n\n ⟨N\|GaμνGaμν\|N⟩=−8π9αs⟨N\|mN−∑q=u,d,smq¯qq\|N⟩. (30)\n\nThe DM-nucleon SI scattering cross section for Model M1 can be expressed as Zheng:2010js\n\n σSIχN=4πμ2χNG2S,N, (31)\n\nwhile that for Models S and V is Yu:2011by\n\n σSIχN=μ2χNπm2χG2S,N. (32)\n\nHere for these three models,\n\n GS,N=−4πk3gχmN9αsΛm2ϕ(1−∑q=u,d,sfNq), (33)\n\nwhere are the nucleon form factors, whose values are adopted from Ref. Ellis:2000ds . In the calculation, the value of should be taken at the scale of D'Eramo:2016mgv . We obtain through RGE running with an input value of Agashe:2014kda .\n\nWe demonstrate the SI scattering cross section in Fig. 7. Also shown are the current 90% CL upper limit from LUX Akerib:2013tjd and the projected sensitivity of XENON1T Aprile:2015uzo . A large portion of the parameter points with in Model S have been excluded by the LUX result. This is because these points basically correspond to large . In Models M1 and V, the constraints would be much weaker. Nevertheless, the parameter points with a correct relic density and a broad decay width can be further tested by XENON1T. In Model V, some points in the resonant annihilation region may remain undetectable in future direct detection experiments.\n\nFinally, we show the viable parameters to explain the diphoton excess in the - plane in Fig. 8, with the color scale indicating . All the parameter points are required to satisfy the LHC constraints. The parameter points represented by black circles satisfy the correct relic density as well as all the DM direct and indirect constraints. For the Fermi-LAT line-spectrum gamma-ray constraints, we take the limit from the region R41 optimized for a NFW profile with . We can see that requires . As the production cross section depends on a factor of , the 8 TeV LHC dijet search constrains below , which has been illustrated in Fig. 1. For , is further constrained below due to the monojet search. Just a few points with and in Models M2, S, and V evade the constraint from Fermi line-spectrum search, which, however, cannot constrain Model M1. Note that if the stricter Fermi limit based on the region R3 for a NFW profile with is adopted, there would be much less black circles.\n\n## Iv Conclusion and discussion\n\nThe recent 750 GeV diphoton excess found in LHC Run 2 data has stimulated great interests. If this signal is confirmed in future LHC searches, it will open a new era of new physics beyond the Standard Model. In this work we attempt to explain the data in the framework of effective field theory. We find that the spin-0 resonance at 750 GeV with a broad width of requires quite large couplings to gluons and photons. Imposing the constraints from the dijet, , , and resonance searches in LHC Run 1, we find that the resonance should have a significant branching ratio into an invisible decay mode, which has also been constrained by the monojet searches in Run 1.\n\nIf the final states of the invisible decay mode are DM particles, the 750 GeV scalar coupling to DM would also be constrained by DM detection experiments. Thus we consider the requirement from DM relic density and the constraints from direct and indirect detection. We find that the Fermi-LAT line-spectrum gamma-ray searches provide strong constraints on the model parameter space. A large parameter region that can explain the diphoton excess has been excluded. The Fermi-LAT dwarf galaxy continuous spectrum gamma-ray observations and the AMS-02 cosmic-ray measurement of the ratio can also constrain the parameter space, but their limits are weaker than those from the line-spectrum gamma-ray searches. Direct detection experiments could constrain the scalar coupling to gluons and DM particles. The future XENON1T experiment is expected to explore a large parameter region accounting for the diphoton excess.\n\nIt is interesting to note that for Majorana fermionic DM coupled to a CP-odd scalar resonance, the DM-nuclei elastic scattering cross section is momentum-suppressed and spin-dependent, and hence could not be constrained by direct detection experiments. However, such a scenario can be deeply explored in line-spectrum gamma-ray searches. In the case of Majorana fermionic DM coupled to a CP-even scalar resonance, annihilation processes cannot be observed due to velocity suppression, but DM-nuclei SI scatterings could be probed in direct searches. Therefore direct and indirect searches are complementary to each other.\n\nIn summary, the LHC Run 2 diphoton excess may reveal the tip of the new physics iceberg and may have close connection to dark matter in the Universe. We note that the corresponding effective couplings of this resonance to SM gauge fields and DM may be quite large. It seems not trivial to explain such large couplings in a UV-complete model. It would be very interesting to further clarify the properties of the resonance and its potential coupling to DM particles by combining more upcoming LHC Run 2 data, Fermi-LAT searches, AMS-02 data, and XENON1T searches.\n\n###### Acknowledgements.\nThis work is supported by the NSFC under Grant Nos. 11475191, 11135009, 11475189, 11175251, and the 973 Program of China under Grant No. 2013CB837000, and by the Strategic Priority Research Program “The Emergence of Cosmological Structures” of the Chinese Academy of Sciences under Grant No. XDB09000000. ZHY is supported by the Australian Research Council.\n\n## Appendix A Partial widths in ϕ decay channels\n\nThis appendix lists the partial widths for decay channels, except for those have already been listed in the main text.\n\nIn the case of a CP-even , we have the following partial width expressions:\n\n Γ(ϕ→ZZ) = k2ZZm3ϕ4πΛ2ηZ(1−4ξ2Z+6ξ4Z), (34) Γ(ϕ→γZ) = k2AZm3ϕ8πΛ2(1−ξ2Z)3, (35) Γ(ϕ→W+W−) = k22m3ϕ2πΛ2ηW(1−4ξ2W+6ξ4W). (36)\n\nwhere and .\n\nIn the case of a CP-odd , we can obtain\n\n Γ(ϕ→ZZ) = k2ZZm3ϕ4πΛ2η3Z, (37) Γ(ϕ→γZ) = k2AZm3ϕ8πΛ2(1−ξ2Z)3, (38) Γ(ϕ→W+W−) = k22m3ϕ2πΛ2η3W. (39)\n\nBelow we write down the partial widths for in the four simplified models. In Model M1,\n\n Γ(ϕ→χχ)=η3χg2χmϕ16π. (40)\n\nIn Model M2,\n\n Γ(ϕ→χχ)=ηχg2χmϕ16π. (41)\n\nIn Model S,\n\n Γ(ϕ→χχ)=ηχg2χ32πmϕ. (42)\n\nIn Model V,\n\n Γ(ϕ→χχ)=ηχg2χm3ϕ128πm4χ(1−4ξ2χ+12ξ4χ). (43)\n\n## Appendix B DM Relic density calculation and annihilation cross sections\n\nBy solving the Boltzmann equation, DM relic density can be expressed as Kolb:1990vq ; Jungman:1995df\n\n Ωχh2≃1.04×109 GeV−1(T0/2.725 K)3xfMpl√g⋆(xf)(a+3b/xf), (44)\n\nwhere with denoting the DM freeze-out temperature. is the effectively relativistic degrees of freedom at the freeze-out epoch. is the Planck mass and is the present CMB temperature. and is the coefficients in the velocity expansion of annihilation cross section , including all open channels. In order to derive the predicted relic density in the four simplified models, we should firstly calculate the and coefficients in various annihilation channels.\n\nIn Model M1, the leading contribution to is of -wave and the coefficient in every channel vanishes. For ,\n\n b=k2AAg2χm4χπΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ]. (45)\n\nFor ,\n\n b=k2ZZg2χρZ(8m4χ−8m2χm2Z+3m4Z)8πΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ], (46)\n\nwhere . For ,\n\n b=k2AZg2χ(4m2χ−m2Z)3128πΛ2m2χ[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ]. (47)\n\nFor ,\n\n b=k22g2χρW(8m4χ−8m2χm2W+3m4W)4πΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ], (48)\n\nwhere . For ,\n\n b=8k23g2χm4χπΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ]. (49)\n\nFor ,\n\n b=g4χm2χρϕ(9m4χ−8m2χm2ϕ+2m4ϕ)24π(2m2χ−m2ϕ)4, (50)\n\nwhere .\n\nIn Model M2, for ,\n\n a = 4k2AAg2χm4χπΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ], (51) b = k2AAg2χm4χ(m4ϕ+m2ϕΓ2ϕ−16m4χ)πΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ]2. (52)\n\nFor ,\n\n a = 4k2ZZg2χm4χρ3ZπΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ], (53) b = k2ZZg2χm2χρZ2πΛ2[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ]2 (54) ×[2m2χ(m4ϕ+m2ϕΓ2ϕ−16m4χ)+m2Z(m4ϕ−24m2ϕm2χ+m2ϕΓ2ϕ+80m4χ)].\n\nFor ,\n\n a = k2AZg2χ(4m2χ−m2Z)332πΛ2m2χ[(m2ϕ−4m2χ)2+m2ϕΓ2ϕ], (55) b =" ]	[ null, "https://media.arxiv-vanity.com/render-output/4770565/x1.png", null, "https://media.arxiv-vanity.com/render-output/4770565/x3.png", null, "https://media.arxiv-vanity.com/render-output/4770565/x4.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8902153,"math_prob":0.92073315,"size":30100,"snap":"2022-05-2022-21","text_gpt3_token_len":6847,"char_repetition_ratio":0.13726076,"word_repetition_ratio":0.026437012,"special_character_ratio":0.2255814,"punctuation_ratio":0.1278826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9627918,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T08:29:39Z\",\"WARC-Record-ID\":\"<urn:uuid:48e60181-2d2f-45b1-a3c7-27a1a728e5cd>\",\"Content-Length\":\"1049510\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5506c799-96b9-429e-9cd5-78a844512c72>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba795641-aeed-4089-84de-6c678d1848c0>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1512.06787/\",\"WARC-Payload-Digest\":\"sha1:VWQGON2JL7VD7XJMLOOENQCGFDR4M2HQ\",\"WARC-Block-Digest\":\"sha1:UUNGNSUQEXIOMLHG3IAASMYA4EROHANV\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663048462.97_warc_CC-MAIN-20220529072915-20220529102915-00305.warc.gz\"}"}
https://mathoverflow.net/questions/225073/is-there-an-odd-number-which-has-no-prime-to-prime-matchings-when-compared-with	[ "Is there an odd number which has no prime to prime matchings when compared with its reverse order? [closed]", null, "For example look at the number 9. It has prime-prime matching at 3,5, and 7.\n\nFor example the sequence of 13 has matchings at 1,3,7,11,13.\n\nFor example 15 has the matchings(crossings) at 3,5,11,13.\n\nIs there an odd number for which the prime prime matchings do not occur?\n\nclosed as off-topic by Wolfgang, Seva, Joonas Ilmavirta, Chris Wuthrich, Gerry MyersonDec 2 '15 at 11:47\n\nThis question appears to be off-topic. The users who voted to close gave this specific reason:\n\n• \"MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics.\" – Seva, Joonas Ilmavirta, Chris Wuthrich\nIf this question can be reworded to fit the rules in the help center, please edit the question.\n\n• So you want $p+1$ to be the sum of two primes. Google for \"Goldbach conjecture\". – Wolfgang Dec 2 '15 at 8:48\n• write natural numbers in order and then below them write those natural numbers in reverse order. and then match, intersect, or cross the primes in the two rows. if both rows has matching prime highlight it as done in the picture. – user42094 Dec 2 '15 at 8:54\n• @wolfgang how is that linked with goldbach conjecture? – user42094 Dec 2 '15 at 8:56\n• Have you looked at it? It conjectures there is no such odd number. So answering your question would be equivalent to solving it :) – Wolfgang Dec 2 '15 at 9:49\n• There is a meta discussion related to this question: meta.mathoverflow.net/q/2627/55893 – Joonas Ilmavirta Dec 2 '15 at 14:11" ]	[ null, "https://i.stack.imgur.com/oqYAv.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89336777,"math_prob":0.7752598,"size":268,"snap":"2019-43-2019-47","text_gpt3_token_len":85,"char_repetition_ratio":0.18560606,"word_repetition_ratio":0.0,"special_character_ratio":0.31343284,"punctuation_ratio":0.1971831,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96743304,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T23:56:46Z\",\"WARC-Record-ID\":\"<urn:uuid:26a8cccf-c359-4152-b8bc-70bf0ed3db09>\",\"Content-Length\":\"96476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff6c6858-49a2-42fd-8296-e2cb0bda73ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:0678caae-5b13-4318-98de-fd35da273c15>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/225073/is-there-an-odd-number-which-has-no-prime-to-prime-matchings-when-compared-with\",\"WARC-Payload-Digest\":\"sha1:AFOTQ7YLA77JJCD6YJITQN2MJGEONHH2\",\"WARC-Block-Digest\":\"sha1:SQVSOKJDEEGE75DUSLBTWEKSTQS37ZHW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987795253.70_warc_CC-MAIN-20191021221245-20191022004745-00259.warc.gz\"}"}
https://www.jpost.com/international/at-least-12-dead-in-polish-roof-collapse	[ "(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] \|\| {}, a[c].trigger = a[c].trigger \|\| function () { (a[c].trigger.arg = a[c].trigger.arg \|\| []).push(arguments)}, a[c].on = a[c].on \|\| function () {(a[c].on.arg = a[c].on.arg \|\| []).push(arguments)}, a[c].off = a[c].off \|\| function () {(a[c].off.arg = a[c].off.arg \|\| []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96315444,"math_prob":0.96919554,"size":372,"snap":"2023-14-2023-23","text_gpt3_token_len":86,"char_repetition_ratio":0.097826086,"word_repetition_ratio":0.0,"special_character_ratio":0.21774194,"punctuation_ratio":0.1594203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789482,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T23:07:16Z\",\"WARC-Record-ID\":\"<urn:uuid:e0932373-973b-465d-8321-938d91784366>\",\"Content-Length\":\"74320\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b065af9-5446-472e-8dfa-e832df654369>\",\"WARC-Concurrent-To\":\"<urn:uuid:86451e04-fd28-4bb3-81d9-a392a2518f47>\",\"WARC-IP-Address\":\"159.60.130.79\",\"WARC-Target-URI\":\"https://www.jpost.com/international/at-least-12-dead-in-polish-roof-collapse\",\"WARC-Payload-Digest\":\"sha1:XZK56U2WKOGLKG73NQG55KCVP4MQC7DO\",\"WARC-Block-Digest\":\"sha1:BTQ5KSW34VU2WENRB4UE7IZNHDONVLGE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296944452.97_warc_CC-MAIN-20230322211955-20230323001955-00625.warc.gz\"}"}
https://nrich.maths.org/public/topic.php?code=71&cl=4&cldcmpid=2860	[ "# Resources tagged with: Mathematical reasoning & proof\n\nFilter by: Content type:\nAge range:\nChallenge level:\n\n### There are 174 results\n\nBroad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof", null, "### Pent\n\n##### Age 14 to 18 Challenge Level:\n\nThe diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.", null, "### Matter of Scale\n\n##### Age 14 to 16 Challenge Level:\n\nProve Pythagoras' Theorem using enlargements and scale factors.", null, "### Folding Fractions\n\n##### Age 14 to 16 Challenge Level:\n\nWhat fractions can you divide the diagonal of a square into by simple folding?", null, "### Folding Squares\n\n##### Age 14 to 16 Challenge Level:\n\nThe diagonal of a square intersects the line joining one of the unused corners to the midpoint of the opposite side. What do you notice about the line segments produced?", null, "### Golden Eggs\n\n##### Age 16 to 18 Challenge Level:\n\nFind a connection between the shape of a special ellipse and an infinite string of nested square roots.", null, "### Plus or Minus\n\n##### Age 16 to 18 Challenge Level:\n\nMake and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.", null, "### Pythagoras Proofs\n\n##### Age 14 to 16 Challenge Level:\n\nCan you make sense of these three proofs of Pythagoras' Theorem?", null, "##### Age 11 to 16 Challenge Level:\n\nDraw some quadrilaterals on a 9-point circle and work out the angles. Is there a theorem?", null, "### Fitting In\n\n##### Age 14 to 16 Challenge Level:\n\nThe largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .", null, "### Pythagorean Triples II\n\n##### Age 11 to 16\n\nThis is the second article on right-angled triangles whose edge lengths are whole numbers.", null, "### Encircling\n\n##### Age 14 to 16 Challenge Level:\n\nAn equilateral triangle is sitting on top of a square. What is the radius of the circle that circumscribes this shape?", null, "### Long Short\n\n##### Age 14 to 16 Challenge Level:\n\nWhat can you say about the lengths of the sides of a quadrilateral whose vertices are on a unit circle?", null, "### Pythagorean Triples I\n\n##### Age 11 to 16\n\nThe first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!", null, "### Continued Fractions II\n\n##### Age 16 to 18\n\nIn this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).", null, "### Square Pair Circles\n\n##### Age 16 to 18 Challenge Level:\n\nInvestigate the number of points with integer coordinates on circles with centres at the origin for which the square of the radius is a power of 5.", null, "### Parallel Universe\n\n##### Age 14 to 16 Challenge Level:\n\nAn equilateral triangle is constructed on BC. A line QD is drawn, where Q is the midpoint of AC. Prove that AB // QD.", null, "### Rhombus in Rectangle\n\n##### Age 14 to 16 Challenge Level:\n\nTake any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.", null, "### The Pillar of Chios\n\n##### Age 14 to 16 Challenge Level:\n\nSemicircles are drawn on the sides of a rectangle. Prove that the sum of the areas of the four crescents is equal to the area of the rectangle.", null, "### Target Six\n\n##### Age 16 to 18 Challenge Level:\n\nShow that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.", null, "### Salinon\n\n##### Age 14 to 16 Challenge Level:\n\nThis shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?", null, "### The Golden Ratio, Fibonacci Numbers and Continued Fractions.\n\n##### Age 14 to 16\n\nAn iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.", null, "### Same Length\n\n##### Age 11 to 16 Challenge Level:\n\nConstruct two equilateral triangles on a straight line. There are two lengths that look the same - can you prove it?", null, "### Kite in a Square\n\n##### Age 14 to 16 Challenge Level:\n\nCan you make sense of the three methods to work out the area of the kite in the square?", null, "### Angle Trisection\n\n##### Age 14 to 16 Challenge Level:\n\nIt is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.", null, "### Square Mean\n\n##### Age 14 to 16 Challenge Level:\n\nIs the mean of the squares of two numbers greater than, or less than, the square of their means?", null, "### Cosines Rule\n\n##### Age 14 to 16 Challenge Level:\n\nThree points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.", null, "### Ordered Sums\n\n##### Age 14 to 16 Challenge Level:\n\nLet a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .", null, "### Towering Trapeziums\n\n##### Age 14 to 16 Challenge Level:\n\nCan you find the areas of the trapezia in this sequence?", null, "### Rolling Coins\n\n##### Age 14 to 16 Challenge Level:\n\nA blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .", null, "### Little and Large\n\n##### Age 16 to 18 Challenge Level:\n\nA point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?", null, "### Picturing Pythagorean Triples\n\n##### Age 14 to 18\n\nThis article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.", null, "### Similarly So\n\n##### Age 14 to 16 Challenge Level:\n\nABCD is a square. P is the midpoint of AB and is joined to C. A line from D perpendicular to PC meets the line at the point Q. Prove AQ = AD.", null, "### Round and Round\n\n##### Age 14 to 16 Challenge Level:\n\nProve that the shaded area of the semicircle is equal to the area of the inner circle.", null, "### Pareq Exists\n\n##### Age 14 to 16 Challenge Level:\n\nProve that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.", null, "### Circle Box\n\n##### Age 14 to 16 Challenge Level:\n\nIt is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?", null, "### Calculating with Cosines\n\n##### Age 14 to 18 Challenge Level:\n\nIf I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?", null, "### Proof Sorter - Quadratic Equation\n\n##### Age 14 to 18 Challenge Level:\n\nThis is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.", null, "##### Age 16 to 18 Challenge Level:\n\nFind all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.", null, "### No Right Angle Here\n\n##### Age 14 to 16 Challenge Level:\n\nProve that the internal angle bisectors of a triangle will never be perpendicular to each other.", null, "### Rational Roots\n\n##### Age 16 to 18 Challenge Level:\n\nGiven that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.", null, "##### Age 14 to 16 Challenge Level:\n\nA picture is made by joining five small quadrilaterals together to make a large quadrilateral. Is it possible to draw a similar picture if all the small quadrilaterals are cyclic?", null, "### Three Balls\n\n##### Age 14 to 16 Challenge Level:\n\nA circle has centre O and angle POR = angle QOR. Construct tangents at P and Q meeting at T. Draw a circle with diameter OT. Do P and Q lie inside, or on, or outside this circle?", null, "### Zig Zag\n\n##### Age 14 to 16 Challenge Level:\n\nFour identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?", null, "##### Age 16 to 18 Short Challenge Level:\n\nCan you work out where the blue-and-red brick roads end?", null, "### Lens Angle\n\n##### Age 14 to 16 Challenge Level:\n\nFind the missing angle between the two secants to the circle when the two angles at the centre subtended by the arcs created by the intersections of the secants and the circle are 50 and 120 degrees.", null, "### Pythagorean Golden Means\n\n##### Age 16 to 18 Challenge Level:\n\nShow that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.", null, "### Tetra Inequalities\n\n##### Age 16 to 18 Challenge Level:\n\nProve that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.", null, "##### Age 14 to 18 Challenge Level:\n\nWhich of these roads will satisfy a Munchkin builder?", null, "### Composite Notions\n\n##### Age 14 to 16 Challenge Level:\n\nA composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.", null, "### Mouhefanggai\n\n##### Age 14 to 16\n\nImagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai." ]	[ null, "https://nrich.maths.org/content/01/10/six6/icon.jpg", null, "https://nrich.maths.org/content/01/12/six2/icon.jpg", null, "https://nrich.maths.org/content/id/6473/icon.png", null, "https://nrich.maths.org/content/00/11/six6/icon.jpg", null, "https://nrich.maths.org/content/id/2686/icon.jpg", null, "https://nrich.maths.org/content/id/2818/icon.png", null, "https://nrich.maths.org/content/id/6553/icon.png", null, "https://nrich.maths.org/content/id/6624/icon.jpg", null, "https://nrich.maths.org/content/01/10/15plus4/icon.png", null, "https://nrich.maths.org/content/97/04/art1/icon.png", null, "https://nrich.maths.org/content/00/02/six5/icon.jpg", null, "https://nrich.maths.org/content/01/07/15plus3/icon.png", null, "https://nrich.maths.org/content/97/02/art1/icon.png", null, "https://nrich.maths.org/content/99/06/art1/icon.gif", null, "https://nrich.maths.org/content/02/07/six6/icon.gif", null, "https://nrich.maths.org/content/02/09/six5/icon.jpg", null, "https://nrich.maths.org/content/01/03/six5/icon.png", null, "https://nrich.maths.org/content/99/05/six6/icon.jpg", null, "https://nrich.maths.org/content/02/06/15plus3/icon.jpg", null, "https://nrich.maths.org/content/id/2425/icon.png", null, "https://nrich.maths.org/content/id/2737/icon.jpg", null, "https://nrich.maths.org/content/id/11134/icon.png", null, "https://nrich.maths.org/content/id/8301/icon.png", null, "https://nrich.maths.org/content/00/09/six1/icon.jpg", null, "https://nrich.maths.org/content/01/04/15plus1/icon.gif", null, "https://nrich.maths.org/content/01/02/15plus3/icon.gif", null, "https://nrich.maths.org/content/98/04/15plus1/icon.jpg", null, "https://nrich.maths.org/content/id/11449/icon.jpg", null, "https://nrich.maths.org/content/97/12/15plus1/icon.jpg", null, "https://nrich.maths.org/content/00/12/six4/icon.jpg", null, "https://nrich.maths.org/content/98/05/art1/icon.png", null, "https://nrich.maths.org/content/02/05/six5/icon.jpg", null, "https://nrich.maths.org/content/99/03/six5/icon.jpg", null, "https://nrich.maths.org/content/99/06/six6/icon.jpg", null, "https://nrich.maths.org/content/04/02/six5/icon.jpg", null, "https://nrich.maths.org/content/id/7787/icon.png", null, "https://nrich.maths.org/content/01/05/art2/icon.jpg", null, "https://nrich.maths.org/content/03/10/15plus4/icon.jpg", null, "https://nrich.maths.org/content/97/09/six5/icon.png", null, "https://nrich.maths.org/content/00/11/15plus3/icon.jpg", null, "https://nrich.maths.org/content/00/06/six6/icon.gif", null, "https://nrich.maths.org/content/00/07/six6/icon.jpg", null, "https://nrich.maths.org/content/00/06/six3/icon.jpg", null, "https://nrich.maths.org/content/id/5924/icon.jpg", null, "https://nrich.maths.org/content/02/03/six6/icon.jpg", null, "https://nrich.maths.org/content/01/05/15plus2/icon.gif", null, "https://nrich.maths.org/content/02/11/15plus3/icon.png", null, "https://nrich.maths.org/content/id/5921/icon.jpg", null, "https://nrich.maths.org/content/03/11/six2/icon.png", null, "https://nrich.maths.org/content/01/11/art1/icon.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92069507,"math_prob":0.93990207,"size":6597,"snap":"2019-51-2020-05","text_gpt3_token_len":1520,"char_repetition_ratio":0.14727741,"word_repetition_ratio":0.024350649,"special_character_ratio":0.22237381,"punctuation_ratio":0.08443466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99594146,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"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,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,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,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\":\"2019-12-07T10:15:55Z\",\"WARC-Record-ID\":\"<urn:uuid:f14a45eb-1fe9-4339-9447-50f2d5353344>\",\"Content-Length\":\"53507\",\"Content-Type\":\"application/http; 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https://convertoctopus.com/136-grams-to-ounces	[ "## Conversion formula\n\nThe conversion factor from grams to ounces is 0.03527396194958, which means that 1 gram is equal to 0.03527396194958 ounces:\n\n1 g = 0.03527396194958 oz\n\nTo convert 136 grams into ounces we have to multiply 136 by the conversion factor in order to get the mass amount from grams to ounces. We can also form a simple proportion to calculate the result:\n\n1 g → 0.03527396194958 oz\n\n136 g → M(oz)\n\nSolve the above proportion to obtain the mass M in ounces:\n\nM(oz) = 136 g × 0.03527396194958 oz\n\nM(oz) = 4.7972588251429 oz\n\nThe final result is:\n\n136 g → 4.7972588251429 oz\n\nWe conclude that 136 grams is equivalent to 4.7972588251429 ounces:\n\n136 grams = 4.7972588251429 ounces\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 ounce is equal to 0.20845237591912 × 136 grams.\n\nAnother way is saying that 136 grams is equal to 1 ÷ 0.20845237591912 ounces.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one hundred thirty-six grams is approximately four point seven nine seven ounces:\n\n136 g ≅ 4.797 oz\n\nAn alternative is also that one ounce is approximately zero point two zero eight times one hundred thirty-six grams.\n\n## Conversion table\n\n### grams to ounces chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from grams to ounces\n\ngrams (g) ounces (oz)\n137 grams 4.833 ounces\n138 grams 4.868 ounces\n139 grams 4.903 ounces\n140 grams 4.938 ounces\n141 grams 4.974 ounces\n142 grams 5.009 ounces\n143 grams 5.044 ounces\n144 grams 5.079 ounces\n145 grams 5.115 ounces\n146 grams 5.15 ounces" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7753947,"math_prob":0.99467313,"size":1691,"snap":"2020-45-2020-50","text_gpt3_token_len":463,"char_repetition_ratio":0.18020155,"word_repetition_ratio":0.0,"special_character_ratio":0.37433472,"punctuation_ratio":0.10682493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956754,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-23T10:48:41Z\",\"WARC-Record-ID\":\"<urn:uuid:e0f323c7-f87a-4bba-84f8-a1edc10465c2>\",\"Content-Length\":\"27731\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22b93864-b34a-455e-a3e9-b0c79c0e5887>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c6b4599-5556-4871-944f-b8c9e808651e>\",\"WARC-IP-Address\":\"104.27.143.66\",\"WARC-Target-URI\":\"https://convertoctopus.com/136-grams-to-ounces\",\"WARC-Payload-Digest\":\"sha1:6N2YDYTP7WZPDM4YCEJKXKGVWVMPQZ6P\",\"WARC-Block-Digest\":\"sha1:EFY7RYBD4V7L6B4QLPUXYI4JI63UX6SU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107881369.4_warc_CC-MAIN-20201023102435-20201023132435-00353.warc.gz\"}"}
https://www.convertunits.com/info/decinewton	[ "# ››Measurement unit: decinewton\n\nFull name: decinewton\n\nPlural form: decinewtons\n\nSymbol: dN\n\nCategory type: force\n\nScale factor: 0.1\n\n# ››SI unit: newton\n\nThe SI derived unit for force is the newton.\n1 newton is equal to 10 decinewton.\n\n# ››Convert decinewton to another unit\n\nConvert decinewton to\n\nValid units must be of the force type.\nYou can use this form to select from known units:\n\nConvert decinewton to\n\n# ››Definition: Decinewton\n\nThe SI prefix \"deci\" represents a factor of 10-1, or in exponential notation, 1E-1.\n\nSo 1 decinewton = 10-1 newtons.\n\nThe definition of a newton is as follows:\n\nIn physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84532535,"math_prob":0.59806806,"size":976,"snap":"2021-31-2021-39","text_gpt3_token_len":296,"char_repetition_ratio":0.23662551,"word_repetition_ratio":0.0,"special_character_ratio":0.22233607,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9938911,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T11:31:13Z\",\"WARC-Record-ID\":\"<urn:uuid:c240af82-2684-4a68-977d-c091d8481ddd>\",\"Content-Length\":\"46270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f09a785-7194-4057-9e9d-ba9ef86f16bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc622840-94cf-4072-a7a6-7c7e02ddc9a2>\",\"WARC-IP-Address\":\"23.20.181.175\",\"WARC-Target-URI\":\"https://www.convertunits.com/info/decinewton\",\"WARC-Payload-Digest\":\"sha1:VTIH3VH5INL3UGGVIC3TY2N5VUHVNSFV\",\"WARC-Block-Digest\":\"sha1:5OCPKVBKHFDIKLW2ZHGUR2AHV2FRVIFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153709.26_warc_CC-MAIN-20210728092200-20210728122200-00370.warc.gz\"}"}
http://mathandmultimedia.com/tag/parameterization/	[ "## GeoGebra Tutorial 6 – Parameterization of Length and Area\n\nThis is the sixth tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series.\n\nIn this tutorial, we are going to learn the following:\n\n1. use variables in GeoGebra\n2. compute using these variables\n3. use variables as parameters of objects\n\nProblem: Given a rectangle with perimeter 10 units, find the dimension of the rectangle that can be formed that has the largest area.\n\nThis problem can be easily solved algebraically, but we are going to use GeoGebra to parameterize the length and the area of the rectangle to find its maximum area. The output of this tutorial is shown above.\n\nBefore doing the tutorial, let us first solve the problem. We know that the rectangle’s perimeter is constant, so we choose the width w. It follows that the height h will depend on the width. For instance, if w=4 units, then h = (10 – 2*4)/2 which is equal to 1. Hence, h = (10 – 2w)/2. Using this information, we plan the GeoGebra construction.\n\n1. First we make our maximum width 5 (Why?). We will create segment AL with length 5 with A at the origin and L at (5,0)\n2. Next, we create point D on AL. With D = (w,0), AD will be the width of the rectangle.\n3. We compute for h = (10 – 2w)/2, then, and take the value as the height of the rectangle. Then, we create point B with coordinates (0,h).\n4. We create the fourth vertex of the rectangle by getting the intersection of the horizontal line passing through B and a vertical line passing through D.\n5. Lastly, create ABCD using the polygon tool, and then produce point P (w, A_r) where A_r is the area of the rectangle.\n\nInstructions\n\n 1.) Open GeoGebra and be sure that the Algebra & Graphics view is selected in the Perspectives panel.", null, "2.) Select the Segment between Two Points tool, click on (0,0) and click on (5,0) to construct segment AB. Show the label of the points, and rename point B to L.", null, "3.) Create a point on AL. You may not see the segment, so before doing this, hide the axes by clicking the Axes icon in the upper left of the Graphics View. If the icons are not displayed, click the arrow.", null, "4.) Rename the recently created point to D. Move the point and notice that it can only move between A and L. Now, hide point L and display the axes. AD will be the width (lower base) of the rectangle. 5.) We now determine the width and the height of the rectangle. First, we want to determine the AD which is the width. To do this, we get the x-coordinate of D (Why?). To get the x-coordinate of D, type w = x(D) in the Input bar and press the ENTER key. This means that the value of w, a declared variable, will be the x-coordinate of point D which is the same as the length of AD. 6.) Next, we compute for the height h of the rectangle. Type h = (10 – 2w)/2 in the input bar and press the ENTER key. Notice the values of h and w were added to the Algebra view. 7.) Next we create point B with coordinates (0,h). To do this, type B = (0,h) in the Input bar and press the ENTER key. This will be the third point on the rectangle.", null, "8.) Move point D. What do you observe? 9.) Next, we locate the fourth vertex of the rectangle. The fourth vertex C will have the y-coordinate the same as B and x-coordinate the same as D. Therefore, we type C = (x(D), y(B)).", null, "10.) Now, we use the polygon tool to construct rectangle ABCD. Click the Polygon tool and then click the points in following order: point A, point B, point C, point D and, again, point A to close the figure. 11.) Now, let us display the area of the polygon. Right click the interior of the rectangle, then click Object Properties to open the Preferences window. In the Basic tab of the Preferences window, check the Show Label check box and choose Value from the drop down list box. Close the window.", null, "12.) Move point D. What do you observe? What length of AD gives the rectangle the largest area? 13.) Now, we create point P, type P = (w, poly1). Note that poly1 is the name of the rectangle and its value is area of the rectangle (see the Algebra view). 14.) Right click on point P, then click check Trace On. This will trace the path of point P.", null, "15.) Move point D. What do you observe? What can you say about the curve formed by the traces of point P? Explain why your observations are such. 16.) Solve the problem algebraically. What is the relationship between the equation formed from getting the solution of the problem and curve formed by traces of point P?", null, "" ]	[ null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/segmentpng3.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/newpointpng4.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/movepng6.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/movepng6.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/regpolygonpng1.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/movepng6.png", null, "http://mathandmultimedia.com/wp-content/uploads/2009/11/movepng6.png", null, "http://www.linkwithin.com/pixel.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8420153,"math_prob":0.9887794,"size":4547,"snap":"2019-51-2020-05","text_gpt3_token_len":1118,"char_repetition_ratio":0.16288796,"word_repetition_ratio":0.0413318,"special_character_ratio":0.26369035,"punctuation_ratio":0.13855422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985119,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T03:36:01Z\",\"WARC-Record-ID\":\"<urn:uuid:177addf6-0e12-4022-bbf8-405d551f122e>\",\"Content-Length\":\"41305\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e832a0d-ff7a-477d-bf7e-5924c2472e2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:76d0b262-1a6f-42cd-8b7b-47355d537dcc>\",\"WARC-IP-Address\":\"184.168.164.1\",\"WARC-Target-URI\":\"http://mathandmultimedia.com/tag/parameterization/\",\"WARC-Payload-Digest\":\"sha1:H4U2POIXWU5MKFICP3USLHJK3WXZJ6KG\",\"WARC-Block-Digest\":\"sha1:F47XPFURQ7R2VDUYBX42KDFF7V73TY5M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251783621.89_warc_CC-MAIN-20200129010251-20200129040251-00252.warc.gz\"}"}
https://stats.stackexchange.com/questions/92984/test-the-randomness-uniformly-distributed-on-a-64-bit-float-random-generator	[ "# Test the randomness (uniformly distributed) on a 64 bit float random generator\n\nWe have a random number generator which is supposed to generate 64 bit floats, uniformly.\n\nWe want to test whether it is a good uniformly random.\n\nI am not asking the general way to test it, as it was asked here https://stackoverflow.com/questions/22916519/test-the-randomness-of-a-black-box-that-outputs-random-64-bit-floats.\n\nAs I understand, the first step for the test is to somehow divide the whole range into equal sizes like bins, then we keep generating and see the number of \"balls\" falling into each bin.\n\nHow should I split the whole range of 64 bits float?\n\nThe range is between -1.79769313486231571e+308 and 1.79769313486231571e+308 and it is huge.\n\nMy idea is to utilise the 64 bits. Can I design the equal size bins like this:\n\n1. Every float has 64 bits, so we have 64 bins.\n2. For every float generated, we read all those bits, if a bit is 1, then we increased the according bin's number by 1\n3. After N samplings, the expected number of each bin should be N/2.\n4. Then we carry on pearson's chi-square test, etc.\n\nAs it stands, this is not a good way to test whether floating point numbers are uniformly distributed. Like Aksakal, I wondered about whether the bits of the exponent part of the floating point representation would be uniformly distributed. The answer to this is that they aren't uniformly distributed, because there are very many more numbers with large exponents than there are numbers with small exponents.\n\nI wrote a small test program that confirms this. It generates $N = 1 \\text{ million}$ uniformly distributed random floating point numbers, and as a control, $N$ random integers. (There were various problems generating 64 bit floating point numbers, see e.g. here, and 32 bits seems sufficient for demonstration purposes.)\n\nFirst, the control case. The plot of the bins of bits for integers is just as you suggested, with each bin $\\approx N/2$.", null, "Now for the floating point numbers. A plot of the sorted numbers is a straight line, indicating that they would pass the Kolmogorov–Smirnov test for uniformity.", null, "But the bins are definitely not uniform.", null, "If you plot only bins 1 to 23 together with bin 32, you do get bins $\\approx N/2$, but bins 24 to 31 show a clear increasing pattern. These bits corresponds precisely with the bits for the exponent in 32 bit floating point numbers. The IEEE single precision floating point definition stipulates\n\n• the least significant 23 bits are for the mantissa\n• the next 8 bits are for the exponent\n• the most significant bit is for the sign\n\nAnother way to see this is to consider a simpler example. Think about generating numbers in base 10 between 0 and $10^7$, with a base 10 exponent. Numbers between 0 and 1 would have an exponent of 0. Numbers between 1 and 10 would have an exponent of 1, numbers between 10 and 100 an exponent of 2, ..., and numbers between $10^6$ and $10^7$ an exponent of 7. The numbers $10^4$ to $10^7$ are $(10^7-10^4)/10^7=99.9\\%$ of the range and in binary their exponents range from 001 to 111, so you'd expect the most significant bit to occur 99.9% of the time, not 50% of the time.\n\nIt would be possible, with some care, to use an approach like this to get the expected frequencies for each bin in the binary exponent of a floating point number, and use this in a $\\chi^2$ test, but Kolmogorov–Smirnov is a better approach in theory and easy to implement. Nevertheless a test like this could pick up distributional biases in the implementation of a random number generation that Kolmogorov–Smirnov might not. For example, when I first tried generating 64 bit double precision floating point random numbers in C++, I forgot to change to a 64 bit Mersenne Twister engine. The sorted numbers give a straight line plot, but you can see from the plots of the bins of the bits that the 64 bit Mersenne Twister engine is superior to the 32 bit one (as you would expect).", null, "(Note in both cases that the last bit, the sign bit, is zero, due to the difficulties of generating random numbers across the whole range.)\n\nhave you looked at A Statistical Test Suite for Random and Pseudorandom Number Generators for Cryptographic Applications by NIST?\n\nI think it's a great place to start your analysis.\n\n• Hi. I understand that there are a number of ways and good ways to really test randomness. But in this question, I particularly know whether my idea is correct or not – Jackson Tale Apr 8 '14 at 13:28\n• the reason i suggest not to make up your own tests, is that it's a tricky business. so many people are using RNGs for so long that there's got to be a test tried and proven to work, such as \"Monobit\" or \"Frequency Test within a Block\", which are suspiciously similar to what you seem to be trying to do (see the NIST suite) – Aksakal Apr 8 '14 at 13:41\n• Hi,sorry, this question is from a telephone interview for a developer position. I think they want to ask for general stats knowledge, combined with CS methods. that's why I focus on bits – Jackson Tale Apr 8 '14 at 13:43\n• the main flaw in your idea is that 64bit float is probably represented as IEEE 754 double precision floating point number, look it up in interweb. as you'll see it has matissa (fraction) and exponent parts. my gut feeling's that your scheme will not work for this format. – Aksakal Apr 8 '14 at 13:48" ]	[ null, "https://i.stack.imgur.com/yWRXp.png", null, "https://i.stack.imgur.com/0LQWz.png", null, "https://i.stack.imgur.com/sgoIJ.png", null, "https://i.stack.imgur.com/6rgMv.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92073524,"math_prob":0.9644463,"size":2981,"snap":"2019-26-2019-30","text_gpt3_token_len":712,"char_repetition_ratio":0.1477998,"word_repetition_ratio":0.0038610038,"special_character_ratio":0.24521972,"punctuation_ratio":0.09747899,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959508,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-16T20:28:40Z\",\"WARC-Record-ID\":\"<urn:uuid:2d7b8f9b-7a73-4efb-8566-4249cfa35397>\",\"Content-Length\":\"153479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a02109e-4f5a-4fae-a233-95658e9f0553>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9ae01ae-44e9-4fc8-9dc9-908b57a1d9e2>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/92984/test-the-randomness-uniformly-distributed-on-a-64-bit-float-random-generator\",\"WARC-Payload-Digest\":\"sha1:WLPB2X5INE7SUAW4RWBOJ7X24DQV7R5K\",\"WARC-Block-Digest\":\"sha1:C7RLU6UCPAIOLSG372ZCP3OOWQTOQVAE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195524879.8_warc_CC-MAIN-20190716201412-20190716223412-00502.warc.gz\"}"}
https://www.tug.org/pipermail/texhax/2007-April/008194.html	[ "# [texhax] \\labal{\\lowercase}\n\nTue Apr 10 15:43:38 CEST 2007\n\nHello list,\n\nI've got a problem which I'm struggling with for a week. I would like\nto write a macro that creates a (LaTeX) \\label of the lowercase text\nof its argument:\n\n\\def\\MyLabel#1{\\label{\\lowercase{#1}}\n\nObviously it doesn't work. My conception was that the problem is\n\\lowercase, which is processed \"in the stomach\" of the interpreter\n(according to the TRM). So I began to write macros that aviod\ncalling both \\lowercase and \\lccode, and I've come up with this:\n(I'm just getting into virgin tex; nevertheless the macros are\nintended for LaTeX usage)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Returns the lower case counterpart of character #1 if it is a\n% capital letter, or #1 otherwise.\n\\def\\LowChar#1{%\n\\ifnum#1=A a\\else\\ifnum#1=B b\\else\\ifnum#1=C c\\else\n\\ifnum#1=D d\\else\\ifnum#1=E e\\else\\ifnum#1=F f\\else\n\\ifnum#1=G g\\else\\ifnum#1=H h\\else\\ifnum#1=I i\\else\n\\ifnum#1=J j\\else\\ifnum#1=K k\\else\\ifnum#1=L l\\else\n\\ifnum#1=M m\\else\\ifnum#1=N n\\else\\ifnum#1=O o\\else\n\\ifnum#1=P p\\else\\ifnum#1=Q q\\else\\ifnum#1=R r\\else\n\\ifnum#1=S s\\else\\ifnum#1=T t\\else\\ifnum#1=U u\\else\n\\ifnum#1=V v\\else\\ifnum#1=W w\\else\\ifnum#1=X x\\else\n\\ifnum#1=Y y\\else\\ifnum#1=Z z\\else #1%\n\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\n\\fi\\fi\\fi\\fi\\fi\\fi\\fi}\n\n% Cycle through the characters of the string via tail recursion.\n\\def\\lowit at b#1#2\\@{%\n\\def\\lowit at tmp{#1}%\n\\expandafter\\ifx\\lowit at tmp\\empty\\else\n\\edef\\lowit at buf{\\lowit at buf \\LowChar#1}%\n\\lowit at b#2\\@\n\\fi}\n\n% The lowercase string is stored in \\lowit at buf temporarily.\n\\def\\lowit at a#1{%\n\\def\\lowit at buf{}%\n\\expandafter\\lowit at b#1\\empty\\empty\\@%\n\\lowit at buf\\egroup}\n\n% This is the entry point. It takes a string and expands to its lowercase.\n% The \\catcode manipulations are necessarry so \\lowit at b won't ignore whitespace.\n\\def\\LowIt{%\n\\bgroup%\n\\def\\whitespace{ }%\n\\catcode\\ =\\active\\lccode\\~\\ \\lowercase{\\let~\\whitespace}%\n\\catcode\\^^I=\\active\\lccode\\~\\^^I\\lowercase{\\let~\\whitespace}%\n\\catcode\\^^M=\\active\\lccode\\~\\^^M\\lowercase{\\let~\\whitespace}%\n\\lowit at a}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nNow, LowIt{BLA BLA BLA} is typeset to bla bla bla'' correctly; I was\nvery happy to see that. However, when I try to \\typeout{\\LowIt{BLA BLA}\nthe whole thing blows up because \\typeout (and \\label for that matter)\nseem to \\edef \\LowIt; then tex starts complaining about \\whitespace being\nundefined. Ouch.\n\nEarlier someone (Uwe?) posted a code fragment that looked primising:\n\\newcommand{\\foo}{%\n\\stepcounter{foo}%\n\\edef\\thelabel{\\noexpand\\label{bar:\\thefoo}}%\n\\thelabel}\n\nWhile it does magic with \\label, I could not apply the idea to my problem.\nCan you give me a hand?\n\nbit,\n`" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7446217,"math_prob":0.91398174,"size":2854,"snap":"2022-27-2022-33","text_gpt3_token_len":1056,"char_repetition_ratio":0.2322807,"word_repetition_ratio":0.0,"special_character_ratio":0.32025227,"punctuation_ratio":0.0626118,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99376005,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T14:25:25Z\",\"WARC-Record-ID\":\"<urn:uuid:94d23614-ebba-46dd-bf0e-11714f116d69>\",\"Content-Length\":\"5972\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4661b0c1-1956-42d0-8ffc-b1ad7219e170>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf9ec0b4-23d8-4345-986f-bfa39b33fa47>\",\"WARC-IP-Address\":\"94.23.251.76\",\"WARC-Target-URI\":\"https://www.tug.org/pipermail/texhax/2007-April/008194.html\",\"WARC-Payload-Digest\":\"sha1:TML7GGA4JPNNLV37SCI4EUW3QS37M2NI\",\"WARC-Block-Digest\":\"sha1:T6ZKES4N5DH5E5HDILHOSSBAT7GTJP74\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571190.0_warc_CC-MAIN-20220810131127-20220810161127-00221.warc.gz\"}"}
https://www.projecteuclid.org/journals/electronic-journal-of-probability/volume-6/issue-none/On-Disagreement-Percolation-and-Maximality-of-the-Critical-Value-for/10.1214/EJP.v6-88.full	[ "Translator Disclaimer\n2001 On Disagreement Percolation and Maximality of the Critical Value for iid Percolation\nJohan Jonasson\nElectron. J. Probab. 6: 1-13 (2001). DOI: 10.1214/EJP.v6-88\n\n## Abstract\n\nTwo different problems are studied:\n\n• 1. For an infinite locally finite connected graph $G$, let $p_c(G)$ be the critical value for the existence of an infinite cluster in iid bond percolation on $G$ and let $P_c = \\sup\\{p_c(G): G \\text{ transitive }, p_c(G) \\lt 1\\}$. Is $P_c \\lt 1$?\n\n• 2. Let $G$ be transitive with $p_c(G) \\lt 1$, take $p \\in [0,1]$ and let $X$ and $Y$ be iid bond percolations on $G$ with retention parameters $(1+p)/2$ and $(1-p)/2$ respectively. Is there a $q \\lt 1$ such that $p \\gt q$ implies that for any monotone coupling $(X',Y')$ of $X$ and $Y$ the edges for which $X'$ and $Y'$ disagree form infinite connected component(s) with positive probability? Let $p_d(G)$ be the infimum of such $q$'s (including $q=1$) and let $P_d = \\sup\\{p_d(G): G \\text{ transitive }, p_c(G) \\lt 1\\}$. Is the stronger statement $P_d \\lt 1$ true? On the other hand: Is it always true that $p_d(G) \\gt p_c (G)$?\n\nIt is shown that if one restricts attention to biregular planar graphs then these two problems can be treated in a similar way and all the above questions are positively answered. We also give examples to show that if one drops the assumption of transitivity, then the answer to the above two questions is no. Furthermore it is shown that for any bounded-degree bipartite graph $G$ with $p_c(G) \\lt 1$ one has $p_c(G) \\lt p_d(G)$. Problem (2) arises naturally from where an example is given of a coupling of the distinct plus- and minus measures for the Ising model on a quasi-transitive graph at super-critical inverse temperature. We give an example of such a coupling on the $r$-regular tree, ${\\bf T}_r$, for $r \\gt 1$.\n\n## Citation\n\nJohan Jonasson. \"On Disagreement Percolation and Maximality of the Critical Value for iid Percolation.\" Electron. J. Probab. 6 1 - 13, 2001. https://doi.org/10.1214/EJP.v6-88\n\n## Information\n\nAccepted: 15 June 2001; Published: 2001\nFirst available in Project Euclid: 19 April 2016\n\nzbMATH: 0987.60099\nMathSciNet: MR1873292\nDigital Object Identifier: 10.1214/EJP.v6-88\n\nSubjects:\nPrimary: 60K35\nSecondary: 82B20 , 82B26 , 82B43\n\nKeywords: coupling , Ising model , planar graph , Random-cluster model , transitive graph\n\nJOURNAL ARTICLE\n13 PAGES", null, "SHARE\nVol.6 • 2001", null, "" ]	[ null, "https://www.projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png", null, "https://www.projecteuclid.org/images/journals/cover_ejp.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80621594,"math_prob":0.9951409,"size":1845,"snap":"2022-27-2022-33","text_gpt3_token_len":559,"char_repetition_ratio":0.109179795,"word_repetition_ratio":0.025974026,"special_character_ratio":0.30460703,"punctuation_ratio":0.10026385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99942374,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T23:58:21Z\",\"WARC-Record-ID\":\"<urn:uuid:3a926dd2-5ee1-42db-9aad-cd9014f03049>\",\"Content-Length\":\"150808\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db0871d8-4f96-4c89-b38d-fa9a1e9bd70b>\",\"WARC-Concurrent-To\":\"<urn:uuid:6bbe22eb-568a-4a22-b89b-1be49dcc3db9>\",\"WARC-IP-Address\":\"107.154.79.145\",\"WARC-Target-URI\":\"https://www.projecteuclid.org/journals/electronic-journal-of-probability/volume-6/issue-none/On-Disagreement-Percolation-and-Maximality-of-the-Critical-Value-for/10.1214/EJP.v6-88.full\",\"WARC-Payload-Digest\":\"sha1:LOX2RGUZSTISRSYIKARPTIQGB2Y7EK6O\",\"WARC-Block-Digest\":\"sha1:FTDYIIQX63YXKD6SK4FHK54R65PGFXMD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571090.80_warc_CC-MAIN-20220809215803-20220810005803-00331.warc.gz\"}"}
https://xiaokangstudynotes.com/2018/06/	[ "# 《自私的基因》书评\n\n1. 基因是进化的基本单位，所谓基因是控制某个生物形状或特性的染色体片段。\n个体或者群体都不是进化的基本单位。\n2. 基因进化的竞争对手是控制同一个性状或特性的等位基因。\n3. 基因的进化的目标是为了提高自己在整个等位基因库中的比例。\n4. 基因进化的一切目标都是以此为出发点，是完全利己的。\n5. 基因进化主要通过控制个体的形状或特性来实现。\n6. 不同的形状或者遗传特性最终会为个体带来不同的繁殖机会，从而影响控制这一特性的基因在后代中的比例。\n7. 基因和等位基因之间的竞争会最终达到一个稳定状态，各个等位基因在基因库中所占的比例保持稳定。\n8. 带来这个稳定状态的策略是进化稳定策略（evolution stable strategy），稳定策略是不同策略博弈的产物。\n9. 基因的竞争并不是单一的，而是相互影响的，最后可能造成不同基因相互配合的效果。\n10. 生物的行为包括利他的行为都可以用基因的自私性来解释。\n\n# Python Notes: Global Interpreter Lock\n\n### Why Python use GIL\n\nPython uses reference count for memory management. Each objects in python has a reference count variables that keep track of the number of reference to the object. When the reference count goes back to 0, Python would release this object from memory.\n\nHowever, in multi threading scenario, multiple thread might access the same object, and object reference count could be changed incorrectly in race conditions. Then objects that should be released could still stay in memory and worst case, objects that should not be release are incorrectly released.\n\nTo solve this problem, python introduced GIL, which is a global lock in python interpreter level. The rules is that, any python code has to acquire this lock to be executed. You might ask why not add one lock to each objects? This could result in deadlock.\n\nIn this way, python code guarantees that only one thread would be able to change the object reference count.\n\n### Problem of GIL\n\nThe GIL solution, however has problems in that Python code would not be able to utilize multi CPUs. If your code is CPU intensive, python multi-thread would not help you at all. However, if you program is not CPU intensive, but I/O intensive, for example, network application, Python thread is still a good choice.\n\n### Solution to this problem?\n\nAre there solutions to the problem? Yes, there are. Python community has tries many times to solve this problem. Python GIL is not really tie to python language it self, it ties to Python interpreter it self. So, as long as we change the underlying python interpreter, python could support multithread. For example, Jython, is implemented in Java.\n\nHowever, another important reason why Python GIL is not removed is that python has many extended libraries that are writing in C. Those libraries works well with Python in that they don’t need to worry about multi-thread models, the GIL model is really easy to integrate. Moving those libraries to other interpreters are hard works.\n\nAnother solution is to use `multiprocess` instead of `multithread`. Python has good designed libraries that supports multiprocess. However, process management would have more overhead than thread management for operating system, which means the performance of multiprocess programs are worse than multithreads.\n\n# Python Notes: Decorator\n\nBy definition, a decorator is a function that takes another function and extends the behavior of the latter function without explicitly modifying it. Built-in decorators such as @staticmethod, @classmethod, and @property work in the same way.\n\nHow decorator works? Let's take a look at the following example:\n\n``````def my_decorator(some_func):\ndef wrapper():\nsome_func()\nprint(\"Some function is being called.\")\n\nreturn wrapper\n\ndef another_func():\nprint(\"Another function is being called.\")\n\nfoo = my_decorator(another_func)\n\nfoo()\n``````\n\nYou would see the following print on the screen:\n\n``````\"Another function is being called.\"\n\"Some function is being called.\"\n``````\n\nAs you can see, we pass `some_func` into a closure, and do something before or after calling this function without modifying its original behavior, and we `return the function`. As we already learned, python functions are just like other python objects, they are first class objects. The returned function could be called just as any other functions.\n\n### @decorator\n\nThe above example is already very similar to decorator. The difference is that decorator often comes with a `@` symbol. This is an example of python syntax sugar, which often refers to syntax in a programming language that aims to make the things easy to read or to express. For example, the following is an example of a decorator:\n\n``````@my_decorator\ndef another_func():\nprint(\"Another function is being called.\")\n``````\n\n### Decorator that takes any argument\n\nIn python, we can use `args` and `kargs` to represent arbitrary arguments. The following example shows how to take arbitrary arguments in a decorator:\n\n``````def proxy(func):\ndef wrapper(args, *kargs):\nreturn func(args, *kargs)\nreturn wrapper\n``````\n\n### Decorator with parameters\n\nSometimes we want the decorator to take parameters, for example, we should implement it in this way:\n\n``````def decorator(argument):\ndef real_decorator(function):\ndef wrapper(args, *kwargs):\ndo_something_with_argument(argument)\nresult = function(args, *kwargs)\nreturn wrapper\nreturn real_decorator\n``````\n\n### Decorator tips\n\nOne practical tips when defining decorator is to use the `functoolss.wraps`, this function would keep all the meta data information of the original functions, including the function signature and docstring information.\n\n``````import functools\n\ndef uppercase(func):\n@functools.wraps(func)\ndef warpper():\nreturn func()\nreturn wrapper\n``````\n\n# Python Notes: function as first class object\n\nPer history of python blog, everything in python are first class objects, that means `all objects that could be named in the language (e.g., integers, strings, functions, classes, modules, methods, etc.) to have equal status. Th:at is, they can be assigned to variables, placed in lists, stored in dictionaries, passed as arguments, and so forth.`.\n\nEssentially, functions return a value based on the given arguments. In Python, functions are first-class objects as well. This means that functions can be passed around, and used as arguments, just like any other value (e.g, string, int, float).\n\nInternally, python use a common C data structure that are used everywhere in the interpreter to represent all objects, either it is a python function or a integer.\n\nHowever, when it comes to python function as first class, there are subtle things to think about when doing design.\n\nThink about the following function definition:\n\n``````class A:\ndef __init__(self, x):\nself.x = x\n\ndef foo(self, bar):\nprint self.x, bar\n``````\n\nWhat would happen if you assign `A.foo` to a variable: `b = A.foo`. The first argument of the function would have to be the instance itself. To handle this problem, python 2 returns a `unbound method`, which is a warper around the original function, but it restrict that the first argument of the function has to be the object instance: `a = A(), b(a)`. In python 3, however, this restriction is removed as the author found this is not very useful.\n\nLet’s think about the second condition, when you have a instance of a class: `a = A(1), b = a.foo`. In this case, python would return a `bound method` which is a thin wrapper around the original function. Bound method stores the instance as a internal object and this object would be the default first argument when calling this function.\n\n# 马尔萨斯和《人口原理》\n\n### 马尔萨斯陷阱\n\n`马尔萨斯人口陷阱(Malthusian Trap)`是说：\n\n• 在没有限制的情况下，人口会呈现指数型增长\n• 食物只会呈现现行增长\n• 人口的增长一定会超过食物增长，从而导致食物不足\n\n• 有意识的晚婚晚育。\n• 缩减人类寿命的时间，比如战争，瘟疫，饥荒等。\n\n# Python Notes: Iterator, Generator and Co-routine\n\n### Iteration\n\nPython support iteration, for example, iterating over a list:\n\n``````for elem in [1, 2, 3]:\nprint elem\n``````\n\nIterating over a dict:\n\n``````for key in {'Google': 'G',\n'Yahoo': 'Y',\n'Microsoft': 'M'}:\nprint key\n``````\n\nIterating over a file:\n\n``````with open('path/to/file.csv', 'r') as file:\nfor line in file:\nprint line\n``````\n\nWe use iterable objects in many ways, for example, reductions: sum(s), min(s), constructors: list(s), in operators: item in s.\n\nThe reason why we can iterate over iterable is because of iterable protocols: any objects that supports `iter()` and `next()` is an itterable. For example, we can define one itterable object in the following way:\n\n``````class count:\n\ndef __init__(self, start):\nself.count = start\n\ndef __iter__(self):\nreturn self\n\nDef next(self):\nif self.count < 0:\nraise StopIteration\nr = self.count\nself.count -=1\nreturn r\n``````\n\nWe can use the above example in this way:\n\n``````c = count(5)\nfor i in c:\nprint I\n# 5, 4, 3, 2, 1\n``````\n\n### Generator\n\nSo what is a generator? By definition: a generator is a `function` that produces `a sequence of results` instead of a single value.\n\nSo generator is a function, it is different from other functions that it generates a sequence of results instead of a single value. Generator function is very different from normal function, calling the generator function will create one generator, but would not execute it, until `next()` is called. The following is an example of generator:\n\n``````def count(n):\nwhile n > 0:\nyield n\nn -= 1\n\nc = count(5)\n``````\n\nNote that when we first initiate count, it won't execute. Until the first time we call, `c.next()` the generator would start to execute. But it will suspend on the `yield` command, until next time it executes.\n\nSo to speak, a generator is a convenient way of writing an iterator, and you don't have to worry about iterator protocols.\n\nExcept for `yield` based generator function, python also supports generator expression:\n\n``````a = [1, 2, 3, 4]\nb = [x2 for x in a]\nc = (x*2 for x in a)\n``````\n\n`b` is still a regular list, while c is a generator.\n\n### Co-routine\n\nPython coroutine is very similar to generator. Think about the following pattern:\n\n``````def receive_count():\ntry:\nwhile True:\nn = (yield) # Yield expression\nprint \"T-minues \", n\nexcept GeneratorExit:\nprint \"Exit from generator.\"\n``````\n\nThe above form of generator is called coroutine. Coroutine is different from generator in that it receives data instead of generates data. Think of it as a `consumer` or `receiver`.\n\nTo use python co-routine, you need to call `next()` first so that the function executes to the `yield` field part, then you can use send to send the value to the function. For example:\n\n`````` c = receive_count()\nc.next() # trigger to yield function\nc.send(1) # sending 1 to the co-routine.\n# prints \"T-minus 1\"\n``````\n\nPython provided a decorator called `@consumer` to execute the next() function part. With the `consumer` decorator, the co-routine can be used directly.\n\nThen the question is: why don't we just declare co-routine as a regular function where you can send the value to it directly instead of relying on the yield expression? Using coroutine in the given examples doesn't fully justify it's value. More often, people use co-routine to implement a application level multiple threading. I will introduce more about this later.\n\n# Python Notes: Context management\n\nPython supports context management. Which often used when handling resources, for example, file, network connection. With statement helps make sure the resources are cleaned up or released. There are two major functions for context management: `__enter__` and `__exit__`.\n\n`__enter__` is triggered when the `with` statement is first triggered, while `__exit__` statement is triggered when the statement finishes execution.\n\nOne very common usage of with statement is when we open up files:\n\n`````` with open('/path/to/file', 'r') as file:\nfor line in file():\nprint(line)\n\n``````\n\nThe `with` statement on this example will automatically close the file descriptor no matter how this with block exits.\n\nPython with statement also supports nesting, for example:\n\n`````` with open('/open/to/file', 'r') as infile:\nwith open('write/to/file', 'w') as outfile:\nfor line in infile:\nif line.startswith('Test'):\noutfile.write(line)\n``````\n\nIf you want your code to support with statement based context management, just override the two functions on your code, for example:\n\n``````class Resource:\ndef __init__(self, res):\nself.res = res\n\ndef __enter__(self):\n\ndef __exit__(self, exc_type, exc_value, traceback):\n\n``````\n\nThere is another way to support context management, which is to use the `contextlib.contextmanager`, we will introduce this later.\n\n# Python Notes: Closure\n\nThe following is an example of python closure.\n\n``````def outer_function(outter_arg):\nclosure_var = outter_arg\ndef inner_function(inner_arg):\nprint(\"{} {}\".format(closure_var, inner_arg))\n\nreturn inner_function\n\n# Usage of a python closure\nclosure_func = outer_function(\"X\")\nclosure_func(\"Y1\") # print \"Y1 X\"\nclosure_funct(\"Y2\") # output \"Y2 X\"\n\n``````\n\n### Variables and nested function\n\nPython has two types of variables: local variable and global variable. Variables defined within a function has a local scope, while variables defined outside a function has global scope.\n\nWhen a function is defined within another function, the function is called nested function. The nested function, however, can access the outer function’s variables. In the example above, outer function defined a variable called closure_var, and this variable would be initiated by the input variable. The inner function is able to access this variable and use this variable in its own function definition.\n\nThere are two steps of using a closure function: initiate the closure function and assign to a local variable, then call the variable with parameter to invoke the inner function.\n\n### When to use closure?\n\n• Reduce the use of global variables\n\nClosure could hide variables inside the function, in this way we reduce the use of global variables.\n\n• Simplify single function classes.\nFor example, the above example could be converted into a single function class:\n``````class OuterClass:\ndef __init__(outer_arg):\nself.arg = outer_arg\n\ndef inner_function(self, inner_arg):\nprint(\"{} {}\".format(self.arg, inner_arg))\n``````\n\nIn general, closure runs faster than instance function calls." ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.98567766,"math_prob":0.83234084,"size":2274,"snap":"2020-34-2020-40","text_gpt3_token_len":2545,"char_repetition_ratio":0.047577094,"word_repetition_ratio":0.0,"special_character_ratio":0.097185574,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95608133,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T05:11:47Z\",\"WARC-Record-ID\":\"<urn:uuid:5d2a69ed-c3bd-4e02-bd48-c681e9299f47>\",\"Content-Length\":\"95926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1b68a4d-8d19-42f5-a35a-e8eaab1dee76>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c3bfadd-b66a-48e2-b52f-54afa153bbc8>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://xiaokangstudynotes.com/2018/06/\",\"WARC-Payload-Digest\":\"sha1:LAJAX5FT2SSPBIAHSMCNUQFWDXMN6PEN\",\"WARC-Block-Digest\":\"sha1:O52VS3ZGBMGRZAEDFCF7BU3EQC5P37DU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401585213.82_warc_CC-MAIN-20200928041630-20200928071630-00037.warc.gz\"}"}
http://www.knowledgeadda.com/2010/03/	[ "## VLSI Design - 7th ND09 EC1401\n\nB.E/B.Tech Degree Examination, November/December 2009.\nSeventh Semester - 7th\nElectronics and Communication Engineering\nEC1401 - VLSI DESIGN\n(Common to B.E.(Part-Time) Sixth Semester Regulation 2005)\n(Regulation 2004)\n\nProgramming Questions \| Travel Tips \| Mobile Review \| Placement Papers\n\nFor More Question paper on ECECLICK HERE\n\nPart A-(102=20 marks)\n\n1.What are the different MOS layers?\n2.What are the two types of layout design rules?\n3.Define rise time and fall time.\n4.What is a pull down device?\n5.What are the difference between task and function?\n6.What is the difference between === and == ?\n7.What is CBIC ?\n8.Draw an assert high switch condition if input = 0 and input =1.\n9.What do you mean by DFT?\n10.Draw the boundary scan input logic diagram.\n\nPart B - (516=80 marks)\n\n11.a) Discuss the steps involved in IC fabrication process.(16)\nOr\nb) Describe n-well process in detail.(16)\n\n12.a)i)Explain the DC characteristics of CMOS inverter with neat sketch.(8)\nii)Explain channel length modulation and body effect.(8)\nOr\nb)i)Explain the different regions of operation in a MOS transistor.(10)\nii)Write a note on MOS models.(6)\n\n13.a)Explain in detail any five operators used in HDL .(16)\nOr\nb)i)Write the verilog code for 4 bit ripple carry full adder.(10)\nii)Give the structural description for priority encoder using verilog.(6)\n\n14.a)Explain in detail the sequence of steps to design an ASIC.(16)\nOr\nb)Describe in detail the chip with programmable logic structures.(16)\n\n15.a)Explain in detail Scan Based Test Techniques.(16)\nOr\nb)Discuss the three main design strategies for testability.(16)\n\n## Electronic Circuits I 3rd ND09 EC9202\n\nANNA UNIVERSITY – CHENNAI\nB.E.\\B.TECH DEGREE EXAMINATION DECEMBER 2009\n3rd - THIRD SEMESTER\nELECTRONICS and COMMUNICATIONS ENGINEERING.\nEC9202 ELECTRONIC CIRCUITS - I\n\nProgramming Questions \| Travel Tips \| Mobile Review \| Placement Papers\n\nFor More Question paper on ECECLICK HERE\n\nPART A-(210=20)\n\n1. Define stability factor.\n2. Draw the fixed bias single stage transistor circuit.\n3. Define CMRR.\n4. Draw the small signal equivalent circuit of FET.\n5. Two amplifiers having gain 20db and 40db are cascaded. Find the overall gain in db.\n6. Define bandwidth.\n7. What is theoretical maximum conversion efficiency of class A power amplifier.\n8. What is distortion in power amplifiers.\n9. Draw the full wave bridge rectifier circuit.\n10. What are the advantages of SMPS over conventional regulators.\n\nPART B-(516=80)\n\n11. A) (i) For the transistor circuit in fig. find the Q-point. Vcc=15v, B=100, VBE=0.7V\n(ii) Calculate the stability factor for a fixed bias circuit.\n(Or)\nB) Discuss the various techniques of stabilization of Q-point in a transistor.\n\n12. A) For the CC transistor amplifier circuit, find the expressions for input impedance and voltage gain. Assume suitable model for transistor.\n(Or)\nB) (i) Discuss the working of a basic emitter coupled differential amplifier circuit (8)\n(ii) Compare CB, CE and CC amplifiers. (8)\n\n13. A) (i) Discuss the frequency response of multistage amplifiers. Calculate the overall upper and lower cutoff frequencies. (10)\n(ii) Discuss the terms rise time and sag. (6) (Or)\nB) Discuss the high frequency equivalent circuit of FET and hence derive gain bandwidth product for any one configuration.\n\n14. A) (i) Derive the theoretical max conversion efficiency of class B power amplifier. (10)\n(ii) Write short notes on power MOSFET amplifier. (6) (Or)\nB) Describe the distortion in power amplifier and the methods to eliminate the same.\n\n15. A) Explain the circuit of voltage regulator and also discuss the short circuit protection mechanism. (Or)\nB) (i) Explain the power control method using SCR.\n(ii) Design zener regulator for following specification Vin=8v to 12v; Vo=10v, RL=10kO. Assume that zener diode is ideal.\n\n## Control System - ND09 5th EC9254\n\nANNA UNIVERSITY – COIMBATORE\nB.E.\\B.TECH DEGREE EXAMINATION DECEMBER 2009\nFIFTH SEMESTER - 5th\nELECTRONICS and COMMUNICATION ENGG.\nCONTROL SYSTEMS - EC9254\n\nProgramming Questions \| Travel Tips \| Mobile Review \| Placement Papers\n\nFor More Question paper on ECECLICK HERE\n\nPART A-(220=40)\n\n1. What is the use of mason’s gain formula.\n2. Give two examples for open loop and closed loop systems.\n3. What do you meant by analogous systems.\n4. What are the frequency domain specifications.\n5. What are the advantages and disadvantages of feedback control systems.\n6. How can we classify second order system based on damping ratio.\n7. Define gain margin.\n8. Define resonant frequency.\n9. State Routh stability criterion.\n10. Define Nyquist stability criterion.\n11. What are the difference between state space analysis and Transfer function analysis.\n12. What are root loci.\n13. What is servomechanism.\n14. What are the different types of controller.\n15. What is synchro.\n16. Name the test signals used in control systems.\n17. What is polar plot.\n18. Give examples for frequency response plot.\n19. Write down state mode and output model of state space systems.\n20. What do you mean by decomposition of transfer function.\n\nPART B-(512=60)\n\n21. A) A unity feedback system has an open loop transfer function G(s)=k/s(s+10). If the damping ratio is 0.5 determine (i) the value of k, (ii) peak overshoot, (iii) time to peak overshoot, (iv) settling time. (8) B) For a unity feedback whose G(s)=1/s(s+1) the input signal is r(t)= 4+6t+2t3. Find the generalized error coefficients. (4)\n\n22. For the given block diagram find corresponding signal flow graph and evaluate closed loop transfer function relating to output and input.\n\n23. Determine the value of k for a unity feedback control system have open loop transfer function G(s)H(s)=k/s(s+2)(s+4) such that (i)gain margin =20db (ii) phase margin =600.\n\n24. Investigate the F(s) for stability using RH criterion (i) F(s) =s4+ks2+(k+1)s+2 (ii) F(s)=s4+s3+3s2+s+6 (iii) F(s)=s5+s4+2s3+2s2+6s+6\n\n25. Obtain the state space model of the system with transfer function C(s)=s2+3s+2 in phase Variable form. R(s) s2+7s+12\n\n26. A system characterized by the following state equation\nFind (i) Transfer function of the system (ii) State transition matrix.\n\n27. Sketch the root locus for the system with characteristic equation 1+G(s)H(s)=K(s+2)(s+3)(s+1)(s-1)\n\n28. Find transfer function of field controlled DC servo motor.\n\n## Digital Signal Processing - ND09 5th EC9304\n\nANNA UNIVERSITY – COIMBATORE\nB.E.\\B.TECH DEGREE EXAMINATION - DECEMBER 2009\nFIFTH SEMESTER - 5th\nELECTRONICS and COMMUNICATION ENGG.\nDIGITAL SIGNAL PROCESSING - EC9304\n\nProgramming Questions \| Travel Tips \| Mobile Review \| Placement Papers\n\nFor More Question paper on ECECLICK HERE\n\nPART A-(220=40)\n\n1. Define twiddle factor of FFT.\n2. The first five DFT coefficients of a sequence x(n) are X(0)=20, X(1)=5+j2, X(2)=0, X(3)=0.2+j0.4, X(4)=0. Determine the remaining DFT coefficients.\n3. Calculate the number of multiplications needed in the calculation of DFT and FFT with 64-point sequence.\n4. What is zero padding. What are its uses.\n5. What are the desirable and undesirable features of FIR filters.\n6. What is the necessary and sufficient condition for linear phase characteristics in FIR filter.\n7. Define Gibb’s phenomenon.\n8. Draw the direct form I structure for the second order system function\nH(z)= b0+b1z-1+b2z-2\n1+a1z-1+a2z-2\n9. Find the digital filter transfer function H(z) by using impulse invariant method.\nfor the analog transfer function H(s)=1/(S+2).\n10. Give any two properties of Butterworth filter.\n11. What is warping effect. What is the effect on magnitude and phase responses.\n12. Mention any two procedures for digitizing the transfer function of an analog filter.\n13. What are the quantization errors due to finite word length registers in digital filters.\n14. What is overflow oscillations. What are the methods to prevent it.\n15. Define a Deadband of a filter.\n16. The filter coefficient H=-0.673 is represented by sign magnitude fixed point arithmetic. If the word length is 6 bits, compute the quantization error due to truncation.\n17. What are the addressing modes of TMS320C50.\n18. What is the advantage of Harvard architecture of TMS320 series.\n19. What is the different buses of TMS320C50.\n20. What are the arithmetic instructions of C50.\n\nPART B-(512=60)\n\n21. A) Find the 8 point DFT of the given sequence x(n)={0,1,2,3,4,5,6,7} using Dif radix-2 FFT algorithm.(8)\nB) Perform the circular convolution of the following two sequences using matrix method. X1(n)={2,1,2,1} , X2(n)={1,2,3,4} (4)\n\n22. A) Design an idea Hilbert transformer having frequency response\nH(ej?)=j for -p=?=0.\n= - j for 0=?=p. Using rectangular window for N=11. (8)\nB) Realize the following system function using minimum number of multipliers (4)\nH(z)=1+1z-1+1z-2+1z-3+1z-4+z-5\n3 4 4 3\n\n23. Design a digital Butter-worth filter satisfying the constraints.\n0.707=¦H(ej?)¦= for 0=?= p/2.\n¦H(ej?)¦=0.2 for 3p/4=?= p.\nWith T=1 sec using bi-linear transformation.\n\n24. A) The input to the system y(n)=0.999y(n-1)+x(n) is applied to an ADC. What is the power produced by the quantization noise at the output of the filter if the input is quantized to (i) 8 bits (ii) 16 bits. (8)\nB) Compare fixed point and floating point arithmetic. (4)\n\n25. A) With a neat block diagram explain in detail the architecture of TMS320C50. (8)\nB) Write short notes on pipelining. (4)\n\n26. Design a high pass filter with Hamming window with a cut-off frequency of 1.2 radians/sec and N=9.\n\n27. Consider the transfer function H(z)=H1(z) where H1(z)=1/(1-a1z-1) and H2(z)=1/(1-a2z-1). Assume a1=0.5 and a2=0.6 and find the output round off noise power.\n\n28. A) Calculate the DFT of the sequence X(n) = {1,1,-2,-2}. (6)\nB) Find the output y(n) of a filter whose impulse response is h(n)= {1,1,1} and input signal x(n)={3,-1,0,1,3,2,0,1,2,1} using overlap save method.\n\n## Principles Of Management - ND09 5th MG1351\n\nANNA UNIVERSITY – COIMBATORE\nB.E\\B.TECH DEGREE EXAMINATION DECEMBER 2009\nFIFTH SEMESTER - 5th\nELECTRONICS and COMMUNICATION ENGG.\nPRINCIPLES OF MANAGEMENT - MG1351\n\nFor More Question paper on ECECLICK HERE\n\nPART A-(220=40)\n\n1. What are the streams related to Neo-classical theory.\n2. What is a system.\n3. Who are all the major contributors to contingency approach of management.\n4. What are the functions of management.\n5. What are two views on strategy.\n6. What are various types of plans based on time.\n7. Define policy.\n8. What is meant by good alternative.\n9. What is meant by formal organization.\n10. What are the benefits of staff authority.\n11. What is meant by delegation of authority.\n12. Define staffing.\n14. Define creativity.\n15. What is meant by job enrichment.\n16. What are the barriers of communication.\n17. State few control techniques.\n18. What are the types of managerial control.\n19. Define productivity.\n20. Define budget.\n\nPART B-(512=60)\n\n21. A) Explain the challenges that are faced by the management. (4)\nB) Describe Henry Fayol’s principles of Management. (8)\n\n22. A) Explain the various steps involved in planning. (8)\nB) Explain the elements of most effective MBO systems.\n\n23. A) Explain the common problems and difficulties faced in decision making process. (6)\nB) Describe the four approaches of selection process. (6)\n\n24. A) Describe the basis for departmentation. (8)\nB) Explain the significance of selection process (4)\n\n25. A) Explain the significance of staffing function. (6)\nB) Describe various style of leadership. (6)\n\n26. A) Explain the role of Non-Financial incentives in Motivation (8)\nB) What are the advantages of job enrichment.\n\n27. A) What are the characteristics features of controlling. Explain. (4)\nB) Explain the process f communication.\n\n28. A) Explain the essential elements of control system (4)\nB) Describe the essential requirements for good control system. (8).\n\n## High Speed Networks - ND09 7th EC1008\n\nB.E/B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2009\nSeventh Semester - 7th\n(Regulation 2004)\nElectronics and Communication Engineering\nEC 1008 – HIGH SPEED NETWORKS\n(Common to B.E (Part Time) Sixth Semester – ECE – Regulation 2005)\n\nFor More Question paper on ECECLICK HERE\n\nPART A – (10 * 2 = 20 marks)\n\n1.What does the term 'asynchronous' indicate ATM networks?\n2.Mention the encoding techniques followed in Fast Ethernet.\n3.Mention any four important parameters that are associated with a single server queue.\n4.Write short notes on Committed Information Rate in Traffic Rate Management.\n5.Write down the expression by which the RTT is computed in the original TCP implementation (i.e., without any refinements).\n6.What is meant by silly window syndrome?\n7.Define Quality of Service in a data network.\n8.What is meant by Active Queue Management Scheme?\n9.Is RTP an application layer protocol or transport layer protocol? Justify your answer.\n10.What is meant by Forward Equivalence Class in MPLS networks?\n\nPART B - ( 5 * 16 = 80 marks )\n\n11. (a) (i) Explain the various cabling schemes and the encoding scheme followed in Gigabyte Ethernet. (8)\n(ii) Discuss the relevance of CSMA/CD in Gigabyte Ethernet. (8)\nOR\n(b) (i) Explain the action that take place in the receiver upon receiving the HEC field in ATM networks. (8)\n(ii) Explain how the GFC is used to provide the flow control at the UNI of ATM networks. (8)\n\n12.(a) (i) An engineering firm provides each of its analyst with a personal computer, all of which are hooked up over a LAN to a database server. In addition, there is an expensive, standalone graphics workstation that is used for special purpose design tasks. During the course of a typical 8-hour day, 10 engineers will make the use of the workstation and spend an average of 30 minutes at a session. Manager is satisfied with this arrangement since the utilization factor of the work station is only 5 hours out of 8. The engineers complain that the wait time for using the work station is long, often an hour or more, and are asking for more workstations. Explain the queuing analysis that should be done by the engineers to convince the manager.(10)\n(ii) Explain the choke packet congestion control mechanism. (6)\nOR\n(b) (i) Consider a LAN with 100 personal computers and a server that matintains a common database for a query application. The average time for the server to respond to a query is 0.6 seconds, and the standard deviation is estimated to equal the mean. At peak times, the query rate over the LAN reaches 20 queries per minute. Answer the following questions.\n(1) What is the average response time ignoring line overhead? (2)\n(2) If a 1.5 second response time is considered the maximum acceptable, what percent growth in message load can occur before the maximum is reached? (3)\n(3) If 20% more utilization is experienced, will response time increase by more or less than 20%? (3)\n(ii) Discuss the issues of fairness. Quality of Service and reservation in traffic management. (8)\n\n13. (a) (i) Explain the various congestion control mechanisms followed in ATM networks. (8)\n(ii) Discuss the ABR and GFR service categories in ATM networks. (8)\nOR\n(b) (i) Explain the Additive Increase Multiplicative Decrease behavior of TCP congestion control algorithm.(8)\n(ii) Explain the basic TCP flow control mechanism in detail. (8)\n\n14. (a) (i) Discuss the motivation and design goals of Random Early Detection scheduling policy.\n(ii) Explain the Random Early Detection algorithm in terms of the estimation of queue size and determining packet discard.\nOR\n(b) (i) Describe the functional modules that should be present in ingress and egress routers of Differentiated Services Networks. (8)\n(ii) Describe the components of a core router of Differentiated Services Network. (8)\n\n15. (a) (i) Explain the soft state based receiver initiated reservation in RSVP. (8)\n(ii) Discuss the flow specifications and filter specifications in RSVP. (8)\nOR\n(b) (i) Discuss the labeling and label switched path of MPLS networks. (8)\n(ii) Explain the stacking rules followed in the label switched path of MPLS networks.(8)\n\n## Object Oriented Analysis and Design ND09 7th CS1402\n\nB.E/B.Tech.DEGREE EXAMINATION, NOVEMBER/DECEMBER 2009\nSeventh Semester\nComputer Science and Engineering\nCS1402-OBJECT ORIENTED ANALYSIS AND DESIGN\n(Common to B.E (Part-Time) Sixth Semester Regulation 2005)\n(Regulation 2004)\n\nFor More Question paper on CSECLICK HERE\n\nPART A- (10 x 2 =20 marks)\n\n1. Give the characteristics of object oriented system.\n2. What is an object? Give an example.\n3. Give a note on patterns and its necessity.\n4. Mention the models in Object Modelling Techniques in Rambaugh methodology and its role for describing the system.\n5. List out the steps for finding the attributes of a class?\n6. Give the hint to identify the attributes of a class?\n7. Define axiom along with its types.\n8. For the schema employee (emp-id, emp-name, street, city) give the class representation along with the attribute types.\n9. Mention the purpose of view layer interface.\n10. What are client/server computing? Give two applications which work on this basis?\n\nPART B- (5 X 16 = 80 marks)\n\n11. (a) Explain and develop the payroll system using the steps of Object oriented approach. [ Marks 16 ]\nOr\n(b) Explain the following\n(i) Object Modeling Technique [ Marks 8 ]\n(ii) Compare Aggregation and Composition with a suitable example. [ Marks 8]\n\n12. (a) Explain the relationships that are possible among the classes in the UML representation with your example. [Marks 16 ]\nOr\n(b) What are the various diagrams that are used in analysis and design steps Of Booch Methodology? Explain with your own example. [Marks 16 ]\n\n13. (a) Explain the method of identifying the classes using the common class approach with an example. [ Marks 16 ]\nOr\n(b) Consider the Hospital Management System application with the Following requirements\n• System should handle the in-patient, out-patient information through receptionist.\n• Doctors are allowed to view the patient history and give their prescription.\n• There should be a information system to provide the required information.\nGive the use case, class and object diagrams. [ Marks 4+8+4 ]\n\n14.(a) With a suitable example explain how to design a class. Give all possible representation in a class (name, attribute, visibility, methods, responsibilities)[Marks 16 ]\nOR\n(b) Design the access layer for the Students information management which includes personal, fees and mark details. [ Marks 16 ]\n\n15.(a) (i) Explain the various testing strategies. [ Marks 12 ]\n(ii) Give the use cases that can be used to generate the test cases for the Bank ATM application. [ Marks 4]\nor\n(b) (i) How will you measure the user satisfaction? Describe. [ Marks 6 ]\n(ii) Perform the satisfaction test for any client/server application. [ Marks 10]\n\n## Analog and Digital communications ND09 3rd CS2204\n\nB.E/B.Tech EXAMINATIONS,NOVEMBER/DECEMBER 2009\nTHIRD SEMESTER\nCOMPUTER SCIENCE AND ENGINEERING\nCS 2204-ANALOG AND DIGITAL COMMUNICATIONS\n(REGULATIONS 2008)\n\nFor More Question paper on CSECLICK HERE\n\nPART A-(10X2=20 marks)\n\n1.A carrier wave is represented by equation s(t)=12sinwt.Draw the wave form of an AM wave for depth of modulation of 0.5.\n2.Compare FM with AM\n3.What is coherent detection?\n4.Why is binary ASK called on-off keying?\n5.What are the errors in DM?\n6.Define companding\n7. What are the two methods of error detection and correction?\n8.What do you mean by signaling rate?\n9.Define processing gain\n10.What is CDMA?\n\nPART B-(5X16=80 marks)\n\n11.(a)(i)The output voltage of a transmitter is given by 500(1+0.4sin3140t)cos6.28X10^7t\nThis voltage is fed to a load of 600 ohms.Determine\n1.Carrier frequency\n2.Modulating frequency\n3.Carrier power (9)\n(ii)Explain i detail about super heterodyne receiver(7)\nOr\n(b)(i)Carrier frequency modulated with a sinusoidal signal of 2kHz resulting in a maximum frequency deviation of 5kHz.Find\n1.Modulating index\n2.Bandwidth of modulated signal (4)\n(ii)Explain the method of generating FM signal using indirect method(12)\n\n12.(a)(i)Explain the coherent binary FSK system with a neat diagram of transmitter and receiver(12)\n(ii)Enumerate the advantages and disadvantages of FSK over PSK system(4)\nOr\n(b)(i)Explain the generation and detection of coherent QPSK system in detail(12)\n(ii)What is DPSK?Explain its bandwidth requirements.\n\n13.(a)(i)Explain in detail about DPCM with suitable diagram(10)\n(ii)1kHz signal is sampled by 8kHz sampling signal and the samples are encoded with 12 bit PCM system.Find\n1.Required bandwidth for PCM system\n2.Total number of bits in the digital output signal in 10 cycles(6)\nOr\n(b)(i)Write short notes on ISI(6)\n(ii)What is eye pattern?Explain how is the performance of a base-band pulse transmission system measured with this?(10)\n\n14(a)(i)Write short notes on error correcting codes\n(ii)Find the generator polynomial of (7,4) cyclic code and find the codeword for the message 1001(12)\nOr\n(b)Explain in detail about the serial interface with its control signals and timing information(16)\n\n15(a)(i)What are pseudo noise sequences?How are they generated?(6)\n(ii)Explain direct sequence spread spectrum system in detail(10)\nOr\n(b)(i)Explain 2 types of FH spread spectrum systems with suitable diagrams(16)\n\n## Fundamental of computing and programming JF10 GE2112\n\nB.E./B.TECH. DEGREE EXAMINATIONS, JANUARY 2010\nREGULATIONS 2008\nFIRST SEMESTER\nCOMMON TO ALL BRANCHES\nGE2112 FUNDAMENTAL OF COMPUTING AND PROGRAMMING\n\nPART A-(102=20MARKS)\n\n1.Distinguish between Analog and Digital computer.\n2.Convert 0.4375 decimal to binary system.\n3.Distinguish between compiler and interpreters.\n4.What is web server?\n5.What is an algorithm?\n6.Give the importance of a graphic packages.\n7.Write a code segment using while statement to print numbers from 10 down to 1.\n8.Write a c program for the following expressions.\n(a)a=5 <= 8 && 6!=5\n(b)a = b++ + ++b where b = 50\n9.How strings are represented in c language?\n10.What are the advantages of unions over structures?\n\nPART B-(516=80 MARKS)\n\n11.(a)(i)What are the characteristic of a computer? Discuss\n(ii)Briefly explain the various generations of computers.\n(iii)Convert the decimal number 59.8125 into binary and octal.\n(OR)\n11(b)Explain the different components of a computer system with block diagram.\n\n12.(a)(i)Describe the different types of software with examples.\n(ii)List the different software development steps and explain.\n(OR)\n12.(b)(i)Explain the common types of internet access.\n(ii)Write short notes on web browser.\n(iii)Explain a typical structure URL.\n\n13. (a)Draw and explain the various symbols of flowchart and also draw the flowchar to add an array\nof N elements\n(OR)\n13.(b)(i)Explain the features of Power Point package.\n(ii)List and explain the features supported by spreadsheet package.\n(iii)Briefly write about Desktop Publishing Software.\n\n14.(a)(i)What are the different operators available in C? Explain with examples.\n(ii)Differentiate between signed and unsigned integer.\n(OR)\n14.(b)(i)Explain the following conditional statements.\n1.nested if-else statement\n2.switch-case statement\n(ii)Write a C program that reads a number and display whether the number is prime or not.\n\n15.(a)(i)Write a C program to reverse a given string.\n(ii)Differentiate pass by value and pass by address in c.\n(OR)\n15.(b)Write a C program that gets and display the report of n students with their personal\nand academic details using structures.\n\n## Engineering Physics I - 1st JF10 PH2111\n\nB.E./B.TECH. DEGREE EXAMINATION, JANUARY 2010\nFIRST SEMESTER\nPH 2111-ENGINEERING PHYSICS-I\n(REGULATIONS 2008)\n\nPART A-(102=20MARKS)\n\n1.What is cavitation?\n2.What is sonogram?\n3.What are different methods of achieving population inversion?\n4.Distinguish between homo junction and hetero-junction semiconductor lasers.\n5.A signal of 100 mW is injected into fiber. The out coming signal from the other end is 40 mW.\nFind the loss in dB?\n6.What is meant by splicing in fiber optics?\n7.Calculate the equivalent wavelength of electrons moving with velocity of 310^7 m/s.\n8.What is Compton effect?Write an expression for the Compton wavelength.\n9.For a cubic crystal draw the planes with Miller indies(110) and(001).\n10.What are Frenkly Schhottky imperfections?\n\nPART-B--(516=80)MARKS\n\n11.(a)(i)Explain how ultrasonic can be produced by using magnetostriction method.(12)\n(ii)Write any four applications of ultrasonic waves (4)\n(or)\n(b)In ultrasonic NDT what are the three different scan displays in common use? Explain (10+6)\n\n12.(a)For atomic transitions, derive Einstein relations and hence deduce the expression for the ratio of spontaneous emission rate to the stimulated emission rate.\n(or)\n(b)What is holography?Describe the construction and reconstruction methods of a hologram.(4+6+6)\n\n13.(a)(i)How are fibers classified?Explain the classification in detail.(8)\n(ii)Explain double crucible method of fiber manufacturing.(8)\n(or)\n(b)(i)Explain the construction and working of displacement and temperature fiber optic sensors (5+5)\n(ii)Explain the construction and working of fiber-optic medical endoscope.(6)\n\n14.(a)(i)Write a note on black body radiation.\n(ii)Derive Planck's law radiation.\n(or)\n(b)(i)What is the principle of electron microscopy? Compare it wroth optical microscope. (4)\n(ii)With schematic diagram explain the construction and working of scanning electron microscope. (12)\n\n15.(a)(i)Explain the terms:atomic radius, coordination number and packing factor.(6)\n(ii)Show that the packing factor for Face Centered Cubic and Hexagonal Closed Packed structures are equal . (10)\n(or)\n(b)(i)What are Miller in dices?Explain. (4)\n(ii)Derive an expression for the inter planar spacing for(hkl) planes of a cubic structure.(10)\n(iii)Calculate the inter planar for the spacing for(101) plane in a simple cubic crystal whose lattice constant is 0.42nm. (2)\n\n## Principles Of Communication ND09 IT2202\n\nAnna University 2009\nB.Tech Information Technology(IT)\nIT 2202-Principles Of Communication\n3rd Semester - Question Paper\nRegulation 2008\n\nFor more question paper of IT\n\nPART-A\n\n1. Define amplitude modulation.\n2. What is modulation index?\n3. Why is ASK called as ON-OFF keying ?\n4. What is the difference between QASK and QPSK ?\n5. Define sampling rate.\n6. What is ISI ?\n7. Define Pseudo noise sequence.\n8. What are the different types of multiple access techniques ?\n9. Define Kepler's three laws of planetary motion.\n10. What are the losses encountered by optical fiber ?\n\nPART-B\n\n11. (a) (i)Write short notes on\n(1) AM voltage distribution. (4)\n(2) AM power distribution. (4)\n(ii) An audio frequency signal 10sin2p500t is used to amplitude modulate a carrier of 50 sin2p*10^5t.Calculate.\n(1) Modulation index. (2)\n(2)Side band frequencies. (2)\n(3)BW required. (2)\n(4)Total power delivered to the load of 600O. (2)\n(OR)\n(b) (i)Compare FM and AM.(12)\n(ii) The phase deviation constant in a phase modulation system is K=0.01rad/v.Calculate the maximum phase deviation when the modulating signal of 10V is applied. (4)\n\n12. (a) (i)Explain the principle of FSK transmitter and receiver (10)\n(ii)Write short note on spectrum and bandwidth of FSK. (6)\n(OR)\n(b) (i)Compare the various types of digital modulation techniques. (8)\n(ii)Explain the eye pattern in base band digital transmission with neat diagram. (8)\n\n13. (a) (i)Explain the elements of PCM system with a neat block diagram. (12)\n(ii)What is companding ? (4)\n(OR)\n(b) (i)Find the signal amplitude for minimum quantization error in a delta modulation system if step size is 1 volt having repetition period 1 ms.The information signal operates at 100Hz. (4)\n(ii) Describe the operations of DPCM system with relevant diagram. (12)\n\n14. (a) (i)Write short notes on Frequency hop spread spectrum. (10)\n(ii)Explain the applications of spread spectrum techniques. (6)\n(OR)\n(b) (i)Give a detailed account of multiple access techniques. (10)\n(ii)Compare TDMA and CDMA. (6)\n\n15. (a) (i)Discuss briefly the basic satellite communication system . (8)\n(ii)Write short notes on LEO and GEO orbits. (8)\n(OR)\n(b) Enumerate the elements of an optical fiber transmission link. (16)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87137413,"math_prob":0.93165815,"size":4235,"snap":"2019-13-2019-22","text_gpt3_token_len":979,"char_repetition_ratio":0.12219334,"word_repetition_ratio":0.014430014,"special_character_ratio":0.23754427,"punctuation_ratio":0.10152284,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9860716,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T16:28:16Z\",\"WARC-Record-ID\":\"<urn:uuid:cf5d4e02-35a0-4952-b09c-864b22f3061d>\",\"Content-Length\":\"216745\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7446304-974d-48f1-b5d6-5cea3e2f4f4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8de75b72-24e8-40f9-96e7-1b199f98a117>\",\"WARC-IP-Address\":\"172.217.164.179\",\"WARC-Target-URI\":\"http://www.knowledgeadda.com/2010/03/\",\"WARC-Payload-Digest\":\"sha1:AHAZRWI25QL5O7Z5RXF75K6AN2X3WY33\",\"WARC-Block-Digest\":\"sha1:GWL5CTZHT2PRWPGZZQIATM5TIMLA6URY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203462.50_warc_CC-MAIN-20190324145706-20190324171706-00464.warc.gz\"}"}
https://www.2outdoors.com/what-will-a-200-watt-solar-system-run/	[ "Tue. Aug 9th, 2022\n\n# What will a 200 watt solar system run\n\nMay 30, 2022\n\n### What will a 200 watt solar system run?\n\nA 200-watt solar panel system is perfect for small appliances. You can use a 200W solar panel to charge a battery to power small appliances. These appliances include coffee makers, laptops, LED lights, LCD TVs, a radio, a mini projector, and a microwave.\n\n### Is 200 watts enough for RV?\n\nHow much power does a 200 watt solar panel produce? For an average irradiance value of 4kWh/m2 a 200 watt solar panel can generate 0.8 kilowatt-hours (kWh) of energy each day, or 292kWh per annum.\n\n### How many batteries can a 200-watt solar panel charge?\n\nA 200-watt solar panel that can produce 1 amp of current takes between 5 and 8 hours to charge a 12-volt car battery fully. The solar panel must be kept oriented perpendicular to the sunlight as the solar panel position plays a vital role in the efficacy of charging and can impact a panel’s charging rate.\n\n### How long does it take a 200W solar panel to charge a 12V battery?\n\nThe short answer is that a 200-watt solar panel that generates 1 amp of current takes between 5 to 8 hours to completely charge a 12-volt car battery.\n\n### How many amp hours does a 200W solar panel produce?\n\nA 200-watt solar panel will produce 10 – 12 amps of power per hour on average. Assuming there are 6 hours of sunlight during the day, this would amount to 60 – 70 amp-hours over 24 hours.\n\n### What size solar panel do I need to charge my RV battery?\n\nBy the rule of thumb, a 100 watt solar panel inputs 30 amp-hours per day into your batteries. So you would need 1.33 100 watt panels, or one 133 watt panel to match your solar power needs.\n\n### What can a 200 watt generator run?\n\nBut how many solar panels do you need for an RV? On average, an RV will need between two and four 200 watt monocrystalline solar panels to offset its energy consumption.\n\n### Will a 200W solar panel run a fridge?\n\nSo with a 200 Watt panel and 120Ah battery you could run your fridge and lights for: 120Ah / 7.5Ah = just over 16 days without any other form of charge.\n\n### What size solar controller do I need for 200W solar panel?\n\nThe solar charge controller needed for a 200-Watt solar panel now depends on the voltage of the battery that you would be using. If you use a 12V battery in this case, then that would be 200/12 which is roughly equal to 16.7. As such, the solar charge controller needed should be capable of handling 16.7A.\n\n### How big is a 200-watt solar panel?\n\nActually, “a 500Wh [12 Volts, 40Ah] is considered the best match for a 200W solar panel.” It is highly recommended to use lithium batteries for solar panels because of their extended life span, excellent power output, robustness, and reliable performance.\n\n### What will a 300 watt solar panel run?\n\nA 300 watt solar panel with full irradiance will run a constant AC load of 270 watts, taking into account inverter losses of 10%. This includes appliances such as blenders, desktop PCs, vacuum cleaners and treadmills. A 300 watt solar panel will also run a small fridge with 120Ah lithium battery.\n\n### How many solar panels do I need to run a TV?\n\nAs a general rule, a 200Ah lead-acid deep-cycle battery would need a 300 watt solar panel to fully recharge from 50% Depth of Discharge (DOD) assuming 4 peak-sun-hours per day.\n\n### Will a solar panel charge a flat battery?\n\nIf a battery is completely drained, a panel can typically charge the battery within five to eight hours. The total charging time will vary depending on the state of a battery. If a battery is totally drained, a solar panel can energize the cells within five to eight hours.\n\n### Can I directly charge battery from solar panel?\n\nA solar panel can be connected directly to a 12 volt car battery, but must be monitored if it’s more than 5 watts. Solar panels rated higher than 5 watts must not be connected directly to a battery, but only through a solar charge controller to protect against over-charging.\n\n### How many amps is 200 watts?\n\nYou can charge 12V batteries with a 100-watt solar panel. The time this would take depends on the capacity of the battery and sunlight exposure. A rough estimate would be that it can take between 10 – 14 hours to fully charge the battery.\n\n### What can I run on a 400 watt solar system?\n\nA 400 watt solar panel with full irradiance will run an AC load (constant) of 360 watts. This value takes into account 10% inverter losses. This includes a combination of appliances like televisions, laptops, slow cookers and ceiling fans. A 400 watt solar panel can run a small fridge using a 120Ah battery.\n\n### How long will a 100 watt solar panel take to charge a 12V battery?\n\nAs a general rule, a typical size 12v 50Ah auto battery at 20% discharge will need 2 hours to fully recharge with a 100 watt solar panel. A lead-acid deep-cycle 12v 50Ah battery at 50% discharge will take about 4 hours to fully recharge using a 100 watt solar panel.\n\n### How many amp hours will a 100 watt solar panel produce?\n\nA 100 watt panel produces an average of about 6 amps per peak sun hour, or about 30 amp-hours per day. Given the above example, you would need three 100 watt solar panels to fully recharge on the average day (80 / 30 ≈ 3).\n\n### Can a 100 watt solar panel run a refrigerator?\n\nAs a general rule, 100 watt solar panel can run a refrigerator for a short time only and would also need a battery. 100 watts of solar panels can generate on average 400 watt-hours of energy per day. A refrigerator with combined freezer needs 2000 watt-hours/day.\n\n### How many watts do you need to run a refrigerator?\n\nA 5,000-watt generator should provide more than enough power to run a refrigerator, with power left to spare. Most home refrigerators need around 2,000 starting watts of power.\n\n### How many watts does a fridge use?\n\nConventional refrigerators typically have a starting wattage of 800-1200 watt-hours/day, and a running wattage of around 150-watt hours/day. Refrigerators are reactive devices that require additional power to start because they contain an electric motor, but significantly fewer watts to run as they remain on.\n\n### How many batteries do I need for a 2000 watt inverter?\n\nTypically two batteries are needed for a 2,000 watt inverter like the part # 34278156 that you referenced.\n\n### How much solar power do you need for an RV fridge?\n\nIn order to power that fridge using solar power, you would need about two to three solar panels. Average solar panels produce approximately 250 to 400 Watts of power. But you are not using an average refrigerator in your RV. Most likely, you need to power a 12V fridge, which is smaller.\n\n### Will a 100 watt solar panel run a camper?\n\nIf it is constantly sunny with no cloud cover, 100-watt solar panels will produce quite a lot of electricity and may be sufficient to supply your RV. However, a 100-watt solar panel output on a cloudy day is very minimal and will not be nearly enough to supply your RV.\n\n### What size solar panel do I need to run a 12V fridge?\n\nBATTERY & SOLAR PANEL SIZING CHART FOR BUSHMAN 12V FRIDGES\n\nAs a rule of thumb, we recommend slightly less than double the Ah of battery storage in watts of solar. So a 100Ah battery would have approximately 200 watts of solar.\n\n### What size solar panel do I need to charge a 100Ah battery?\n\nIn general, a 100Ah deep-cycle lead-acid battery would require 180 watts of solar panel to fully recharge from 50% Depth of Discharge (DOD) assuming 4.2 peak-sun-hours per day.\n\n### How much solar power do I need for camping?\n\nIf you have a single 12 volt battery, at about 100 AH, you should have 300 watts of solar panels, minimum. With two 12 volt batteries, or two 6 golf cart volt batteries, with between 200-250 AH, you should have 400 watts of solar panels, minimum.\n\n### How many watts does a 12 volt fridge use?\n\n12-volt refrigerators generally range from 40 to 100 watts, although there are both smaller and larger units. A 12-volt refrigerator that features 60 running watts draws on ~5 Amps from the 12V battery.\n\n### What size MPPT controller do I need for 200 watt solar panel?\n\nSolar Charge controller Sizing (A)\n\nFor example: if we have 2 x 200W solar panels and a 12V battery, then the maximum current = 400W/12V = 33Amps. In this example, we could use either a 30A or 35A MPPT solar charge controller.\n\n### How many watts can a 100 amp charge controller handle?\n\nWith a maximum of 100 amps output, a single charge controller can handle array sizes up to 6,000 watts on a 48 volt battery bank.\n\n### How many watts can a 15 amp charge controller handle?\n\nThis 15-AMP digital charge controller can handle systems up to 270 watts and is compatible with Lithium and Lead-Acid Batteries.\n\n### How heavy is a 250w solar panel?\n\nHow Big Is a 300w Solar Panel? 300 Watt solar panels are considered typical rooftop panels as they are big enough to generate enough power for full home use. This means they are typically the same size as your standard residential solar panel at approximately 5-5.5 feet long by 3-3.5 feet wide.\n\n### How many amps does a 250 watt solar panel produce?\n\nA solar panel’s output power depends on the panel’s size and the efficiency of the PV cells. Solar panel efficiency, in turn, is affected by insolation, temperature, shading, and orientation. In ideal circumstances, a 250- to 400-watt solar panel can produce between 14 and 24 amps.\n\n### What size solar panel do I need to charge 120Ah battery?\n\nIn general, a 120Ah deep-cycle lead-acid battery would require a 200 watt solar panel to recharge from 50% Depth of Discharge (DOD) if we assume 4 peak-sun-hours per day. It would take one day to fully recharge assuming clear skies.\n\n### How many batteries do you need for a solar system?\n\nIf you want to save the most money possible, you’ll need enough battery storage to cover your energy usage when your solar panels aren’t producing – somewhere around 2-3 batteries. If you want to keep the power on when the grid is down, you’ll usually just need one solar battery.\n\n### What can a 150 watt solar panel run?\n\nA 150W solar panel will easily power most of your camp lights, torches and their rechargeable batteries, especially if those lights and torches are LED. These draw very little power from your solar panel battery and are known to be longer-lasting than incandescent bulbs.\n\n### Can you run an RV air conditioner with solar power?\n\nIf your system is big enough, you can run RV A/C with solar power. Yes, it’s technically possible to power an RV air conditioner with solar panel. But to generate enough power, a large amount of solar panels and upgrades to the electrical system are required.\n\n### How fast will a 300 watt solar panel charge a battery?\n\nCharging your battery would take 10 hours using one 300-watt solar panel, assuming perfect conditions.\n\n### Will a 200 watt solar panel run a TV?\n\nDifferent types of TVs require different amounts of power. The Department of Energy provides a handy home appliance energy use calculator, which says modern TVs use anywhere from 150 watts (LCD or LED TVs under 40 inches) to 300 watts (plasma TVs)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8965841,"math_prob":0.7968894,"size":12867,"snap":"2022-27-2022-33","text_gpt3_token_len":3161,"char_repetition_ratio":0.22288735,"word_repetition_ratio":0.21434529,"special_character_ratio":0.26556307,"punctuation_ratio":0.09131736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9516464,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T11:25:41Z\",\"WARC-Record-ID\":\"<urn:uuid:b964e530-ebc5-4c3a-ba4f-1674fe4bb8ad>\",\"Content-Length\":\"109171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5060c97f-3014-441d-b7cc-22305a92620d>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d7149f4-27cb-4298-8f8e-128170919799>\",\"WARC-IP-Address\":\"172.67.186.248\",\"WARC-Target-URI\":\"https://www.2outdoors.com/what-will-a-200-watt-solar-system-run/\",\"WARC-Payload-Digest\":\"sha1:GXYTPZTGSL4YF6W44DOCL5H6ZEBCUCJ5\",\"WARC-Block-Digest\":\"sha1:KXUFUIMTXP5ZVORYTXHY5AICZTIN6K5R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570921.9_warc_CC-MAIN-20220809094531-20220809124531-00774.warc.gz\"}"}
https://help.simetrix.co.uk/8.4/simplis/simplis_mdm/topics/2_3_initial_inductor_design.htm	[ "# 2.3 Design an Inductor Using MDM\n\nIn this section of the tutorial you create an initial inductor design for the Buck converter. Then you will perform post-processing in MDM after simulating the schematic to obtain a detailed inductor loss estimate.\n\nIn this topic:\n\n## Key Concepts\n\nThis topic addresses the following key concepts:\n\n• The Level 2 model only adds a DC winding resistance (ESR) to the inductor symbol inside the schematic. In order to obtain more accurate winding and core losses, post-processing must be performed in MDM.\n• MDM post-processing occurs after the circuit simulation in SIMPLIS, and must be set up properly to take in a portion of the resulting circuit simulation waveforms.\n• MDM provides a calculation of the average power loss in the inductor over the simulation period provided. Therefore, for meaningful results, you must make sure that you are providing MDM with steady-state converter waveforms over an integer number of steady-state cycles (preferably one).\n\n## What You Will Learn\n\nIn this topic, you will learn the following:\n\n• How to design an inductor using SIMPLIS MDM that meets your design requirements.\n• How to properly set-up and perform post-processing in MDM to obtain detailed inductor loss estimates.\n• How to analyze the results MDM provides.\n\n## 2.3.1 Design the Inductor\n\nRemember, you want an inductor with an inductance in between 600 and 700 nH. To create an initial inductor design for the Buck converter, follow these steps:\n\n1. Double-click the symbol L1.\n2. In the resulting dialog, switch the model to Level 2.\n3. Check the Write MDM generated model parameters to all lower model levels? checkbox.\n\n4. Click the Create with SIMPLIS MDM... button.\n5. In the MDM main window, go to the Core tab, if it is not already showing.\n6. In the core geometry selection menu tree, select EE 3 air gaps > EPCOS > E32. You have selected the E32 core with three variable air gaps for this design.\n7. In the core material selection menu tree, select Ferrite > EPCOS > N87+. The E32 core is commercially available made from the N87 material.\n8. In the lower right panel, click", null, "to re-center the inductor visualization.\n9. In the middle panel showing the core dimensions, scroll down until you see the input field for Air Gap Size (gap): and set it to 1.\nResult: The core tab should now look like this:", null, "10. Open the MDM Status Window. Move it to the side so you can see both it and the entire MDM main window.\n11. Switch to the Winding tab.\n12. Scroll down in the left panel until you see the third unnamed sub-panel. In the wire selection tree menu, select Predefined Windings > AWG10.\n13. Increase the number of turns to 3.\nResult: In the status window, an inductance of 878 nH for this inductor is displayed.\n14. Since the inductance is too large with just three turns, try a smaller core. Go back to the Core tab.\n15. Select a smaller core: EE 3 air gaps > EPCOS > E30.\n16. Increase the air gap of the core to 1.1 and press Enter.\nResult: In the status window, an inductance of 615.6 nH for this inductor is now displayed:", null, "17. You now have an acceptable initial inductance. Go back to the Winding tab.\n18. With only three turns, the required inductance has been reached. However there is still a lot of free space in the core's winding window. More copper can therefore be added, but there is no need for extra turns. To add more copper without adding turns, you can make the winding multi-filar. MDM allows you to create a bifilar, trifilar, ..., n-filar winding. On the panel next to the wire selection menu tree, find the spinner named Filar:.\n19. From the drop-down box next to this spinner, select Y.\n20. Increase the Filar spinner to 2.\nResult: The winding tab should now look like this:", null, "21. Close the MDM Status Window.\n22. In the upper right corner of the main MDM window, click Finish.\n\nYou have now saved an MDM physical model of the inductor to the symbol L1. From this physical model, a Level 2 model for L1 has been extracted. To see this, double-click on L1. You should see the following:", null, "You can see that the model status box is green and that model status is Ok: A Model Has Been Created For This Inductor.\n\nMake sure that the checkbox Power probe calculates detailed losses is unchecked. Then click OK to close the dialog.\n\n## 2.3.2 Add a Power Probe\n\nYou can see now that the ESR of the Level 2 model generated by MDM is displayed on the schematic above L1. In order to measure losses, you must add a power probe, as is standard procedure in SIMPLIS when you want to measure the power loss in a component. To do so,\n\n1. In the Part Selector, double-click on Probes > Power Probe.\n2. Put the power probe on the right pin of L1:", null, "3. Double click the probe you just placed.\n4. Make sure that in the Edit Probe dialog that appears, under Analyses the boxes Transient and POP are checked:", null, "5. This will ensure that the power loss of the inductor is displayed in the Waveform Viewer when the simulation is run. Now under Axis type select Use dedicated grid. Click OK.\n\n## 2.3.3 Add Mean Measurement to Power Probe\n\nNext you will add the mean measurement to the power probe. This measurement will be performed after every simulation, with the mean value for the power dissipation output to the waveform viewer.\n\nTo add the mean measurement to the Power(L1) probe, follow these steps:\n\n1. Select the power probe Power(L1).\n3. Select the Mean measurement in the drop down list.", null, "4. Click Ok.\n\n## 2.3.3 Run the Simulation Without MDM Post-Processing\n\nNow you will run the simulation using the power probe to calculate inductor losses based on the Level 2 model, but without using MDM for post-processing. This was previously the only way to calculate losses in SIMPLIS. In the previous section you added a mean value measurement to the power probe. In this section you will see that this measurement only measures the loss due to the RMS current in the inductor.\n\n1. Press F9 to run the simulation.\nResult: The results appear in the Waveform Viewer, including now the power loss of the inductor.\n2. Click on the Waveform Viewer.\n3. In the measurement pane in the lower right corner of the Waveform Viewer, scroll until you find the measurement for Power(L1) Mean:", null, "You will see that the power probe has measured an average loss in the inductor of 7.39mW.\n\nHowever, it is very important to note that the ESR of the Level 2 model that is shown on the schematic is based on the DC resistance of the inductor winding ONLY. Therefore, the loss measurement which you just performed does NOT include core losses, AC winding losses, and losses due to proximity effect.\n\nIn order to obtain those detailed losses, you must run MDM post-processing after the circuit simulation.\n\n## 2.3.4 Run the Simulation With MDM Post-Processing\n\nNow you will run set up and run MDM post-processing after the circuit simulation to obtain a detailed calculation of inductor losses - including core losses, AC winding losses, and proximity effects, as well as the inductor temperature, none of which you can obtain by just using a regular SIMPLIS power probe as in the previous sub-section.\n\nIn post-processing, MDM calculates average inductor loss over a specified section of the circuit simulation waveform. An instantaneous power loss waveform, as given by the regular power probe, is not provided. Therefore, for the results to be meaningful, you want to make sure that MDM uses an integer multiple of the converter's steady-state cycle for the calculation. The more cycles are included, the longer the calculation will take, so the ideal number of cycles is one. Calculating average power loss over one steady-state cycle will give you the average power loss in the inductor when the converter is in steady-state.\n\nNote that MDM assumes that you have set-up the post-processing to produce meaningful results. It will calculate the power over a quarter- or half-cycle, over 3.37 cycles or during converter startup if you so instruct it to. While average power loss in steady-state is what you are usually most interested in, if you run post-processing on a different type of waveform, it's up to you to make sense of the results properly.\n\nFor the inductor, MDM needs both the current and voltage waveform of the inductor. You do not need to necessarily plot these explicitly to the Waveform Viewer in SIMPLIS. In this example, you will notice that while the inductor current is plotted, the inductor voltage is not. MDM will automatically acquire the waveforms it needs regardless of what you plot to the Waveform Viewer.\n\nTo set up MDM post-processing for the inductor in the Buck converter,\n\n1. Double-click L1\nResult: The Edit Multi-Level PWL Inductor dialog opens, showing the Level 2 model.\n2. Check the Power probe calculates detailed losses (Core, Winding, etc.) and temperature using MDM post processing checkbox.\nResult: The two fields below the checkbox are now enabled.\n3. The Fundamental frequency field determines how much of the circuit simulation waveform is sent for post-processing in MDM. MDM will take the last fundamental period (calculated as the inverse of the entered fundamental frequency) of the simulation waveforms to use for the post-processing calculations. As noted earlier, it is best if this is a single steady-state converter cycle. In a Buck converter, a steady-state cycle is a switching cycle. Since the switching frequency of this converter is 500kHz, for Fundamental frequency enter 500k.\n4. MDM must also be aware of the shape of the waveform. This is needed for the accurate calculation of core losses. You must specify this in the Wave shape drop-down menu. Since a rectangular voltage is applied to an inductor in a buck converter causing it to conduct a triangular current, set the Wave shape to Piecewise linear (PWL). The dialog should then look like this:", null, "5. Click OK.\nResult: The label MDM POSTPROCESSING appears above L1:", null, "6. Press F9 to run the simulation.\nResult: After the circuit simulation completes, a window indicating that MDM post-processing appears, displaying \"Calculating losses for L1...\".\n7. Once MDM post-processing is finished, the MDM Results window will appear:", null, "By default, the MDM Results window is open to the Results Overview tab. To the left, it shows the initial inductance, a textual breakdown of the different types of losses in the inductor, and also the core and winding temperatures, as well as the boxed volume of the inductor. Note that the DC Bias Current Losses are given as 0.007W - the same as you previously measured with the power probe. However the actual total losses are much higher - 0.044W. You can see that MDM post-processing can give you a much better insight into what the inductor losses will be than you can obtain by just setting an ESR in the schematic.\n\nYou might ask yourself if you could not obtain the same results by just setting the ESR of L1, in a Level 1 model, to the required value to obtain a loss of 44mW. While this is of course possible, there are two limitations with this approach. First, this would not provide a detailed loss analysis which you can use to improve your inductor design. Second, you do not know this value of ESR ahead of time. While it is easy to calculate the DC winding resistance once a physical model of the inductor is created in MDM, the AC winding losses and the core losses depend on many factors that are operating-point dependent: frequency, flux density inside the core, temperature, the duty ratio of the waveforms, and DC current. Thus the \"ESR\" for each operating point would be different. MDM post-processing will give you accurate loss estimates for different operating points without the need to adjust the ESR of the inductor in the schematic.\n\nThere are some other important things to note:\n\n• The MDM Results window does not block SIMetrix/SIMPLIS like the main MDM window does. You can close it at any time, or just minimize it or leave it in the background. In the MDM Beta version, you cannot bring back a Results window you have closed. Once you close it, the only way to regenerate it is to re-run the simulation.\n• Checking the post-processing checkbox in the Edit Multi-Level PWL Inductor dialog converts an \"ordinary\" power probe into an \"MDM post-processing\" power probe. Note that it not sufficient for just the box to be checked to perform MDM post-processing: both a power probe must be placed on the symbol and the checkbox must be checked.\n• With the MDM POSTPROCESSING label above the inductor symbol, you can easily see by looking at a schematic for which symbols MDM post-processing is selected, and by extension, which symbols have a valid MDM Level 2 model defined.\n• In the title bar of the MDM Results window you can see which symbol the results are for and to which SIMPLIS simulation group they refer. In this example the title is MDM Results for L1 (simplis_pop3). Therefore when you have several windows open, you can refer to this to figure out for which symbol and which simulation run the results are.\n\n## 2.3.5 Analyze the MDM Results\n\nNow you can analyze the MDM inductor results in more depth. You can see that the DC winding losses and core losses are quite low at about 7mW and 5mW, respectively. However the total winding losses are quite high, with proximity losses dominating at 19mW. To gain more insight into why proximity losses are so high, switch to the Losses By Winding tab of the MDM Results window:", null, "Here the loss density of each turn (in W/cm3) of the winding is displayed. The more red the turn, the higher the losses inside it. Note that the scale to the right is relative and differs for each inductor and operating point. The most intensely red turn is the one with the most losses relative to the others - it does not necessarily mean that it's very hot.\n\nYou can see that the turns closest to the air gap have the highest loss density. This is the result of fringing flux field produced by the air gaps of the core. As the magnetic flux in the core travels across the air gap, it fringes and a field is generated inside the winding window which creates proximity losses in the winding. So despite using a large bifilar wire to reduce the winding resistance, the way the turns are arranged relative to the air gap produces significant winding losses.\n\nNow switch to the Waveforms tab of the MDM Results window. Here the flux density inside the inductor core (in Tesla), the inductor current (in Amperes) and the inductor voltage (in Volts) are shown:", null, "Note that the peak flux density is about 25mT. If you were to examine the N87+ material used for this inductor inside MagDB, you would see that the peak flux density of this material is given as 360mT. Also note that the peak current in the inductor is about 6.5A.\n\nNow switch to the L vs. Current tab. Here the inductance of the inductor as a function of the current is shown:", null, "The inductance starts to drop off only after 48A, with another significant drop after 75A, and the inductor saturates only above 105A. This is many times higher than the peak inductor current in this converter. You can conclude from this, as well as from the peak flux density, that this inductor is using a core which is too large for this application. Indeed, if you go back to the Results Overview tab, you will see that the boxed volume is 13.12 cm3. A 7A, 600nH inductor can be much smaller. Additionally, you can see by looking at the core and winding temperatures that the inductor is quite cool, heating up less than 3 degrees above the ambient temperature of 25 degrees.\n\nTherefore, there is a lot of room to improve this inductor design. You will do this in the next section: 2.4 Refine the Inductor Design Using MDM\n\nMinimize or close the MDM Results window. Save the schematic as 2_my_buck_E30_core.sxsch.\n\nA complete schematic with the inductor design developed in this section, set up for MDM post-processing, is available as 2.3_SIMPLIS_MDM_tutorial_buck_converter_E30core.sxsch in the zip archive of schematic files:", null, "" ]	[ null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/icon_one.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-1_core_tab.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-1_status_window.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-1_winding_tab.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-1_level2_model_nopostprocess.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-2_power_probe.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-2_power_probe_dialog.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-2-3_power_probe_mean_measurement.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-3_waveform_viewer_power_probe.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-4_level2_model_postprocess.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-4_post_process_label.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-4_mdm_results.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-5_results_by_turn.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-5_results_waveforms.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-5_results_ind_vs_curr.png", null, "https://help.simetrix.co.uk/8.4/simplis/library/images/simplis_mdm/2-3-5_buck_schematic_with_mdm.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9350589,"math_prob":0.8920795,"size":8849,"snap":"2020-45-2020-50","text_gpt3_token_len":1945,"char_repetition_ratio":0.16359524,"word_repetition_ratio":0.026710099,"special_character_ratio":0.20702904,"punctuation_ratio":0.09630911,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9558398,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T13:39:57Z\",\"WARC-Record-ID\":\"<urn:uuid:0eef8eef-8610-486e-bdf7-6f9b9ecb13fa>\",\"Content-Length\":\"280206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90f5989d-d706-4ec0-b519-8d9538582cc7>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8e432f4-7bdc-4346-885e-057d4b8744ec>\",\"WARC-IP-Address\":\"78.109.168.89\",\"WARC-Target-URI\":\"https://help.simetrix.co.uk/8.4/simplis/simplis_mdm/topics/2_3_initial_inductor_design.htm\",\"WARC-Payload-Digest\":\"sha1:63SFG3WIUUIDEGN64A743KQMNAEG5M23\",\"WARC-Block-Digest\":\"sha1:CFKT3QSW5YOGW2FBVYSZNTMWYZUXMIGL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141198409.43_warc_CC-MAIN-20201129123729-20201129153729-00418.warc.gz\"}"}
https://answers.search.yahoo.com/search?p=ACP&b=31&pz=10&xargs=0&ei=UTF-8&flt=cat%3AMathematics	[ "# Yahoo Web Search\n\nrelated to ACP\n\n1. Sort by\n\n1. ### Help with a Logic Problem!?!?\n\n1. Y ∙ ∙ ∙ ∙ ∙ ∙ ∙ /~Y → (P → ~T) 2. Y v (P → ~T) ∙ ∙ ∙ ∙ ∙ ∙ ∙ 1, Addition 3. ~Y → (P → ~T) ∙ ∙ ∙ ∙ ∙ 2, Material Implication nothing else is needed.\n\n1 Answers · Science & Mathematics · 14/04/2011\n\n2. ### What is the smallest equilateral triangle, given that ...?\n\n...http://farm5.static.flickr.com/4092/5038287825_17374d81cc.jpg Rotating the triangle ACP (#1') or ABP (#1\") about A at ±60° and applying the...\n\n6 Answers · Science & Mathematics · 04/10/2010\n\n3. ### AP STAT OR HONORS PHYSICS?\n\nMmmmm...sounds like a difficult choice. I took a heavy load of AP classes all throughout highschool and was able to graduate in 3.5 years. The method in which you structure your questions makes it seem as though you don't...\n\n3 Answers · Science & Mathematics · 08/09/2011\n\n4. ### How to solve Systems of Linear Equation with variables?\n\nIt's not really six variables. You have three variables x, y, z and three unknown constants a, b, c. Note: I assume the last equation should be 2ax - by + cz = 14. The main thing we have to be careful about is not dividing by 0. So...\n\n1 Answers · Science & Mathematics · 10/04/2014\n\n...)(cp+ds) bd=+1 so as with question 1 we find b=d=-1 : (ap-s)(cp-s) = acp^2 -(a+c)ps + s^2 ac=9 so we could have 1x9 or 3x3. only a=c=3 gives...\n\n3 Answers · Science & Mathematics · 21/08/2009\n\n6. ### Solve this geometry problem.....?\n\n... at 60° about A to position A'B'P' ≡ ACP'. The triangle APP' is isosceles: \|AP\| = \|AP...\n\n2 Answers · Science & Mathematics · 22/12/2010\n\n7. ### Philosophy/Symbolic Logic Help needed! (Solving a proof!)?\n\n...3. E v B............../A -> C \|4. A..................ACP \|5. ~(E v F).........2,4, MP \|6. ~E & ~F........5...\n\n3 Answers · Science & Mathematics · 26/10/2008\n\n8. ### Proving geometric theorems?\n\n...common tangent at P. As before by alternate segment theorem, <ACP = <APX; as well <BNP = <...\n\n1 Answers · Science & Mathematics · 25/02/2014\n\n9. ### Trigonometry. Is this an ambiguous triangle?\n\n.... AP = 9*sin(40°) = 5.79 cm Now apply Pythagoras to ACP. 6^2 = 5.79^2 + CP^2 CP = 1.59 cm (this would have to be 4 to make...\n\n1 Answers · Science & Mathematics · 16/09/2017\n\n10. ### The traingles ABC and DEF are similar. The area of ABC is 20cm and the area of DEF is 40cm. if AB= a cm find D?\n\n...altitude FQ from F on to DE In two triangles, ACP and DFQ, <A = <D (from 3) &...\n\n3 Answers · Science & Mathematics · 22/01/2010" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81217694,"math_prob":0.8843947,"size":2028,"snap":"2019-43-2019-47","text_gpt3_token_len":713,"char_repetition_ratio":0.19367589,"word_repetition_ratio":0.1056701,"special_character_ratio":0.4289941,"punctuation_ratio":0.28294572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99058056,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-11T20:34:34Z\",\"WARC-Record-ID\":\"<urn:uuid:37f2a9c9-f380-446f-9c9c-e3eda79fade0>\",\"Content-Length\":\"121692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:227eef5a-18b1-4f0a-af20-31baa5b586f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a94641c-02ea-4a10-900d-f38306f04349>\",\"WARC-IP-Address\":\"66.218.84.137\",\"WARC-Target-URI\":\"https://answers.search.yahoo.com/search?p=ACP&b=31&pz=10&xargs=0&ei=UTF-8&flt=cat%3AMathematics\",\"WARC-Payload-Digest\":\"sha1:BJANFOFJP2ONM3UFWG4OZDF54EOXC73I\",\"WARC-Block-Digest\":\"sha1:NIPCEMZ6LRH5MNJW56JRLQSO722VBVIE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664437.49_warc_CC-MAIN-20191111191704-20191111215704-00297.warc.gz\"}"}
https://simple.wikipedia.org/wiki/Numerical_methods_for_partial_differential_equations	[ "# Numerical methods for partial differential equations\n\nNumerical methods for partial differential equations are computational schemes to obtain approximate solutions of partial differential equations (PDEs).\n\n## Scientific Background\n\n### Motivation of this area\n\nMany PDEs appeared for the study of physics and other areas in science. Therefore, many mathematicians have challenged to make methods to solve them, but there is no method to mathematically solve PDEs except the Hirota direct method and the inverse scattering method. This is why numerical methods for PDEs are needed.\n\n### The Finite Difference Method (FDM) and its problems\n\nOne of the most basic PDE solver is the finite difference method (FDM). This method approximates derivatives as differences:\n\n$f^{\\prime }(x)\\simeq {\\frac {f(x+h)-f(x)}{h}},\\quad h\\ll 1.$", null, "This method works for easy problems. But it is powerless to some equations (such as the Navier–Stokes equations) because they are non-linear. Since this difficulty appeared, numerical analysts started to study other methods (just like the finite element method, FEM). On the other hand, some experts started to consider improvements for FDM.\n\n### Evolution of the FDM\n\nExperts have discovered difference methods which preserves the property of the given PDE.\n\n#### Integrable Difference Schemes\n\nRyogo Hirota, Mark Ablowitz and others have made methods that preserves the integrability (important mathematical property in the theory of dynamical systems) of PDEs. These methods are known to have better accuaracy than the original FDM.\n\n#### Structure Preserving Numerical Methods\n\nMany PDEs have appeared from physics. So we can think about difference methods preserving physical properties. These difference methods are known as structure preserving numerical methods. The following list is the examples of them:\n\nSome experts are studying their relation between numerical linear algebra.\n\n#### Others\n\nThe difference mthods in above have high accuracy, but their usage is limited because they depend on the behaviour of the given PDEs. This is why new types of FDM are still studied. For example, the following methods are studied:\n\n## Validated Numerics for PDEs\n\nNot only approximate solvers, but the study to \"verify the existence of solution by computers\" is also active. This study is needed because numerically obtained solutions could be phantom solutions (fake solutions). This kind of incident is already reported.\n\n## Journal\n\nThe scientific journal \"Numerical Methods for Partial Differential Equations\" is published to promote the studies of this area.\n\n## Related Software\n\nChebfun is one of the most famous software in this field. They are also many libraries based on the finite element method such as:" ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/773af7d9567b87801966747aaef4ae586ddfa46a", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.641816,"math_prob":0.83523065,"size":17725,"snap":"2023-14-2023-23","text_gpt3_token_len":5055,"char_repetition_ratio":0.15298234,"word_repetition_ratio":0.09192308,"special_character_ratio":0.30087447,"punctuation_ratio":0.29270977,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99424946,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T16:42:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a55b60fc-652b-495d-9643-0dc7c48c132c>\",\"Content-Length\":\"104805\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:46e85530-368e-4deb-9efe-2a1d2b6c4bfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:15743a05-083c-48d4-b701-200aca86e9ca>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://simple.wikipedia.org/wiki/Numerical_methods_for_partial_differential_equations\",\"WARC-Payload-Digest\":\"sha1:HK6HNDBGHAICVOUA63LY6PYNHDWEEXT3\",\"WARC-Block-Digest\":\"sha1:TWTB342T4K4CAIWKJPAP57CQZGJBSSPZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652959.43_warc_CC-MAIN-20230606150510-20230606180510-00060.warc.gz\"}"}
https://cstheory.stackexchange.com/questions/9527/permutation-game-redux	[ "# Permutation game redux\n\nThis is a restatement of an earlier question.\n\nConsider the following impartial perfect information game between two players, Alice and Bob. The players are given a permutation of the integers 1 through n. At each turn, if the current permutation is increasing, the current player loses and the other player wins; otherwise, the current player removes one of the numbers, and play passes to the other player. Alice plays first. For example:\n\n• (1,2,3,4) — Bob wins immediately, by definition.\n\n• (4,3,2,1) — Alice wins after three turns, no matter how anyone plays.\n\n• (2,4,1,3) — Bob can win on his first turn, no matter how Alice plays.\n\n• (1,3,2,4) — Alice wins immediately by removing the 2 or the 3; otherwise, Bob can win on his first turn by removing the 2 or the 3.\n\n• (1,4,3,2) — Alice eventually wins if she takes the 1 on her first turn; otherwise, Bob can win on his first turn by not removing the 1.\n\nIs there a polynomial-time algorithm to determine which player wins this game from a given starting permutation, assuming perfect play? More generally, because this is a standard impartial game, every permutation has a Sprague–Grundy value; for example, (1,2,4,3) has value 1 and (1,3,2) has value 2. How hard is it to compute this value?\n\nThe obvious backtracking algorithm runs in O(n!) time, although this can be reduced to $O(2^n poly(n))$ time via dynamic programming.\n\n• Seems to me that the naive algorithm runs in O(2^n⋅poly(n)) time. – Tsuyoshi Ito Dec 29 '11 at 13:36\n• From your examples, it is obvious that Alice always wins if the sequence is descending and Bob always wins if the sequence is ascending. This problem reminds me of analyzing sorting algorithms, which have been extensively studied and allow you to use a wide arsenal of tools. – chazisop Dec 29 '11 at 13:53\n• @chazisop: “Alice always wins if the sequence is descending”: That is the case if and only if n is even. – Tsuyoshi Ito Dec 29 '11 at 14:09\n• @JɛﬀE in case 3, how does Bob win on his first turn ? – Suresh Venkat Dec 29 '11 at 23:23\n• @Suresh: In the case of (2,4,1,3), the graph representation is the linear graph on 4 vertices (2-1-4-3). If Alice removes an end node, this leaves the linear graph on 3 vertices; Bob wins by removing the center vertex (so 3 is answered by 1, and 2 is answered by 4). If Alice removes an interior node, this leaves two connected vertices and an isolated node; Bob wins by removing either of the two connected vertices (so 1 is answered by 3 or 4, and 4 is answered by 1 or 2). – mjqxxxx Dec 30 '11 at 1:32\n\nThe \"permutation game\" is isomorphic to the following game:\n\nDisconnect. Players alternately remove vertices from a graph $G$. The player that produces a fully disconnected graph (i.e., a graph with no edges) is the winner.\n\nThe graph $G_{\\pi}$ corresponding to a particular initial permutation $\\pi\\in S_n$ contains just those edges $(i,j)$ for which $i-j$ and $\\pi(i)-\\pi(j)$ have opposite signs. That is, each pair of numbers in the wrong order in the permutation is associated with an edge. Clearly the allowed moves are isomorphic to those in the permutation game (remove a number = remove a node), and the winning conditions are isomorphic as well (no pairs in descending order = no edges remaining).\n\nA complementary view is obtained by considering playing a \"dual\" game on the graph complement $G^{c}_\\pi = G_{R(\\pi)}$, which contains those edges $(i,j)$ for which $i$ and $j$ are in the correct order in the permutation. The dual game to Disconnect is:\n\nReconnect. Players alternately remove vertices from a graph $G$. The player that produces a complete graph is the winner.\n\nDepending on the particular permutation, one of these games may seem simpler than the other to analyze. The advantage of graph representation is that it is clear that disconnected components of the graph are separate games, and so one hopes for some reduction in complexity. It also makes the symmetries of the position more apparent. Unfortunately, the winning conditions are non-standard... the permutation game will always end before all moves are used up, giving it something of a misère character. In particular, the nim-value cannot be calculated as the nim-sum (binary XOR) of the nim-values of the disconnected components.\n\nFor Disconnect, it is not hard to see that for any graph $G$ and any even $n$, the game $G \\cup \\bar{K}_n$ is equivalent to $G$ (where $\\bar{K}_n$ is the edgeless graph on $n$ vertices). To prove it, we need to show that the disjunctive sum $G + G\\cup\\bar{K}_n$ is a second-player win. The proof is by induction on $\|G\|+n$. If $G$ is edgeless, then the first player loses immediately (both games are over). Otherwise, the first player can move in either $G$, and the second player can copy his move in the other one (reducing to $G' + G'\\cup \\bar{K_n}$ with $\|G'\|=\|G\|-1$); or, if $n\\ge 2$, the first player can move in the disconnected piece, and the second player can do the same (reducing to $G + G\\cup\\bar{K}_{n-2}$).\n\nThis shows that any graph $G$ is equivalent to $H \\cup K_p$, where $H$ is the part of $G$ with no disconnected vertices, and $p=0$ or $1$ is the parity of the number of disconnected vertices in $G$. All games in an equivalence class have the same nim-value, and moreover, the equivalence relation respects the union operation: if $G \\sim H \\cup K_p$ and $G' \\sim H' \\cup K_{p'}$ then $G \\cup G' \\sim (H \\cup H')\\cup K_{p\\oplus p'}$. Moreover, one can see that the games in $[H \\cup K_0]$ and $[H \\cup K_1]$ have different nim-values unless $H$ is the null graph: when playing $H + H \\cup K_1$, the first player can take the isolated vertex, leaving $H+H$, and then copy the second player's moves thereafter.\n\nI do not know any related decomposition results for Reconnect.\n\nTwo special types of permutations correspond to particularly simple heap games.\n\n1. The first is an ascending run of descents, e.g., $32165487$. When $\\pi$ takes this form, the graph $G_{\\pi}$ is a union of disjoint cliques, and the game of Disconnect reduces to a game on heaps: players alternately remove a single bean from a heap until all heaps have size $1$.\n2. The second is a descending run of ascents, e.g., $78456123$. When $\\pi$ takes this form, the graph $G^{c}_{\\pi}$ is a union of disjoint cliques, and the game of Reconnect reduces to a game on heaps: players alternately remove a single bean from a heap until there is only one heap left.\n\nA little thought shows that these two different games on heaps (we can call them 1-Heaps and One-Heap, at some risk of confusion) are, in fact, themselves isomorphic. Both can be represented by a game on a Young diagram (as initially proposed by @domotorp) in which players alternate removing a lower-right square until only a single row is left. This is obviously the same game as 1-Heaps when columns correspond to heaps, and the same game as One-Heap when rows correspond to heaps.\n\nA key element of this game, which extends to Disconnect and Reconnect, is that the duration is related to the final game state in a simple way. When it is your turn, you will win if the game has an odd number of moves remaining, including the one you're about to make. Since a single square is removed each move, this means you want the number of squares remaining at the end of the game to have the opposite parity that it has now. Moreover, the number of squares will have the same parity on all of your turns; so you know from the outset what parity you want the final count to have. We can call the two players Eve and Otto, according to whether the final count must be even or odd for them to win. Eve always moves in states with odd parity and produces states with even parity, and Otto is the opposite.\n\nIn his answer, @PeterShor gives a complete analysis of One-Heap. Without repeating the proof, the upshot is the following:\n\n• Otto likes $1$-heaps and $2$-heaps, and can tolerate a single larger heap. He wins if he can make all heap sizes except one $\\le 2$, at least without giving Eve an immediate win of the form $(1,n)$. An optimal strategy for Otto is to always take from the second-largest heap except when the state is $(1,1,n>1)$, when he should take from the $n$. Otto will lose if there are too many beans in big heaps to start with.\n• Eve dislikes $1$-heaps. She wins if she can make all heap sizes $\\ge 2$. An optimal strategy for Eve is to always take from a $1$-heap, if there are any, and never take from a $2$-heap. Eve will lose if there are too many $1$-heaps to start with.\n\nAs noted, this gives optimal strategies for 1-Heaps as well, although they are somewhat more awkward to phrase (and I may well be making an error in primary-to-dual \"translation\"). In the game of 1-Heaps:\n\n• Otto likes one or two large heaps, and can tolerate any number of $1$-heaps. He wins if he can make all but the two largest heaps be $1$-heaps, at least without giving Eve an immediate win of the form $(1,1,\\dots,1,2)$. An optimal strategy for Otto is to always take from the third-largest heap, or from the smaller heap when there are only two heaps.\n• Eve dislikes a gap between the largest and second-largest heaps. She wins if she can make the two largest heaps the same size. An optimal strategy for Eve is to always take from the largest heap, if it is unique, and never if there are exactly two of the largest size.\n\nAs @PeterShor notes, it isn't clear how (or if) these analyses could be extended to the more general games of Disconnect and Reconnect.\n\n• I think that this kind of games are collectively referred to as “vertex deletion games.” But I agree with you that the winning condition is quite nonstandard in that it refers to the global property of the graph instead of local properties such as the degree of a vertex. – Tsuyoshi Ito Dec 30 '11 at 1:43\n• The constructed graph is called a permutation graph (en.wikipedia.org/wiki/Permutation_graph) in the literature. Some structural properties might help. – Yoshio Okamoto Dec 30 '11 at 4:41\n• @Yoshio: That's a good point. The permutation game is isomorphic to the graph game, but the starting graphs aren't arbitrary. So even if the general graph game is hard to analyze, it's possible that when restricted to this subclass of graphs, it becomes simpler. – mjqxxxx Dec 30 '11 at 4:52\n• On the other hand, the more general formulation might be easier to prove hard. Variants of vertex-deletion games are known to be PSPACE-hard, for example: emis.ams.org/journals/INTEGERS/papers/a31int2005/a31int2005.pdf – Jeffε Dec 30 '11 at 11:35\n• I've added a question on this kind of game specifically over at math.SE (math.stackexchange.com/questions/95895/…). Incidentally, since permutation graphs are circle graphs, an alternate formulation is the following: Players take turns removing chords from an initial set; the player that leaves a non-intersecting set of chords is the winner. – mjqxxxx Jan 2 '12 at 20:44\n\nIn his answer, domotorp suggests analyzing a special case of the game. This special case arises when the permutation is a series of increasing sequences, each of which is larger than the following one, such as (8,9,5,6,7,4,1,2,3). In this game, you start with a collection of heaps of stones, and players alternately remove one stones from a heap. The player who leaves a single heap wins. We will say the $i$th heap has $h_i$ stones in it, and assume that the $h_i$ are given in decreasing order. For example, for the permutation above, the $h_i$ are 3,3,2,1. I tried giving the analysis of this game in the comments to domotorp's answer, but (a) I got it wrong and (b) there isn't enough room in comments to give a real proof.\n\nTo analyze this game, we need to compare two quantities: $s$, the number of heaps containing single stones and $t=\\sum_{i\\geq 2, h_i>2\\,} h_i -2$; note that we ignore the largest heap in the sum. This is the number of stones you would have to remove to ensure that all heaps but one contain no more than two stones. We claim that the losing positions are as follows:\n\n1. Positions where $t \\leq s-2$ containing an odd number of stones.\n\n2. Positions where $t \\geq s$ containing an even number of stones.\n\nIt is easy to show that from a losing position, you must go to a winning position, since $t - s$ can only change by at most 1 each turn, and the number of stones goes down by 1 each move.\n\nTo finish showing that this is correct, we need to show that from any position which is not in category (1) or (2), the first player can in one move either reach a position in category (1) or (2), or win directly.\n\nThere are two cases:\n\n1. Positions where $t \\geq s-1$ containing an odd number of stones. Here, if $s>0$, remove a stone from a heap with a single stone. If there's only one heap left, we've won. Otherwise, we now have $t \\geq s$. If there are no heaps with a single stone, remove a stone from a heap with at least three stones. (Since there were an odd number of stones, this is possible). Since $s = 0$, we have $t \\geq s$.\n\n2. Positions where $t \\leq s-1$ containing an even number of stones. Here, if there are any heaps with at least two stones other than the largest heap, remove a stone from one of them. If this heap has three or more stones in it, $t$ decreases by one. If it has exactly two stones in it, $s$ increases by one. We now have $t \\leq s-2$. The last case is when all the heaps except one consist of single stones; in this case, it is easy to check the first player wins if there are an even number of stones.\n\nI've tried generalizing this strategy to the original game, and haven't figured out how to do it.\n\n• In my answer, I noted that having a solution to this special case also solves the special case with an increasing series of decreasing runs, by playing in the \"dual\" position obtained by transposing the Young diagram. In particular, Eve's optimal strategy becomes \"take from the largest heap, unless there are exactly two of that size\", and Otto's optimal strategy becomes \"take from the smallest heap\". – mjqxxxx Jan 9 '12 at 20:39\n• I am sure that this approach will lead to a perfect solution, but at the moment there still is a minor mistake, e.g. (3,1) is not losing and (3,1,1) is. The problem is that the definition of 2. should exclude this case, since we can reach a one heap position in one step. But I think this is the only problem with 2. and hopefully it is not hard to correct it. – domotorp Jan 10 '12 at 6:09\n• @domotorp: For (3,1), t = 0 and s = 1, so t $\\not\\geq$ s, and criterion (2) says that it is not a losing position. For (3,1,1), t = 0 and s = 2, so t $\\leq$ s $-$ 2, and criterion (1) says that it is a losing position. I think you missed that in the definition of t, you ignore the largest heap. – Peter Shor Jan 10 '12 at 18:17\n• Of course, I forgot that part at the end... Then this game is solved! – domotorp Jan 11 '12 at 6:35\n• Not a complete answer, but still worth the bounty. – Jeffε Jan 11 '12 at 10:19\n\nI've implemented an $O(2^n n)$ solution for quick hypothesis checking. Feel free to play with it. If you don't have C++ compiler locally, you can run it on different inputs remotely using \"upload with new input\" link.\n\n@JɛﬀE It happened that (1,4,3,2) has value 1, not 2 as you suggested.\n\n• Oops, my mistake. Fixed the question: g(1,3,2) = mex{g(1,3), g(1,2), g(3,2)} = mex{0, 0, 1} = 2. – Jeffε Jan 7 '12 at 18:28\n• @JɛﬀE It's interesting that for $n \\le 10$ SG-value of any position is not greater than 2. I'm trying to prove that now for arbitrary $n$, although I dunno how will that help. – Dmytro Korduban Jan 7 '12 at 18:51\n• @maldini: it gives hope that the game has some nice properties, which might make it tractable. I wonder what happens to the game generalized to graphs, or the game just generalized to perfect graphs. – Peter Shor Jan 17 '12 at 20:37\n\nEdit 5th of Jan: In fact the One Heap Game described below is a special case of the problem, i.e. when the numbers follow each other in a specific way such that the first group is bigger than the second group which is bigger than the third etc. and the numbers in each group are increasing. E.g. 8, 9, 4, 5, 6, 7, 2, 3, 1 is such a permutation. So I propose to solve this special case first.\n\nDisclaimer: I no longer claim that the below proof is correct, see e.g. the comment of Tsuyoshi which shows that deleting a number from a permutation will give a diagram not obtainable by deleting a square from the diagram of the permutation. I left the answer here to show that this approach does not work, plus since it contains another simple game.\n\nThe game has a very simple other formulation thanks to Young Tableaux. I am sure that it can be analyzed from there as other games and it will yield a linear time algorithm.\n\nFirst define the following game on Young Diagrams: At each turn, if the current diagram is horizontal (all squares in one line), the current player loses and the other player wins; otherwise, the current player removes one of the bottom-right squares, and play passes to the other player.\n\nNow order the sequence of numbers into a Young Tableaux. The main claim is that the winner of the original game is the same as the winner as the diagram game starting with this shape. To see this, notice that whenever the players delete a number, the diagram of the new sequence can be achieved by deleting a bottom-right square of the diagram. Moreover, any such diagram can be achieved by deleting the number from the respective bottom-right square. These statements follow from standard Young Tableaux theory.\n\nAlthough this diagram game is simple enough, it is trivially equivalent to the following game, which seems more standard:\n\nOne Heap Game: The players are given some heaps with some pebbles in each. At each turn, if their is only one heap left, the current player loses and the other player wins; otherwise, the current player removes a pebble from a heap, and play passes to the other player.\n\nIf there is a simple solution to the heap game (and I strongly believe there is one) we also get a solution to the original game: Just put the sequence in a Young Tableaux, and transform its diagram into heaps.\n\nUnfortunately I do not see which heap positions are winning/how to determine the Sprague–Grundy values. I checked a few cases by hand, and the following are the losing positions with at most 6 pebbles:\n\none heap; (1,1,1); (2,2); (3,1,1); (2,1,1,1); (1,1,1,1,1); (4,2); (3,3); (2,2,2).\n\nAnyone can solve this game?\n\nEdit: Peter Shor can, see his answer!\n\n• Can you give at least one example showing how a particular permutation is turned into a Young Tableau and how the same game (number removal until an ascending sequence is reached) is played on the Tableau? In particular I don't understand what it means to remove \"one of the bottom-right squares\". – mjqxxxx Dec 30 '11 at 13:32\n• Here is a counterexample to a weaker claim that removing a number from a permutation corresponds to a removing one of the bottom-right cells from the corresponding Young diagram (instead of Young tableau). Let n=5, and consider a position specified by the permutation [4,1,3,5,2] (that is, σ(1)=4, σ(2)=1, and so on), and remove 3 from it. The corresponding Young diagram before the move is 5=3+1+1, but the corresponding Young diagram after the move is 4=2+2, which is not obtained from removing one cell from 3+1+1. – Tsuyoshi Ito Dec 31 '11 at 3:10\n• And the permutation [5,4,1,2,3] has the same Young diagram as [4,1,3,5,2], but you cannot reach the Young diagram 4=2+2 from it. So the game depends on more than the shape of the Young tableau. – Peter Shor Dec 31 '11 at 11:32\n• Hooray for constructive misunderstanding! – Jeffε Jan 1 '12 at 10:37\n• @JɛﬀE: Yeah, this is much more useful than a proof of mere existence of misunderstanding. – Tsuyoshi Ito Jan 3 '12 at 12:28" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9429102,"math_prob":0.9886458,"size":6907,"snap":"2020-34-2020-40","text_gpt3_token_len":1720,"char_repetition_ratio":0.118788935,"word_repetition_ratio":0.104872,"special_character_ratio":0.25119445,"punctuation_ratio":0.10084034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99200356,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T07:29:49Z\",\"WARC-Record-ID\":\"<urn:uuid:ccedfba5-3f2c-472b-b20e-2a700c2bf015>\",\"Content-Length\":\"215410\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37a7032a-1fac-488c-889e-004b8ff84e72>\",\"WARC-Concurrent-To\":\"<urn:uuid:460ac81e-f5f9-45da-b634-09bfdb3d0259>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://cstheory.stackexchange.com/questions/9527/permutation-game-redux\",\"WARC-Payload-Digest\":\"sha1:A3BDGHCSK7PELHGLRXTYVA75DZC53MGG\",\"WARC-Block-Digest\":\"sha1:SAU33QXBYGWWD2GX3QWPLNSSPICCILEE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400265461.58_warc_CC-MAIN-20200927054550-20200927084550-00498.warc.gz\"}"}
http://www.sfu.ca/personal/archives/richards/Zen/show9/ch9.html	[ "Inferential Statistics\n\nStart Here\n\nContents\n\n1. Title page\n\n2. Descriptive vs Inferential statistics\n\n3. Inferential statistics is a leap into the unknown\n\n4. Remember the study of students' drinking habits?\n\n5. Your sample's mean was 10.54\n\n6. You found that other people got different results\n\n7. If you could get enough samples from that population ...\n\n8. The Sampling Distribution of Sample Means\n\n9. It has some useful properties\n\n10. The third useful property\n\n11. Estimate the Standard Error of the Mean\n\n12. You use information about sample statistics to estimates of population parameters\n\n13. The key to inferential statistics is the sampling distribution\n\n14. Four important assuimptions\n\n15. The mean and standard deviation of the sampling distribution\n\n16. Standard errors are measures of sampling variability\n\n17. A large standard error means that ....\n\n18. Only two things influence the standard error\n\n19. Standard errors and level of confidence\n\n20. How close are you? How confident are you?\n\n21. You know four things.\n\n22. A picture showing how 95% of sample means are within +/- 1.96 SEM of the population mean\n\n23. A picture showing how you can be 95% certain that .....\n\n24. The Standard Error of Proportions\n\n25. The Standard Error of Differences Between Means\n\n26. Standard errors measure sampling variability\n\n27. Two things affect sampling variability\n\n28. Blank Slide", null, "" ]	[ null, "http://www.sfu.ca/personal/archives/richards/Zen/show9/Images/Logo.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75109714,"math_prob":0.89959204,"size":1450,"snap":"2019-43-2019-47","text_gpt3_token_len":342,"char_repetition_ratio":0.12586446,"word_repetition_ratio":0.016877636,"special_character_ratio":0.22137931,"punctuation_ratio":0.098712444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.970299,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T08:29:48Z\",\"WARC-Record-ID\":\"<urn:uuid:f49ba9bb-f7c6-4740-8d8e-52a8d1f8e9c2>\",\"Content-Length\":\"6582\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cab03e7-7313-4384-bf04-bec2da01b65f>\",\"WARC-Concurrent-To\":\"<urn:uuid:98c3a8b8-0921-490f-a8f6-8b8b5371fc4d>\",\"WARC-IP-Address\":\"142.58.102.68\",\"WARC-Target-URI\":\"http://www.sfu.ca/personal/archives/richards/Zen/show9/ch9.html\",\"WARC-Payload-Digest\":\"sha1:KTJE267UINB2LOTTOSZQMYKUZOT4BH6E\",\"WARC-Block-Digest\":\"sha1:3LMMGWCYSG427HCKQEZBI4NOT6BSLMR6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987813307.73_warc_CC-MAIN-20191022081307-20191022104807-00539.warc.gz\"}"}
https://forum.ansys.com/forums/reply/283408/	[ "", null, "KK\nSubscriber\nIs it not possible to implement a simple least squares objective function in Optislang?\nJ(p) = zWz or after linearisation (GWG)p = GWz\nz = resiude vector\nW = weighting matrix\nG = weighting matrix (partial derivative of the residuals after the paramers)\np = parameter change\n\nI would like to know if it is possible to implement a least squares objective function using Optislang, or if there are simpler approaches that can take my weighting matrices directly from Optislang. Furthermore, I would be interested to know which approach is considered the best or simplest to realise the alignment between test and simulation. Are there any proven methods that are particularly recommended and what factors should be taken into account when choosing?", null, "" ]	[ null, "https://secure.gravatar.com/avatar/c9ad06b1e52cb65e7841d7d17fad75b6", null, "https://forum.ansys.com/wp-content/themes/ansysbbpress/assets/images/loading.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9114545,"math_prob":0.9839035,"size":744,"snap":"2023-14-2023-23","text_gpt3_token_len":152,"char_repetition_ratio":0.10135135,"word_repetition_ratio":0.0,"special_character_ratio":0.18682796,"punctuation_ratio":0.048387095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9711086,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T11:50:21Z\",\"WARC-Record-ID\":\"<urn:uuid:1b3b246e-40b5-43b2-b8e3-1804820b6a9f>\",\"Content-Length\":\"106523\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:459b9364-705f-4b4d-9e7c-1373bb7e84a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:4823c930-27ac-43d7-bcac-77dc31df9e30>\",\"WARC-IP-Address\":\"23.212.251.26\",\"WARC-Target-URI\":\"https://forum.ansys.com/forums/reply/283408/\",\"WARC-Payload-Digest\":\"sha1:X62ZAWOF5JMPDFQGCGKSUCX2PE72ZO7G\",\"WARC-Block-Digest\":\"sha1:74NKPAN6WLHIYUK5CD66JMAPGE3YUEUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654871.97_warc_CC-MAIN-20230608103815-20230608133815-00343.warc.gz\"}"}
https://www.indiabix.com/c-programming/bitwise-operators/discussion-503	[ "# C Programming - Bitwise Operators - Discussion\n\n### Discussion :: Bitwise Operators - Find Output of Program (Q.No.3)\n\n3.\n\nAssuming a integer 2-bytes, What will be the output of the program?\n\n``````#include<stdio.h>\n\nint main()\n{\nprintf(\"%x\\n\", -1<<3);\nreturn 0;\n}\n``````\n\n [A]. ffff [B]. fff8 [C]. 0 [D]. -1\n\nExplanation:\n\nThe system will treat negative numbers in 2's complement method.\n\nExample:\n\nAssume the size of int is 2-bytes(16 bits). The integer value 1 is represented as given below:\n\nBinary of 1: 00000000 00000001 (this is for positive value of 1)\n\n1's complement of binary 1: 11111111 11111110\n2's complement of binary 1: 11111111 11111111\n\nThy system will store '11111111 11111111' in memory to represent '-1'.\n\nIf we do left shift (3 bits) on 11111111 11111111 it will become as given below:\n\n11111111 11111111 ---(left shift 3 times)---> 11111111 11111000.\n\nSo, 11111111 11111000 ---(binary to hex)---> FF F8. (Required Answer)\n\nNote:\n\nHow is the negative number obtained from 2's complement value?\n\nAs stated above, -1 is represented as '11111111 11111111' in memory.\n\nSo, the system will take 2's complement of '11111111 11111111' to the get the original negative value back.\n\nExample:\n\nBit Representation of -1: 11111111 11111111\n\nSince the left most bit is 1, it is a negative number. Then the value is\n\n1's complement: 00000000 00000000\n2's complement: 00000000 00000001 (Add 1 to the above result)\n\nTherefore, '00000000 00000001' = 1 and the sign is negative.\n\nHence the value is -1.\n\n Sanju said: (Aug 17, 2013) You have given: Fill with 1s in the left side for right shift for negative numbers. Then why this?\n\n Madhuri Agrawal said: (Oct 5, 2015) As 32 will be written as 100000 in its binary format preceded by all 0's. So when we do negation of the same, so it will give the output as: 011111 preceded by any number of 1's. Now coming to converting the same bit string in its hexadecimal format. So it will result in 'df' preceded by as many number of 'f' as we want.\n\n Vamsi said: (Dec 17, 2015) I didn't understand any about the -1 in the given answer.\n\n Manjula said: (Oct 4, 2016) Here, why we have to take left shift?\n\n Dedeepya said: (Sep 17, 2017) Here in the program bit wise operator \"<<\" indicates left shift operator." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89083683,"math_prob":0.9479859,"size":929,"snap":"2022-27-2022-33","text_gpt3_token_len":270,"char_repetition_ratio":0.104864866,"word_repetition_ratio":0.0,"special_character_ratio":0.3078579,"punctuation_ratio":0.1573604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99488276,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T17:04:26Z\",\"WARC-Record-ID\":\"<urn:uuid:68f82e59-9054-4eb3-a51b-338773276fbd>\",\"Content-Length\":\"24134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3688e16a-8adc-4bec-92e4-24c887c04889>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b48e52e-b2ad-48c0-aa8c-f5750de6ad07>\",\"WARC-IP-Address\":\"35.244.11.196\",\"WARC-Target-URI\":\"https://www.indiabix.com/c-programming/bitwise-operators/discussion-503\",\"WARC-Payload-Digest\":\"sha1:NYYJDOAX5TZYQL6DDDSMVBQPM6NPQO7R\",\"WARC-Block-Digest\":\"sha1:3C4X3DKZKSXIQTXIIU5X42VW4WSH6VP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103943339.53_warc_CC-MAIN-20220701155803-20220701185803-00554.warc.gz\"}"}
http://epirecip.es/epicookbook/chapters/blackross2015/r	[ "# Final size for SIR model\n\nAuthor: Sangeeta Bhatia @sangeetabhatia03\n\nDate: 2018-10-03\n\nThis implementation is tested against the MATLAB implementation provided by the authors here. The following distribution was generated using their code.\n\n## Test data generated using authors' code.\nN <- 10\nbeta <- 2/(N-1);\ngamma1 <- 1;\nfinal_size1 <- c(0.00000, 0.33333, 0.08640, 0.04908, 0.03814, 0.03638, 0.04079, 0.05256, 0.07614, 0.11862, 0.16854)\n\n\nAlthough most of the code uses base R functions, we load libraries for testing and plotting.\n\nlibrary(ggplot2)\nlibrary(testthat)\n\n## Some more test data; extreme cases.\n## What if the recovery rate is 0?\ngamma2 <- 0\nfinal_size2 <- c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1)\n\noutbreak_size_distr <- function(N, beta, gamma) {\n## total number of possible states\nnstates <- (N + 1) * (N + 2) / 2\n## empty vector to store outbreak size distribution\ndistr <- rep(0, N + 1)\n## probability distribution of states\np_states <- rep(0, nstates)\n## Start with 1 infection event. This could be\n## customised. Note that state 2 corresponds to\n## state (1, 0)\np_states <- 1\n## counter to index over states\nstate <- 1\nfor (z2 in 0:N) {\ndistr[z2 + 1] <- p_states[state];\nstate <- state + 1;\nif ((z2 + 1) > (N - 1)) {\nnext\n}\nfor (z1 in (z2 + 1): (N - 1)) {\np <- 1 / ( 1 + gamma/(beta(N - z1)))\np_states[state + 1] <- p_states[state + 1] + p_states[state] p\np_states[state + N - z2] <- p_states[state + N - z2] + p_states[state] * (1 - p)\nstate <- state + 1\n}\nif (z2 < N) {\np_states[state + N - z2] <- p_states[state + N - z2] + p_states[state]\nstate <- state + 1\n}\n\n}\ndistr\n}\n\ndistr <- outbreak_size_distr(N = 10, beta = 2/9, gamma = 1)\ndistr <- round(distr, 5)\n\ntestthat::expect_equal(distr, final_size1)\n\ndistr <- outbreak_size_distr(N = 10, beta = 2/9, gamma = 0)\ndistr <- round(distr, 5)\n\ntestthat::expect_equal(distr, final_size2)\n\n\nThe above assertion checks that everything works as expected. For a given value of N and gamma, how does beta influence the outbreak size distribution?\n\nN <- 100\nbeta <- seq(from = 10, to = 100, by = 50) / (N - 1)\ngamma <- 1\nfinal_distr <- lapply(beta, function(b) outbreak_size_distr(N, beta = b, gamma))\n\nresults <- data.frame()\nfor (i in seq_along(beta)) {\nresults <- rbind(results, data.frame(infection_rate = rep(beta[i], N + 1),\ndistr = final_distr[[i]]))\n}\nresults$infection_rate <- factor(results$infection_rate)\n\n\nThus the probability of the final size of the outbreak being very small or very large is high and close to 0 for anything in between. What if we change the recovery rate?\n\nN <- 100\nbeta <- 99 / (N - 1)\ngamma <- beta / c(1, 100)\nfinal_distr <- lapply(gamma, function(g) outbreak_size_distr(N, beta, gamma = g))\nresults <- data.frame()\nfor (i in seq_along(gamma)) {\nresults <- rbind(results, data.frame(recovery_rate = rep(gamma[i], N + 1),\ndistr = final_distr[[i]]))\n}\nresults$recovery_rate <- factor(results$recovery_rate)\nx <- rep(0:N, length(gamma))\nggplot(results, aes(x, distr, fill = recovery_rate)) +\ngeom_col(position = \"dodge\")", null, "" ]	[ null, "http://epirecip.es/epicookbook/images/chapters/blackross2015/r_15_1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6480695,"math_prob":0.9986128,"size":2996,"snap":"2021-04-2021-17","text_gpt3_token_len":992,"char_repetition_ratio":0.14739305,"word_repetition_ratio":0.11377245,"special_character_ratio":0.38618156,"punctuation_ratio":0.16964285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99934715,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T01:43:01Z\",\"WARC-Record-ID\":\"<urn:uuid:275b4e11-9319-4774-af8d-c5269b609e4e>\",\"Content-Length\":\"59708\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43dbb094-59e3-4af9-a0a2-851eb3b4008e>\",\"WARC-Concurrent-To\":\"<urn:uuid:fbd3ad16-c434-4874-a5e0-ef5f702d448f>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"http://epirecip.es/epicookbook/chapters/blackross2015/r\",\"WARC-Payload-Digest\":\"sha1:DO7ARML4JW3QOVH5X4TM4UJ5P74SYKB2\",\"WARC-Block-Digest\":\"sha1:Z4LJCFENAK5ETGFVBIXOIZKI3RRJ3FWI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039626288.96_warc_CC-MAIN-20210423011010-20210423041010-00245.warc.gz\"}"}
https://www.digicamdb.com/specs/canon_eos-7d-mark-ii/	[ "# Canon EOS 7D Mark II\n\n### Specs", null, "compare »\n Brand: Canon Model: EOS 7D Mark II Megapixels: 20.20 Sensor: 22.4 x 15 mm Price: check here »\n\n## Sensor info\n\nCanon 7D Mark II comes with a 22.4 x 15 mm CMOS sensor, which has a diagonal of 26.96 mm (1.06\") and a surface area of 336.00 mm².\nDiagonal\n26.96 mm\nSurface area\n336 mm²\nPixel pitch\n4.08 µm\nPixel area\n16.65 µm²\nPixel density\n6 MP/cm²\n\n## Actual sensor size\n\nNote: Actual size is set to screen → change »\nThis is the actual size of the 7D Mark II sensor: 22.4 x 15 mm\nThe sensor has a surface area of 336 mm². There are approx. 20,200,000 photosites (pixels) on this area. Pixel pitch, which is a measure of the distance between pixels, is 4.08 µm. Pixel pitch tells you the distance from the center of one pixel (photosite) to the center of the next.\n\nPixel or photosite area is 16.65 µm². The larger the photosite, the more light it can capture and the more information can be recorded.\n\nPixel density tells you how many million pixels fit or would fit in one square cm of the sensor. Canon 7D Mark II has a pixel density of 6 MP/cm².\n\nThese numbers are important in terms of assessing the overall quality of a digital camera. Generally, the bigger (and newer) the sensor, pixel pitch and photosite area, and the smaller the pixel density, the better the camera. If you want to see how 7D Mark II compares to other cameras, click here.\n\n## Specifications\n\n Brand: Canon Model: EOS 7D Mark II Effective megapixels: 20.20 Total megapixels: 20.90 Sensor size: 22.4 x 15 mm Sensor type: CMOS Sensor resolution: 5486 x 3682 Max. image resolution: 5472 x 3648 Crop factor: 1.6 Optical zoom: Digital zoom: ISO: Auto, 100-16000 (expandable to 51200) RAW support: Manual focus: Normal focus range: Macro focus range: Focal length (35mm equiv.): Aperture priority: Yes Max aperture: Max. aperture (35mm equiv.): n/a Depth of field: simulate → Metering: Evaluative, Partial, Center-weighted, Spot Exposure Compensation: ±5 EV (in 1/3 EV, 1/2 EV steps) Shutter priority: Yes Min. shutter speed: 30 sec Max. shutter speed: 1/8000 sec Built-in flash: External flash: Viewfinder: Optical (pentaprism) White balance presets: 8 Screen size: 3\" Screen resolution: 1,040,000 dots Video capture: Max. video resolution: 1920x1080 (60p/50p/30p/25p/24p) Storage types: SD/SDHC/SDXC/Type I CompactFlash USB: USB 3.0 (5 GBit/sec) HDMI: Wireless: GPS: Battery: Battery Pack LP-E6N (or LP-E6) Weight: 910 g Dimensions: 148.6 x 112.4 x 78.2 mm Year: 2014\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### Canon 7D Mark II diagonal:\n\nw = 22.40 mm\nh = 15.00 mm\n Diagonal = √ 22.40² + 15.00² = 26.96 mm\n\n## Surface area\n\nSurface area is calculated by multiplying the width and the height of a sensor.\n\nWidth = 22.40 mm\nHeight = 15.00 mm\n\nSurface area = 22.40 × 15.00 = 336.00 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### Canon 7D Mark II pixel pitch:\n\nSensor width = 22.40 mm\nSensor resolution width = 5486 pixels\n Pixel pitch = 22.40 × 1000 = 4.08 µm 5486\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### Canon 7D Mark II pixel area:\n\nPixel pitch = 4.08 µm\n\nPixel area = 4.08² = 16.65 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\nYou could also use this formula:\n Pixel density = effective megapixels × 1000000 / 10000 sensor surface area in mm²\n\n### Canon 7D Mark II pixel density:\n\nSensor resolution width = 5486 pixels\nSensor width = 2.24 cm\n\nPixel density = (5486 / 2.24)² / 1000000 = 6 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### Canon EOS 7D Mark II sensor resolution:\n\nSensor width = 22.40 mm\nSensor height = 15.00 mm\nEffective megapixels = 20.20\nr = 22.40/15.00 = 1.49\n X = √ 20.20 × 1000000 = 3682 1.49\nResolution horizontal: X × r = 3682 × 1.49 = 5486\nResolution vertical: X = 3682\n\nSensor resolution = 5486 x 3682\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### Canon 7D Mark II crop factor:\n\nSensor diagonal = 26.96 mm\n Crop factor = 43.27 = 1.6 26.96\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### Canon EOS 7D Mark II 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 Canon EOS 7D Mark II, take the aperture of the lens you're using and multiply it with crop factor.\n\nCrop factor for Canon 7D Mark II is 1.6\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/specs/canon_eos-7d-mark-ii/images/cameras/canon_eos-7d-mark2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78312826,"math_prob":0.97446597,"size":8767,"snap":"2021-43-2021-49","text_gpt3_token_len":2624,"char_repetition_ratio":0.18737875,"word_repetition_ratio":0.3919598,"special_character_ratio":0.31755447,"punctuation_ratio":0.14216739,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9707754,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T05:00:45Z\",\"WARC-Record-ID\":\"<urn:uuid:2633121e-d593-4f2d-8846-3482c4833d6f>\",\"Content-Length\":\"33873\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a268cdcd-de32-480e-a264-02d5347b040a>\",\"WARC-Concurrent-To\":\"<urn:uuid:36b96b45-babc-4ca6-8e7b-1ec4184bb007>\",\"WARC-IP-Address\":\"173.248.187.251\",\"WARC-Target-URI\":\"https://www.digicamdb.com/specs/canon_eos-7d-mark-ii/\",\"WARC-Payload-Digest\":\"sha1:6RNSTDS4VSF7S567KXKT36L75D4UZAK2\",\"WARC-Block-Digest\":\"sha1:ONA5V723UDVP6GG62EY3WQVSZ5E5OL3U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587623.1_warc_CC-MAIN-20211025030510-20211025060510-00311.warc.gz\"}"}
https://answers.everydaycalculation.com/lcm/3-252	[ "Solutions by everydaycalculation.com\n\n## What is the LCM of 3 and 252?\n\nThe lcm of 3 and 252 is 252.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 3\n3 = 3\n2. Find the prime factorization of 252\n252 = 2 × 2 × 3 × 3 × 7\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 2 × 2 × 3 × 3 × 7\n4. LCM = 252\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.731256,"math_prob":0.9992824,"size":465,"snap":"2019-43-2019-47","text_gpt3_token_len":156,"char_repetition_ratio":0.12581345,"word_repetition_ratio":0.0,"special_character_ratio":0.41075268,"punctuation_ratio":0.08421053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99620676,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T17:19:05Z\",\"WARC-Record-ID\":\"<urn:uuid:e3c351dd-3dc0-4795-880b-c3770cba6402>\",\"Content-Length\":\"5778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30dd466d-0bcb-48f2-b37c-5c0725d352e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e92ad32-c3bd-429d-a4a4-20f14f0e0f4c>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/lcm/3-252\",\"WARC-Payload-Digest\":\"sha1:G4Q56WXFPFIZCELL3PFK6ELEHOLE42A6\",\"WARC-Block-Digest\":\"sha1:HZNCCEIZZZYGRZYSV7A7G6VFBD5456AL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670597.74_warc_CC-MAIN-20191120162215-20191120190215-00175.warc.gz\"}"}
https://studysoup.com/tsg/13548/introductory-chemistry-5-edition-chapter-6-problem-96p	[ "×\nGet Full Access to Introductory Chemistry - 5 Edition - Chapter 6 - Problem 96p\nGet Full Access to Introductory Chemistry - 5 Edition - Chapter 6 - Problem 96p\n\n×\n\n# A 2.241-g sample of nickel reacts with oxygen to form", null, "ISBN: 9780321910295 34\n\n## Solution for problem 96P Chapter 6\n\nIntroductory Chemistry \| 5th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Introductory Chemistry \| 5th Edition\n\n4 5 1 277 Reviews\n25\n2\nProblem 96P\n\nA 2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide. Calculate the empirical formula of the oxide.\n\nStep-by-Step Solution:\nStep 1 of 3\n\nSolution 96P\n\nThe empirical formula can be found using the table given below.The number of moles of each elements is obtained and the simplest whole number ratio in which the elements are present in the compound is found.\n\n Elements Mass of the elements Atomic weight Number of moles Ni 2.241 59", null, "=0.03 O 0.611 16", null, "=0.03\n\nThe simplest ratio in which the elements Nickel and oxygen is present is 1:1.So the empirical formula should be NiO\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780321910295\n\nThe answer to “?A 2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide. Calculate the empirical formula of the oxide.” is broken down into a number of easy to follow steps, and 23 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 96P from 6 chapter was answered, more than 705 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 96P from chapter: 6 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: oxide, nickel, form, formula, metal. This expansive textbook survival guide covers 19 chapters, and 2046 solutions.\n\n## Discover and learn what students are asking\n\nCalculus: Early Transcendental Functions : Integration Techniques, LHpitals Rule, and Improper Integrals\n?Verify the reduction formula $$\\int(\\ln x)^{n} d x=x(\\ln x)^{n}-n \\int(\\ln x)^{n-1} d x$$.\n\nCalculus: Early Transcendental Functions : Basic Differentiation Rules and Rates of Change\n?Finding a Value In Exercises 65–70, find k such that the line is tangent to the graph of the function. Function\n\nCalculus: Early Transcendental Functions : Hyperbolic Functions\n?In Exercises 7-14, verify the identity. $$\\tanh ^{2} x+\\operatorname{sech}^{2} x=1$$\n\nStatistics: Informed Decisions Using Data : Estimating a Population Standard Deviation\n?True or False: The chi-square distribution is symmetric.\n\nStatistics: Informed Decisions Using Data : Testing the Significance of the Least-Squares Regression Model\n?In Problems 5–10, use the results of Problems 7–12, respectively, from Section 4.2 to answer the following questions: (a) What are the estima\n\nChemistry: The Central Science : The Chemistry of Life: Organic and Biological Chemistry\n?Identify the functional groups in each of the following compounds:\n\nUnlock Textbook Solution" ]	[ null, "https://studysoup.com/cdn/24cover_2610068", null, "https://studysoup.com/cdn/24cover_2610068", null, "https://chart.googleapis.com/chart", null, "https://chart.googleapis.com/chart", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8808639,"math_prob":0.8892287,"size":583,"snap":"2022-27-2022-33","text_gpt3_token_len":154,"char_repetition_ratio":0.13471502,"word_repetition_ratio":0.018691588,"special_character_ratio":0.2864494,"punctuation_ratio":0.09917355,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982158,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T02:24:00Z\",\"WARC-Record-ID\":\"<urn:uuid:0254051c-11e8-4e98-803a-688f066a8158>\",\"Content-Length\":\"99469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:afe1dd03-d957-48a9-a8f6-24248060baf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b8ffd0a-affc-4f0a-98fa-890fec48ff18>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/13548/introductory-chemistry-5-edition-chapter-6-problem-96p\",\"WARC-Payload-Digest\":\"sha1:7DE6MXYMPOYZEXX26N724GQZ45WH7OWU\",\"WARC-Block-Digest\":\"sha1:B2XPQQA5GUQZIYIIPFAUFWLSN2AEXXOR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103646990.40_warc_CC-MAIN-20220630001553-20220630031553-00004.warc.gz\"}"}
https://acp.copernicus.org/articles/18/10025/2018/	[ "Atmos. Chem. Phys., 18, 10025–10038, 2018\nhttps://doi.org/10.5194/acp-18-10025-2018\nAtmos. Chem. Phys., 18, 10025–10038, 2018\nhttps://doi.org/10.5194/acp-18-10025-2018\n\nResearch article 16 Jul 2018\n\nResearch article \| 16 Jul 2018", null, "# Turbulent transport of energy across a forest and a semiarid shrubland\n\nTurbulent transport of energy across a forest and a semiarid shrubland\nTirtha Banerjee1,a, Peter Brugger1, Frederik De Roo1, Konstantin Kröniger1, Dan Yakir2, Eyal Rotenberg2, and Matthias Mauder1 Tirtha Banerjee et al.\n• 1Karlsruhe Institute of Technology (KIT), Institute of Meteorology and Climate Research, Atmospheric Environmental Research (IMK-IFU), 82467 Garmisch-Partenkirchen, Germany\n• 2Department of Earth and Planetary Sciences (EPS), The Weizmann Institute of Science, Rehovot 76100, Israel\n• acurrent address: Earth and Environmental Sciences Division, Los Alamos National Laboratory, Los Alamos, New Mexico, USA\n\nCorrespondence: Tirtha Banerjee (tirtha.banerjee@lanl.gov)\n\nAbstract\n\nThe role of secondary circulations has recently been studied in the context of well-defined surface heterogeneity in a semiarid ecosystem where it was found that energy balance closure over a desert–forest system and the structure of the boundary layer was impacted by advection and flux divergence. As a part of the CliFF (“Climate feedbacks and benefits of semi-arid forests”, a collaboration between KIT, Germany, and the Weizmann Institute, Israel) campaign, we studied the boundary layer dynamics and turbulent transport of energy corresponding to this effect in Yatir Forest situated in the Negev Desert in Israel. The forest surrounded by small shrubs presents a distinct feature of surface heterogeneity, allowing us to study the differences between their interactions with the atmosphere above by conducting measurements with two eddy covariance (EC) stations and two Doppler lidars. As expected, the turbulence intensity and vertical fluxes of momentum and sensible heat are found to be higher above the forest compared to the shrubland. Turbulent statistics indicative of nonlocal motions are also found to differ over the forest and shrubland and also display a strong diurnal cycle. The production of turbulent kinetic energy (TKE) over the forest is strongly mechanical, while buoyancy effects generate most of the TKE over the shrubland. Overall TKE production is much higher above the forest compared to the shrubland. The forest is also found to be more efficient in dissipating TKE. The TKE budget appears to be balanced on average both for the forest and shrubland, although the imbalance of the TKE budget, which includes the role of TKE transport, is found to be quite different in terms of diurnal cycles for the forest and shrubland. The difference in turbulent quantities and the relationships between the components of TKE budget are used to infer the characteristics of the turbulent transport of energy between the desert and the forest.\n\nShare\n1 Introduction\n\nUnderstanding the interaction between vegetation canopies and atmosphere is a crucial component in the quantification of biosphere–atmosphere exchange of heat, carbon dioxide, water and trace gas fluxes. It is also important for the development of numerical weather and climate models where the fluxes in the canopy surface layer (CSL) and the atmospheric surface layer (ASL) are parameterized through bulk exchange coefficients of momentum and scalar. However, idealizations of the forest canopies as horizontally homogeneous momentum sinks and scalar sources introduces uncertainties in flux estimations and estimating diffusion coefficients. The presence of heterogeneities such as roughness transitions, complex topography and mesoscale circulations are common sources of such uncertainties that give rise to nonlocal motions and secondary circulations. These secondary circulations not only occur in forests but are also generic characteristics of boundary layer flows over natural and man-made landscapes with discongruity of land use types, surface moisture, temperature, etc. . Different types of land cover such as agricultural lands or urban areas can affect local energy balance closure and the structure of the overlying boundary layer as well as cloud formation and regional weather . Strong differences in surface properties and large swaths of such surface patches are known to induce secondary circulations . Recent works by , and have suggested that non-closure of the energy balance is also related to advection and flux divergence due to secondary circulations . The non-closure of the energy balance refers to the fact that the available energy RnG is often higher than the turbulent energy H+LE at micrometeorological sites, where Rn is net radiation, G is soil heat flux, H is sensible heat flux and LE is latent heat flux. Thus, it is established that studies involving surface heterogeneities such as a difference in roughness characteristics and albedo are crucial for the advancements of our understanding of biosphere–atmosphere interaction since the quasi-universal scaling laws of turbulent moments and simple parametrizations of exchange coefficients are disturbed and rendered nonoperational.\n\nSeveral studies have attempted to study the nature of turbulence across a roughness transition such as a grassland and a forest canopy by means of experimental and numerical methods and documented several length scales associated with the roughness transitions, recirculation zones and the nature of the turbulent momentum budget. However, all of these studies are concerned with the flow adjustment in the immediate vicinity of the roughness transition (edges or gaps). have studied the dynamics of the convective boundary layer over a well-defined surface heterogeneity – namely Yatir Forest and the shrubland surrounding it, which are located in the northern part of the Negev Desert in Israel. Eddy covariance (EC) and Doppler lidar measurements were conducted by at two sites approximately 6.5 km apart: one in the forest and one in the desert. The forest has a darker surface and consequently lower albedo (12.5 %) than the desert (33.7 %). Moreover, the higher surface roughness of the forest results in higher turbulence intensity, which leads to more efficient heat transfer above the forest, a phenomenon called canopy convector effect . The region being very dry, there is very little latent heat flux (Bowen ratio >10 over the summer), resulting in a spatial difference in surface buoyancy flux of 220–290 W m−2 between the desert and the forest. Furthermore, the length scale of surface heterogeneities (6–10 km) is larger than the minimal length scale needed for the development of secondary circulations: ${L}_{\\mathrm{rau}}={C}_{\\mathrm{Rau}}U/{w}_{}\\approx \\mathrm{2}$–5 km , where U is mean wind speed, w is the convective velocity scale and CRau=0.8 is an empirical parameter, so that it is possible for secondary circulations to develop.\n\nThe present work is an attempt to examine this hypothesis of secondary circulations in more detail. We use eddy covariance and Doppler lidar measurements at two sites 4.3 km apart over the shrubland and Yatir Forest, where the shrubland is upwind of the forest in the path of the principal wind direction (during the summer, there exists a heat-induced low-pressure system to the east, resulting in the main wind direction from the northwest). We investigate the individual components of the turbulent kinetic energy budget, as well as the nature of advection and turbulent transport over the forest and the desert and determine if there is a relationship between them. Not many instances were found in the literature where the nature of turbulent transport was studied across large-scale surface roughness heterogeneities, except for and . However, only studied turbulent production and the turbulent velocity fluctuations in the presence of a complex topography – so the nature of turbulent transport via secondary circulations was not highlighted. studied the decay of turbulence over different land surface types. Hence, the difference in turbulence production and simultaneous transport across different land use types was not studied, which determines the scope of the current work.\n\n2 Method\n\n## 2.1 Theory\n\nThe turbulent kinetic energy (TKE) budget is given by without invoking any special assumption:\n\n$\\begin{array}{ll}\\text{(1)}& & \\frac{\\partial e}{\\partial t}+{U}_{j}\\frac{\\partial e}{\\partial {x}_{j}}={\\mathit{\\delta }}_{i\\mathrm{3}}\\frac{g}{T}\\left(\\stackrel{\\mathrm{‾}}{{u}_{i}^{\\prime }{T}^{\\prime }}\\right)-\\stackrel{\\mathrm{‾}}{{u}_{i}^{\\prime }{u}_{j}^{\\prime }}\\frac{\\partial {U}_{i}}{\\partial {x}_{j}}-\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}_{j}^{\\prime }e}\\right)}{\\partial {x}_{j}}& -\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}_{i}^{\\prime }{p}^{\\prime }}\\right)}{\\partial {x}_{i}}-\\mathit{ϵ},\\end{array}$\n\nwhere i and j are the usual tensor indices, which can take the values of 1, 2 and 3 to indicate x, y and z directions, respectively, and δi3 is the Kronecker delta. $e=\\left(\\mathrm{1}/\\mathrm{2}\\right)\\left({\\mathit{\\sigma }}_{u}^{\\mathrm{2}}+{\\mathit{\\sigma }}_{v}^{\\mathrm{2}}+{\\mathit{\\sigma }}_{w}^{\\mathrm{2}}\\right)=\\left(\\mathrm{1}/\\mathrm{2}\\right)\\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime \\mathrm{2}}}+\\stackrel{\\mathrm{‾}}{{v}^{\\prime \\mathrm{2}}}+\\stackrel{\\mathrm{‾}}{{w}^{\\prime \\mathrm{2}}}\\right)$ is the TKE, U denotes mean longitudinal velocity; u, v and w denote the fluctuations from mean for the longitudinal, transverse and vertical velocity components; g is acceleration due to gravity; T denotes mean potential temperature; T is the potential temperature fluctuation; p is the dynamic pressure perturbation; ρ is density of air. The first term on the left-hand side (LHS) denotes storage or TKE tendency. The second term on the LHS indicates advection of TKE by mean wind flow. The first term on the right-hand side (RHS) denotes buoyant production/destruction of TKE. The second term on the RHS denotes mechanical/shear production of TKE. The third term on RHS denotes turbulent transport of TKE and can also be called turbulent flux divergence. The fourth term on RHS denotes transport of TKE by pressure velocity correlation. ϵ is the dissipation of TKE.\n\nExpanding the equations in terms of x, y and z coordinates, the full TKE budget can be written as Eq. (A1) as shown in Appendix A. Since it is difficult to keep track of the full equation due to the large number of terms, it would be easier to use a simple form of the TKE budget (Stull2012)\n\n$\\begin{array}{}\\text{(2)}& \\mathrm{0}=-\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}\\frac{\\mathrm{d}U}{\\mathrm{d}z}+\\frac{g}{T}\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}-\\mathit{ϵ}-\\text{Imbalance},\\end{array}$\n\nwhere the “imbalance” is defined in Eq. (A2). Note that $\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}$ and $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}$ denote vertical momentum flux and sensible heat flux, respectively. Also note that if the term imbalance is set to zero, one recovers the TKE budget for an idealized surface layer where the coordinate system is aligned with the mean wind, and a planar, homogeneous flow with zero subsidence is assumed. Since our objective in the current problem is to study the effect of heterogeneity, we cannot make these assumptions. Moreover, we are also constrained by being able to measure only at two single points in space quite far apart. Single point eddy covariance measurements cannot compute spatial gradients, and the pressure perturbations are not measured either. Thus, explicit computations of the imbalance terms are not possible. Due to the three-dimensional nature of the problem, it is also difficult to anticipate what degrees of assumptions are sufficient, so that some of the terms can be ignored safely.\n\nUnder these constraints, a strategy is needed to evaluate the TKE budget. The dominant mechanical production term, the buoyant production/destruction term and the dissipation term will be evaluated directly from the data. The residual of the TKE budget will be described as the imbalance as per Eq. (3) which would contain the effects of advection and transport terms. The advantage of using this strategy is that since the original TKE budget equation has to be closed, the errors in computing the production and dissipation terms can also be assumed to be inside the imbalance term.\n\n$\\begin{array}{}\\text{(3)}& \\text{Imbalance}=-\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}\\frac{\\mathrm{d}U}{\\mathrm{d}z}+\\frac{g}{T}\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}-\\mathit{ϵ}.\\end{array}$\n\nTo compute the mechanical production term, we momentarily assume that the TKE budget is well balanced and Monin–Obukhov similarity theory (MOST) is valid . This allows us to write\n\n$\\begin{array}{}\\text{(4)}& \\frac{\\mathrm{d}U}{\\mathrm{d}z}={\\mathit{\\varphi }}_{\\mathrm{m}}\\left(\\mathit{\\zeta }\\right)\\frac{{u}_{}}{\\mathit{\\kappa }z},\\end{array}$\n\nwhere ϕm is the stability correction function for momentum which varies with the stability parameter $\\mathit{\\zeta }=\\left(z-d\\right)/L$ and κ=0.4, the von Kármán constant. ${u}_{}=\\sqrt{{\\left\|\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}\\right\|}^{\\mathrm{2}}+{\\left\|\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{w}^{\\prime }}\\right\|}^{\\mathrm{2}}}$ is the friction velocity, z is the measurement height, and $L=-{u}_{}^{\\mathrm{3}}/\\left(\\mathit{\\kappa }\\left(g/T\\right)\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}\\right)$ is the Obukhov length; d is zero plane displacement height, taken as 2∕3 of canopy height. The standard MOST scaling relations for ϕm are used, i.e., ${\\mathit{\\varphi }}_{\\mathrm{m}}=\\mathrm{0.74}+\\mathrm{4.7}\\mathit{\\zeta }$ for stable (ζ>0) and ${\\mathit{\\varphi }}_{\\mathrm{m}}=\\left(\\mathrm{1}-\\mathrm{16}\\mathit{\\zeta }{\\right)}^{-\\mathrm{1}/\\mathrm{4}}$ for unstable (ζ<0) stratification .\n\nEquation (4) allows us to compute the mechanical production term in Eq. (2) as\n\n$\\begin{array}{}\\text{(5)}& -\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}\\frac{\\mathrm{d}U}{\\mathrm{d}z}={\\mathit{\\varphi }}_{\\mathrm{m}}\\frac{{u}_{}^{\\mathrm{3}}}{\\left(\\mathit{\\kappa }z\\right)}.\\end{array}$\n\nThe buoyancy term can directly be computed from the EC measurements as well. To compute the dissipation term ϵ, we use the scaling relation of second-order structure function ${D}_{uu}=\\stackrel{\\mathrm{‾}}{{\\left[u\\left(x+r\\right)-u\\left(x\\right)\\right]}^{\\mathrm{2}}}$ in the inertial subrange\n\n$\\begin{array}{}\\text{(6)}& {D}_{uu}\\left(r\\right)={C}_{u}{\\mathit{ϵ}}^{\\mathrm{2}/\\mathrm{3}}{r}^{\\mathrm{2}/\\mathrm{3}},\\end{array}$\n\nwhere Cu≈2 (Stull2012) and r is the spatial lag in the longitudinal direction, which can be computed by multiplying the sampling time interval with the mean longitudinal velocity, assuming that Taylor's frozen turbulence hypothesis is valid (r=\|u\|Δt). The range of r where this relation is valid is found to be between 0.2 to 2 m, and ϵ is found by regression of Eq. (6). Note that the computation of ϵ is independent of any assumptions used to compute the production terms.\n\n## 2.2 Research site", null, "Figure 1Map of Yatir Forest in Israel and locations of the measurement stations. Insets: snapshots of measurement setups. Bottom panel: topography map of Yatir Forest from Google maps. The blue arrow indicates north.\n\nThe measurements were conducted in Yatir Forest and the surrounding shrubland in Israel between 18 and 30 August 2015 as part of the “Climate feedbacks and benefits of semi-arid forests” (CliFF) campaign, a joint collaboration between Karlsruhe Institute of Technology (KIT), Germany, and the Weizmann Institute, Israel. Figure 1 gives an idea of the locations of the EC towers. Tower 1 (lat 31.375728, long 35.024262) was located in the semiarid shrubland 620 m above sea level and tower 2 (lat 31.345315, long 35.052224) was located inside the forest 660 m above sea level. The linear distance between the two locations was measured to be 4.3 km, and as can be observed from Fig. 1, there is a distinct surface heterogeneity between the two sites. The climate of the area is in between Mediterranean and semiarid, with a mean annual precipitation of about 285 mm . Note that the measurement sites reported in this work are different from those in . The trees in the forest were mostly Aleppo pine (Pinus halepensis), with an average height of 10 m with negligible height variation. The surrounding land was sparsely populated by small shrubs, and in the dry season, when the measurements were conducted, was mostly free of vegetation. Thus, it is referred to as “desert” for easy distinction . The measurement height for the forest was 19 m above ground (9 m above the canopy height). Note that with this height selection, the measurements were conducted above the roughness sub-layer, which ends at approximately 2 times the canopy height . A mast was used over the desert and the measurement height was 9 m until 23 August, after which it was changed to 15 m for the remaining period. In this zone of the atmospheric surface layer, the longitudinal and crosswise velocity variances decrease logarithmically with height and the vertical velocity variance shows independence from height . High-frequency turbulent data were collected at 20 Hz and 30 min averaging periods were used for both sites. After conducting quality control of the data following , a planar fit coordinate rotation is applied to the velocity components since the data are collected on a sloped ground. The coordinate rotation following ensures that the cross stream velocity component v is zero and corrects the tilting of the anemometer with respect to the local streamlines. Moreover, a different set of coordinate rotation is applied for the desert data after 23 August.\n\nIn addition, two Doppler lidars were used at the two locations which measured vertical velocities . The Doppler lidars used were StreamLine systems from HaloPhotonics. They were operated in a vertical stare mode most of the time (interrupted every half hour for less than 90 s). Technical specifications and instrument settings of the Doppler lidars are given in Table 1. The Doppler lidar at tower 1 was not working from 19 August 2015, 15:00 UTC, until 21 August 2015, 10:30 UTC and very briefly on the 23 August 2015 around 10:00 UTC due to power cuts.\n\nTable 1Instrument specification and settings of the Doppler lidars. From top to bottom: serial number of the forest and desert lidar, pulse length of the laser pulse at full width at half maximum, range gate length, pulse repetition frequency, number of averaged pulses for a backscatter coefficient profile, and the wavelength of the emitted laser pulse (short wavelength infrared).", null, "", null, "Figure 2Time series of half-hourly averages of mean speed (ms−1), mean vertical velocity (ms−1), friction velocity (ms−1) and mean potential temperature (K) for the measurement period. The black line indicates desert, and the red line indicates forest.", null, "Figure 3Time series of longitudinal velocity variance (m2 s−2), vertical velocity variance (m2 s−2), momentum flux (m2 s−2) and sensible heat flux (K ms−1) for the measurement period. The black line indicates desert, and the red line indicates forest.\n\n3 Results and discussion\n\n## 3.1 Time series of turbulence statistics\n\nTime series of mean speed (ms−1), mean vertical velocity (W, ms−1) (after applying coordinate rotation), friction velocity (u, ms−1) and mean near-surface air (potential) temperature (T, K) for the measurement period are shown in Fig. 2. Figure 3 shows time series of longitudinal velocity variance ($\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{u}^{\\prime }}$, m2 s−2), vertical velocity variance ($\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$, m2 s−2), momentum flux ($\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}$, m2 s−2) and sensible heat flux ($\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}$, K ms−1). The black line indicates desert, and the red line indicates forest. As noted, the desert is associated with a higher wind speed because of a lower amount of friction on the desert surface. The higher vertical velocity over the desert indicates the presence of stronger updrafts, which would be explained by higher buoyancy-driven turbulence. The friction velocity (u) over the forest is much higher compared to the desert, especially in the daytime, which is expected because of higher surface roughness over the forest. u, above both the forest and the desert, shows a strong diurnal cycle. However, there seems to be a prominent increase in u over the desert after 23 August. This can be attributed to the raising of the tower height. Moreover, the gentle topography around the desert could result in the strong vertical updrafts above the desert. Interestingly, the near-surface air temperatures over both the forest and the desert show a strong diurnal cycle and their differences are about 5 K on average during daytime and almost zero at night.\n\nThe longitudinal velocity variance $\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{u}^{\\prime }}$ over the forest and the desert show similar variations over time. The vertical velocity variance $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$ over the forest is higher than its desert counterpart; however, after 23 August, the levels of $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$ over the desert increase as well and become similar to the forest. This is due to changing the tower height. As the vertical profiles of $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$ are different between the desert and the forest (due to roughness length differences), the observed differences between $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$ are a function of observation height. At 15 m above the desert and 19 m above the forest floor, high enough to be in the “constant flux layer”, the vertical profiles of TKE ($\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{u}^{\\prime }}+\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$) converge. However, when observed at a lower elevation and below the constant flux layer, the data show clear differences in $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}$.\n\nThe vertical momentum flux $\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}$ over the forest is much higher compared to the desert, which is also expected because of the higher surface roughness of the forest, making it a much more efficient momentum sink compared to the desert. Note that the shear transport of momentum flux is still much more effective over the forest compared to the desert because of roughness effects even though the mean quantities can be higher over the desert. The sensible heat flux $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}$ over the forest is also higher, as discussed before, due to the canopy convector effect.", null, "Figure 4Time series of mechanical production of TKE (m2 s−3), buoyant production of TKE (m2 s−3), full TKE production (m2 s−3), dissipation of TKE (m2 s−3) and imbalance of TKE (m2 s−3). The black line indicates desert, and the red line indicates forest.\n\n## 3.2 Nature of TKE budget\n\nFigure 4 shows the time series of the components of the TKE budget as discussed in Sect. 2.1. The first row shows mechanical production of TKE (PMech, m2 s−3); the second row shows buoyant production of TKE (PBuoy, m2 s−3); the third row shows full TKE production (PTKE, m2 s−3), which is the sum of mechanical and buoyant TKE production. The fourth row shows dissipation of TKE (ϵ, m2 s−3) and the fifth row shows an imbalance of TKE (Imb, m2 s−3). The black line indicates desert, and the red line indicates forest. As noted in Fig. 4, the production of turbulence is mostly by mechanical or shear forcing because of the roughness of the forest, whereas mechanical production of TKE over the desert is very small and does not have a strong diurnal cycle like the forest, although it increases slightly after 23 August. On the other hand, TKE production over the desert is mostly carried by buoyancy. Buoyant TKE production is slightly larger over the forest. The buoyant TKE production over the desert is also higher after 23 August. Given the moderate temperature difference between the desert and the forest, the difference in their corresponding buoyant TKE production is interesting. It also indicates that mechanical forcing and not buoyancy makes a difference (mechanical production is higher by approximately 1 order of magnitude than buoyant production) in the turbulence generation over the desert and the forest. The diurnal cycle of the TKE dissipation ϵ is interesting as well. The dissipation of TKE seems to be higher above the forest as well compared to the desert.\n\nA smaller TKE dissipation is recorded when the measurement location is further from the ground and above the roughness sub-layer. One strong argument for observed changes after 23 August being tower-height effects rather than a change in any large-scale forcing is that changes in the desert are observed only after the 23 August, while the forest observations maintain rather consistent dynamics.\n\nThe diurnal cycles of the TKE imbalance computed by Eq. (3) are also very interesting. The imbalance over the forest is often positive over the daytime, while over the desert it is often negative, highlighting the difference in turbulent transport and advection over the two different regimes. Also note that the positive imbalance for the forest and negative imbalance for the desert almost have a phase (anti-)synchronization, indicting that the turbulence above the forest and the desert are responsive to one another and that they are part of a coupled system, indicating again the role of the secondary circulations.\n\n## 3.3 Transport of TKE over the desert and the forest\n\nFigure 5 is used to better understand the nature of turbulent transport between the desert and the forest. Panel (a) depicts the TKE imbalance over the desert vs. the net production of TKE over the forest. As observed, there is a significant correlation (0.5) between them, indicating that the advection and transport of TKE by flux divergence and pressure fluctuations reach downstream by means of the secondary circulations and produce TKE over the forest. On the other hand, the reverse is not true, as observed in panel (b) of Fig. 5. There is little correlation between the imbalance of TKE over the forest and the production of TKE over the desert (0.14). As observed in panel (c), the production over the desert is also well correlated with the production over the forest (0.3) as both the desert and the forest are subject to the same forcing. However, the TKE production over the desert is not that well correlated with the TKE imbalance over the desert as seen in panel (d). Thus, while there should be some cross correlation in panel (a) because of desert production, that is not the only effect. The nonlocal large-scale motions contribute to the transport over the desert (without significantly altering TKE production over the desert) which in turn cause TKE production above the forest because of the higher mechanical forcing.\n\nThus, it can be stated that at least in the canopy sub-layer and in the atmospheric surface layer, the effects of secondary circulations are transported from over the desert towards the forest following the background wind direction, and it is not the other way around. It is worth noting here that the term “secondary circulation” has been used somewhat loosely here and contain the effects of horizontal transport as well, since partitioning the imbalance term is not possible within the scope of this campaign. In the case of transport from the forest towards the desert, it is more likely that horizontal advection is the main mechanism. The nature of the full extent of the secondary circulation are a part of a much larger flow pattern and are not fully captured by the eddy covariance towers, which only capture the fine-scale turbulence. To reveal the full nature of the secondary circulations, one can look at lidar observations as shown in Appendix B as well as large eddy simulations .", null, "Figure 5(a) TKE imbalance for the desert vs. TKE production for the forest. (b) TKE imbalance for the forest vs. TKE production for the desert. (c) TKE production for the desert vs. TKE production for the forest. (d) TKE production for the desert vs. TKE imbalance for the desert. Significance: 0.05 level.\n\n## 3.4 Effect of nonlocal motions\n\nFigure 6 shows the time series of the triple moments $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{u}^{\\prime }}$, $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{w}^{\\prime }}$ and $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{T}^{\\prime }}$ in the first three rows. The vertical velocity skewness term $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{w}^{\\prime }}$ (second row) is of importance as it appears in the transport term of the TKE budget (Eq. 2) and is a measure of non-Gaussian turbulence, which indicates the presence of nonlocal coherent motions such as sweeps and ejections. Note that the vertical velocity skewness is often negative above the canopy, which is consistent with the generic feature of canopy turbulence . The daytime vertical velocity skewness over the desert is often positive, indicating again the presence of nonlocal coherent structures active over the desert. The measure of skewness increases over the desert after 23 August, which is also due to the height change. The other two terms $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{u}^{\\prime }}$ and $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{T}^{\\prime }}$ are also associated with turbulent transport of momentum and heat as evident from their respective budget equations .\n\n$\\begin{array}{ll}\\text{(7)}& & \\frac{\\partial \\stackrel{\\mathrm{‾}}{{w}^{\\prime }{u}^{\\prime }}}{\\partial t}=\\mathrm{0}=-\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}\\frac{\\partial \\stackrel{\\mathrm{‾}}{U}}{\\partial z}-\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{u}^{\\prime }}\\right)}{\\partial z}+\\frac{g}{\\stackrel{\\mathrm{‾}}{T}}\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{T}^{\\prime }}& -\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime }\\frac{\\partial {p}^{\\prime }}{\\partial z}}\\right),\\end{array}$\n\nand\n\n$\\begin{array}{ll}\\text{(8)}& & \\frac{\\partial \\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}}{\\partial t}=\\mathrm{0}=-\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}\\frac{\\partial \\stackrel{\\mathrm{‾}}{T}}{\\partial z}-\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{T}^{\\prime }}\\right)}{\\partial z}+\\frac{g}{\\stackrel{\\mathrm{‾}}{T}}\\stackrel{\\mathrm{‾}}{{T}^{\\prime }{T}^{\\prime }}& -\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\left(\\stackrel{\\mathrm{‾}}{{T}^{\\prime }\\frac{\\partial {p}^{\\prime }}{\\partial z}}\\right).\\end{array}$", null, "Figure 6Top three panels: time series of triple moments $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{u}^{\\prime }}$, $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{w}^{\\prime }}$ (m3 s−3) and $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{T}^{\\prime }}$ (K m2 s−2). Bottom two panels show the integral timescales of horizontal (Inu) and vertical velocities (Inw) in seconds.\n\nMoreover, the triple moments have been shown to be directly correlated with the relative contributions of nonlocal events such as sweeps and ejections . Note that momentum transport term $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{u}^{\\prime }}$ is also opposite in sign for the desert and the forest, and it shows a strong diurnal cycle. After 23 August, an increase in momentum transport is noted for the desert. However, the diurnal cycle of the heat transport term $\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }{T}^{\\prime }}$ is not as strong as its momentum counterpart, but it is often found to be larger over the desert compared to the forest, consistent with the findings from the TKE budget that show heat is transported from over the desert towards the forest. It is, however, important to note that the structures are representative of the fine-scale turbulence and not directly representative of the large-scale circulation structures spanning the whole boundary layer. The fourth and fifth rows of Fig. 6 show the time series of the integral timescale of horizontal (Inu) and vertical (Inw) velocity components in seconds. Inu and Inw for every half hour time period are computed by integrating the normalized autocorrelation function of u and w until the first zero crossing . They can be interpreted as the characteristic timescale of the most energetic eddies in each direction. As noted in Fig. 6, timescales in the horizontal directions are larger compared to the vertical direction. More interesting is the observation that the integral timescales for the eddies above the desert are larger than those above the forest, which also increase after 23 August. This is another indicator of buoyant production of turbulence, which generates larger eddies than shear production.\n\n4 Conclusions\n\nWe studied the nature of turbulent transport over a well-defined surface heterogeneity, comprising a desert and forest in the Yatir semiarid area in Israel. Eddy covariance and Doppler lidar measurements were conducted for 12 days between 18 and 31 August 2015 over two locations in the forest and the shrubland (referred to as “desert” because of the almost complete lack of vegetation during the observation period). Earlier campaigns in this area focused on energy balance closure and hypothesized that there are secondary circulations because of surface heterogeneity. The present work was aimed to study the nature of turbulent transport over the forest and the desert in more detail to address the following questions:\n\n1. How does Yatir Forest affect the boundary layer dynamics such as eddy size distribution, boundary layer height and diurnal variations in turbulent statistics and fluxes compared to the surrounding desert?\n\n2. Can the existence of secondary circulation be confirmed?\n\n3. Is there any horizontal energy transport between the forest and the desert and how does it vary with time?\n\nTo answer the abovementioned questions, we computed half hour average turbulent statistics for both the desert and the forest and looked at their diurnal variations. We also computed individual components of the turbulent kinetic energy (TKE) budget and argued that the turbulent transport of energy should be contained in the imbalance of the TKE budget, which consists of the effects of advection, transport by turbulent flux divergence and pressure velocity interactions, since we could not compute those terms explicitly. Moreover, we also computed triple moments, which are associated with nonlocal motions and coherent structures, and integral timescales, which are associated with the most energetic eddies. The findings to the questions are listed below.\n\n1. The forest is found to be associated with a higher level of turbulent intensity because of higher roughness although the desert had higher mean speeds and vertical updrafts, possibly due to the presence of secondary circulations. Gentle topography around the desert might contribute to the updrafts over the desert as well. The higher roughness of the desert is also responsible for higher wind speeds above the desert. There is little air temperature difference between the desert and the forest, although the mean velocities and temperature have strong diurnal cycles. Momentum and heat flux are also found to be stronger above the forest. The presence of the secondary circulation enhances the turbulent fluxes as well as the turbulent intensity above the desert.\n\n2. The role of secondary circulations can be better understood once the components of the TKE budget are studied. Over the forest, the production of turbulence is mechanical, while over the desert, TKE production is mostly carried by buoyancy. The forest is more efficient in dissipating TKE as well. The imbalance of TKE is taken as the indicator of TKE transport and is found to vary diurnally almost anti-synchronously over the desert and the forest, confirming the role of a secondary circulation. The TKE budget is closed better over the forest compared to the desert. Turbulent triple moments, which are indicators of nonlocal motions and coherent structures, also show strong variability over the desert and are opposite in signs also confirming the role of secondary circulations. The integral timescales are found to be greater over the desert compared to the forest. This suggests that the secondary circulations that transport energy are more active over the desert – however, they cannot produce much turbulence over the desert since they only rely on buoyancy-driven turbulence as mechanical forcing is missing over the desert. This is also highlighted by the fact that mean velocities are higher above the desert while turbulent fluctuations are higher above the forest.\n\n3. To elucidate the role of horizontal transport between the desert and the forest, we studied the correlation between the TKE imbalance over the desert and the TKE production over the forest. The moderately high correlation suggests that the secondary circulation is transported from over the desert towards the forest, enhancing TKE production over the forest, at least in the canopy sub-layer and the atmospheric surface layer. The low correlation between the TKE imbalance over the forest and TKE production over the desert confirms the directionality of this horizontal exchange, which is from the desert towards the forest and not the other way around.\n\nTo summarize, we have examined the existence and role of secondary circulations that exists because of large-scale surface heterogeneities and possible due to some topography effects between the desert and the forest by looking at proxy quantities computed from turbulence measurements. Although the campaign was conducted at a particular site, the conclusions drawn are fairly general and can be extended to other scenarios involving surface heterogeneities, such as urban landscapes and agricultural fields. Future work will attempt to highlight a more spatially detailed picture of the turbulent structure under the interesting scenario of secondary circulations and horizontal energy transport.\n\nData availability\n\nWe suggest contacting the principal investigator Matthias Mauder (matthias.mauder@kit.edu) if readers are interested in obtaining the data used in the paper.\n\nAppendix A: Full form of the TKE budget\n$\\begin{array}{ll}\\text{(A1)}& & \\frac{\\partial e}{\\partial t}+U\\frac{\\partial e}{\\partial x}+V\\frac{\\partial e}{\\partial y}+W\\frac{\\partial e}{\\partial z}=\\frac{g}{T}\\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{T}^{\\prime }}\\right)& -\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{u}^{\\prime }}\\frac{\\partial U}{\\partial x}-\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{u}^{\\prime }}\\frac{\\partial V}{\\partial x}-\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{u}^{\\prime }}\\frac{\\partial W}{\\partial x}\\\\ & -\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{v}^{\\prime }}\\frac{\\partial U}{\\partial y}-\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{v}^{\\prime }}\\frac{\\partial V}{\\partial y}-\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{v}^{\\prime }}\\frac{\\partial W}{\\partial y}\\\\ & -\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{w}^{\\prime }}\\frac{\\partial U}{\\partial z}-\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{w}^{\\prime }}\\frac{\\partial V}{\\partial z}-\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}\\frac{\\partial W}{\\partial z}\\\\ & -\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime }e}\\right)}{\\partial x}-\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{v}^{\\prime }e}\\right)}{\\partial y}-\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }e}\\right)}{\\partial z}-\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{p}^{\\prime }}\\right)}{\\partial x}-\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{p}^{\\prime }}\\right)}{\\partial y}\\\\ & -\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{p}^{\\prime }}\\right)}{\\partial z}-\\mathit{ϵ}.\\end{array}$\n\nThus, to be consistent, with Eq. (2), all the terms in Eq. (A1) that cannot be evaluated using one-point measurements can be clubbed in the imbalance term, which can be described by\n\n$\\begin{array}{ll}\\text{(A2)}& & \\mathrm{Imbalance}=\\frac{\\partial e}{\\partial t}+U\\frac{\\partial e}{\\partial x}+V\\frac{\\partial e}{\\partial y}+W\\frac{\\partial e}{\\partial z}& +\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{u}^{\\prime }}\\frac{\\partial U}{\\partial x}+\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{u}^{\\prime }}\\frac{\\partial V}{\\partial x}+\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{u}^{\\prime }}\\frac{\\partial W}{\\partial x}\\\\ & +\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{v}^{\\prime }}\\frac{\\partial U}{\\partial y}+\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{v}^{\\prime }}\\frac{\\partial V}{\\partial y}+\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{v}^{\\prime }}\\frac{\\partial W}{\\partial y}\\\\ & +\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{w}^{\\prime }}\\frac{\\partial V}{\\partial z}+\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{w}^{\\prime }}\\frac{\\partial W}{\\partial z}\\\\ & +\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime }e}\\right)}{\\partial x}+\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{v}^{\\prime }e}\\right)}{\\partial y}+\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }e}\\right)}{\\partial z}+\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{u}^{\\prime }{p}^{\\prime }}\\right)}{\\partial x}\\\\ & +\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{v}^{\\prime }{p}^{\\prime }}\\right)}{\\partial y}+\\frac{\\mathrm{1}}{\\mathit{\\rho }}\\frac{\\partial \\left(\\stackrel{\\mathrm{‾}}{{w}^{\\prime }{p}^{\\prime }}\\right)}{\\partial z}.\\end{array}$", null, "Figure A1Vertical mean velocity profile averaged from 18 to 29 August (only times with both instruments simultaneously online and the nearest three range gates are discarded). Left to right: 4 h window centered on noon, daytime (sunrise to sundown) and nighttime (sundown to sunrise). The forest is shown as a solid red line, the desert as solid black line and a vertical line at $\\stackrel{\\mathrm{‾}}{w}=\\mathrm{0}$ as a black dashed line. Note that near the surface, the desert always has larger w, but only during the noontime with the updrafts of the forest is there a change in sign.\n\nThus, if no assumptions or idealizations are invoked, the imbalance of the commonly used operational TKE budget (Eq. 2) consists of TKE tendency, advection, shear production, TKE flux divergence and pressure velocity interactions. Using an array of sonics in each direction will enable determination of all these terms. However, as evident from the myriad of terms contributing to the imbalance, it is difficult to determine what degree of assumptions of homogeneity in which direction are sufficient so that certain terms can be ignored. Thus, unless all terms in Eq. (A2) can be determined, it is easier to stick to the most idealized form of Eq. (2) and treat all other terms as imbalances. Future work will try to determine the partitioning of advection, flux divergence and the other shear production terms contributing to TKE budget imbalances in the presence of heterogeneities.\n\nAppendix B: Further evidence of secondary circulation\n\nFigure A1 shows mean vertical velocity W above the forest and the desert averaged over all observations using the Doppler lidars. Secondary circulation cannot be thought of as a single large rotational system spanning the desert and the forest; rather, it is a much more complex and three-dimensional structure. Close to the surface layer and the canopy sub-layer, the transport of energy is indeed from the desert to the forest (Fig. 5). Further, we observe that the desert has more updrafts and the forest has more downdrafts close to the surface. However, as we go up above roughly 100 m, this behavior flips. Lastly, found in his simulations that large rotational systems developed at specific locations connected to surface features. Therefore, we conclude that the bulk transport in the convective mixed layer by a secondary circulation is from the forest to the desert, but it is advected with the mean wind and heavily influenced by surface features on a smaller scale than the forest itself.\n\nAuthor contributions\n\nTB did the data analysis and wrote the paper. PB collected the data, FDR, and KK helped in the interpretation. MM, DY and EY oversaw the whole project and was involved in the entire workflow, starting from data collection to data analysis and interpretation.\n\nCompeting interests\n\nThe authors declare that they have no conflict of interest.\n\nAcknowledgements\n\nThis research was supported by the German Research Foundation (DFG) as part of the project “Climate feedbacks and benefits of semi-arid forests” (CliFF) and the project “Capturing all relevant scales of biosphere–atmosphere exchange – the enigmatic energy balance closure problem”, which is funded by the Helmholtz-Association through the President's Initiative and Networking Fund and by KIT.\n\nThe article processing charges for this open-access\npublication were covered by a Research\nCentre of the Helmholtz Association.\n\nEdited by: Thomas Karl\nReviewed by: Gil Bohrer and one anonymous referee\n\nReferences\n\nBanerjee, T. and Katul, G.: Logarithmic scaling in the longitudinal velocity variance explained by a spectral budget, Phys. 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D.: The Effects of Canopy Leaf Area Index on Airflow across Foest Edges: Large-Eddy Simulation and Analytical Results, Bound.-Lay. Meteorol., 126, 433–460, 2008. a\n\nCava, D., Katul, G., Scrimieri, A., Poggi, D., Cescatti, A., and Giostra, U.: Buoyancy and the sensible heat flux budget within dense canopies, Bound.-Lay. Meteorol., 118, 217–240, 2006. a, b\n\nChamecki, M.: Persistence of velocity fluctuations in non-Gaussian turbulence within and above plant canopies, Phys. Fluids (1994–present), 25, 115110, https://doi.org/10.1063/1.4832955, 2013. a\n\nChatziefstratiou, E. K., Velissariou, V., and Bohrer, G.: Resolving the effects of aperture and volume restriction of the flow by semi-porous barriers using large-eddy simulations, Bound.-Lay. Meteorol., 152, 329–348, 2014. a\n\nCourault, D., Drobinski, P., Brunet, Y., Lacarrere, P., and Talbot, C.: Impact of surface heterogeneity on a buoyancy-driven convective boundary layer in light winds, Bound.-Lay. Meteorol., 124, 383–403, 2007. a\n\nDalpe, B. and Masson, C.: Numerical Simulation of Wind Flow near a Forest Edge, J. Wind Eng. Ind. Aerod., 97, 228–241, 2009. a\n\nDalu, G. and Pielke, R.: Vertical heat fluxes generated by mesoscale atmospheric flow induced by thermal inhomogeneities in the PBL, J. Atmos. Sci., 50, 919–926, 1993. a\n\nDetto, M., Katul, G. G., Siqueira, M., Juang, J. Y., and Stoy, P.: The Structure of Turbulence near a Tall Forest Edge: The Backward-Facing Step Flow Analogy Revisited, Ecological Applications, 18, 1420–1435, 2008. a\n\nDias-Junior, C. Q., Marques Filho, E. P., and Sá, L. D.: A large eddy simulation model applied to analyze the turbulent flow above Amazon forest, J. Wind Eng. Ind. Aerod., 147, 143–153, 2015. a\n\nDixon, N., Parker, D., Taylor, C., Garcia-Carreras, L., Harris, P., Marsham, J., Polcher, J., and Woolley, A.: The effect of background wind on mesoscale circulations above variable soil moisture in the Sahel, Q. J. Roy. Meteorol. Soc., 139, 1009–1024, 2013. a\n\nDupont, S. and Brunet, Y.: Coherent structures in canopy edge flow: a large-eddy simulation study, J. Fluid Mech., 630, 93–128, 2009. a\n\nDyer, A.: A review of flux-profile relationships, Bound.-Lay. Meteorol., 7, 363–372, 1974. a\n\nEder, F., De Roo, F., Kohnert, K., Desjardins, R. L., Schmid, H. P., and Mauder, M.: Evaluation of two energy balance closure parametrizations, Bound.-Lay. Meteorol., 151, 195–219, 2014. a\n\nEder, F., De Roo, F., Rotenberg, E., Yakir, D., Schmid, H. P., and Mauder, M.: Secondary circulations at a solitary forest surrounded by semi-arid shrubland and their impact on eddy-covariance measurements, Agr. Forest Meteorol., 211, 115–127, 2015. a, b, c, d, e, f, g, h, i\n\nFesquet, C., Dupont, S., Drobinski, P., Dubos, T., and Barthlott, C.: Impact of Terrain Heterogeneity on Coherent Structure Properties: Numerical Approach, Bound.-Lay. Meteorol., 133, 71–92, 2009. a\n\nFoken, T.: The energy balance closure problem: an overview, Ecol. Appl., 18, 1351–1367, 2008. a\n\nFuentes, J. D., Chamecki, M., Nascimento dos Santos, R. M., Von Randow, C., Stoy, P. C., Katul, G., Fitzjarrald, D., Manzi, A., Gerken, T., Trowbridge, A., Souza Freire, L., Ruiz-Plancarte, J., Furtunato Maia, J. M., Tóta, J., Dias, N., Fisch, G., Schumacher, C., Acevedo, O., Rezende Mercer, J., and Yañez-Serrano, A. M.: Linking meteorology, turbulence, and air chemistry in the Amazon rainforest, B. Am. Meteorol. Soc., 97, 2329–2342, 2016. a\n\nGarcia-Carreras, L., Parker, D. J., Taylor, C. M., Reeves, C. E., and Murphy, J. G.: Impact of mesoscale vegetation heterogeneities on the dynamical and thermodynamic properties of the planetary boundary layer, J. Geophys. 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K.: The Evolution of Turbulence across a Forest Edge, Bound.-Lay. Meteorol., 84, 467–496, 1997. a\n\nKaimal, J. C. and Finnigan, J. J.: Atmospheric boundary layer flows: their structure and measurement, Oxford University Press, 1994. a, b, c\n\nKanani-Sühring, F. and Raasch, S.: Spatial variability of scalar concentrations and fluxes downstream of a clearing-to-forest transition: a large-eddy simulation study, Bound.-Lay. Meteorol., 155, 1–27, 2015. a\n\nKanda, M., Inagaki, A., Letzel, M. O., Raasch, S., and Watanabe, T.: LES study of the energy imbalance problem with eddy covariance fluxes, Bound.-Lay. Meteorol., 110, 381–404, 2004. a\n\nKang, S.-L. and Lenschow, D. H.: Temporal evolution of low-level winds induced by two-dimensional mesoscale surface heat-flux heterogeneity, Bound.-Lay. Meteorol., 151, 501–529, 2014. a\n\nKatul, G. G., Cava, D., Siqueira, M., and Poggi, D.: Scalar turbulence within the canopy sublayer, Coherent Flow Structures at Earth's Surface, chap. 6, 73–95, 2013. a, b\n\nKröniger, K., Banerjee, T., De Roo, F., and Mauder, M.: Flow adjustment inside homogeneous canopies after a leading edge–An analytical approach backed by LES, Agr. Forest Meteorol., 255, 17–30, https://doi.org/10.1016/j.agrformet.2017.09.019, 2017. a\n\nKröniger, K., , DeRoo, F., Brugger, P., Huq, S., Banerjee, T., Zinsser, J., Rotenberg, E., Yakir, D., Rohatyn, S., and Mauder, M.: Effect of Secondary Circulations on the Surface–Atmosphere Exchange of Energy at an Isolated Semi-arid Forest, Bound.-Lay. Meteorol., https://doi.org/10.1007/s10546-018-0370-6, 2018. a, b, c\n\nLi, D., Salesky, S. T., and Banerjee, T.: Connections between the Ozmidov scale and mean velocity profile in stably stratified atmospheric surface layers, J. Fluid Mech., 797, https://doi.org/10.1017/jfm.2016.311, 2016. a, b\n\nLi, Z. J., Lin, J. D., and Miller, D. R.: Air-Flow over and through a Forest Edge – a Steady-State Numerical-Simulation, Bound.-Lay. Meteorol., 51, 179–197, 1990. a\n\nMahfouf, J.-F., Richard, E., and Mascart, P.: The influence of soil and vegetation on the development of mesoscale circulations, J. Appl. Meteorol. Climatol., 26, 1483–1495, 1987. a\n\nMarkfort, C., Porté-Agel, F., and Stefan, H.: Canopy-wake dynamics and wind sheltering effects on Earth surface fluxes, Environ. Fluid Mech., 14, 663–697, 2014. a\n\nMarusic, I., Monty, J. P., Hultmark, M., and Smits, A. J.: On the logarithmic region in wall turbulence, J. Fluid Mech., 716, https://doi.org/10.1017/jfm.2012.511, 2013. a\n\nMauder, M., Jegede, O., Okogbue, E., Wimmer, F., and Foken, T.: Surface energy balance measurements at a tropical site in West Africa during the transition from dry to wet season, Theor. Appl. 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Fluid Mech., 119, 173–217, 1982. a\n\nQueck, R., Bernhofer, C., Bienert, A., and Schlegel, F.: The TurbEFA Field Experiment – Measuring the Influence of a Forest Clearing on the Turbulent Wind Field, Bound.-Lay. Meteorol., 160, 397–423, 2016. a\n\nRaupach, M. and Finnigan, J.: Scale issues in boundary-layer meteorology: Surface energy balances in heterogeneous terrain, Hydrol. Process., 9, 589–612, 1995. a, b\n\nRaupach, M., Coppin, P., and Legg, B.: Experiments on scalar dispersion within a model plant canopy part I: The turbulence structure, Bound.-Lay. Meteorol., 35, 21–52, 1986. a, b\n\nRominger, J. T. and Nepf, H. M.: Flow Adjustment and Interior Flow Associated with a Rectangular Porous Obstruction, J. Fluid Mech., 680, 636–659, 2011. a\n\nRotenberg, E. and Yakir, D.: Distinct patterns of changes in surface energy budget associated with forestation in the semiarid region, Glob. Change Biol., 17, 1536–1548, 2011. a\n\nSalesky, S. T., Katul, G. G., and Chamecki, M.: Buoyancy effects on the integral lengthscales and mean velocity profile in atmospheric surface layer flows, Phys. Fluids (1994–present), 25, 105101, https://doi.org/10.1063/1.4823747, 2013. a\n\nSchlegel, F., Stiller, J., Bienert, A., Maas, H. G., Queck, R., and Bernhofer, C.: Large-Eddy Simulation of Inhomogeneous Canopy Flows Using High Resolution Terrestrial Laser Scanning Data, Bound.-Lay. Meteorol., 142, 223–243, 2012. a\n\nStoy, P. C., Mauder, M., Foken, T., Marcolla, B., Boegh, E., Ibrom, A., Arain, M. A., Arneth, A., Aurela, M., Bernhofer, C., et al.: A data-driven analysis of energy balance closure across FLUXNET research sites: The role of landscape scale heterogeneity, Agr. Forest Meteorol., 171, 137–152, 2013. a\n\nStull, R. B.: An introduction to boundary layer meteorology, vol. 13, Springer Science & Business Media, 2012. a, b, c\n\nSühring, M. and Raasch, S.: Heterogeneity-induced heat-flux patterns in the convective boundary layer: can they be detected from observations and is there a blending height? – a large-eddy simulation study for the LITFASS-2003 experiment, Bound.-Lay. Meteorol., 148, 309–331, 2013. a\n\nTownsend, A. A.: The structure of turbulent shear flow, Cambridge University Press, 1976. a\n\nvan Heerwaarden, C. C. and Guerau de Arellano, J. V.: Relative humidity as an indicator for cloud formation over heterogeneous land surfaces, J. Atmos. Sci., 65, 3263–3277, 2008. a\n\nVan Heerwaarden, C. C., Mellado, J. P., and De Lozar, A.: Scaling laws for the heterogeneously heated free convective boundary layer, J. Atmos. Sci., 71, 3975–4000, 2014. a\n\nWilczak, J. M., Oncley, S. P., and Stage, S. A.: Sonic anemometer tilt correction algorithms, Bound.-Lay. Meteorol., 99, 127–150, 2001. a\n\nYang, B., Raupach, M. R., Shaw, R. H., Tha, K., Paw, U., and Morse, A. P.: Large-Eddy Simulation of Turbulent Flow across a Forest Edge. Part I: Flow Statistics, Bound.-Lay. Meteorol., 120, 377–412, 2006. a\n\nYue, P., Zhang, Q., Wang, R., Li, Y., and Wang, S.: Turbulence intensity and turbulent kinetic energy parameters over a heterogeneous terrain of Loess Plateau, Adv. Atmos. Sci., 32, 1291–1302, 2015. a, b\n\nZhuang, Y. and Amiro, B.: Pressure fluctuations during coherent motions and their effects on the budgets of turbulent kinetic energy and momentum flux within a forest canopy, J. Appl. Meteorol., 33, 704–711, 1994. a" ]	[ null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-avatar-thumb150.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f01-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-t01-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f02-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f03-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f04-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f05-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f06-thumb.png", null, "https://acp.copernicus.org/articles/18/10025/2018/acp-18-10025-2018-f07-thumb.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8819929,"math_prob":0.971789,"size":49286,"snap":"2022-40-2023-06","text_gpt3_token_len":12291,"char_repetition_ratio":0.16387323,"word_repetition_ratio":0.04008222,"special_character_ratio":0.24599278,"punctuation_ratio":0.21238421,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9843413,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T23:02:10Z\",\"WARC-Record-ID\":\"<urn:uuid:9a0c5b86-bfd4-4be4-a293-d61c0118557f>\",\"Content-Length\":\"314954\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:513d0d7d-d186-49a6-8315-04d45eeb7786>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9cd70e2-7ee2-40a2-b691-cded91040264>\",\"WARC-IP-Address\":\"81.3.21.103\",\"WARC-Target-URI\":\"https://acp.copernicus.org/articles/18/10025/2018/\",\"WARC-Payload-Digest\":\"sha1:J7PHDDCU2AX4SXH5L4YVMP4MEINOVHGO\",\"WARC-Block-Digest\":\"sha1:7DITKGDNSUVBLUH2PUVXE3CWLWFSRHOV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499891.42_warc_CC-MAIN-20230131222253-20230201012253-00709.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/25-42-minus-12-10	[ "Solutions by everydaycalculation.com\n\n## Subtract 12/10 from 25/42\n\n1st number: 25/42, 2nd number: 1 2/10\n\n25/42 - 12/10 is -127/210.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 42 and 10 is 210\n2. For the 1st fraction, since 42 × 5 = 210,\n25/42 = 25 × 5/42 × 5 = 125/210\n3. Likewise, for the 2nd fraction, since 10 × 21 = 210,\n12/10 = 12 × 21/10 × 21 = 252/210\n4. Subtract the two fractions:\n125/210 - 252/210 = 125 - 252/210 = -127/210\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7725077,"math_prob":0.99533486,"size":390,"snap":"2020-34-2020-40","text_gpt3_token_len":163,"char_repetition_ratio":0.2253886,"word_repetition_ratio":0.0,"special_character_ratio":0.53846157,"punctuation_ratio":0.086021505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9978907,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T18:39:29Z\",\"WARC-Record-ID\":\"<urn:uuid:c6546507-3555-4cce-bd04-a29060ce2c94>\",\"Content-Length\":\"8025\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8949806-cdcb-4594-a165-785787641843>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe27c7b5-38fa-4ef4-afd4-8381a0645121>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/25-42-minus-12-10\",\"WARC-Payload-Digest\":\"sha1:OVPKIPQ44WUUKNMQCFCIWZCGK3HTXTJN\",\"WARC-Block-Digest\":\"sha1:NMZZMG2BS5CTLZDKPSAISXELYWQ7LT7Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738819.78_warc_CC-MAIN-20200811180239-20200811210239-00399.warc.gz\"}"}
https://curvyeditor.com/documentation/splines/cache	[ "# Caching\n\nThe key to achieve the highest possible performance when working with splines is precalculation. In Curvy the following spline data is precomputed:\n\n• Positions\n• Distances between positions\n• Tangent (Direction) for each position\n• Up Vector (Orientation) for each position\n\nAs you remember (perhaps read here again), a spline by nature has no resolution. You enter a F value and get a position in return. But of course the step width should be limited to reasonable values. At some point a lower stepsize makes no sense due float inaccuracy. We've seen spline solutions using doubles, but transforms and visual presentation in Unity are float based, so calculating doubles would produce a precision that can't be visualized.\n\nTo sum it up, calculating spline curves is about querying a formula continuously with a given stepsize and retrieving approximated points. Those points connected form the spline you see on screen. The number of sample points a spline uses is calculated automatically based on it's length, but can be influenced by you. In general, sample points (we call them Approximation Points or cache points) are spread at equal distance over a spline segment (in fact the segment length is roughly guessed at calculation time by taking the distance between the two involved Control Points).\n\nAny API methods ending with “Fast” means that cache data is being used by that method\n\nThe sample points number for a spline is defined by two parameters:\n\n• Max Points Per Unit: This defines the number of sample points per world distance unit.\n• Cache Density: This is a value between 1 and 100. The higher the number, the more sample points are per distance unit. When it is equal to 100, the spline will have Max Points Per Unit sample points per unit. For example, if you set Cache Density to 100 and Max Points Per Unit to 8, a spline segment with a length of 20 units will calculate 160 cache points. The Cache Density's default value of 50 is a good compromise between performance and precision most of the time.\n\nIn general, having more sample points will result in a higher degree of accuracy. As a downside, precomputing takes longer and more memory is used. You can (but don't necessarily need to) tweak the above two settings if you need more accurate values or a faster calculation. Which values work best for you depends on your setup.\n\nCurvy uses a sophisticated caching system. If you alter a spline, only the necessary values are being recalculated (usually on the next call to Update). On top of that you can enable threading to speed up calculation. Threading comes with a management overhead, so it's most useful for very large splines that are continuously changed." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9102654,"math_prob":0.9911421,"size":2683,"snap":"2020-10-2020-16","text_gpt3_token_len":550,"char_repetition_ratio":0.12989922,"word_repetition_ratio":0.0,"special_character_ratio":0.20089452,"punctuation_ratio":0.0877551,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.990154,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-10T18:15:50Z\",\"WARC-Record-ID\":\"<urn:uuid:da337b7d-0fba-409d-a192-5de1cb86c445>\",\"Content-Length\":\"26722\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a880c07c-bcf4-4b62-86bf-bc5bec3c4426>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5d75287-2f5c-48ed-9a7c-93493fa5c6d1>\",\"WARC-IP-Address\":\"160.153.78.3\",\"WARC-Target-URI\":\"https://curvyeditor.com/documentation/splines/cache\",\"WARC-Payload-Digest\":\"sha1:34Y7RIVZQJN4ALNAE3DT4BKIWOL5T3FZ\",\"WARC-Block-Digest\":\"sha1:JG4FOWA7B63HO3Y7E5ZVS5AEMRAJ5QOO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370511408.40_warc_CC-MAIN-20200410173109-20200410203609-00020.warc.gz\"}"}
http://ecowebsupport.org/anna-white-paf/ce3b0f-universal-law-of-gravitation-formula	[ "Newton's law of universal gravitation can be written as a vector equation to account for the direction of the gravitational force as well as its magnitude. We can now determine why this is so. If the two masses are m1 and m2 and the distance between them is r, the magnitude of the force (F) […] Substituting mg for $$F$$ in Newton’s universal law of gravitation gives The theorem tells us how different parts of the mass distribution affect the gravitational force measured at a point located a distance $\\text{r}_0$ from the center of the mass distribution: As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere. CC licensed content, Specific attribution, http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation, http://en.wikipedia.org/wiki/Isaac_Newton%23Apple_incident, http://en.wikipedia.org/wiki/Shell_theorem, http://en.wikipedia.org/wiki/Center_of_mass, http://en.wikipedia.org/wiki/center%20of%20mass, https://commons.wikimedia.org/wiki/File:Shell-diag-1.png, http://en.wikipedia.org/wiki/Law_of_universal_gravitation, http://cnx.org/content/m42073/latest/?collection=col11406/1.7, http://en.wikipedia.org/wiki/Gravitational_constant, http://en.wiktionary.org/wiki/gravitational_force, http://upload.wikimedia.org/wikipedia/commons/4/43/Earth-G-force.png. What is Difference Between Heat and Temperature? It occurred to Newton that if … Also Read: Important Gravitation Formulas for JEE. This potential energy formula contains a constant, G, which is called the \"universal gravitational constant\". The portion of the mass that is located at radii $\\text{r}<\\text{r}_0$ causes the same force at $\\text{r}_0$ as if all of the mass enclosed within a sphere of radius $\\text{r}_0$ was concentrated at the center of the mass distribution (as noted above). Universal Gravitation Formula The law of universal gravitation states that any two objects in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. m/s2. The value for Universal law of gravitation is: G = 6.673 × 10 -11 Nm² / kg² This value is used for solving numericals based on Newton’s law of universal gravitation. Gravity for astronauts in orbit. State the universal law of gravitation. The universal constant G must not be confused with the g that is the acceleration of a body arising from the earth’s gravity. This is a general physical law derived from empirical observations by what Isaac Newton called inductive reasoning. Since force is a vector quantity, the vector summation of all parts of the shell contribute to the net force, and this net force is the equivalent of one force measurement taken from the sphere’s midpoint, or center of mass (COM). That is because shells at a greater radius than the one at which the object is, do not contribute a force to an object inside of them (Statement 2 of theorem). So when finding the force of gravity exerted on a ball of 10 kg, the distance measured from the ball is taken from the ball’s center of mass to the earth’s center of mass. Newton’s law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Derive its mathematical formula. Sir Isaac Newton’s inspiration for the Law of Universal Gravitation was from the dropping of an apple from a tree. The gravitational force on an object within a hollow spherical shell is zero. State the Universal Law of Gravitation Write the equation to calculate gravitational force and state the meaning of its variables Explain how gravity is an inverse-square law The value of g. Another way to calculate the acceleration due to gravity g is by using Newton's Law of Universal Gravitation. what is Newton's universal law of gravitation? As previously noted, the universal gravitational constant G is determined experimentally. It wasn’t until Henry Cavendish’s verification of the gravitational constant that the Law of Universal Gravitation received its final algebraic form: $\\displaystyle \\text{F} = \\text{G}\\frac{\\text{Mm}}{\\text{r}^2}$. Newton’s law of universal gravitation – problems and solutions. gravitation and the universal law of gravitation. Forces on two masses: All masses are attracted to each other. So, the gravitational force acting upon point mass $\\text{m}$ is: $\\displaystyle \\text{F}=\\frac{\\text{GmM}_{<\\text{d}}}{\\text{d}^2}$, where it can be shown that $\\displaystyle \\text{M}_{<\\text{d}}=\\frac{4}{3}\\pi \\text{d}^3 \\rho$, ($\\rho$ is the mass density of the sphere and we are assuming that it does not depend on the radius. It is called the universal constant of gravitation. Save my name, email, and website in this browser for the next time I comment. In modern language, the law states the following: Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. where F gravity is the gravitational force between two objects, M 1 and M 2 are the masses of the two objects, and R is their separation. The weight of an object mg is the gravitational force between it and Earth. Any two objects with mass are attracted to each other by gravity. Diagram used in the proof of the Shell Theorem: This diagram outlines the geometry considered when proving The Shell Theorem. These forces are equal in magnitude but opposite in direction. This value is used for solving numericals based on Newton’s law of universal gravitation. Sir Isaac Newton came up with one of the heavyweight laws in physics for you: the law of universal gravitation. According to Newton’s law of gravitation, every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Given that a sphere can be thought of as a collection of infinitesimally thin, concentric, spherical shells (like the layers of an onion), then it can be shown that a corollary of the Shell Theorem is that the force exerted in an object inside of a solid sphere is only dependent on the mass of the sphere inside of the radius at which the object is. Alok Jha . Total force is the force of gravity or Fg. If you can improve it, please do. Observe how the force of gravity is directly proportional to the product of the two masses and inversely proportional to the square of the distance of separation. Chec That is, the sphere’s mass is uniformly distributed.). How does acceleration due to gravity vary? Newton's law of universal gravitation states that everybody of nonzero mass attracts every other object in the universe. F\\propto \\frac {m_ {1}m_ {2}} {r^2} \\Rightarrow F=G\\frac {m_ {1}m_ {2}} {r^2} Sir Isaac Newton put forward the universal law of gravitation in 1687 and used it to explain the observed motions of the planets and moons. The law of universal gravitation was formulated by Isaac Newton $$\\left(1643-1727\\right)$$ and published in $$1687.$$ Figure 1. Review the key concepts, equations, and skills for Newton's law of gravity, including how to find the gravitational field strength. The distance between the centers of masses is r. According to the law of gravitation, the gravitational force of attraction F with which the two masses m 1 and m2 separated by a … Newton’s universal law of gravitation equation. Universal Law of Gravitation Statement Therefore, combining the above two equations we get: $\\text{F}=\\frac{4}{3} \\pi \\text{Gm} \\rho \\text{d}$. It exists between all objects, even though it may seem ridiculous. The Universal law of gravitation can be summed by this gravitational force formula F G = (G.m1.m2)/ d 2 G is a constant which is discussed later in this post. By his dynamical and gravitational theories, he explained Kepler’s laws and established the modern quantitative science of gravitation. When the bodies have spatial extent, gravitational force is calculated by summing the contributions of point masses which constitute them. Sun 13 Oct 2013 03.00 EDT. See Also : Difference between g and G, Consider two bodies of masses m1 and m2 . (Note: The proof of the theorem is not presented here. Newton’s Law of Universal Gravitation; Newton’s Law of Universal Gravitation. In other words, the Earth attracts objects near its surface to itself. The proportionalities expressed by Newton's universal law of gravitation is represented graphically by the following illustration. where $\\text{F}$ represents the force in Newtons, $\\text{M}$ and $\\text{m}$ represent the two masses in kilograms, and $\\text{r}$ represents the separation in meters. Newton’s Law of Gravitation Gravitational force is a attractive force between two masses m 1 and m 2 separated by a distance r. The gravitational force acting between two point objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. OpenStax College, College Physics. Describe how gravitational force is calculated for the bodies with spatial extent. Thus he proposed his law of universal gravitation. Newton’s Law of Universal Gravitation – Page 2. We shall also discuss the conditions for objects to float in liquids. Newton’s law of gravitation can be stated as:”Everybody in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.” Newton went a step further in his analysis of gravity. The universal gravitation equation thus takes the form. Introduction to gravity. G is gravitational constant, m 1 , m 2 are the masses of two bodies separated by a distance d, then give the statement of Newton's law of gravitation. Science Physics Newton's Law of Gravity. The net gravitational force that a spherical shell of mass $\\text{M}$ exerts on a body outside of it, is the vector sum of the gravitational forces acted by each part of the shell on the outside object, which add up to a net force acting as if mass $\\text{M}$ is concentrated on a point at the center of the sphere (Statement 1 of Shell Theorem). Calculate by the gravitational force formula Newton's law of gravitation. Google Classroom Facebook Twitter. This law says that every mass exerts an attractive force on every other mass. Since the mass of the earth is very large, it attracts nearby objects with a significant force. So, Fg (gravity force pulling on object) ∝ object’s mass (m) This force is also known as the gravitational force F g. Why do all objects attract downwards? Analyze - why two people sitting next to each other don't feel gravitational force? It describes the gravitational interaction between bodies endowed with mass, and establishes a proportional relationship of the force with which those bodies attract the each other. But it's important to realize that the distance between the two objects, especially when we're talking about the universal law of gravitation, is the distance between their center of masses. By equating Newton’s second law with his law of universal gravitation, and inputting for the acceleration a the experimentally verified value of 9.8 $\\text{m/}\\text{s}^2$, the mass of earth is calculated to be $5.96 \\cdot 1024$ kg, making the earth’s weight calculable given any gravitational field. In accordance with this law, two point masses attract each other with a force that is directly proportional to the masses of these bodies $${m_1}$$ and $${m_2},$$ and inversely proportional to the square of the distance between them: State the universal law of gravitation. In SI units its value is 6.673 × 10-11 Nm2kg-2. The mathematical formula for gravitational force is $\\text{F} = \\text{G}\\frac{\\text{Mm}}{\\text{r}^2}$ where $\\text{G}$ is the gravitational constant. The Law of Universal Gravitation states that the gravitational force between two points of mass is proportional to the magnitudes of their masses and the inverse-square of their separation, $\\text{d}$: $\\displaystyle \\text{F}=\\frac{\\text{GmM}}{\\text{d}^2}$. The portion of the mass that is located at radii $\\text{r}>\\text{r}_0$ exerts no net gravitational force at the distance $\\text{r}_0$ from the center. The value for Universal law of gravitation is: G = 6.673 × 10-11 Nm² / kg². More on Newton's Law of Universal Gravitation Prehistoric man realized a long time ago that when objects are released near the surface of the Earth, they always fall down to the ground. As in the case of hollow spherical shells, the net gravitational force that a solid sphere of uniformly distributed mass $\\text{M}$ exerts on a body outside of it, is the vector sum of the gravitational forces acted by each shell of the sphere on the outside object. The law of universal gravitation is indirectly derived from Kepler's law and the equation of motion. The weight of an object mg is the gravitational force between it and Earth. Its value is = … It's probably a little bit lower than that, actually. According to the universal law of gravitation, all material bodies attract each other, while the attractive force does not depend on the physical or chemical properties of the bodies. Gravitation: Each object in the universe attracts every other object with a force, which is called the force of gravitation.. We are giving a detailed and clear sheet on all Physics Notes that are very useful to understand the Basic Physics Concepts.. Newton’s Law of Gravitation \| Definition, Formula – Gravitation This law states that any two objects pull on each other with force gravity. Gravitational Force formula derivation from the Universal Law of Gravitation. But Newton's law of universal gravitation extends gravity beyond earth. 10.1 Gravitation We know that the moon goes around the earth. Newton's Universal Law of Gravitation: 'a simple equation, but devastatingly effective'. Required fields are marked *. Newton's law of universal gravitation is about the universality of gravity. It is important to note that the gravitational forces between two particles are an action-reaction pair. Why do we use mass and weight interchangeably in day-to-day life? Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. (adsbygoogle = window.adsbygoogle \|\| []).push({}); Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance. The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation This attractive force always points inward, from one point to the other. Express the Law of Universal Gravitation in mathematical form. Let’s see a video about universal law of gravitation. This video explains the concept of the Universal Law of Gravitation. Its value is the same everywhere. where $\\text{F}$ is the force between the masses, $\\text{G}$ is the gravitational constant, $\\text{m}_1$ is the first mass, $\\text{m}_2$ is the second mass and $\\text{r}$ is the distance between the centers of the masses. Finding the gravitational force between three-dimensional objects requires treating them as points in space. universal law of gravitation+formulae+unit 2 See answers Brainlybikesh Brainlybikesh Explanation:-⇒ universal law of gravitation is also know as Newton Law of Gravitation ⇒ Every particles in the universe attracts every other particles with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them . The Law of Universal Gravitation is one of the physical laws formulated by Isaac Newton in his book Philosophiae Naturalis Principia Mathematica of 1687. How many Types of Multivibrators Are There? Thus, if a spherically symmetric body has a uniform core and a uniform mantle with a density that is less than $\\frac{2}{3}$ of that of the core, then the gravity initially decreases outwardly beyond the boundary, and if the sphere is large enough, further outward the gravity increases again, and eventually it exceeds the gravity at the core/mantle boundary. The weight of an object on the earth is the result of the gravitational force of attraction between the earth and the object. More generally, this result is true even if the mass $\\text{M}$ is not uniformly distributed, but its density varies radially (as is the case for planets). Coulomb’s law Vs Gravitational law. If you want to learn Brief differences b/w law of Electrostatic and Universal law of gravitation or gravitational law, then you are at the right place.. Keep reading.. The old lore goes something like this- Newton had taken a year off from the Cambridge University due to a plague epidemic there. This equation gives us the expression of the gravitational force. The weight of an object mg is the gravitational force between it and Earth. Give its si unit and numerical value. Substituting mg for $$F$$ in Newton’s universal law of gravitation gives Understanding Newton’s Universal Law of Gravitation. Leave a Comment / Physics. For all general purposes, my center of mass, maybe it's like three feet above the ground, because I'm not that tall. Recall that the Force of gravity around the surface of the Earth is calculated with the formula F g = m x g. But we can also use Newton's Law of Universal Gravitation … The value of force F g is the same for both the masses m 1 as well as m 2. That is, the individual gravitational forces exerted by the elements of the sphere out there, on the point at $\\text{r}_0$, cancel each other out. This attractive force is called gravity. This video explains the concept of the Universal Law of Gravitation. Newton’s law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. More on Newton's Law of Universal Gravitation. What are the SI units of the proportionality constant G? Solve Numericals. Substituting mg for $$F$$ in Newton’s universal law of gravitation gives Statue of Isaac Newton in the chapel of Trinity College, Cambridge. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them: $\\displaystyle \\text{F} = \\text{G}\\frac{\\text{m}_{1}\\text{m}_{2}}{\\text{r}^{2}}$. If you want to learn Brief differences b/w law of Electrostatic and Universal law of gravitation or gravitational law, then you are at the right place.. Keep reading.. The law is part of classical mechanics. (The law was by Newton). Coulomb’s law Vs Gravitational law. Isaac Newton proved the Shell Theorem, which states that: Since force is a vector quantity, the vector summation of all parts of the shell/sphere contribute to the net force, and this net force is the equivalent of one force measurement taken from the sphere’s midpoint, or center of mass (COM). Consider two bodies of masses m 1 and m 2. While an apple might not have struck Sir Isaac Newton’s head as myth suggests, the falling of one did inspire Newton to one of the great discoveries in mechanics: The Law of Universal Gravitation. which shows that mass $\\text{m}$ feels a force that is linearly proportional to its distance, $\\text{d}$, from the sphere’s center of mass. Newton’s Law of Gravitation. Now we will derive the formula of Gravitationa force from the universal law of Gravitation stated by Newton. Also, the motion of the planets around the earth is explained. It is talked about in Isaac Newton's Philosophiae Naturalis Principia Mathematica. In the limit, as the component point masses become “infinitely small”, this entails integrating the force (in vector form, see below) over the extents of the two bodies. We can now determine why this is so. Newton's law of gravitation review. Law of gravitation states that every object in this universe attract each other by mutual force of attraction which is directly proportional to product of their mass and inversely proportional to the square of the distance between them, measured from their centre. It wasn’t until Henry Cavendish’s verification of the gravitational constant that the Law of Universal Gravitation received its final algebraic form: F =GMm r2 F = G Mm r 2. where F F represents the force in Newtons, M M and m m represent the two masses in kilograms, and r r … Including a dramatization of The Cavendish Experiment and force visualization via qualitative examples. ALLobjects attract each other with a force of gravitational attraction. It explains the motion of the moon around the earth. Derivation of Universal Law of Gravitation. b) F = d 2 G m 1 m 2 is the mathematical form of Newton's law of gravitation. Inputs: object 1 mass (m 1) ... Change Equation Select to solve for a different unknown Newton's law of gravity. Calculate by the gravitational force formula Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric. The difference between Coulomb’s law and gravitational law is provided here. The Law of Universal Gravitation is one of the physical laws formulated by Isaac Newton in his book Philosophiae Naturalis Principia Mathematica of 1687. This led him to believe that it is also the gravitational force that acts between the Sun and each of the planets to keep them in their orbits. The resulting net gravitational force acts as if mass $\\text{M}$ is concentrated on a point at the center of the sphere, which is the center of mass, or COM (Statement 1 of Shell Theorem). Newton's universal law of gravitation is a physical law that describes the attraction between two objects with mass. In this formula, quantities in bold represent vectors. Formulate the Shell Theorem for spherically symmetric objects. In other words, the Earth attracts objects near its surface to itself. The second situation we will examine is for a solid, uniform sphere of mass $\\text{M}$ and radius $\\text{R}$, exerting a force on a body of mass $\\text{m}$ at a radius $\\text{d}$ inside of it (that is, $\\text{d}< \\text{R}$). This law says that every mass exerts an attractive force on every other mass. Likewise, the second particle exerts a force on the first particle that is directed toward the second particle along the line joining them. Prehistoric man realized a long time ago that when objects are released near the surface of the Earth, they always fall down to the ground. The Universal Law of Gravitation is a law devised by Issac Newton, the father of classical physics. Newton's place in the Gravity Hall of Fame is not due to his discovery of gravity, but rather due to his discovery that gravitation is universal. , Your email address will not be published. What do you mean by Thermal conductivity? Mathematically it can be represented as, F = Gm 1 m 2 /r 2 Where, F is the Gravitational … In particular, in this case a spherical shell of mass $\\text{M}$ (left side of figure) exerts a force on mass $\\text{m}$ (right side of the figure) outside of it. Nearby objects with masses, big or small the magnitude of the gravitational force between it and.... Gravity beyond earth - why two people sitting next to each other do n't feel gravitational force F g. do! 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https://encyclopediaofmath.org:443/wiki/Laplace-Beltrami_equation	[ "# Laplace-Beltrami equation\n\nBeltrami equation\n\nA generalization of the Laplace equation for functions in a plane to the case of functions $u$ on an arbitrary two-dimensional Riemannian manifold $R$ of class $C ^ {2}$. For a surface $R$ with local coordinates $\\xi , \\eta$ and first fundamental form\n\n$$d s ^ {2} = E d \\xi ^ {2} + 2 F d \\xi d \\eta + G d \\eta ^ {2} ,$$\n\nthe Laplace–Beltrami equation has the form\n\n$$\\tag{* } \\Delta u \\equiv \\ \\frac \\partial {\\partial \\xi } \\left ( \\frac{F \\frac{\\partial u }{\\partial \\eta } - G \\frac{\\partial u }{\\partial \\xi } }{\\sqrt {E G - F ^ { 2 } } } \\right ) + \\frac \\partial {\\partial \\eta } \\left ( \\frac{F \\frac{\\partial u }{\\partial \\xi } - E \\frac{\\partial u }{\\partial \\eta } }{\\sqrt {E G - F ^ { 2 } } } \\right ) = 0 .$$\n\nFor $E = G$ and $F = 0$, that is, when $( \\xi , \\eta )$ are isothermal coordinates on $R$, equation () becomes the Laplace equation. The Laplace–Beltrami equation was introduced by E. Beltrami in 1864–1865 (see ).\n\nThe left-hand side of equation () divided by $\\sqrt {E G - F ^ { 2 } }$ is called the second Beltrami differential parameter.\n\nRegular solutions $u$ of the Laplace–Beltrami equation are generalizations of harmonic functions and are usually called harmonic functions on the surface $R$( cf. also Harmonic function). These solutions are interpreted physically like the usual harmonic functions, e.g. as the velocity potential of the flow of an incompressible liquid flowing over the surface $R$, or as the potential of an electrostatic field on $R$, etc. Harmonic functions on a surface retain the properties of ordinary harmonic functions. A generalization of the Dirichlet principle is valid for them: Among all functions $v$ of class $C ^ {2} ( G) \\cap C ( \\overline{G}\\; )$ in a domain $G \\subset R$ that take the same values on the boundary $\\partial G$ as a harmonic function $v \\in C ( \\overline{G}\\; )$, the latter gives the minimum of the Dirichlet integral\n\n$$D ( v) = {\\int\\limits \\int\\limits } _ { G } \\nabla v \\cdot \\sqrt {E G - F ^ { 2 } } \\ d \\xi d \\eta ,$$\n\nwhere\n\n$$\\nabla v = \\ \\frac{E \\left ( \\frac{\\partial v }{\\partial \\eta } \\right ) ^ {2} - 2 F \\frac{\\partial v }{\\partial \\xi } \\frac{\\partial v }{\\partial \\eta } + G \\left ( \\frac{\\partial v }{\\partial \\xi } \\right ) ^ {2} }{E G - F ^ { 2 } }$$\n\nis the first Beltrami differential parameter, which is a generalization of the square of the gradient $\\mathop{\\rm grad} ^ {2} u$ to the case of functions on a surface.\n\nFor generalizations of the Laplace–Beltrami equation to Riemannian manifolds of higher dimensions see Laplace operator.\n\nHow to Cite This Entry:\nLaplace-Beltrami equation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Laplace-Beltrami_equation&oldid=47576\nThis article was adapted from an original article by E.D. SolomentsevE.V. Shikin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6848572,"math_prob":0.9998024,"size":3005,"snap":"2022-27-2022-33","text_gpt3_token_len":908,"char_repetition_ratio":0.17394201,"word_repetition_ratio":0.109540634,"special_character_ratio":0.33344427,"punctuation_ratio":0.0967118,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999958,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T16:01:06Z\",\"WARC-Record-ID\":\"<urn:uuid:11674860-a8f7-472b-bb46-eb3d204c67bb>\",\"Content-Length\":\"17584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ebecbb52-5f26-455f-829b-7dba4690ce7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7eb0853d-3585-45b4-9cea-4bd56873fcde>\",\"WARC-IP-Address\":\"34.96.94.55\",\"WARC-Target-URI\":\"https://encyclopediaofmath.org:443/wiki/Laplace-Beltrami_equation\",\"WARC-Payload-Digest\":\"sha1:TUENPQS4Q4KINEVQHNVUEKKJC3UMCYIR\",\"WARC-Block-Digest\":\"sha1:HQZYLQOXPDPZCOG22CWENQYDP7FRVD3O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104675818.94_warc_CC-MAIN-20220706151618-20220706181618-00044.warc.gz\"}"}