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https://www.r-bloggers.com/2009/09/power-analysis-for-mixed-effect-models-in-r/	[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nThe power of a statistical test is the probability that a null hypothesis will be rejected when the alternative hypothesis is true. In lay terms, power is your ability to refine or “prove” your expectations from the data you collect. The most frequent motivation for estimating the power of a study is to figure out what sample size will be needed to observe a treatment effect. Given a set of pilot data or some other estimate of the variation in a sample, we can use power analysis to inform how much additional data we should collect.\n\nI recently did a power analysis on a set of pilot data for a long-term monitoring study of the US National Park Service. I thought I would share some of the things I learned and a bit of R code for others that might need to do something like this. If you aren’t into power analysis, the code below may still be useful as examples of how to use the error handling functions in R (`withCallingHandlers`, `withRestarts`), parallel programming using the `snow` package, and linear mixed effect regression using `nlme`. If you have any suggestions for improvement or if I got something wrong on the analysis, I’d love to hear from you.\n\n## 1 The Study\n\nThe study system was cobblebars along the Cumberland river in Big South Fork National Park (Kentucky and Tennessee, United States). Cobblebars are typically dominated by grassy vegetation that include disjunct tall-grass prairie species. It is hypothesized that woody species will encroach onto cobblebars if they are not seasonally scoured by floods. The purpose of the NPS sampling was to observe changes in woody cover through time. The study design consisted of two-stages of clustering: the first being cobblebars, and the second being transects within cobblebars. The response variable was the percentage of the transect that was woody vegetation. Because of the clustered design, the inferential model for this study design has mixed-effects. I used a simple varying intercept model:", null, "where y is the percent of each transect in woody vegetation sampled n times within J cobblebars, each with K transects. The parameter of inference for the purpose of monitoring change in woody vegetation through time is β, the rate at which cover changes as a function of time. α, γ, σ2γ, and σ2y are hyper-parameters that describe the hierarchical variance structure inherent in the clustered sampling design.\n\nBelow is the function code used I used to regress the pilot data. It should be noted that with this function you can log or logit transform the response variable (percentage of transect that is woody). I put this in because the responses are proportions (0,1) and errors should technically follow a beta-distribution. Log and logit transforms with Gaussian errors could approximate this. I ran all the models with transformed and untransformed response, and the results did not vary at all. So, I stuck with untransformed responses:\n\n```Model <- function(x = cobblebars,\ntype = c(\"normal\",\"log\",\"logit\")){\n## Transforms\nif (type == \"log\")\nx\\$prop.woody <- log(x\\$prop.woody)\nelse if (type == \"logit\")\nx\\$prop.woody <- log(x\\$prop.woody / (1 - x\\$prop.woody))\n\nmod <- lme(prop.woody ~ year,\ndata = x,\nrandom = ~ 1 \| cobblebar/transect,\nna.action = na.omit,\ncontrol = lmeControl(opt = \"optim\",\nmaxIter = 800, msMaxIter = 800)\n)\nmod\\$type <- type\n\nreturn(mod)\n}\n```\n\nHere are the results from this regression of the pilot data:\n\n```Linear mixed-effects model fit by REML\nData: x\nAIC BIC logLik\n-134.4319 -124.1297 72.21595\n\nRandom effects:\nFormula: ~1 \| cobblebar\n(Intercept)\nStdDev: 0.03668416\n\nFormula: ~1 \| transect %in% cobblebar\n(Intercept) Residual\nStdDev: 0.02625062 0.05663784\n\nFixed effects: prop.woody ~ year\nValue Std.Error DF t-value p-value\n(Intercept) 0.12966667 0.01881983 29 6.889896 0.0000\nyear -0.00704598 0.01462383 29 -0.481815 0.6336\nCorrelation:\n(Intr)\nyear -0.389\n\nNumber of Observations: 60\nNumber of Groups:\ncobblebar transect %in% cobblebar\n6 30\n```\n\n## 2 We don't learn about power analysis and complex models\n\nWhen I decided upon the inferential model the first thing that occurred to me was that I never learned in any statistics course I had taken how to do such a power analysis on a multi-level model. I've taken more statistics courses than I'd like to count and taught my own statistics courses for undergrads and graduate students, and the only exposure to power analysis that I had was in the context of simple t-tests or ANOVA. You learn about it in your first 2 statistics courses, then it rarely if ever comes up again until you actually need it.\n\nI was, however, able to find a great resource on power analysis from a Bayesian perspective in the excellent book \"Data Analysis Using Regression and Multilevel/Hierarchical Models\" by Andrew Gelman and Jennifer Hill. Andrew Gelman has thought and debated about power analysis and you can get more from his blog. The approach in the book is a simulation-based one and I have adopted it for this analysis.\n\n## 3 Analysis Procedure\n\nFor the current analysis we needed to know three things: effect size, sample size, and estimates of population variance. We set effect size beforehand. In this context, the parameter of interest is the rate of change in woody cover through time β, and effect size is simply how large or small a value of β you want to distinguish with a regression. Sample size is also set a priori. In the analysis we want to vary sample size by varying the number of cobblebars, the number of transects per cobblebar or the number of years the study is conducted.\n\nThe population variance cannot be known precisely, and this is where the pilot data come in. By regressing the pilot data using the model we can obtain estimates of all the different components of the variance (cobblebars, transects within cobblebars, and the residual variance). Below is the R function that will return all the hyperparameters (and β) from the regression:\n\n```GetHyperparam<-function(x,b=NULL){\n## Get the hyperparameters from the mixed effect model\nfe <- fixef(x)\n\nif(is.null(b))\nb<-fe # use the data effect size if not supplied\n\nmu.a <- fe\n\nvc <- VarCorr(x)\nsigma.y <- as.numeric(vc[5, 2]) # Residual StdDev\nsigma.a <- as.numeric(vc[2, 2]) # Cobblebar StdDev\nsigma.g <- as.numeric(vc[4, 2]) # Cobblebar:transect StdDev\n\nhp<-c(b, mu.a, sigma.y, sigma.a, sigma.g)\nnames(hp)<-c(\"b\", \"mu.a\", \"sigma.y\", \"sigma.a\", \"sigma.g\")\nreturn(hp)\n}\n```\n\nTo calculate power we to regress the simulated data in the same way we did the pilot data, and check for a significant β. Since optimization is done using numeric methods there is always the chance that the optimization will not work. So, we make sure the regression on the fake data catches and recovers from all errors. The solution for error recovery is to simply try the regression on a new set of fake data. This function is a pretty good example of using the R error handling function `withCallingHandlers` and `withRestarts`.\n\n```fakeModWithRestarts <- function(m.o, n = 100, ...){\n## A Fake Model\nwithCallingHandlers({\ni <- 0\nmod <- NULL\nwhile (i < n & is.null(mod)){\nmod <- withRestarts({\nf <- fake(m.orig = m.o, transform = F, ...)\nreturn(update(m.o, data = f))\n},\nrs = function(){\ni <<- i + 1\nreturn(NULL)\n})\n}\nif(is.null(mod))\nwarning(\"ExceededIterations\")\nreturn(mod)\n},\nerror = function(e){\ninvokeRestart(\"rs\")\n},\nwarning = function(w){\nif(w\\$message == \"ExceededIterations\")\ncat(\"\\n\", w\\$message, \"\\n\")\nelse\ninvokeRestart(\"rs\")\n})\n}\n```\n\nTo calculate the power of a particular design we run `fakeModWithRestarts` 1000 times and look at the proportion of significant β values:\n\n```dt.power <- function (m, n.sims = 1000, alpha=0.05, ...){\n## Calculate power for a particular sampling design\nsignif<-rep(NA, n.sims)\nfor(i in 1:n.sims){\nlme.power <- fakeModWithRestarts(m.o = m, ...)\nif(!is.null(lme.power))\nsignif[i] <- summary(lme.power)\\$tTable[2, 5] < alpha\n}\npower <- mean(signif, na.rm = T)\nreturn(power)\n}\n```\n\nFinally, we want to perform this analysis on many different sampling designs. In my case I did all combinations of set of effect sizes, cobblebars, transects, and years. So, I generated the appropriate designs:\n\n```factoredDesign <- function(Elevs = 0.2/c(1,5,10,20),\nNlevs = seq(2, 10, by = 2),\nJlevs = seq(4, 10, by = 2),\nKlevs = c(3, 5, 7), ...){\n## Generates factored series of sampling designs for simulation\n## of data that follow a particular model.\n## Inputs:\n## Elevs - vector of effect sizes for the slope parameter.\n## Nlevs - vector of number of years to sample.\n## Jlevs - vector of number of cobblebars to sample.\n## Klevs - vector of number of transects to sample.\n## Results:\n## Data frame with where columns are the factors and\n## rows are the designs.\n\n# Level lengths\nlE <- length(Elevs)\nlN <- length(Nlevs)\nlJ <- length(Jlevs)\nlK <- length(Klevs)\n\n# Generate repeated vectors for each factor\nE <- rep(Elevs, each = lNlJlK)\nN <- rep(rep(Nlevs, each = lJlK), times = lE)\nJ <- rep(rep(Jlevs, each = lK), times = lElN)\nK <- rep(Klevs, times = lElNlJ)\n\nreturn(data.frame(E, N, J, K))\n}\n```\n\nOnce we know our effect sizes, the different sample sizes we want, and the estimates of population variance we can generate simulated dataset that are similar to the pilot data. To calculate power we simply simulate a large number of dataset and calculate the proportion of slopes, β that are significantly different from zero (p-value < 0.05). This procedure is repeated for all the effect sizes and sample sizes of interest. Here is the code for generating a simulated dataset. It also does the work of doing the inverse transform of the response variables if necessary.\n\n```fake <- function(N = 2, J = 6, K = 5, b = NULL, m.orig = mod,\ntransform = TRUE, ...){\n## Simulated Data for power analysis\n## N = Number of years\n## J = Number of cobblebars\n## K = Number of transects within cobblebars\nyear <- rep(0:(N-1), each = JK)\ncobblebar <- factor(rep(rep(1:J, each = K), times = N))\ntransect <- factor(rep(1:K, times = NJ))\n\n## Simulated parameters\nhp<-GetHyperparam(x=m.orig)\nif(is.null(b))\nb <- hp['b']\ng <- rnorm(JK, 0, hp['sigma.g'])\na <- rnorm(JK, hp['mu.a'] + g, hp['sigma.a'])\n\n## Simulated responses\neta <- rnorm(JKN, a + b * year, hp['sigma.y'])\nif (transform){\nif (m.orig\\$type == \"normal\"){\ny <- eta\ny[y > 1] <- 1 # Fix any boundary problems.\ny[y < 0] <- 0\n}\nelse if (m.orig\\$type == \"log\"){\ny <- exp(eta)\ny[y > 1] <- 1\n}\nelse if (m.orig\\$type == \"logit\")\ny <- exp(eta) / (1 + exp(eta))\n}\nelse{\ny <- eta\n}\n\nreturn(data.frame(prop.woody = y, year, transect, cobblebar))\n}\n```\n\nThen I performed the power calculations on each of these designs. This could take a long time, so I set this procedure to use parallel processing if needed. Note that I had to re-~source~ the file with all the necessary functions for each processor.\n\n```powerAnalysis <- function(parallel = T, ...){\n## Full Power Analysis\n\n## Parallel\nif(parallel){\ncloseAllConnections()\ncl <- makeCluster(7, type = \"SOCK\")\non.exit(closeAllConnections())\nclusterEvalQ(cl, source(\"cobblebars2.r\"))\n}\n\n## The simulations\ndat <- factoredDesign(...)\n\nif (parallel){\ndat\\$power <- parRapply(cl, dat, function(x,...){\ndt.power(N = x, J = x, K = x, b = x, ...)\n}, ...)\n} else {\ndat\\$power <- apply(dat, 1, function(x, ...){\ndt.power(N = x, J = x, K = x, b = x, ...)\n}, ...)\n}\n\nreturn(dat)\n}\n```\n\nThe output of the `powerAnalysis` function is a data frame with columns for the power and all the sample design settings. So, I wrote a custom plotting function for this data frame:\n\n```plotPower <- function(dt){\nxyplot(power~N\|J*K, data = dt, groups = E,\npanel = function(...){panel.xyplot(...)\npanel.abline(h = 0.8, lty = 2)},\ntype = c(\"p\", \"l\"),\nxlab = \"sampling years\",\nylab = \"power\",\nstrip = strip.custom(var.name = c(\"C\", \"T\"),\nstrip.levels = c(T, T)),\nauto.key = T\n)\n}\n```\n\nBelow is the figure for the cobblebar power analysis. I won't go into detail on what the results mean since I am concerned here with illustrating the technique and the R code. Obviously, as the number of cobblebars and transects per year increase, so does power. And, as the effect size increases, observing it with a test is easier.", null, "Date: 2009-09-18 Fri\n\nHTML generated by org-mode 6.30trans in emacs 22" ]	[ null, "https://i2.wp.com/toddjobe.blogspot.com/2009/09/ltxpng/powerAnalysis_0001.png", null, "https://i2.wp.com/toddjobe.blogspot.com/2009/09/ltxpng/powerAnalysis_0002.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8126074,"math_prob":0.98227674,"size":12511,"snap":"2022-05-2022-21","text_gpt3_token_len":3264,"char_repetition_ratio":0.11809387,"word_repetition_ratio":0.031538088,"special_character_ratio":0.27759573,"punctuation_ratio":0.1543332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973056,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T12:29:18Z\",\"WARC-Record-ID\":\"<urn:uuid:ce60ff46-f74a-46ea-98ff-90dc818395ce>\",\"Content-Length\":\"111180\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cbf9754d-4385-4b99-b29b-f56382028778>\",\"WARC-Concurrent-To\":\"<urn:uuid:2091afd8-9ae6-430b-a6d4-f5dc5e6a9be6>\",\"WARC-IP-Address\":\"172.64.104.18\",\"WARC-Target-URI\":\"https://www.r-bloggers.com/2009/09/power-analysis-for-mixed-effect-models-in-r/\",\"WARC-Payload-Digest\":\"sha1:VKYJ6CVGJRMQL2DKK3BF7VHMIBOICW5U\",\"WARC-Block-Digest\":\"sha1:LVJQEWZ7VZ7RCNKVP47GIFH3GN3RPSVB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662647086.91_warc_CC-MAIN-20220527112418-20220527142418-00374.warc.gz\"}"}
https://russianpatents.com/patent/221/2214680.html	[ "Way to normalize the metric values of the component decoder in a mobile communication system and device for its implementation\n\nThe invention relates to communication systems and can be used in means of mobile communication. The technical result is to use changes in the set of metrics on the set of time segments. The decoder contains a decision tree, which generates a signal decisions when all metric values exceed the predetermined value. MyCitadel subtracts the predetermined value from the metric values in response to the signal decision to normalize the value of metrics. The decision tree contains many storage devices that remember the appropriate value metrics. 3 S. and 3 C.p. f-crystals, 13 ill., table 1.\n\nTechnical field the Present invention relates to a device and method of iterative decoding for mobile communication systems and, in particular, relates to devices and method for normalizing metric values stored in the component decoder is an iterative decoder in a mobile communication system.\n\nThe prior art In General, iterative decoding is used in such mobile communication systems, as IMT-2000 (or systems multiple access, code-division multiplexing mdcr-2000 and UMTS), which is applied that is a similar connection, using cascaded convolutional codes, cascading block codes or composite codes. The technical scope of iterative decoding is associated with the so-called \"soft\" (not defined) solutions and optimal characteristics-correcting code errors.\n\nIn Fig.1 shows a known iterative decoder, comprising two component decoder. According Fig.1 the first component decoder 101 receives signals Xtosystematic code, the first signal parity Y1Kreceived from the demultiplexer 107 (which demuxes input signals parity Ytoand first external information signal. The first component decoder 101 decodes the received signals, giving primary decoded signal associated with the decoding results. This signal consists of component Xtosignals a systematic code and the second external information component. The interleaver 103 performs interleaving of primary decoded signal. The second component decoder 105 receives the primary decoded signal coming from the output of the interleaver 103, and the second signal parity Y2Kreceived from demultiply parity Y2Kby issuing a second decoded signal is converted interleaver 111. Next, the second component decoder 105 via directed interleaver 109 delivers the external information component to the first component decoder 101.\n\nAs shown in Fig.2, the first component decoder includes unit 113 measurements of branching (WWII) to calculate metrics branching and block 115 summation-compare-select (CERs) to calculate metrics and perform the comparison in each state to choose the path with fewer errors.\n\nIn the General case, the iterative decoder computes the metric value Mtaccording to the following equation (1).", null, "where Mt- the accumulated value of the metric at time t; Ut- the code word for the systematic bits, the code word for each bit Xto; xt,j- codeword redundancy bits; yt,j- the resulting value for the channel (systematic + excess); Lc is the reliability of the channel, and L(ut) is the a priori value of reliability at time t.\n\nFrom equation (1) implies that at each calculation of the metric the metric value Mtcontinuously growing by the second, third and fourth members. When perepolneny. However, the main purpose of the iterative decoder is performing iterative decoding to improve the characteristics of the decoding (i.e. error rate bit (hospital has no facilities) or error rate for personnel (PSCS)). Thus, during execution of the iterative decoder's function after a number of consecutive iterations metric values can increase and grow beyond the specified range. Therefore, if you are developing hardware decoder assumes the task of a certain range for the values of the metrics, the value metric may exceed the specified range, and there is a problem of overflow.\n\nSummary of the invention Therefore, the object of the present invention is to provide a device and method for normalizing metric values of the component decoder, and exceeding all accumulated metric values for the current state of a certain threshold, these accumulated metric values are normalized to a specific level after subtracting from them the specified value.\n\nTo achieve the above result, it is proposed a decoder that uses the change in the set of metrics on the set of time segments. The decoder includes therefore the TES value. MyCitadel it subtracts a predetermined value from the metric values in response to the signal decision to normalize the metric values. The decision tree includes multiple storage devices for storing the corresponding values of the metrics with a predetermined number of bits. The logical element AND-NOT to generate a signal (\"1\" or high signal) when all values most bits (PRS) provided in the respective storage device is equal to \"1\" (high level). MyCitadel sets to zero the PRS in each storage device, when the logical element AND-NOT outputs of higher-level decisions, resulting from each of the metric values is subtracted preset value.\n\nBrief description of drawings\nThe above and other objectives, features and advantages of the present invention are explained in the following detailed description, illustrated by the drawings, which represent the following:\nFig. 1 is a block diagram showing the iterative decoder containing two-component decoder;\nFig.2 is a detailed block diagram showing the component decoders of Fig. 1;\nFig. 3 is a diagram illustrating the operation of CERs komponentov is but first embodiment of the present invention;\nFig. 4 is a flowchart showing the procedure of normalizing metric values according to the first embodiment of the present invention;\nFig. 5 is a diagram illustrating the operation of the CER component decoder, which has the facility to normalize the metric values in the block CER component decoder according to the second variant of the present invention;\nFig. 6 is a diagram showing the format of the storage device for the values of metrics to normalize the metric values according to the second variant of the present invention;\nFig. 7 is a flowchart illustrating the procedure of normalizing metric values according to the second variant of the present invention;\nFig. 8A and 8B is a diagram illustrating the right way, a wrong way and a difference of ways, and the quantization scheme for the code symbols;\nFig. 9A-9C is a diagram illustrating a right way and wrong way in accordance with the signal-to-noise ratio; and\nFig. 10 is a graph showing the value of", null, "maxin the saturation state, depending on the relationship of energy (signal) to the noise power Eb/No.\n\nDetailed description the preferred option of carrying out the invention\nCERs for a component decoder in accordance with the present itaut the threshold value.\n\nThere are two ways to normalize the accumulated metric values in accordance with the present invention. According to the first method the accumulated metric values are normalized using the minimum accumulated metric values when one of the accumulated metric values of the respective States exceeds the threshold value. According to the second method accumulated metric values are normalized using the pre-set value, when all the accumulated metric values exceed the threshold value.\n\nNormalization for CERs of the present invention can be used for the normalization to unit CERs 115 iterative decoder 101 described above in connection with Fig.2.\n\nA. the First option\nNext with reference to Fig.3 describes a first variant embodiment of the invention. In Fig.3 shows the block structure of CERs, with the device normalizing metric values for the code restriction K=3 according to the first embodiment of the present invention.\n\nBelow with reference to Fig.3 describes a device normalizing metric values. In Fig.3 shows four \"current state\", each of which has a metric value. When K=3 the number of shift registers for the values of the metrics is the La of each state. When all defined values of the metrics exceeds a threshold value, the comparator 117 outputs the specified value to the adders 125 - 125d, and each adder connected between one's current state and one by the following conditions. Then the adders 125 - 125d of the accumulated metric values for the current state subtracts the specified value, and the resulting values are given in the following States. In this description, the term \"cumulative metric values of the current state\" is used instead of the term \"metric values of the current state and on the contrary, to emphasize the fact that the values of the metrics for the current state of the sequential computation of the metrics are accumulated.\n\nIn Fig. 4 shows the procedure for normalizing metric values according to the first embodiment of the present invention. According Fig.4, the comparator 117 at step 401 determines the metric values for the four current States. After determining the values of the metrics comparator 117 at step 403 checks whether at least one of the specific accumulated metric values in the threshold value. If none of the accumulated metric values exceeds the threshold value, the comparator 117 passes to step 407 for performing, the comparator 117 at step 405 outputs the adders 125-125d minimum of four detected accumulated metric values. Then the adders 125-125d from all four of the accumulated metric values are subtracted minimum accumulated metric value, and then proceeds in the following States. After that, as shown in step 407, the decoder goes to the normal operation of CERs.\n\nC. the Second option\nThe following describes a second variant embodiment of the invention.\n\nIn Fig.5 shows the block structure normalization of CERs according to the second variant of the present invention. According Fig.5, the comparator includes many storage devices 130, 132, 134 and 136 for storing the accumulated metric values of the respective States, the logical element And 121 to determine whether all the accumulated metric values stored in the storage devices 130, 132, 134 and 136, exceeds a threshold value, and the inverter 119 for setting to zero the high order bit (PRS) of the respective storage devices 130, 132, 134 and 136 in response to a high signal issued by the logical element And 121.\n\nThe format of the memory devices described with reference to Fig.6. Here it is assumed that each accumulated Zn is with one additional bits to prevent overflow of the accumulated metric values. Thus, the accumulated value of the metric is only 9 bits per sample. As shown in Fig.5, the logical element 121 And receives the ninth bit, which is the highest bit (PRS) storage devices 130, 132, 134 and 136, and generates an output signal of high level when all the input signals is \"1\". That is, when none of the PRS storage devices 130, 132, 134 and 136 is not equal to \"1\", the logical element 121 And does not generate an output signal (low level signal). When all of the PRS of the storage devices have a high level or \"1\", the logical element 121 And generates a high signal. When the logical element 121 And outputs a signal of high level, the inverter 119 outputs a signal installation in the zero bit of the PRS storage devices 130, 132, 134, 136, thereby establishing the PRS bits to zero. This is equivalent to subtraction of 256 from each of the accumulated metric values, which allows to Express the accumulated metric values using 8 bits.\n\nSuppose, further, that the difference between the accumulated values of the metrics of the two States is", null, "k= (uki-ukj-)", null, "", null, "maxwhere i and j represent one of the values 0, 1, 2 and 3, a k - p the IR for the two States is", null, "max= 255 = 28-1. Finally, suppose that u1k- minimum metric value, a u3k- maximum metric value, as shown in Fig.6.\n\nAs for overflow, if all bits of the PRS metric values at time k is equal to \"1\", the minimum value is 256.\n\nAccording to the above assumptions, if the PRS for u3kis \"1\", then the PRS other States will be equal to \"0\" or \"1\". Until all bits of the PRS for uik(where 0", null, "i", null, "3) becomes equal to \"1\", the output signal of the transfer will not appear even if the PRS u3k(and maybe one or two other metric values) is equal to \"1\". That is, until all bits of the PRS becomes equal to \"1\", the output signal of the transfer on the ninth bit does not occur for any of them. This means that", null, "kdoes not exceed", null, "max.\nIn Fig.7 shows a flowchart illustrating the procedure of normalizing metric values according to the second variant. According Fig.5 and 7, at step 501, the comparator 117 determines the accumulated metric values in specific units, the logical element 121 And the comparator 117a determines (or receives) PRS bits of the accumulated values ametryplene metric values corresponding to the current state exceeds a threshold value. That is, the logical element 121 of the comparator 117 determines whether all bits of the PRS is \"1\", as shown in decision block 503. If none of the PRS is not equal to \"1\", the comparator 117 proceeds to step 507 to perform common operations CERs. If all bits of the PRS accumulated metric values are equal to \"1\", the comparator 117a goes to step 505, where each metric value is subtracted threshold value. That is, all bits of the PRS are set to zero. This corresponds to the logical element 121 And applying a high signal to the inverter 119, which in response to this signal generates a setting signal to zero bits PRS respective accumulated metric values, thereby establishing the PRS to zero. After installing PRS bits to zero comparator 117a, as shown in step 507, performs the normal operation of CERs.\n\nNext with reference to Fig. with 8A 10 description", null, "maxdefined above. When", null, "k<", null, "maxoverflow does not occur.", null, "maxhas a lower value at low value of the ratio Eb/No and has a higher value at high value of the ratio Eb/No. That is, the difference between the values of the metrics has the m, that noise at low Eb/No increases, resulting in reduction of the aforementioned difference, at high Eb/No noise is extremely small, which increases the difference", null, "maxbetween the values of the metrics. Therefore, it is very important, which is set to", null, "maxat high Eb/No. In the first case, you can just assume that", null, "maxhas an infinite value at infinite value of Eb/No. However, for example, the Viterbi algorithm with weakly defined (\"soft\") output (SOVA algorithm) the difference between the metrics is limited to a constant, defined asfree.\n\nFor example, suppose you have 4 bits per sample, the rate of the code R= 1/3, K= 9 convolutional code transmits a code word with all zeros \"000\". In this case, when the high value of the ratio Eb/No most of the errors you receive during the comparison/selection between the way with all zeros and by dfreeas shown in Fig.8A. Here, the value of the branch metric and the metric value of the path is calculated by the following equations (2) and (3) respectively.", null, "", null, "where 1=0, 1, 2, and 3, Ck,j- code word y(i)k,j- received signal, r is the PTO is sup>ki= us,ki-uc,ki", null, "", null, "maxwhere \"s\" denotes the selected path, and \"C\" indicates a competitive way. You must calculate", null, "maxat high Eb/No, when", null, "kihas a maximum value. This means that if", null, "kiless", null, "maxat high Eb/No, when", null, "kihas a maximum value, then the difference between the values of the metrics does not exceed", null, "max.\nIn this state of \"i\" there is a difference metric between two paths: the path with all zeros and the path dfree. In Fig.8B shows that the difference between the two paths depends on the code symbol dfree.\n\nIn other words, the metric of the selected path is a value obtained by summing the metric of the path for the zero path with a metric value in the first state in the previous time, and the metric competitive path is a value obtained by adding the path metric corresponding to a competitive way, with the metric value in the second state in PA more than the metric of the path between the first state and the time of the comparison, the difference", null, "maxequal to or greater than", null, "ki. Therefore, when the conditions are satisfied for", null, "maxalso satisfied the terms and", null, "ki. The fact that the difference does not exceed", null, "maxthat means that the difference in metric values between the two conditions in the above-mentioned point in time does not exceed", null, "The difference metric is given by the following expression:\n(dfree(convolutional encoder)=18 for K=9, R=1/3)", null, "where M denotes the value of the metric at the point of branching of the chosen path and competitive way. Therefore, when a condition is met", null, "max", null, "270, the value of the difference between the corresponding States does not exceed", null, "max. Because it was assumed that the sample has 4 bits, the number of storage devices for storing the metric values is 8, and as to prevent overflow is added 1-bit storage device,/p> In Fig. 9A shows the value of", null, "maxat high signal-to-noise ratio, the value of", null, "maxis calculated by the formula", null, "max= dfree", null, "Max(Q[ctot]) ... (4)\nwhere Q denotes the level of quantization, a Max(Q[.]) denotes the distance between \"0\" and \"1\". For example, 4 bits per sample Q=16 and Max(Q[.])= 15, and 3 bits per sample Q=8, a Max(Q[.])=7.\n\nIn Fig.9B shows the value of", null, "maxwith an average signal-to-noise ratio, where a value of", null, "maxat this point, is calculated by the formula", null, "max= (dfree+", null, ")", null, "Max(Q[.]) ... Ur.5\nwhere the value of", null, "due to noise, a very small value, and it is less than or equal to 2 x dfreex Max(Q[.]) in the convolutional encoder (SC). However, this is not the case when", null, "summed as shown in equation (5).\n\nIn Fig. 9C shows the value of", null, "maxat low signal-to-noise, while at this point, the value of", null, "maxis calculated by the formula", null, "max=0, it should be noted that the value of", null, "maxgradually increases with increasing Eb/No, and starting from a certain point saturation occurs. If", null, "maxsatisfies equation (5), equation (6) is also satisfied.\n\nThe following describes the characteristics of the convolutional encoder in the system of the CDMA-2000.\n\nFor K=9 and R=1/2, dfree=12, and dfree", null, "14, 16, 18, 20.\n\nFor K=9 and R=1/3, dfree=18, and the next dfree", null, "20, 22.\n\nFor K=9 and R=1/4, dfree=24, and the next dfree", null, "26, 18.\n\nThe table shows the value of", null, "maxin the convolutional encoder (SC).\n\nTherefore, the number of bits that are added to prevent overflow for 8 bits per sample, which are assigned for metric values, is defined as follows.\n\nFor R=1/2 the number of bits equal to 1, since 28=256 180<256; for R= 1/3 the number of bits is 2, because 29=512, and 270<512; for R=1/4 the number of bits is 2, because 29=512, and 360<512. In other words, because the rate of the code R=1/2 requires 8 bits to prevent overflow, you need to add only 1 bit. In addition, because scene, adding to the number of bits required for a given speed of your code, only 1 bit.\n\nAs described above, the new device can prevent errors due to overflow by normalizing the accumulated metric values for decoding, resulting in more efficient memory usage.\n\nAlthough the invention has been shown and described with reference to specific preferred implementation, specialists in the art it is obvious that it can be made various changes in form and detail, not beyond being and scope of the invention defined in the claims.\n\nClaims\n\n1. The decoder uses the change in the set of metrics on the set of time segments, containing the decision tree, which generates a signal decisions when all metric values exceed the predetermined value, and myCitadel, which subtracts the predetermined value from the metric values in response to a signal solutions.\n\n2. The decoder under item 1, characterized in that the decision tree contains many storage devices having a pre-set E, which as the input signal gets values most bits (PRS) supplied from the respective storage devices, and the output of the logical element AND-NOT given the signal of the higher-level decisions, when all values of the PRS have a high level.\n\n3. The decoder under item 2, characterized in that myCitadel contains an inverter that has an input signal, which is combined with the output signal of the logical element AND-NOT, and the output signal, which mates with the PRS storage devices, and the signal of the higher-level decisions issued by the logical element AND-NOT on the inverter generates an output signal of the inverter, which sets to zero the value of the PRS of the respective storage devices.\n\n4. The method of normalizing metric values in the decoder, using the edit multiple values of the metrics on the set of time segments, comprising the steps of determining whether all of the metric values exceed the preset level, and subtracting a predetermined value from the values of the metrics for transition to the next state when all metric values exceeded this predetermined value.\n\n5. The method according to p. 4, characterized in that the subtraction pre-installed is achene metrics.\n\n6. The method of normalizing metric values in the decoder, using the edit multiple values of the metrics on the set of temporal segments and having multiple storage devices for storing the metric values with a predetermined number of bits, comprising the steps of determining whether all the values of the PRS of the respective storage devices is equal to 1, and setting to zero the values of the PRS, when all values of the PRS is equal to 1.\n\nSame patents:", null, "The invention relates to communication systems and can be used in the quantization means", null, "The invention relates to the field of radio communications and computing, and more particularly to methods and devices for data transmission in the computer network by radio with pseudorandom change the operating frequency", null, "The invention relates to electrical engineering and can be used in communication systems with pseudorandom change the operating frequency", null, "The invention relates to electrical engineering and can be used in the communication system with signal transmission in a wide range, in particular to the actions of search honeycomb performed by the mobile station, and to receive specific honeycomb long code used in the communication system in a wide range", null, "The invention relates to a communication system, multiple access, code-division multiplexing (mdcr)", null, "SUBSTANCE: proposed device depends for its operation on comparison of read-out signal with two thresholds, probability of exceeding these thresholds being enhanced during search interval with the result that search is continued. This broadband signal search device has linear part 1, matched filter 2, clock generator 19, channel selection control unit 13, inverter 12, fourth adder 15, two detectors 8, 17, two threshold comparison units 9, 18, NOT gates 16, as well as AND gate 14. Matched filter has pre-filter 3, delay line 4, n attenuators, n phase shifters, and three adders 7, 10, 11.\n\nEFFECT: enhanced noise immunity under structural noise impact.\n\n1 cl, 3 dwg", null, "SUBSTANCE: proposed automatically tunable band filter has series-connected limiting amplifier 1, tunable band filter 2 in the form of first series-tuned circuit with capacitor whose value varies depending on voltage applied to control input, first buffer amplifier 3, parametric correcting unit 4 in the form of second series-tuned circuit incorporating variable capacitor, second buffer amplifier 5, first differential unit 6, first amplitude detector 7, first integrating device 9, and subtraction unit 9. Inverting input of subtraction unit 9 is connected to reference-voltage generator 10 and output, to control input of variable capacitors 2 and 4. Automatically tunable band filter also has series-connected second amplitude detector 11, second integrating unit 12, and threshold unit 13. Synchronous operation of this filter during reception and processing of finite-length radio pulses is ensured by synchronizer 14 whose output is connected to units 10, 8, and 12. This automatically tunable band filter also has second differential unit whose input is connected to output of buffer amplifier 3 and output, to second control input of variable capacitor of band filter 2.\n\nEFFECT: enhanced noise immunity due to maintaining device characteristics within wide frequency range.\n\n1 cl, 1 dwg", null, "FIELD: radio communications engineering; mobile ground- and satellite-based communication systems.\n\nSUBSTANCE: proposed modulator that incorporates provision for operation in single-channel mode with selected frequency modulation index m = 0.5 or m = 1.5, or in dual-channel mode at minimal frequency shift and without open-phase fault has phase-shifting voltage analyzer 1, continuous periodic signal train and clock train shaping unit 2, control voltage shaping unit 3 for switch unit 3, switch unit 3, switch unit 4, two amplitude-phase modulators 5, 6, phase shifter 7, carrier oscillator 8, and adder 9.\n\nEFFECT: enlarged functional capabilities.\n\n1 cl, 15 dwg", null, "FIELD: electronic engineering.\n\nSUBSTANCE: device has data processing circuit, transmitter, commutation unit, endec, receiver, computation unit, and control unit.\n\nEFFECT: high reliability in transmitting data via radio channel.\n\n4 dwg", null, "FIELD: electronic engineering.\n\nSUBSTANCE: method involves building unipolar pulses on each current modulating continuous information signal reading of or on each pulse or some continuous pulse sequence of modulating continuous information code group. The number of pulses, their duration, amplitude and time relations are selected from permissible approximation error of given spectral value and formed sequence parameters are modulated.\n\nEFFECT: reduced inetrsymbol interference; high data transmission speed.\n\n16 cl, 8 dwg", null, "FIELD: communication system transceivers.\n\nSUBSTANCE: transceiver 80 has digital circuit 86 for converting modulating signals into intermediate-frequency ones. Signal source 114 transmits first periodic reference signal 112 at first frequency. Direct digital synthesizer 84 receives second periodic signal 102 at second frequency from first periodic reference signal. Converter circuit affording frequency increase in digital form functions to convert and raise frequency of modulating signals into intermediate-frequency digital signals using second periodic signal 102. Digital-to-analog converter 82 converts intermediate-frequency digital signals into intermediate-frequency analog signals using first periodic reference signal 112.\n\nEFFECT: reduced power requirement at low noise characteristics.\n\n45 cl, 3 dwg", null, "SUBSTANCE: proposed receiver has multiplier 4, band filter 6, demodulator 8, weighting coefficient unit 5, adding unit 7, analyzing and control unit 10, synchronizing unit 3, n pseudorandom sequence generators 21 through 2n, decoder 1, and switch unit 9. Receiver also has narrow-band noise suppression unit made in the form of transversal filter. Novelty is that this unit is transferred to correlator reference signal channel, reference signal being stationary periodic signal acting in absence of noise and having unmodulated harmonic components that can be rejected by filters of simpler design than those used for rejecting frequency band of input signal and noise mixture. Group of synchronized pseudorandom sequence generators used instead of delay line does not need in-service tuning.\n\nEFFECT: facilitated realization of narrow-band noise suppression unit; simplified design of rejection filters.\n\n1 cl, 8 dwg", null, "SUBSTANCE: proposed method and device are intended to control transmission power levels for plurality of various data streams transferred from at least one base station to mobile one in mobile radio communication system. First and second data streams are transmitted from base station and received by mobile station. Power-control instruction stream is generated in mobile station in compliance with first or second data stream received. Power control signal is shaped in mobile station from first power control instruction stream and transferred to base station. Received power control instruction stream is produced from power control signal received by base station; power transmission levels of first and second data streams coming from base station are controlled in compliance with power control instruction stream received. In this way control is effected of transmission power levels of first data stream transferred from each base station out of first active set to mobile station and of transmission power levels of second data stream which is transferred from each base station out of second active set to mobile station.\n\nEFFECT: enlarged functional capabilities.\n\n80 cl, 21 dwg", null, "SUBSTANCE: proposed method and device designed for fast synchronization of signal in wade-band code-division multiple access (WCDMA) system involve use of accumulations of variable-length samples, testing of decoder estimates for reliability, and concurrent decoding of plurality of sync signals in PERCH channel. Receiver accumulates samples required for reliable estimation of time interval synchronization. As long as time interval synchronization estimates have not passed reliability tests, samples are accumulated for frame synchronization estimates. As long as frame synchronization estimates have not passed reliability tests, samples are analyzed to determine channel pilot signal shift.\n\nEFFECT: reduced time for pulling into synchronism.\n\n13 cl, 9 dwg", null, "FIELD: satellite navigation systems and may be used at construction of imitators of signals of satellite navigational system GLONASS and pseudo-satellites.\n\nSUBSTANCE: for this purpose two oscillators of a lettered frequency and of a fixed frequency are used. Mode includes successive fulfillment of the following operations - generation of a stabilized lettered frequency, its multiplication with an oscillator's fixed frequency and filtration of lateral multipliers with means of filters of L1 and L2 ranges and corresponding option of a fixed and a lettered frequencies.\n\nEFFECT: reduces phase noise and ensures synthesizing of lettered frequencies of L1 and L2 ranges of satellite navigational system from one supporting generator at minimum number of analogous super high frequency units.\n\n3 cl, 1 dwg", null, "" ]	[ null, "https://img.russianpatents.com/img_data/65/658960.gif", null, "https://img.russianpatents.com/chr/916.gif", null, "https://img.russianpatents.com/chr/916.gif", null, "https://img.russianpatents.com/chr/8804.gif", null, "https://img.russianpatents.com/chr/916.gif", null, "https://img.russianpatents.com/chr/916.gif", null, "https://img.russianpatents.com/chr/8804.gif", null, "https://img.russianpatents.com/chr/8804.gif", 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https://itprospt.com/num/20695790/show-that-the-setb-x27-1-1-2-1-1-1-4-3-1-2-0-3-1-2-2-0-basis	[ "1\n\n# Show that the setB' = {(1,-1,2,1), (1,1,-4,3),(1,2,0.3), (1,2,-2,0)}basis for R\"...\n\n## Question\n\n###### Show that the setB' = {(1,-1,2,1), (1,1,-4,3),(1,2,0.3), (1,2,-2,0)}basis for R\"\n\nShow that the set B' = {(1,-1,2,1), (1,1,-4,3),(1,2,0.3), (1,2,-2,0)} basis for R\"", null, "", null, "#### Similar Solved Questions\n\n##### Problem 1Q An object is thrown downward from the top of the building with initial velocity of 50 ftls. Assuming positive direction of y measured downward from the top; derive an expression for (a) the velocity; and the displacement as function of time_ Assume J(0) = 0 .What is the velocityat F052What is the positionat F25?\nProblem 1Q An object is thrown downward from the top of the building with initial velocity of 50 ftls. Assuming positive direction of y measured downward from the top; derive an expression for (a) the velocity; and the displacement as function of time_ Assume J(0) = 0 . What is the velocity at F052 ...\n##### 14 In &udying Ine ortena gaelicchaadoisic the following sanpledata weecbtanulCaxtuisicFrequencIf the assumption that the characteristics occur #ith the same frequency, find the - test stalistic4) 825233300D) 11.0715) Oneway mayasof vaimoisusd letto equdity Inreed moopopuldion varanos oudity = Inrea0r moc eamplcmcay qudity theeo moropopulaion prcportions equaity tnreaor mcre popultion maans\n14 In &udying Ine ortena gaelicchaadoisic the following sanpledata weecbtanul Caxtuisic Frequenc If the assumption that the characteristics occur #ith the same frequency, find the - test stalistic 4) 82 5233 300 D) 11.07 15) Oneway mayasof vaimoisusd letto equdity Inreed moopopuldion varanos oud...\n##### Eanpe Dests2 3 4 5 61 7 8438 : 10 8\nEanpe Dests 2 3 4 5 61 7 8 43 8 : 10 8...\n##### A) 83% 16:) The\" B) 120% elemental% C) )b% (a 58 butane, C4H1o; 2 & 68.241\nA) 83% 16:) The\" B) 120% elemental% C) )b% (a 58 butane, C4H1o; 2 & 68.24 1...\n##### Choose the correct events referring to the synthesis, translocation and transport in the cell of a multipass transmembrane O-glycosylated plasma membrane protein from the list below and order them chronologicallyProtein is glycosylated in the Golgi SRP (Signal recognition particle) recognizes the signal sequence Protein is glycosylated in the ER (endoplasmic reticulum) Translocator opens and translocation starts Protein is transported to the Golgi Vesicles transporting the protein fuse with the\nChoose the correct events referring to the synthesis, translocation and transport in the cell of a multipass transmembrane O-glycosylated plasma membrane protein from the list below and order them chronologically Protein is glycosylated in the Golgi SRP (Signal recognition particle) recognizes the s...\n##### 105. The hydrogen atom contains 1 proton and 1 electron: The radius of the proton is approximately 1.0 fm (femtometer) , and the radius of the hydrogen atom is approximately 53 pm (picometers) Calculate the volume of the nucleus and the volume of theatom for hydrogen. What percentage of the hydrogen atom s volume does the nucleus occupy? (Hint: Convert both given radii t0 m, and then calculate their volumes using the formula for the volume of sphere, which is V = 37r' ,)\n105. The hydrogen atom contains 1 proton and 1 electron: The radius of the proton is approximately 1.0 fm (femtometer) , and the radius of the hydrogen atom is approximately 53 pm (picometers) Calculate the volume of the nucleus and the volume of theatom for hydrogen. What percentage of the hydrogen...\n##### Explore how a technique called centered differencing can be used to extend our understanding of derivatives by accommodating corners, and some cusps. These exercises should be done in order. The graph of $f$ is shown in Figure $4.14$, and the point $P$ is at $left(t_{0}, fleft(t_{0}ight)ight)$.(a) Draw the line segment whose slope is $M(Delta t)=frac{fleft(t_{0}+Delta tight)-fleft(t_{0}-Delta tight)}{2 Delta t}$.(b) Suppose that $f$ is differentiable at $t_{0}$. Verify (analytically, not graphic\nExplore how a technique called centered differencing can be used to extend our understanding of derivatives by accommodating corners, and some cusps. These exercises should be done in order. The graph of $f$ is shown in Figure $4.14$, and the point $P$ is at $left(t_{0}, fleft(t_{0} ight) ight)$. (...\n##### Consider the following differential equation V4) -v\"+a\" + bv' + cy=0where a, b, â‚¬ are constants: Given equation has the following general solution: Kx) = â‚¬1e ~x+ Gzez* C3cos(3x) + c4sin(3x) where â‚¬1, Cz, C3, C4 are arbitrary constants Then, which of the following is true?Yanitiniz:a + b + 0=30a + b+c=20a +b+c= - 200 + b+c= -30None of the above\nConsider the following differential equation V4) -v\"+a\" + bv' + cy=0 where a, b, â‚¬ are constants: Given equation has the following general solution: Kx) = â‚¬1e ~x+ Gzez* C3cos(3x) + c4sin(3x) where â‚¬1, Cz, C3, C4 are arbitrary constants Then, which of the following i...\n##### Determine currents in each of the resistors Rl = 12 }; R2 = 12 02; R3 = 5 (};R4 =6 (1; RS = 6 ( ; R6 = 6\nDetermine currents in each of the resistors Rl = 12 }; R2 = 12 02; R3 = 5 (};R4 =6 (1; RS = 6 ( ; R6 = 6...\n##### Question 4 (8 p) Complete the following reactions providing the structures of (A), (B), (C) and (D): Write reasonable mechanism for the preparation of (D) from (A}:EtoM 'EtOCH;2.B,OC4HjoO1.H,0CH;0OCHi EL,02 , H,0 +C;HpO\nQuestion 4 (8 p) Complete the following reactions providing the structures of (A), (B), (C) and (D): Write reasonable mechanism for the preparation of (D) from (A}: Eto M 'EtO CH; 2.B,O C4HjoO 1.H,0 CH;0 OCHi EL,0 2 , H,0 + C;HpO...\n##### (6 x 5) Trarslate each of these logical expressions into English Let M(-)\" denote 'X is man\" Let W(r) denote \"X = woman\" Let W (I,y) denote works for y\"_ Let - denote Ivan and denote Peter.J(W(s) _ Vy( M(y) = -W(I,y))) Bolucion:VI(M(c) 3y(W(y) W(I,y))) Bolucion:Hi(M(I) Vy(W(I,y) - W(y))) Bolutioc:Hiy(M(c) W(y,I) W(y)) Bolucion:W(i,p) Vr(W(p,r) +-W(r)) Bolucion:\n(6 x 5) Trarslate each of these logical expressions into English Let M(-)\" denote 'X is man\" Let W(r) denote \"X = woman\" Let W (I,y) denote works for y\"_ Let - denote Ivan and denote Peter. J(W(s) _ Vy( M(y) = -W(I,y))) Bolucion: VI(M(c) 3y(W(y) W(I,y))) Bolucion: Hi(M(...\n##### Question 4A set contains fourteen elements. How many subsets does it have? 16384Blank16384Question 5The set contalning all the elements that are common tO both set A and set B Is called:None.The Intersection of set A and set B.The Unlon of set A and set B_\nQuestion 4 A set contains fourteen elements. How many subsets does it have? 16384 Blank 16384 Question 5 The set contalning all the elements that are common tO both set A and set B Is called: None. The Intersection of set A and set B. The Unlon of set A and set B_...\n##### What is the E\" for the following reaction: 2H+ + 2e < > Hzat 1M[P OA-100v OB.0v 0â‚¬ 100 MA 00. 100 V 0 E 100 mA\nWhat is the E\" for the following reaction: 2H+ + 2e < > Hzat 1M[P OA-100v OB.0v 0â‚¬ 100 MA 00. 100 V 0 E 100 mA...\n##### Anenbor child rocelves 59.000 9It towar contcounded quanadyInyasied 21525Honh(Round Vt neltest cenL)\nAnenbor child rocelves 59.000 9It towar contcounded quanady Inyasied 21525 Honh (Round Vt neltest cenL)...\n##### If there is a reaction where the reactants and products are about equal in energy will tend to he reversible or irreversable? and would it be under kinetic or therodynamic control?\nIf there is a reaction where the reactants and products are about equal in energy will tend to he reversible or irreversable? and would it be under kinetic or therodynamic control?...\n##### LlunCCCinvaQuustion 8 ptsLonsidcrthamacnonZHz (e} Nz (c} ZH,0 {El Lab cxperimcnnId conchnon thit thc MNOJ[H-l?Mthc mcacnon undcr ccrln condition: %ath 0.10 MNO and 0.10 MHz, te nte rExcbion @ mneaeuned 64 i0 Mk When INO] k Increxsed 0 40MAhil Ulothermaction prameten renulnthc mnthc Terchon FunJA Cakculate tte exercted ncw ratc(Report Jrsvrer decimal torm with Ixo signifcInt 6 gures JadUnits Ior mtewould be M/sQuestion 3neaction second ordcr wlth rEspect to bcntophenole (lets cullit 82 Whn sturt\nLlunCC Cinva Quustion 8 pts Lonsidcrthamacnon ZHz (e} Nz (c} ZH,0 {El Lab cxperimcnnId conchnon thit thc MNOJ[H-l? Mthc mcacnon undcr ccrln condition: %ath 0.10 MNO and 0.10 MHz, te nte rExcbion @ mneaeuned 64 i0 Mk When INO] k Increxsed 0 40MAhil Ulothermaction prameten renulnthc mnthc Terchon Fu..." ]	[ null, "https://cdn.numerade.com/ask_images/abc14d350605460ba3cd6c4b554c94da.jpg ", null, "https://cdn.numerade.com/previews/1f9a1d05-bbda-4b8e-ac1d-e0fb388c715e_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8163194,"math_prob":0.9813712,"size":10147,"snap":"2022-40-2023-06","text_gpt3_token_len":3120,"char_repetition_ratio":0.11071675,"word_repetition_ratio":0.59598345,"special_character_ratio":0.288164,"punctuation_ratio":0.13225499,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995507,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T13:33:03Z\",\"WARC-Record-ID\":\"<urn:uuid:43705ac2-f548-424a-8bb3-ae6a3fe25d93>\",\"Content-Length\":\"53479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e45d648b-f1db-4adc-ba8a-8d1a7a2c3109>\",\"WARC-Concurrent-To\":\"<urn:uuid:7802d4f7-7c23-4718-bcde-3bed59d7f96f>\",\"WARC-IP-Address\":\"104.26.6.163\",\"WARC-Target-URI\":\"https://itprospt.com/num/20695790/show-that-the-setb-x27-1-1-2-1-1-1-4-3-1-2-0-3-1-2-2-0-basis\",\"WARC-Payload-Digest\":\"sha1:UQLRR4MBPDJXKX62SXKQ7PFQ4PVCXK3Q\",\"WARC-Block-Digest\":\"sha1:LHCZZTXRMPCMA6DGCTZAWWNU2FOAHCCL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337836.93_warc_CC-MAIN-20221006124156-20221006154156-00273.warc.gz\"}"}
https://www.wowebook.com/book/excel-2013-formulas/	[ "# Excel 2013 Formulas", null, "Maximize the power of Excel 2013 formulas with this must-have Excel reference\n\nJohn Walkenbach, known as “Mr. Spreadsheet,” is a master at deciphering complex technical topics and Excel formulas are no exception. This fully updated book delivers more than 800 pages of Excel 2013 tips, tricks, and techniques for creating formulas that calculate, developing custom worksheet functions with VBA, debugging formulas, and much more.\n\n• Demonstrates how to use all the latest features in Excel 2013\n• Shows how to create financial formulas and tap into the power of array formulas\n• Serves as a guide to using various lookup formulas, working with conditional formatting, and developing custom functions\n• Shares proven solutions for handling typical (and not-so-typical) Excel formula challenges\n• Includes links to the “Mr. Spreadsheet” website, which contains all the templates and worksheets used in the book, plus access to John Walkenbach’s award-winning Power Utility Pak.\n\nFrom charts to PivotTables and everything in between, Excel 2013 Formulas is your formula for Excel success.\n\nPart I: Basic Information\nChapter 1. Excel in a Nutshell\nChapter 2. Basic Facts about Formulas\nChapter 3. Working with Names\n\nPart II: Using Functions in Your Formulas\nChapter 4. Introducing Worksheet Functions\nChapter 5. Manipulating Text\nChapter 6. Working with Dates and Times\nChapter 7. Counting and Summing Techniques\nChapter 8. Using Lookup Functions\nChapter 9. Working with Tables and Lists\nChapter 10. Miscellaneous Calculations\n\nPart III: Financial Formulas\nChapter 11. Borrowing and Investing Formulas\nChapter 12. Discounting and Depreciation Formulas\nChapter 13. Financial Schedules\n\nPart IV: Array Formulas\nChapter 14. Introducing Arrays\nChapter 15. Performing Magic with Array Formulas\n\nPart V: Miscellaneous Formula Techniques\nChapter 16. Importing and Cleaning Data\nChapter 17. Charting Techniques\nChapter 18. Pivot Tables\nChapter 19. Conditional Formatting\nChapter 20. Using Data Validation\nChapter 21. Creating Megaformulas\nChapter 22. Tools and Methods for Debugging Formulas\n\nPart VI: Developing Custom Worksheet Functions\nChapter 23. Introducing VBA\nChapter 24. Function Procedure Basics\nChapter 25. VBA Programming Concepts\nChapter 26. VBA Custom Function Examples\n\nPart VII: Appendixes\nAppendix A. Excel Function Reference\nAppendix B. Using Custom Number Formats\n\n### Book Details\n\n• Paperback: 864 pages\n• Publisher: Wiley (April 2013)\n• Language: English\n• ISBN-10: 1118490444\n• ISBN-13: 978-1118490440" ]	[ null, "http://img.wowebook.com/images/2038502170.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67605716,"math_prob":0.4453315,"size":2534,"snap":"2021-21-2021-25","text_gpt3_token_len":590,"char_repetition_ratio":0.186166,"word_repetition_ratio":0.0,"special_character_ratio":0.2190213,"punctuation_ratio":0.13551402,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99001837,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T12:59:01Z\",\"WARC-Record-ID\":\"<urn:uuid:8a850e7b-eeaa-4e9a-9787-99f0b1e6d871>\",\"Content-Length\":\"35846\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:038a577a-be17-4e3d-9afb-6219098da31b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9450147-0ad2-40f2-ab91-a2bc0edafae9>\",\"WARC-IP-Address\":\"172.67.133.191\",\"WARC-Target-URI\":\"https://www.wowebook.com/book/excel-2013-formulas/\",\"WARC-Payload-Digest\":\"sha1:KCY2ESPIA66YVDRE6HUSWURWPHCFCJ5V\",\"WARC-Block-Digest\":\"sha1:JZKZPN6RYVVEUZ3IXZDXJ4NEOGSWMIK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991269.57_warc_CC-MAIN-20210516105746-20210516135746-00107.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/gr-qc/0005086/	[ "# A class of perfect-fluid cosmologies with polarised Gowdy symmetry and a Kasner-like singularity\n\nKeith Anguige\n###### Abstract\n\nWe prove the existence of a class of perfect-fluid cosmologies which have polarised Gowdy symmetry and a Kasner like singularity. These solutions of the Einstein equations depend on four free functions of one space coordinate and are constructed by solving a system of Fuchsian equations.\n\n## 1 Introduction\n\nIn this paper we use the Fuchsian algorithm to construct a family of perfect-fluid cosmologies with polarised Gowdy symmetry and with Kasner like asymptotics at early times. This family depends on the maximum number of free functions for spacetimes within the symmetry class. The technique used is to perturb exact Bianchi I solutions in one space-direction and to solve a Fuchsian system of equations for the perturbation. The results obtained are a generalisation of those in in the sense that the symmetry requirement has been relaxed a little, but here we require the free data for the field equations to be analytic rather than merely .\n\nThe main result is stated as Theorem 6.1 at the end of the paper.\n\n## 2 Exact solutions\n\nThe general Bianchi I solution of the Einstein-perfect fluid equations, as given in is\n\n ds2=−B2(γ−1)dτ2+τ2p1B2q1dx2+τ2p2B2q2dy2+τ2p3B2q3dz2 (1)\n\nwhere\n\n B2−γ=α+m2τ2−γ α≥0, m>0 (2)\n p1+p2+p3=1 , p21+p22+p23=1 , qi=23−pi (3)\n\n(The case is an FRW model and is excluded in what follows.)\n\nThe density of the fluid is given by\n\n μ=4m23τγBγ (4)\n\nWhen we come to write down Einstein’s equations for a metric with polarised Gowdy symmetry we will use conformal coordinates, i.e coordinates for which the metric takes the form\n\n ds2=e2A(t,x)(−dt2+dx2)+R(t,x)eW(t,x)dy2+R(t,x)e−W(t,x)dz2 (5)\n\nThe model solution (1) may be written in conformal coordinates by making the transformation\n\n τ→t=∫τ0Bp1+γ−53(s)s−p1 ds (6)\n\nThen the metric (1) takes the form (5) with\n\n eA=τp1B23−p1 , eW=τp2−p3Bp3−p2 , R=τ1−p1B(p1+13) (7)\n\nand given implicitly as function of by the relation (6).\n\n## 3 The Einstein-perfect fluid equations\n\nWe will assume that spacetime is filled with a polytropic perfect fluid, so that the stress tensor takes the form\n\n Tαβ=(ρ+P)uαuβ+Pgαβ (8)\n\nwith and .\n\nWith this stress tensor the Einstein evolution equations for the metric (5) are\n\n Rtt−Rxx=Re2Aρ(2−γ) (9)\n Wtt−Wxx+R−1(RtWt−RxWx)=0 (10)\n Att−Axx−14R−2(R2t−R2x)+14(W2t−W2x)=−12γe2Aρ (11)\n\n R−1Rxx−14R−2(R2t+R2x)−R−1RtAt−R−1RxAx+14(W2t+W2x)\n =−e2Aρ(1+γ(v1)2) (12)\n −R−1Rtx+R−1(RtAx+AtRx)+12R−2RtRx−12WtWx=−e2Aγρv0v1 (13)\n\nwhere and hence .\n\nThe Euler equations take the following explicit form\n\n γv0(v0ρt+v1ρx)+(1−γ)ρt\n +γρ{v0(2v0t+Axv1)+v1(v1x−v1Ax+Axv0+Atv1)}\n +γρv0{v0At+v1x+v1Ax+R−1(v0Rt+v1Rx)}=0 (14)\n\nand\n\n γv1(v0ρt+v1ρx)+(1−γ)ρx+γρ{v0(v1t+v0Ax)}\n +γρ{v1(v0t+2v1x+v0(2At+R−1Rt)+v1(Ax+R−1Rx)}=0 (15)\n\n## 4 Inhomogeneous perturbations\n\nWe now seek solutions of (9)-(15) which depend on the maximum number of free analytic functions, namely four, and which approach the model forms (1)-(4) for each as .\n\nTo be specific we make the following ansatz\n\n A=p1logτ+(23−p1)logB+t~A (16)\n R=τ1−p1B(p1+13)(1+~R) (17)\n W=c(x)+(p2−p3)logτ+(p3−p2)logB+t~W (18)\n logρ=logμ+t(γ−1−p1)/(1−p1)+ϵϕ (19)\n v1=τγ−1−p1(G(x)+tϵ~ψ) (20)\n\nwhere\n\n B2−γ=α(x)+m2(x)τ2−γ (21)\n G(x)=−34m2γα(p1+13)/(2−γ)× (22)\n {−12(p2−p3)cx+αxα(3γ(1−p1)+2(3p1−1))3(2−γ)+(p1)x(2(1−p1)(2−γ)logα+1)} (23)\n\nand is determined by the relation (6).\n\nand are chosen as strictly positive analytic functions, is an analytic function and the Kasner exponent is subject to the restriction for some arbitrarily small constant . The analytic function is chosen to satisfy\n\n 0<ϵ(x)\n\nWe are looking for analytic solutions of the Einstein-perfect fluid equations for which and tend to zero as tends to zero. The choice of ensures that the momentum constraint is satisfied at .\n\nIt is convenient to write the field equations for and in first order form. To do this we introduce the following new variables:\n\n U=~Rt, Q=(t~A)t, X=~Rx, Y=t~Ax, V=(t~W)t, Z=t~Wx, S=t−1~R\n\nIn terms of these variables the evolution equations (9)-(11) take the form\n\n tXt=tUx (25)\n tYt=tQx (26)\n t~Rt=tU (27)\n t~At+~A−Q=0 (28)\n tZt=tVx (29)\n t~Wt+~W−V=0 (30)\n tSt+S−U=0 (31)\n tUt+2U=tXx+2U(1−((1−p1)B2−γ+m2(p1+13)τ2−γ)B−(p1+13)tτp1−1)\n +tτp1−1B−(p1+13){2X(τ1−p1Bp1+13)x+(1+~R)(τ1−p1Bp1+13)xx}\n +4m23(2−γ)(1+tS)tτ2p1−γB43−2p1−γ(exp(2t~A+tϵ+(γ−1−p1)(1−p1)−1ϕ)−1) (32)\n tQt−12U+12(p2−p3)(1−p1)V=tYx+t{(p1logτ)x+(logB23−p1)xx}\n +14(1+~R)−2tU2−12U(1−tτp1−1B−(p1+13)(1+tS)−1((1−p1)B2−γ+m2(p1+13)τ2−γ))\n +14tτ2(p1−1)B−2(p1+13)(τ1−p1Bp1+13X+(1+~R)(τ1−p1Bp1+13)x)2\n −14tV2+12V(p2−p3)((1−p1)−1−tτp1−1B53+p1−γ+m2tτ1+p1−γB−13−p1)\n +14t(cx+((p2−p3)logτ)x+((p3−p2)logB)x+Z)2\n −2m23γB43−2p1−γtτ2p1−γ(exp(2t~A+tϵ+(γ−1−p1)(1−p1)−1ϕ)−1) (33)\n tVt+V+(p2−p3)(1−p1)U=tZx+t{cxx+((p2−p3)logτ)xx+((p3−p2)logB)xx}\n +tτp1−1B−p1−13(1+~R)(cx+((p2−p3)logτ)x+((p3−p2)logB)x+Z)(τ1−p1Bp1+13X+(1+~R)(τ1−p1Bp1+13)x)\n +U((p2−p3)(1−p1)−(1+tS)−1((p2−p3)tτp1−1B53−p1−γ+m2(p3−p2)tτ1+p1−γB−13−p1+tV))\n +V(1−B−p1−13tτp1−1((1−p1)B2−γ+m2τ2−γ(p1+1/3))) (34)\n\nThe Euler equations , after some rearrangement, may be written\n\n (1+2τ2βψ2v0−2γv0τ2βψ21+γτ2βψ2)(t~ψt+ϵ~ψ)=\n t1−ϵψ(γv0(v0+τβ)1+γτ2βψ2−2)(τβψx+ψ(τβ)x)\n +t1−ϵτ−β(γ−1γ+γ(v0)2τ2βψ21+γτ2βψ2)((m2)x−γ(τx+Bx)+tϵ+β(1−p1)−1(ϕx+ϕ(β(1−p1)−1)xlogt))\n +t1−ϵv0ψ(γ((v0)2+τ2βψ2)1+γτ2βψ2−2)((2/3−p1)B−13−p1m2τ−β+Q)\n −t1−ϵτ−β(1+2τ2βψ2−γτβψv01+γτ2βψ2(3v0τβψ−τ2βψ2))((p1logτ)x+(logB23−p1)x+Y)\n +t1−ϵψ(γ(v0)31+γτ2βψ2−v0)((1+~R)−1U+m2(p1+1/3)B−p1−13τβ)\n +t1−ϵτγ−2ψ2(γ(v0)21+γτ2βψ2−1)B−p1−13(1+~R)−1(τ1−p1Bp1+13X+(1+~R)(τ1−p1Bp1+13)x)\n +t1−ϵτp1−1ψB53−p1−γ{β(1+2γv0τ2βψ21+γτ2βψ2−1+2τ2βψ2v0)\n +p1(γ(v0((v0)2+τ2βψ2)1+γτ2βψ2−1)+2(1−v0))+(1−p1)(γ((v0)21+γτ2βψ2−1)+1−v0)} (35)\n\nand\n\n (1+γτ2βψ2−2γ(v0)2τ2βψ21+2τ2βψ2)tϕt+(ϵ+β1−p1)ϕ=\n =γ{(ϵ+β(1−p1)−1)ϕ(1γ+2(v0)2τ2βψ21+2τ2βψ2−1+γτ2βψ2γ)\n +γt1+βp1−1−ϵτp1−1τ2βψ2(1−2(v0)21+2τ2βψ2)(B53−p1−γ+m2τ2−γB−13−p1)\n −(tϕx+tϕlogt(ϵ+β(1−p1)−1)x+t1+βp1−1−ϵ((m2)x−γ(τx+Bx)))v0τβψ\n ×(1+21+2τ2βψ2(γ−1γ−τ2βψ2))\n +t1+βp1−1−ϵ(4τ2βψ2v01+2τ2βψ2−v0−τβψ)((τβG)x+τβtϵ~ψx+(τβtϵ)x~ψ)\n +t1+βp1−1−ϵ(4(v0)2τ2βψ21+2τ2βψ2−1−2τ2βψ2)Q\n +t1+βp1−1−ϵτ2βψ2(4(v0)21+2τ2βψ2−2)(p1τp1−1B53−p1−γ+((2/3)−p1)B−13−p1m2τ−β)\n +t1+βp1−1−ϵ(τβψ(τβψ−v0)((p1logτ)x+(logB23−p1)x+Y)+(v0τβψ−(v0)2)1+~RU)\n +t1+βp1−1−ϵτ2βψ2(2(v0)21+2τ2βψ2−1)τp1−1B−p1−13((1−p1)B2−γ+m2(p1+1/3)τ2−γ)\n +t1+βp1−1−ϵτγ−2ψ(2v0τ2βψ21+2τ2βψ2−v0)B−p1−131+~R(τ1−p1Bp1+13X+(1+~R)(τ1−p1Bp1+13)x)⎫⎬⎭ (36)\n\n## 5 Existence and uniqueness of solutions\n\nA careful inspection of the field equations (25)-(36) shows that they may be written in the form\n\n t∂tu+N(x)u=tδH(t,x,u,ux) (37)\n\nwhere stands for is a strictly positive constant and is continuous in and analytic in its other arguments. The matrix has positive eigenvalues.\n\nIt follows by that (37) has a unique analytic solution with\n\n## 6 The constraints\n\nDefine constraint quantities by\n\n C0=R−1Rxx−14R−2(R2t+R2x)−R−1RtAt−R−1RxAx+14(W2t+W2x)\n +e2Aρ(1+γ(v1)2) (38)\n C1=−R−1Rtx+R−1(RtAx+AtRx)+12R−2RtRx−12WtWx+e2Aγρv0v1 (39)\n\nIf the evolution equations (25)-(36) are satisfied then a calculation shows that the following hold\n\n ∂tC0=−∂xC1−RtRC0−RxRC1 (40)\n ∂tC1=−∂xC0−RtRC1−RtRC0 (41)\n\nOne also calculates that the quantities tend to zero as tends to zero. These quantities satisfy the following\n\n t∂t~C0+(tRtR−1)~C0=−t∂x~C1−tRxR~C1 (42)\n t∂t~C1+(tRtR−1)~C1=−t∂x~C0−tRxR~C0 (43)\n\nNow is and for some . It thus follows that and are identically zero and thus the constraints are satisfied.\n\nSummarising the results of sections (4)-(6) we have proved the following\n\nTheorem 6.1 Given two strictly positive analytic functions , an analytic function and an analytic function satisfying\n\n −13≤p1(x)<γ−1−k (44)\n\nfor some small , there exists a unique solution of the Einstein equations coupled to a -law perfect fluid on satisfying\n\n ds2=e2A(t,x)(−dt2+dx2)+R(t,x)eW(t,x)(dy)2+R(t,x)e−W(t,x)(dz)2\n A=p1logt+logB23−p1+O(t)\n R=τ1−p1Bp1+13(1+O(t))\n W=c(x)+(p2−p3)logt+(p3−p2)logB+O(t)\n logρ=log4m23τγBγ+O(t(γ−1−p1)(1−p1)−1)\n v1=O(t(γ−1−p1)(1−p1)−1)\n\nwhere is implicitly given by\n\n t=∫τ0Bp1+γ−53(s)s−p1 ds." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.874589,"math_prob":0.9996031,"size":4659,"snap":"2022-40-2023-06","text_gpt3_token_len":1088,"char_repetition_ratio":0.13834587,"word_repetition_ratio":0.026143791,"special_character_ratio":0.25241467,"punctuation_ratio":0.055622734,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976473,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T16:08:00Z\",\"WARC-Record-ID\":\"<urn:uuid:63162cf6-bdb6-4681-9dc1-d5e210d7e145>\",\"Content-Length\":\"925843\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6db5e8e0-0530-42a5-9ff0-b7fb9ced2f19>\",\"WARC-Concurrent-To\":\"<urn:uuid:80bef897-013c-47c6-835c-ba34ca4e0e50>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/gr-qc/0005086/\",\"WARC-Payload-Digest\":\"sha1:X3NHTYMLD4CRL4JXBYOW7N3NMT6DTQLY\",\"WARC-Block-Digest\":\"sha1:Z62AB4BE24S326HDSA75CITGHB6GV3DV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00328.warc.gz\"}"}
https://www.colorhexa.com/5b698e	[ "# #5b698e Color Information\n\nIn a RGB color space, hex #5b698e is composed of 35.7% red, 41.2% green and 55.7% blue. Whereas in a CMYK color space, it is composed of 35.9% cyan, 26.1% magenta, 0% yellow and 44.3% black. It has a hue angle of 223.5 degrees, a saturation of 21.9% and a lightness of 45.7%. #5b698e color hex could be obtained by blending #b6d2ff with #00001d. Closest websafe color is: #666699.\n\n• R 36\n• G 41\n• B 56\nRGB color chart\n• C 36\n• M 26\n• Y 0\n• K 44\nCMYK color chart\n\n#5b698e color description : Mostly desaturated dark blue.\n\n# #5b698e Color Conversion\n\nThe hexadecimal color #5b698e has RGB values of R:91, G:105, B:142 and CMYK values of C:0.36, M:0.26, Y:0, K:0.44. Its decimal value is 5990798.\n\nHex triplet RGB Decimal 5b698e `#5b698e` 91, 105, 142 `rgb(91,105,142)` 35.7, 41.2, 55.7 `rgb(35.7%,41.2%,55.7%)` 36, 26, 0, 44 223.5°, 21.9, 45.7 `hsl(223.5,21.9%,45.7%)` 223.5°, 35.9, 55.7 666699 `#666699`\nCIE-LAB 44.632, 4.261, -22.03 14.247, 14.28, 27.595 0.254, 0.254, 14.28 44.632, 22.438, 280.948 44.632, -8.546, -32.143 37.789, 1.169, -16.845 01011011, 01101001, 10001110\n\n# Color Schemes with #5b698e\n\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #8e805b\n``#8e805b` `rgb(142,128,91)``\nComplementary Color\n• #5b838e\n``#5b838e` `rgb(91,131,142)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #675b8e\n``#675b8e` `rgb(103,91,142)``\nAnalogous Color\n• #838e5b\n``#838e5b` `rgb(131,142,91)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #8e675b\n``#8e675b` `rgb(142,103,91)``\nSplit Complementary Color\n• #698e5b\n``#698e5b` `rgb(105,142,91)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #8e5b69\n``#8e5b69` `rgb(142,91,105)``\n• #5b8e80\n``#5b8e80` `rgb(91,142,128)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #8e5b69\n``#8e5b69` `rgb(142,91,105)``\n• #8e805b\n``#8e805b` `rgb(142,128,91)``\n• #3d475f\n``#3d475f` `rgb(61,71,95)``\n• #47526f\n``#47526f` `rgb(71,82,111)``\n• #515e7e\n``#515e7e` `rgb(81,94,126)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #66759d\n``#66759d` `rgb(102,117,157)``\n• #7583a7\n``#7583a7` `rgb(117,131,167)``\n• #8591b1\n``#8591b1` `rgb(133,145,177)``\nMonochromatic Color\n\n# Alternatives to #5b698e\n\nBelow, you can see some colors close to #5b698e. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #5b768e\n``#5b768e` `rgb(91,118,142)``\n• #5b728e\n``#5b728e` `rgb(91,114,142)``\n• #5b6d8e\n``#5b6d8e` `rgb(91,109,142)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #5b658e\n``#5b658e` `rgb(91,101,142)``\n• #5b618e\n``#5b618e` `rgb(91,97,142)``\n• #5b5c8e\n``#5b5c8e` `rgb(91,92,142)``\nSimilar Colors\n\n# #5b698e Preview\n\nThis text has a font color of #5b698e.\n\n``<span style=\"color:#5b698e;\">Text here</span>``\n#5b698e background color\n\nThis paragraph has a background color of #5b698e.\n\n``<p style=\"background-color:#5b698e;\">Content here</p>``\n#5b698e border color\n\nThis element has a border color of #5b698e.\n\n``<div style=\"border:1px solid #5b698e;\">Content here</div>``\nCSS codes\n``.text {color:#5b698e;}``\n``.background {background-color:#5b698e;}``\n``.border {border:1px solid #5b698e;}``\n\n# Shades and Tints of #5b698e\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #07080b is the darkest color, while #fefefe is the lightest one.\n\n• #07080b\n``#07080b` `rgb(7,8,11)``\n• #0e1116\n``#0e1116` `rgb(14,17,22)``\n• #161922\n``#161922` `rgb(22,25,34)``\n• #1e222e\n``#1e222e` `rgb(30,34,46)``\n• #252b3a\n``#252b3a` `rgb(37,43,58)``\n• #2d3446\n``#2d3446` `rgb(45,52,70)``\n• #353d52\n``#353d52` `rgb(53,61,82)``\n• #3c465e\n``#3c465e` `rgb(60,70,94)``\n• #444e6a\n``#444e6a` `rgb(68,78,106)``\n• #4c5776\n``#4c5776` `rgb(76,87,118)``\n• #536082\n``#536082` `rgb(83,96,130)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #63729a\n``#63729a` `rgb(99,114,154)``\n• #6e7ca2\n``#6e7ca2` `rgb(110,124,162)``\n• #7a87aa\n``#7a87aa` `rgb(122,135,170)``\n• #8692b1\n``#8692b1` `rgb(134,146,177)``\n• #929db9\n``#929db9` `rgb(146,157,185)``\n• #9ea7c1\n``#9ea7c1` `rgb(158,167,193)``\n• #aab2c8\n``#aab2c8` `rgb(170,178,200)``\n• #b6bdd0\n``#b6bdd0` `rgb(182,189,208)``\n• #c2c8d8\n``#c2c8d8` `rgb(194,200,216)``\n• #ced3df\n``#ced3df` `rgb(206,211,223)``\n``#dadde7` `rgb(218,221,231)``\n• #e6e8ef\n``#e6e8ef` `rgb(230,232,239)``\n• #f2f3f6\n``#f2f3f6` `rgb(242,243,246)``\n• #fefefe\n``#fefefe` `rgb(254,254,254)``\nTint Color Variation\n\n# Tones of #5b698e\n\nA tone is produced by adding gray to any pure hue. In this case, #6d717c is the less saturated color, while #0141e8 is the most saturated one.\n\n• #6d717c\n``#6d717c` `rgb(109,113,124)``\n• #646d85\n``#646d85` `rgb(100,109,133)``\n• #5b698e\n``#5b698e` `rgb(91,105,142)``\n• #526597\n``#526597` `rgb(82,101,151)``\n• #4961a0\n``#4961a0` `rgb(73,97,160)``\n• #405da9\n``#405da9` `rgb(64,93,169)``\n• #3759b2\n``#3759b2` `rgb(55,89,178)``\n• #2e55bb\n``#2e55bb` `rgb(46,85,187)``\n• #2551c4\n``#2551c4` `rgb(37,81,196)``\n• #1c4dcd\n``#1c4dcd` `rgb(28,77,205)``\n• #1349d6\n``#1349d6` `rgb(19,73,214)``\n• #0a45df\n``#0a45df` `rgb(10,69,223)``\n• #0141e8\n``#0141e8` `rgb(1,65,232)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #5b698e is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50437844,"math_prob":0.6499177,"size":3719,"snap":"2021-43-2021-49","text_gpt3_token_len":1683,"char_repetition_ratio":0.12113055,"word_repetition_ratio":0.011070111,"special_character_ratio":0.5574079,"punctuation_ratio":0.23756906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9895715,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T20:54:00Z\",\"WARC-Record-ID\":\"<urn:uuid:71aa1e59-95cd-435e-b3eb-8d079ac2826a>\",\"Content-Length\":\"36207\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:267e4882-bcff-456b-8d27-9caff7543672>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a79c3ef-94d2-485a-a177-80c6efeb2d5f>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/5b698e\",\"WARC-Payload-Digest\":\"sha1:B3BLEYQZHOZCFSRISZRJGYUECLTQUMKI\",\"WARC-Block-Digest\":\"sha1:4OEXYKZGOQD6BDIJMRSKIUK73LKJVQKU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363312.79_warc_CC-MAIN-20211206194128-20211206224128-00023.warc.gz\"}"}
https://www.onlinemathlearning.com/area-perimeter-4md3.html	[ "", null, "# Area & Perimeter of Rectangles (Grade 4)\n\nRelated Topics:\nLesson Plans and Worksheets for Grade 4\nLesson Plans and Worksheets for all Grades\n\nVideos, examples, solutions, and lessons to help Grade 4 students learn to apply the area and perimeter formulas for rectangles in real world and mathematical problems.\n\nFor example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor\n.\n\nCommon Core: 4.MD.3\n\n### Suggested Learning Targets\n\n• I can explain the area and perimeter formula.\n• I can use the formulas to solve problems.\nFinding Perimeter & Unknown Measure (4.MD.3) Example:\n1. Mr. Myers is planning to make a flower garden outside our classroom window. He wants the garden to be 16 feet long and 3 feet wide. If he decides to fence in this flower garden, how much fencing would Mr. Myers need?\n2. An outdoor deck is 7 feet wide. The perimeter of the deck is 64 feet. What is the length of the deck?\n3. Fred has 30 yards of border for his rectangular bulletin board. The bulletin board is 6 yards long. How wide is Fred's bulletin board?\nArea Formula for Square or Rectangle\n\nPerimeter and Area of Squares and Rectangles\n• Definitions of perimeter and area.\n• How to find perimeter and area of squares and rectangles.\n• Area formulas for squares and rectangles.\n• Several examples narrated and explained.\n• Each polygon or figure shown with square units. Missing Lengths Given Area or Perimeter - 4.MD.3\nIn this video, students will learn to solve real world problems using the formulas for area and perimeter. This video relates to Common Core Standard 4.MD.3. Students learn to solve for a missing side given the perimeter or area.\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "", null, "" ]	[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85775566,"math_prob":0.92255163,"size":2083,"snap":"2020-45-2020-50","text_gpt3_token_len":458,"char_repetition_ratio":0.12169312,"word_repetition_ratio":0.0056497175,"special_character_ratio":0.21363418,"punctuation_ratio":0.12224939,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9898071,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T23:34:15Z\",\"WARC-Record-ID\":\"<urn:uuid:34eb4d32-c3e0-4195-9215-32916ab519ca>\",\"Content-Length\":\"44516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:227e07a3-881a-482a-ae0b-a1413d044231>\",\"WARC-Concurrent-To\":\"<urn:uuid:e84e8650-7b3b-49e9-a898-cfff09e0d2b9>\",\"WARC-IP-Address\":\"173.247.219.45\",\"WARC-Target-URI\":\"https://www.onlinemathlearning.com/area-perimeter-4md3.html\",\"WARC-Payload-Digest\":\"sha1:UXIHUHOAO6WNPXPNOIQQCAYX23PMCMQJ\",\"WARC-Block-Digest\":\"sha1:NDKJAC6ZKYAA35DHBBDV7Z4Z4BPDQPLT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195929.39_warc_CC-MAIN-20201128214643-20201129004643-00461.warc.gz\"}"}
https://demo.formulasearchengine.com/wiki/Cohomology	[ "# Cohomology\n\nTemplate:More footnotes In mathematics, specifically in homology theory and algebraic topology, cohomology is a general term for a sequence of abelian groups defined from a co-chain complex. That is, cohomology is defined as the abstract study of cochains, cocycles, and coboundaries. Cohomology can be viewed as a method of assigning algebraic invariants to a topological space that has a more refined algebraic structure than does homology. Cohomology arises from the algebraic dualization of the construction of homology. In less abstract language, cochains in the fundamental sense should assign 'quantities' to the chains of homology theory.\n\nFrom its beginning in topology, this idea became a dominant method in the mathematics of the second half of the twentieth century; from the initial idea of homology as a topologically invariant relation on chains, the range of applications of homology and cohomology theories has spread out over geometry and abstract algebra. The terminology tends to mask the fact that in many applications cohomology, a contravariant theory, is more natural than homology. At a basic level this has to do with functions and pullbacks in geometric situations: given spaces X and Y, and some kind of function F on Y, for any mapping f : XY composition with f gives rise to a function F o f on X. Cohomology groups often also have a natural product, the cup product, which gives them a ring structure. Because of this feature, cohomology is a stronger invariant than homology, as it can differentiate between certain algebraic objects that homology cannot.\n\n## Definition\n\nIn algebraic topology, the cohomology groups for spaces can be defined as follows (see Hatcher). Given a topological space X, consider the chain complex\n\n$\\cdots \\rightarrow C_{n}{\\stackrel {\\partial _{n}}{\\rightarrow }}\\ C_{n-1}\\rightarrow \\cdots$", null, "as in the definition of singular homology (or simplicial homology). Here, the Cn are the free abelian groups generated by formal linear combinations of the singular n-simplices in X and ∂n is the nth boundary operator.\n\nNow replace each Cn by its dual space Cn−1 = Hom(Cn, G), and ∂n by its transpose\n\n$\\delta ^{n}:C_{n-1}^{}\\rightarrow C_{n}^{}$", null, "to obtain the cochain complex\n\n$\\cdots \\leftarrow C_{n}^{}{\\stackrel {\\delta ^{n}}{\\leftarrow }}\\ C_{n-1}^{}\\leftarrow \\cdots$", null, "Then the nth cohomology group with coefficients in G is defined to be Ker(δn+1)/Im(δn) and denoted by Hn(C; G). The elements of Cn are called singular n-cochains with coefficients in G , and the δn are referred to as the coboundary operators. Elements of Ker(δn+1), Im(δn) are called cocycles and coboundaries, respectively.\n\nNote that the above definition can be adapted for general chain complexes, and not just the complexes used in singular homology. The study of general cohomology groups was a major motivation for the development of homological algebra, and has since found applications in a wide variety of settings (see below).\n\nGiven an element φ of Cn-1, it follows from the properties of the transpose that $\\delta ^{n}(\\varphi )=\\varphi \\circ \\partial _{n}$", null, "as elements of Cn. We can use this fact to relate the cohomology and homology groups as follows. Every element φ of Ker(δn) has a kernel containing the image of ∂n. So we can restrict φ to Ker(∂n−1) and take the quotient by the image of ∂n to obtain an element h(φ) in Hom(Hn, G). If φ is also contained in the image of δn−1, then h(φ) is zero. So we can take the quotient by Ker(δn), and to obtain a homomorphism\n\n$h:H^{n}(C;G)\\rightarrow {\\text{Hom}}(H_{n}(C),G).$", null, "It can be shown that this map h is surjective, and that we have a short split exact sequence\n\n$0\\rightarrow \\ker h\\rightarrow H^{n}(C;G){\\stackrel {h}{\\rightarrow }}{\\text{Hom}}(H_{n}(C),G)\\rightarrow 0.$", null, "## History\n\nAlthough cohomology is fundamental to modern algebraic topology, its importance was not seen for some 40 years after the development of homology. The concept of dual cell structure, which Henri Poincaré used in his proof of his Poincaré duality theorem, contained the germ of the idea of cohomology, but this was not seen until later.\n\nThere were various precursors to cohomology. In the mid-1920s, J. W. Alexander and Solomon Lefschetz founded the intersection theory of cycles on manifolds. On an n-dimensional manifold M, a p-cycle and a q-cycle with nonempty intersection will, if in general position, have intersection a (p + q − n)-cycle. This enables us to define a multiplication of homology classes\n\nHp(M) × Hq(M) → Hp+qn(M).\n\nAlexander had by 1930 defined a first cochain notion, based on a p-cochain on a space X having relevance to the small neighborhoods of the diagonal in Xp+1.\n\nIn 1931, Georges de Rham related homology and exterior differential forms, proving De Rham's theorem. This result is now understood to be more naturally interpreted in terms of cohomology.\n\nIn 1934, Lev Pontryagin proved the Pontryagin duality theorem; a result on topological groups. This (in rather special cases) provided an interpretation of Poincaré duality and Alexander duality in terms of group characters.\n\nAt a 1935 conference in Moscow, Andrey Kolmogorov and Alexander both introduced cohomology and tried to construct a cohomology product structure.\n\nIn 1936 Norman Steenrod published a paper constructing Čech cohomology by dualizing Čech homology.\n\nFrom 1936 to 1938, Hassler Whitney and Eduard Čech developed the cup product (making cohomology into a graded ring) and cap product, and realized that Poincaré duality can be stated in terms of the cap product. Their theory was still limited to finite cell complexes.\n\nIn 1944, Samuel Eilenberg overcame the technical limitations, and gave the modern definition of singular homology and cohomology.\n\nIn 1945, Eilenberg and Steenrod stated the axioms defining a homology or cohomology theory. In their 1952 book, Foundations of Algebraic Topology, they proved that the existing homology and cohomology theories did indeed satisfy their axioms.\n\nIn 1948 Edwin Spanier, building on work of Alexander and Kolmogorov, developed Alexander–Spanier cohomology.\n\n## Cohomology theories\n\n### Eilenberg–Steenrod theories\n\nA cohomology theory is a family of contravariant functors from the category of pairs of topological spaces and continuous functions (or some subcategory thereof such as the category of CW complexes) to the category of Abelian groups and group homomorphisms that satisfies the Eilenberg–Steenrod axioms.\n\nSome cohomology theories in this sense are:\n\n{{safesubst:#invoke:anchor\|main}}\n\n### Axioms and generalized cohomology theories\n\n{{#invoke:see also\|seealso}} There are various ways to define cohomology groups (for example singular cohomology, Čech cohomology, Alexander–Spanier cohomology or Sheaf cohomology). These give different answers for some exotic spaces, but there is a large class of spaces on which they all agree. This is most easily understood axiomatically: there is a list of properties known as the Eilenberg–Steenrod axioms, and any two constructions that share those properties will agree at least on all finite CW complexes, for example.\n\nOne of the axioms is the so-called dimension axiom: if P is a single point, then Hn(P) = 0 for all n ≠ 0, and H0(P) = Z. We can generalise slightly by allowing an arbitrary abelian group A in dimension zero, but still insisting that the groups in nonzero dimension are trivial. It turns out that there is again an essentially unique system of groups satisfying these axioms, which are denoted by $H_{}(X;A)$", null, ". In the common case where each group Hk(X) is isomorphic to Zrk for some rk in N, we just have $H_{k}(X;A)=A^{r_{k}}$", null, ". In general, the relationship between Hk(X) and $H_{k}(X;A)$", null, "is only a little more complicated, and is again controlled by the Universal coefficient theorem.\n\nMore significantly, we can drop the dimension axiom altogether. There are a number of different ways to define groups satisfying all the other axioms, including the following:\n\nThese are called generalised homology theories; they carry much richer information than ordinary homology, but are often harder to compute. Their study is tightly linked (via the Brown representability theorem) to stable homotopy.\n\nA cohomology theory E is said to be multiplicative if $E^{}(X)$", null, "is a graded ring.\n\n### Other cohomology theories\n\nTheories in a broader sense of cohomology include:" ]	[ null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null, "https://demo.formulasearchengine.com/index.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89422756,"math_prob":0.9269235,"size":10126,"snap":"2020-45-2020-50","text_gpt3_token_len":2565,"char_repetition_ratio":0.1641968,"word_repetition_ratio":0.0012861736,"special_character_ratio":0.21666996,"punctuation_ratio":0.1169045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99720395,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T16:20:47Z\",\"WARC-Record-ID\":\"<urn:uuid:a9cd2249-27aa-4bf3-bd43-3829db0501e7>\",\"Content-Length\":\"120099\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7dad296-37e1-4cf3-b832-21e64080712e>\",\"WARC-Concurrent-To\":\"<urn:uuid:4bed199a-0f93-43c1-831f-f98606133a63>\",\"WARC-IP-Address\":\"132.195.228.228\",\"WARC-Target-URI\":\"https://demo.formulasearchengine.com/wiki/Cohomology\",\"WARC-Payload-Digest\":\"sha1:YCN6FTQA2TLN7LPX5LENZRCFMRAETBMQ\",\"WARC-Block-Digest\":\"sha1:VI4U5TXJGX3QYDATL542KMIWT3KW67VB\",\"WARC-Truncated\":\"disconnect\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107904834.82_warc_CC-MAIN-20201029154446-20201029184446-00082.warc.gz\"}"}
http://codeforces.com/problemset/problem/1144/E	[ "E. Median String\ntime limit per test\n2 seconds\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nYou are given two strings $s$ and $t$, both consisting of exactly $k$ lowercase Latin letters, $s$ is lexicographically less than $t$.\n\nLet's consider list of all strings consisting of exactly $k$ lowercase Latin letters, lexicographically not less than $s$ and not greater than $t$ (including $s$ and $t$) in lexicographical order. For example, for $k=2$, $s=$\"az\" and $t=$\"bf\" the list will be [\"az\", \"ba\", \"bb\", \"bc\", \"bd\", \"be\", \"bf\"].\n\nYour task is to print the median (the middle element) of this list. For the example above this will be \"bc\".\n\nIt is guaranteed that there is an odd number of strings lexicographically not less than $s$ and not greater than $t$.\n\nInput\n\nThe first line of the input contains one integer $k$ ($1 \\le k \\le 2 \\cdot 10^5$) — the length of strings.\n\nThe second line of the input contains one string $s$ consisting of exactly $k$ lowercase Latin letters.\n\nThe third line of the input contains one string $t$ consisting of exactly $k$ lowercase Latin letters.\n\nIt is guaranteed that $s$ is lexicographically less than $t$.\n\nIt is guaranteed that there is an odd number of strings lexicographically not less than $s$ and not greater than $t$.\n\nOutput\n\nPrint one string consisting exactly of $k$ lowercase Latin letters — the median (the middle element) of list of strings of length $k$ lexicographically not less than $s$ and not greater than $t$.\n\nExamples\nInput\n2\naz\nbf\n\nOutput\nbc\n\nInput\n5\nafogk\nasdji\n\nOutput\nalvuw\n\nInput\n6\nnijfvj\ntvqhwp\n\nOutput\nqoztvz" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86700237,"math_prob":0.9977703,"size":1573,"snap":"2020-24-2020-29","text_gpt3_token_len":410,"char_repetition_ratio":0.14850223,"word_repetition_ratio":0.29554656,"special_character_ratio":0.32231405,"punctuation_ratio":0.0866426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993449,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-28T15:44:04Z\",\"WARC-Record-ID\":\"<urn:uuid:485bea23-3fad-4027-95f3-1d3d53ff43dd>\",\"Content-Length\":\"52139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d471701-69f4-4842-ac75-314ad7475bc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:655f6353-1fcd-4f78-98c7-b497b04fbc56>\",\"WARC-IP-Address\":\"81.27.240.126\",\"WARC-Target-URI\":\"http://codeforces.com/problemset/problem/1144/E\",\"WARC-Payload-Digest\":\"sha1:NER273JHJMHERMMJG5P6ZM2MYAKVDX6I\",\"WARC-Block-Digest\":\"sha1:DLDRBLCQ5BQTRUSXXALUTTDMMWOXUIDE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347399820.9_warc_CC-MAIN-20200528135528-20200528165528-00564.warc.gz\"}"}
https://www.storyofmathematics.com/factors/factors-of-388/	[ "", null, "# Factors of 388: Prime Factorization, Methods, and Examples\n\nThe factors of 388 are 1, 2, 4, 97, 194, and 388. These numbers produce a zero remainder when divided by 388. 388 is an even composite number having 6 factors. It has both positive and negative integers as factors.", null, "Further information is provided in the article.\n\n### Factors of 388\n\nHere are the factors of number 388.\n\nFactors of 388: 1, 2, 4, 97, 194, 388\n\n### Negative Factors of 388\n\nThe negative factors of 388 are similar to its positive aspects, just with a negative sign.\n\nNegative Factors of 388: -1, -2, -4, -97, -194, -388\n\n### Prime Factorization of 388\n\nThe prime factorization of 388 is the way of expressing its prime factors in the product form.\n\nPrime Factorization: 22 × 97\n\nIn this article, we will learn about the factors of 388 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.\n\n## What Are the Factors of 388?\n\nThe factors of 388 are 1, 2, 4, 97, 194, 388. These numbers are the factors as they do not leave any remainder when divided by 388.\n\nThe factors of 388 are classified as prime numbers and composite numbers. The prime factors of the number 388 can be determined using the prime factorization technique.\n\n## How To Find the Factors of 388?\n\nYou can find the factors of 388 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.\n\nTo find the factors of 388, create a list containing the numbers that are exactly divisible by 388 with zero remainders. One important thing to note is that 1 and 388 are the 388’s factors as every natural number has 1 and the number itself as its factor.\n\n1 is also called the universal factor of every number. The factors of 388 are determined as follows:\n\n$\\dfrac{388}{1} = 388$\n\n$\\dfrac{388}{2} = 194$\n\n$\\dfrac{388}{4} = 97$\n\n$\\dfrac{388}{388} = 1$\n\nTherefore, 1, 2, 4, 97, 194, 388 are the factors of 388.\n\n### Total Number of Factors of 388\n\nFor 388, there are 6 positive factors and 6 negative ones. So in total, there are 12 factors of 388.\n\nTo find the total number of factors of the given number, follow the procedure mentioned below:\n\n1. Find the factorization/prime factorization of the given number.\n2. Demonstrate the prime factorization of the number in the form of exponent form.\n3. Add 1 to each of the exponents of the prime factor.\n4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.\n\nBy following this procedure, the total number of factors of X is given as:\n\nFactorization of 388 is 1, 2, 4, 97, 194, 388.\n\nThe exponent of 1, 2, 4, 97, 194, 388 is 1.\n\nAdding 1 to each and multiplying them together results in m.\n\nTherefore, the total number of factors of 388 is 12. 6 is positive, and 6 factors are negative.\n\n### Important Notes\n\nHere are some essential points that must be considered while finding the factors of any given number:\n\n• The factor of any given number must be a whole number.\n• The factors of the number cannot be in the form of decimals or fractions.\n• Factors can be positive as well as negative.\n• Negative factors are the additive inverse of the positive factors of a given number.\n• The factor of a number cannot be greater than that number.\n• Every even number has 2 as its prime factor, the smallest prime factor.\n\n## Factors of 388 by Prime Factorization\n\nThe number 388 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.", null, "Before finding the factors of 388 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.\n\nTo start the prime factorization of 388, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.\n\nContinue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 388 can be expressed as:\n\n388 = 22 × 97\n\n## Factors of 388 in Pairs\n\nThe factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.", null, "For 388, the factor pairs can be found as:\n\n1 x 388 = 388\n\n2 x 194 = 388\n\n4 x 97 = 388\n\nThe possible factor pairs of 388 are given as (1, 388), (2, 194), and (4, 97).\n\nAll these numbers in pairs, when multiplied, give 388 as the product.\n\nThe negative factor pairs of 388 are given as:\n\n-1 x -388 = 388\n\n-2 x -194 = 388\n\n-4 x -97 = 388\n\nIt is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -4, -97, -194, -388 are called negative factors of 388.\n\nThe list of all the factors of 388, including positive as well as negative numbers, is given below.\n\nFactor list of 388: 1, -1, 2, -2, 4, -4, 97, -97, 194, -194, 388, and -388\n\n## Factors of 388 Solved Examples\n\nTo better understand the concept of factors, let’s solve some examples.\n\n### Example 1\n\nHow many factors of 388 are there?\n\n### Solution\n\nThe total number of Factors of 388 is 12.\n\nFactors of 388 are 1, 2, 4, 97, 194, and 388.\n\n### Example 2\n\nFind the factors of 388 using prime factorization.\n\n### Solution\n\nThe prime factorization of 388 is given as:\n\n388 $\\div$ 2 = 194\n\n194 $\\div$ 2 = 97\n\n97 $\\div$ 97 = 1\n\nSo the prime factorization of 388 can be written as:\n\n22 × 97 = 388" ]	[ null, "https://www.storyofmathematics.com/wp-content/uploads/2022/02/som-header1.jpg", null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%201200%20630%22%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%201800%20900%22%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%201800%20900%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90164655,"math_prob":0.99565583,"size":5688,"snap":"2023-40-2023-50","text_gpt3_token_len":1494,"char_repetition_ratio":0.2489444,"word_repetition_ratio":0.06621881,"special_character_ratio":0.3043249,"punctuation_ratio":0.13939899,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998771,"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\":\"2023-10-03T08:35:35Z\",\"WARC-Record-ID\":\"<urn:uuid:031d6e6a-929a-4e40-bbc2-742dcef5132a>\",\"Content-Length\":\"192278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e0b65f2-5054-4ec8-9912-ea8dfc22d5b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c9f7602-8316-4e6a-8522-09d4ca2ce54a>\",\"WARC-IP-Address\":\"172.67.190.47\",\"WARC-Target-URI\":\"https://www.storyofmathematics.com/factors/factors-of-388/\",\"WARC-Payload-Digest\":\"sha1:IV47QWNO4NIQIGTLNABWSXKUAE7SRPYZ\",\"WARC-Block-Digest\":\"sha1:34CJ7EA7CRLAVHUYVHCPGRMAFHU3CZQM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511055.59_warc_CC-MAIN-20231003060619-20231003090619-00875.warc.gz\"}"}
https://math.stackexchange.com/questions/4316470/what-is-the-difference-between-formula-and-function	[ "# What is the difference between 'formula' and 'function'?\n\nI hope this question fits in this site. I was asked by my student when I was explaining the concept of relation and function and after thinking, I still can't differentiate those two concepts (formula and function). I've looked for the answers from other sites and I have read these:\n\nThe first link above tells us\n\n'The difference between a formula and function is that a formula is defined as the statement used for the calculation [a]. These formulas could be simple or complex and always start with equal to operator [b]. While function is defined as the code that is designed for the calculations and is used inside the formula [c].'\n\nWe know a function can be used to calculate to find the result of the calculation from the range with the given number taken from the domain [a]. For instance, the exponential function\n\n$$P(t) = P_0e^{kt}$$\n\ncan be used to calculate the population in the region X.\n\nA function also has the equal sign, because if it doesn't, I'd say it's an expression [b] & [c]. And I don't know about the relevancy of the second link since the context is used in Excel. But it says that\n\nFormula(s) is/are nothing but doing some arithmetic operation(s) or calculation(s).\n\nDoes that mean these are formulas?:\n\n$$2+2=4\\quad 5-3=2\\quad 2(3+5)/4=4$$\n\nHow to tell to my students about their difference? Are the exactly the same concept? I have a class tomorrow and I've promised to answer their question that day.\n\n• It is common to use a formula/expression to define a function, but not all functions have a formula that describe them. (In fact, by just about any halfway principled way of comparing, most functions don't). Nov 26, 2021 at 2:41\n• Almost anything is a formula. You pick a language and its words are called formulas. So, yes $2+2=5$, $\\rightarrow\\text{car}$, $\\forall x>0, \\exists y$ can be formulas in some language. The concept of function is more concrete, but even it has two inequivalent definitions that are often used interchangeably in mathematics. I think they mention both here.\n– plop\nNov 26, 2021 at 2:43\n• See Relation for the definition of binary relation, and function is a special type of binary relation as discussed. I believe 'formula' is not something fundamental, but we commonly use it in English to refer symbolic expressions. Nov 26, 2021 at 2:55\n• How could I forget to give $H_2O$ as an example of formula?\n– plop\nNov 26, 2021 at 3:00\n• The Math Educators StackExchange may be a good place for this question.\n– Blue\nNov 26, 2021 at 3:28\n\nIn my estimation, you are asking two different questions here. The one about what might tell your students is easier to answer, but there are lots of different answers that are reasonable. Here is what I would say to, e.g., a college student in any course before linear algebra:\n\nA function is a machine that takes in numbers and spits out other numbers. A formula is a mathematical expression that I can use to calculate. Functions are more general objects than formulas, because not every function can be written as a formula. My favorite example is the floor function, which works like this: it takes in a number, cuts off the decimal, and spits out the whole number that's left over. [examples]. This function makes perfect sense, even though I can't write down a \"formula\" for it like x^2 or $$\\sin(x)$$.\n\nNone of the first three sentences are true (the fourth one is!), even to the extent that they are meaningful. But the lies are mild, and they help simplify the language to get the point across.\n\nThe second question is what the difference actually is. I will not come to any forceful conclusion on this question, but instead frame my thoughts as a series of remarks.\n\nRemark 1. This question has been discussed a lot on MSE, because, as you have observed, it is subtle, and it concerns relatively elementary objects. Exactly which question it is a duplicate of is not totally clear, though.\n\nRemark 2. I can't open the first link, but based on the language used, it kind of sounds like this source is using the programming definition of a 'function', which differs in a subtle way from the mathematical definition. In particular, we usually do not agree with [c]: for us, the function is the encapsulating object, and there are various methods by which a function's values may be given, one of which is a formula (see this Wikipedia quote)\n\nRemark 3. Continuing on this, you write\n\nWe know a function can be used to calculate to find the result of the calculation from the range with the given number taken from the domain [a]...\n\nI know no such thing. A function may encode a calculation (or it may not!) but it is odd to me to say that we use a function to calculate elements in its range. For instance, in your example I would not say I am using the function $$P$$ to calculate $$P(3)$$; I guess I can see why you would say this, but it doesn't reflect how the calculation actually proceeds. When I am actually getting around to doing the computation, what I am using is the expression $$P_0e^{kt}$$ by substituting $$t=3$$. The result of this calculation is certainly in the range of $$P$$; this is guaranteed by the equation $$P(t)=P_0e^{kt}$$... indeed this equation is what convinced me to use the expression to compute $$P(3)$$.\n\nRemark 4. The word \"formula\" did not appear in the previous remark, what's up with that? Well, in casual speech, I would describe both $$P(t)=P_0e^{kt}$$ and $$P_0e^{kt}$$ as formulas for $$P(t)$$. If I am required to be more careful, I would not use the word at all.\n\nRemark 5. Logicians have strong opinions about [read: a definition of] what a \"formula\" is, that does not align well with the common-language meaning, despite being related. (The tension is well-illustrated by this answer.) To answer your question, the three equalities in your question undoubtedly are formulas if we are using the logician's definition. Still, I would not call them formulas because for me a formula needs to be something that I can compute with; those equalities look to me like a \"record\" of a computation that has already been completed.\n\nAs this answer has now been accepted, let me highlight Mauro Allegranza's very useful comment [emphasis mine]:\n\nThe simplest explanation is in the answer to the linked post: \"A formula is a string of symbols, arranged according to mathematical grammar [i.e. an expression of the mathematical language]. A function is a mathematical object.\" The clear distinction is that between the world of (mathematical) objects and the language used to speak of them (exactly like natural language). –\n\n• You forgot to include the link for the examples. Nov 26, 2021 at 6:21\n• No I didn't; sorry for the confusion. I mean something like doing out $\\lfloor 3.482\\rfloor$ on paper, but I didn't feel like boring this crowd with the roleplay. Nov 26, 2021 at 8:38\n\nThe difference between a formula and a function is like the difference between the word \"Earth\" and the planet Earth. A formula is something formed by combining mathematical symbols in a syntactically valid way. A function is a relation mapping each element from the domain to an element in its range. A formula can define a function.\n\nFor example, $$1 + x$$ is a formula because it combines the symbols '1', '+' and '$$x$$' in a valid way; and this formula defines a function which relates each number to the number which is 1 larger than it. On the other hand, $$x + 1$$ is a different formula, because the symbols are combined in a different way; however, this different formula defines the same function because it relates numbers to each other in the same way.\n\nPerhaps a clearer example is the prime-counting function, which is certainly a function, but the phrase \"the prime-counting function\" is not a formula, because it combines English words, not mathematical symbols. In fact, almost all functions are not defined by any formula, because there are uncountably many functions $$: \\mathbb{N} \\rightarrow \\mathbb{N}$$ (just taking $$\\mathbb{N}$$ as an example for the domain and range) but only countably many formulae (each of which is a finite sequence of symbols chosen from a finite alphabet).\n\nA function $$f$$ is a triple, consisting of a domain $$D_f$$ (a set), a set of values $$V_f$$, and a relation $$R_f$$, i.e. a subset of $$D_f \\times V_f$$, which has the additional condition that for any element $$d \\in D_f$$, there is only one element in $$R_f$$ which has this particular element $$d$$ as first component.\n\nThat's the mathematical definition of a function. The intention is:\n\n$$D_f$$ defines all possible inputs; $$V_f$$ is a set containing all possible outputs, and $$R_f$$ contains all pairs of $$(d,v)$$ where $$d$$ is an input and $$f(d) = v$$, i.e. $$v$$ is the output belonging to the input $$d$$. The condition described above means that for any input there is only one output.\n\nThere is no mention at all of any formula, because there is no need for this. As Eric Nathan Stucky has mentioned, there are a lot of functions that cannot be described with a formula.\n\nIf you have a formula, you are in the lucky situation that you have a way of calculating your function values. Although I've never before heard the word \"formula\" in this context; to me the word would be \"function term\".\n\n• What happened to Binet's formula? en.wikipedia.org/wiki/Fibonacci_number Nov 26, 2021 at 17:31\n• @lalala Oops. Learned something, edited my post. Thanks! Nov 30, 2021 at 7:55\n\nBest illustrated by examples.\n\n$$y(x)=ax^2+bx+c$$\n\nis a particuar function among many. Satisfied only by two x values when $$y(0)=0$$\n\n$$\\sin (a+b)=\\sin a\\cos b+\\cos a \\sin b$$\n\nis a formula/ identity satisfied by any values of $$(a,b).$$\n\nThe difference comes from arbitrariness of chosen variables.\n\n• Not sure what you mean by saying that a function is \"satisfied by only two x values\". Also, a logician might say that the formula $\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b$ defines a function $f : \\mathbb{R}^2 \\rightarrow \\{\\top, \\bot\\}$ which is equivalent to $f(a, b) = \\top$. Nov 26, 2021 at 11:19\n• No \"$y(x)=ax^2+bx+c$\" is not a function. It is an equation, that in a context where it is understood that $y$ designates a function $\\Bbb R\\to\\Bbb R$, that $a,b,c$ are fixed real numbers, and that the equation is asserted to hold for all $x\\in\\Bbb R$, can be used to specify the function designated by $y$. And in no way it is only satisfied by two values of $x$. Nov 27, 2021 at 4:28\n• It was just given in simpler terms to contrast with a formula. Also the quadratic y(x)=0 has two roots. Nov 27, 2021 at 15:39\n\n## What to tell your students\n\nIn math class, a function is a restricted kind of relation between two sets. For the sake of simplicity we'll call these sets \"domain\" and \"range\". A relation is a function when, for any given element of the domain, there is at most one related element of the range. (There are other names for these sets in use. These are the names I learned in school.) Also, most functions do not have names, but some few do. An example is cosine.\n\nA formula is a (valid) sequence of symbols in some formal language. Most of the formulae we work with in this class are of the algebraic type, so they involve numbers, variables, operators, and the occasional function. Ordinarily, you'll interpret or manipulate a formula in order to draw some conclusion.\n\nOften a function may be defined by a formula. For example, the function square(x) may be defined as x*x.\n\nIn other contexts, these terms may be used slightly differently. Ask a chemist about functional groups, for example." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94804204,"math_prob":0.99264205,"size":4031,"snap":"2023-40-2023-50","text_gpt3_token_len":904,"char_repetition_ratio":0.11547057,"word_repetition_ratio":0.0,"special_character_ratio":0.22500621,"punctuation_ratio":0.11604938,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997446,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T15:26:01Z\",\"WARC-Record-ID\":\"<urn:uuid:d5c1c84f-6f65-4924-a551-f2497209eda4>\",\"Content-Length\":\"200914\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dad727a5-040d-43c9-be74-a3f9b4e0b2bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:192e5150-b5b7-448b-8e70-2b6c083330a5>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/4316470/what-is-the-difference-between-formula-and-function\",\"WARC-Payload-Digest\":\"sha1:CKWZ3BGEDPRAVFGET4DJJLLZQHCN37PE\",\"WARC-Block-Digest\":\"sha1:MM76TE35ABMDXSFZ6SYPJPF4E2J7XVMR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510300.41_warc_CC-MAIN-20230927135227-20230927165227-00405.warc.gz\"}"}
https://www.geeksforgeeks.org/gate-gate-cs-2006-question-4/	[ "Related Articles\n\n# GATE \| GATE-CS-2006 \| Question 4\n\n• Last Updated : 26 Feb, 2021\n\nA relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v. Then R is:\nThen R is:\n(A) Neither a Partial Order nor an Equivalence Relation\n(B) A Partial Order but not a Total Order\n(C) A Total Order\n(D) An Equivalence Relation\n\nExplanation: An equivalence relation on a set x is a subset of xx, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y” to mean (x,y) is an element of R, and we say “x is related to y,” then the properties are:\n1. Reflexive: a R a for all a Є R,\n2. Symmetric: a R b implies that b R a for all a,b Є R\n3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.\n\nAn partial order relation on a set x is a subset of xx, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y” to mean (x,y) is an element of R, and we say “x is related to y,” then the properties are:\n1. Reflexive: a R a for all a Є R,\n2. Anti-Symmetric: a R b and b R a implies that for all a,b Є R\n3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.\n\nAn total order relation a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y” to mean (x,y) is an element of R, and we say “x is related to y,” then the properties are:\n1. Reflexive: a R a for all a Є R,\n2. Anti-Symmetric: a R b implies that b R a for all a,b Є R\n3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.\n4. Comparability : either a R b or b R a for all a,b Є R.\n\nAs given in question, a relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v , reflexive property is not satisfied here, because there is > or < relationship between (x ,y) pair set and (u,v) pair set . Other way , if there would have been x <= u and y>= v (or x=u and y=v) kind of relation among elements of sets then reflexive property could have been satisfied. Since reflexive property in not satisfied here , so given realtion can not be equivalence, partial orderor total order relation.\n\nSo, option (A) is correct.\n\nThis solution is contributed by Nirmal Bharadwaj.\n\nQuiz of this Question\n\nAttention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.\n\nLearn all GATE CS concepts with Free Live Classes on our youtube channel.\n\nMy Personal Notes arrow_drop_up" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8409971,"math_prob":0.9849987,"size":2468,"snap":"2021-31-2021-39","text_gpt3_token_len":703,"char_repetition_ratio":0.12540585,"word_repetition_ratio":0.5516569,"special_character_ratio":0.28119937,"punctuation_ratio":0.15165877,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995703,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T09:01:05Z\",\"WARC-Record-ID\":\"<urn:uuid:3d5e3e99-334c-48af-b119-2b29e6e1eaca>\",\"Content-Length\":\"97610\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04bc4cdb-a994-4266-885d-5281f453215b>\",\"WARC-Concurrent-To\":\"<urn:uuid:02af864e-5b2d-48f1-b523-33c71cc42956>\",\"WARC-IP-Address\":\"23.46.239.11\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/gate-gate-cs-2006-question-4/\",\"WARC-Payload-Digest\":\"sha1:BGA3EY6QCOZSZ6HE2KD4MQS26Y76QWVE\",\"WARC-Block-Digest\":\"sha1:LRSXHHQJOZTYKGXF3NLOCTKRDBNA6RCG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150134.86_warc_CC-MAIN-20210724063259-20210724093259-00715.warc.gz\"}"}
https://www.studysmarter.us/textbooks/math/precalculus-enhanced-with-graphing-utilities-6th/sequences-induction-the-binomial-theorem/q-76-in-problems-71-82-find-the-sum-of-each-sequence/	[ "", null, "Suggested languages for you:\n\nEurope\n\nAnswers without the blur. Sign up and see all textbooks for free!", null, "Q 76.\n\nExpert-verified", null, "Found in: Page 809", null, "### Precalculus Enhanced with Graphing Utilities\n\nBook edition 6th\nAuthor(s) Sullivan\nPages 1200 pages\nISBN 9780321795465", null, "# In Problems 71-82, find the sum of each sequence.$\\underset{k=1}{\\overset{26}{\\sum \\left(3}}k-7\\right)$\n\nThe sum of this sequence is 871.\n\nSee the step by step solution\n\n## Step 1. Write the given information.\n\nThe sum of sequence:\n\n$\\underset{k=1}{\\overset{26}{\\sum \\left(3k-7\\right)}}$\n\n## Step 2. Use the properties of sequence.\n\nUsing the properties of sequence:\n\n$\\underset{k=1}{\\overset{n}{\\sum \\left({a}_{k}-{b}_{k}\\right)}}=\\underset{k=1}{\\overset{n}{\\sum \\left({a}_{k}\\right)}}-\\underset{k=1}{\\overset{n}{\\sum \\left({b}_{k}\\right)}}\\phantom{\\rule{0ex}{0ex}}&\\phantom{\\rule{0ex}{0ex}}\\underset{k=1}{\\overset{n}{\\sum \\left(c{a}_{k}\\right)}}=c\\underset{k=1}{\\overset{n}{\\sum {a}_{k}}}$\n\nSo,\n\n$\\underset{k=1}{\\overset{26}{\\sum \\left(3k-7\\right)}}=\\underset{k=1}{\\overset{26}{\\sum \\left(3k\\right)}}-\\underset{k=1}{\\overset{26}{\\sum \\left(7\\right)}}$\n\n&\n\n$\\underset{k=1}{\\overset{26}{\\sum \\left(3k\\right)}}=\\underset{k=1}{\\overset{26}{3\\sum \\left(k\\right)}}$\n\n## Step 3. Use the formula for sum of sequences of n real numbers.\n\nUsing formula for summation is:\n\n$\\underset{k=1}{\\overset{n}{\\sum k}}=1+2+...+n\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{n}{\\sum k}}=\\frac{n\\left(n+1\\right)}{2}$\n\nSo,\n\nrole=\"math\" localid=\"1646750162806\" $3\\underset{k=1}{\\overset{26}{\\sum k}}=3×\\frac{26\\left(26+1\\right)}{2}\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{3\\sum k}}=3×\\frac{26\\left(27\\right)}{2}\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{3\\sum k}}=3×351\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{3\\sum k}}=1053$\n\n## Step 4. Use the formula for sum of sequence.\n\nUsing formula for summation is:$\\underset{k=1}{\\overset{n}{\\sum c}}=c+c+...+c\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{n}{\\sum c}}=cn,c-realnumber\\phantom{\\rule{0ex}{0ex}}So,\\phantom{\\rule{0ex}{0ex}}\\underset{k=1}{\\overset{26}{\\sum 7}}=7×26\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{\\sum 7}}=182\\phantom{\\rule{0ex}{0ex}}$\n\n## Step 5. Now, subtract Step 3 from Step 4.\n\nFrom Step 3,\n\n$\\underset{k=1}{\\overset{26}{3\\sum k}}=1053$\n\nFrom Step 4,\n\n$\\underset{k=1}{\\overset{26}{\\sum 7}}=182$\n\nSo,\n\n$\\underset{k=1}{\\overset{26}{\\sum \\left(3k-7\\right)}}=\\underset{k=1}{\\overset{26}{\\sum \\left(3k\\right)}}\\underset{k=1}{\\overset{26}{-\\sum \\left(3\\right)}}\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{\\sum \\left(3k-7\\right)}}=1053-182\\phantom{\\rule{0ex}{0ex}}⇒\\underset{k=1}{\\overset{26}{\\sum \\left(3k-7\\right)}}=871$", null, "### Want to see more solutions like these?", null, "" ]	[ null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/dist/assets/images/header-logo.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/src/assets/images/ab-test/searching-looking.svg", null, "https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-images/Sullivan_Precalculus.jpeg", null, "https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-images/Sullivan_Precalculus.jpeg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/src/assets/images/ab-test/businessman-superhero.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/img/textbook/banner-top.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/img/textbook/cta-icon.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6977346,"math_prob":0.9997594,"size":375,"snap":"2023-14-2023-23","text_gpt3_token_len":105,"char_repetition_ratio":0.14824797,"word_repetition_ratio":0.03508772,"special_character_ratio":0.29333332,"punctuation_ratio":0.175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999243,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T21:03:36Z\",\"WARC-Record-ID\":\"<urn:uuid:a680a25f-566e-4a6e-a97d-8e6e1755bbb4>\",\"Content-Length\":\"158370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1291b38-058e-45c1-a800-b9bb2f70de8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba744501-b4d8-46f8-b4b3-0708dc177260>\",\"WARC-IP-Address\":\"18.194.226.228\",\"WARC-Target-URI\":\"https://www.studysmarter.us/textbooks/math/precalculus-enhanced-with-graphing-utilities-6th/sequences-induction-the-binomial-theorem/q-76-in-problems-71-82-find-the-sum-of-each-sequence/\",\"WARC-Payload-Digest\":\"sha1:WUT55JOQ2OVQ7YHBEEZ2HJSAXKEKXUAR\",\"WARC-Block-Digest\":\"sha1:UFCBYP3YH3IROMFPPWZ6VWMDUB4SFK3W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945372.38_warc_CC-MAIN-20230325191930-20230325221930-00750.warc.gz\"}"}
https://chem.libretexts.org/Courses/City_College_of_San_Francisco/Foundations_-_Review_Source_for_Chem_101A/06%3A_Appendices/7.01%3A_Essential_Mathematics	[ "Skip to main content\n\n# 7.1: Essential Mathematics\n\n## Exponential Arithmetic\n\nExponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:\n\n\\begin{align} 1000&=1×10^3\\\\ 100&=1×10^2\\\\ 10&=1×10^1\\\\ 1&=1×10^0\\\\ 0.1&=1×10^{−1}\\\\ 0.001&=1×10^{−3}\\\\ 2386&=2.386×1000=2.386×10^3\\\\ 0.123&=1.23×0.1=1.23×10^{−1} \\end{align}\n\nThe power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 × 109, and 0.00000000036 = 3.6 × 10−10.\n\n### Addition of Exponentials\n\nConvert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.\n\nExercise $$\\PageIndex{1}$$: Adding Exponentials\n\nAdd 5.00 × 10−5 and 3.00 × 10−3.\n\nSolution\n\n\\begin{align} 3.00×10^{−3}&=300×10^{−5}\\\\ (5.00×10^{−5})+(300×10^{−5})&=305×10^{−5}=3.05×10^{−3} \\end{align}\n\n### Subtraction of Exponentials\n\nConvert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.\n\nExercise $$\\PageIndex{2}$$: Subtracting Exponentials\n\nSubtract 4.0 × 10−7 from 5.0 × 10−6.\n\nSolution\n\n$4.0×10^{−7}=0.40×10^{−6}\\\\ (5.0×10^{−6})−(0.40×10^{−6})=4.6×10^{−6}$\n\n### Multiplication of Exponentials\n\nMultiply the digit terms in the usual way and add the exponents of the exponential terms.\n\nExercise $$\\PageIndex{3}$$: Multiplying Exponentials\n\nMultiply 4.2 × 10−8 by 2.0 × 103.\n\nSolution\n\n$(4.2×10^{−8})×(2.0×10^3)=(4.2×2.0)×10^{(−8)+(+3)}=8.4×10^{−5}$\n\n### Division of Exponentials\n\nDivide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.\n\nExercise $$\\PageIndex{4}$$: Dividing Exponentials\n\nDivide 3.6 × 105 by 6.0 × 10−4.\n\nSolution\n\n$\\dfrac{3.6×10^{−5}}{6.0×10^{−4}}=\\left(\\dfrac{3.6}{6.0}\\right)×10^{(−5)−(−4)}=0.60×10^{−1}=6.0×10^{−2}$\n\n### Squaring of Exponentials\n\nSquare the digit term in the usual way and multiply the exponent of the exponential term by 2.\n\nExercise $$\\PageIndex{5}$$: Squaring Exponentials\n\nSquare the number 4.0 × 10−6.\n\nSolution\n\n$(4.0×10^{−6})^2=4×4×10^{2×(−6)}=16×10^{−12}=1.6×10^{−11}$\n\n### Cubing of Exponentials\n\nCube the digit term in the usual way and multiply the exponent of the exponential term by 3.\n\nExercise $$\\PageIndex{6}$$: Cubing Exponentials\n\nCube the number 2 × 104.\n\nSolution\n\n$(2×10^4)^3=2×2×2×10^{3×4}=8×10^{12}$\n\n### Taking Square Roots of Exponentials\n\nIf necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.\n\nExercise $$\\PageIndex{7}$$: Finding the Square Root of Exponentials\n\nFind the square root of 1.6 × 10−7.\n\nSolution\n\n\\begin{align} 1.6×10^{−7}&=16×10^{−8}\\\\ \\sqrt{16×10^{−8}}=\\sqrt{16}×\\sqrt{10^{−8}}&=\\sqrt{16}×10^{−\\large{\\frac{8}{2}}}=4.0×10^{−4} \\end{align}\n\n## Significant Figures\n\nA beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.\n\nThe importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).\n\nExercise $$\\PageIndex{8}$$: Addition and Subtraction with Significant Figures\n\nAdd 4.383 g and 0.0023 g.\n\nSolution\n\n\\begin{align} &\\mathrm{4.38\\underline{3}\\:g}\\\\ &\\mathrm{\\underline{0.002\\underline{3}\\:g}}\\\\ &\\mathrm{4.38\\underline{5}\\:g} \\end{align}\n\nIn multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.\n\nExercise $$\\PageIndex{9}$$: Multiplication and Division with Significant Figures\n\nMultiply 0.6238 by 6.6.\n\nSolution\n\n$0.623\\underline{8}×6.\\underline{6}=4.\\underline{1}$\n\nWhen rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).\n\n## The Use of Logarithms and Exponential Numbers\n\nThe common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2, because 10 must be raised to the second power to equal 100. Additional examples follow.\n\nLogarithms and Exponential Numbers\nNumber Number Expressed Exponentially Common Logarithm\n1000 103 3\n10 101 1\n1 100 0\n0.1 10−1 −1\n0.001 10−3 −3\n\nWhat is the common logarithm of 60? Because 60 lies between 10 and 100, which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782; that is,\n\n$60=10^{1.7782}$\n\nThe common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is −1.4069, or\n\n$0.03918=10^{-1.4069}=\\dfrac{1}{10^{1.4069}}$\n\nTo obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate 10x (where x is the logarithm of the number).\n\nThe natural logarithm of a number (ln) is the power to which e must be raised to equal the number; e is the constant 2.7182818. For example, the natural logarithm of 10 is 2.303; that is,\n\n$10=e^{2.303}=2.7182818^{2.303}$\n\nTo obtain the natural logarithm of a number, use the ln button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate ex (where x is the natural logarithm of the number).\n\nLogarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving exponents.\n\n1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers. $\\log xy= \\log x + \\log y, \\textrm{ and }\\ln xy=\\ln x + \\ln y$\n2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers. $\\log\\dfrac{x}{y}=\\log x-\\log y,\\textrm{ and } \\ln\\dfrac{x}{y}=\\ln x-\\ln y$\n3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number. $\\log x^n=n\\log x \\textrm{ and }\\ln x^n=n\\ln x$\n\n## The Solution of Quadratic Equations\n\nMathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions.\n\n$ax^2+bx+c=0$\n\nThe solution or roots for any quadratic equation can be calculated using the following formula:\n\n$x=\\dfrac{-b±\\sqrt{b^2−4ac}}{2a}$\n\nSolving Quadratic Equations Solve the quadratic equation 3x2 + 13x − 10 = 0.\n\nSolution Substituting the values a = 3, b = 13, c = −10 in the formula, we obtain\n\n$x=\\dfrac{−13±\\sqrt{(13)^2−4×3×(−10)}}{2×3}$\n$x=\\dfrac{−13±\\sqrt{169+120}}{6}=\\dfrac{−13±\\sqrt{289}}{6}=\\dfrac{−13±17}{6}$\n\nThe two roots are therefore\n\n$x=\\dfrac{−13+17}{6}=\\dfrac{2}{3}\\textrm{ and }x=\\dfrac{−13−17}{6}=−5$\n\nQuadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance.\n\n## Two-Dimensional (x-y) Graphing\n\nThe relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled (x), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured (y).\n\nWhen the value of y is changing as a function of x (that is, different values of x correspond to different values of y), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for (x,y) data pairs.\n\nExercise $$\\PageIndex{10}$$\n\nGraphing the Dependence of y on x\n\nx y\n1 5\n2 10\n3 7\n4 14\n\nThis table contains the following points: (1,5), (2,10), (3,7), and (4,14). Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of y on x.", null, "If the function that describes the dependence of y on x is known, it may be used to compute x,y data pairs that may subsequently be plotted.\n\nExercise $$\\PageIndex{11}$$\n\nPlotting Data Pairs If we know that y = x2 + 2, we can produce a table of a few (x,y) values and then plot the line based on the data shown here.\n\nx y = x2 + 2\n1 3\n2 6\n3 11\n4 18", null, "• Was this article helpful?" ]	[ null, "https://chem.libretexts.org/@api/deki/files/61354/CNX_Chem_00_BB_Dependence_img.jpg", null, "https://chem.libretexts.org/@api/deki/files/61355/CNX_Chem_00_BB_Function_img.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8290494,"math_prob":0.99938685,"size":9110,"snap":"2021-21-2021-25","text_gpt3_token_len":2638,"char_repetition_ratio":0.16571492,"word_repetition_ratio":0.09244884,"special_character_ratio":0.31635565,"punctuation_ratio":0.1131874,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999871,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T07:31:37Z\",\"WARC-Record-ID\":\"<urn:uuid:e153b724-6c97-49a3-9883-9fc622c4e658>\",\"Content-Length\":\"112820\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4cf6139a-50c1-4245-9aee-1dfce54b26c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:c80bd688-bea9-4a1b-a891-7d5a517b268e>\",\"WARC-IP-Address\":\"13.249.43.88\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Courses/City_College_of_San_Francisco/Foundations_-_Review_Source_for_Chem_101A/06%3A_Appendices/7.01%3A_Essential_Mathematics\",\"WARC-Payload-Digest\":\"sha1:HBQZPNFSIASDVYV3MK2GSYZ5RVZQAHO3\",\"WARC-Block-Digest\":\"sha1:W37ELTRFR4WW6Q4L7ARAYWWZ6C2OWLGZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487643703.56_warc_CC-MAIN-20210619051239-20210619081239-00483.warc.gz\"}"}
https://python.engineering/python-check-whether-two-lists-circularly-identical/	[ "# Python \| Check if two lists match in a circle\n\n\| \| \|", null, "Examples :\n\n` Input: list1 = [10, 10, 0, 0, 10] list2 = [10, 10, 10, 0, 0] Output: Yes Explanation: yes they are circularly identical as when we write the list1 last index to second last index, then we find it is circularly same with list1 Input: list1 = [10, 10, 10, 0, 0] list2 = [1, 10, 10, 0, 0] Output: No `\n\nMethod 1. Using list traversal\n\nUsing traversing , we have to double the given list. Check for any x (0 \"= n) for any x + n and compare with list2 to see if list1 and list2 are the same, if both are the same then list2 is cyclically identical. Check this property using two loops. The first loop will run from 0 to len (list1) and then check if the index (x to x + n) is the same as list2, if yes then returns true, otherwise returns false.\n\nPython implementation of the above approach:\n\n ` # Python program to check if two ` ` # lists circularly identical ` ` # using a workaround ` ` # function to check circular identity or not ` ` def ` ` circularly_identical (list1, list2): ` ` # doubling list ` ` list3 ` = ` list1 ` ` * ` ` 2 ` ` # crawl list twice1 ` ` for ` ` x ` ` in ` ` range ` ` (` ` 0 ` `, ` ` len ` ` (list1)): ` ` z ` ` = ` ` 0 ` ` # check if list2 == list1 briefly ` ` for ` ` y ` ` in ` ` range ` ` (x, x ` ` + ` ` len ` ` (list1)): ` ` if ` ` list2 [z] ` ` = ` ` = ` ` list3 [y]: ` ` z ` ` + ` ` = ` ` 1 ` ` else ` `: ` ` break ` ` # if all n elements are the same in a circle ` ` if ` ` z ` ` = ` ` = ` ` len ` ` (list1): ` ` return ` ` True ` ` return ` ` False ` ` ` ` ` ` ` ` # driver code ` ` list1 ` ` = ` ` [` ` 10 ` `, ` ` 10 ` `, ` ` 0 ` `, ` ` 0 ` `, ` ` 10 ] `` list2 = [ 10 , 10 , 10 , 0 , 0 ] list3 = [ 1 , 10 , 10 , 0 , 0 ] # check list 1 and list 2 if < / code> (circularly_identical (list1, list2)): print \" Yes \" else : print \"No\" # check list 2 and list 3 if (circularly_identical (list2, list3)): print \" Yes \" else : print \"No\" `\n\nExit :\n\n` Yes No `\n\nC tutorial 1: using the map () function\n\nUsing the Python built-in function map (), we can do this in one step where we need to map list2 to a string and then see if it exists when mapping a double from list1 (2 * list1) to on another line.\n\nBelow is the Python implementation of the above approach:\n\n ` # Python program to check if two ` ` # the lists are circularly identical ` ` # using the map function ` ` # function to check circular identity or not ` ` def ` ` circularly_identical (list1, list2): ` ` ` ` ` ` return ` ` ( ` ` ’’ ` `. join (` ` map ` ` (` ` str ` `, list2)) ` ` in ` ` ’’ ` `. join (` ` map ` ` (` ` str ` `, list1 ` ` * ` ` 2 ` `))) ` ` # driver code ` ` list1 ` ` = ` ` [` ` 10 ` ` , ` ` 10 ` `, < / code> 0 , 0 , 10 ] `` list2 = [ 10 , 10 , 10 , 0 , 0 ] list3 = [ 1 , 10 , 10 , 0 , 0 ] # check list 1 and list 2 if (circularly_identical (list1, list2)): print \"Yes\" else : print \"No\" # check list 2 and list 3 if (circularly_identical (list2, list3)): print \"Yes\" else : `` print < code class = \"string\"> \"No\" `\n\nExit:\n\n` Yes No `\n\n## Shop", null, "Best laptop for Fortnite\n\n\\$", null, "Best laptop for Excel\n\n\\$", null, "Best laptop for Solidworks\n\n\\$", null, "Best laptop for Roblox\n\n\\$", null, "Best computer for crypto mining\n\n\\$", null, "Best laptop for Sims 4\n\n\\$", null, "Best laptop for Zoom\n\n\\$499", null, "Best laptop for Minecraft\n\n\\$590\n\nLatest questions\n\nNUMPYNUMPY\n\npsycopg2: insert multiple rows with one query\n\nNUMPYNUMPY\n\nHow to convert Nonetype to int or string?\n\nNUMPYNUMPY\n\nHow to specify multiple return types using type-hints\n\nNUMPYNUMPY\n\nJavascript Error: IPython is not defined in JupyterLab\n\n## Wiki\n\nPython OpenCV \| cv2.putText () method\n\nnumpy.arctan2 () in Python\n\nPython \| os.path.realpath () method\n\nPython OpenCV \| cv2.circle () method\n\nPython OpenCV cv2.cvtColor () method\n\nPython - Move item to the end of the list\n\ntime.perf_counter () function in Python\n\nCheck if one list is a subset of another in Python\n\nPython os.path.join () method" ]	[ null, "http://espressocode.top/images/paichildbusverthhandre372881.jpg", null, "https://python.engineering/python-check-whether-two-lists-circularly-identical/", null, "https://python.engineering/python-check-whether-two-lists-circularly-identical/", null, "https://python.engineering/python-check-whether-two-lists-circularly-identical/", null, "https://python.engineering/python-check-whether-two-lists-circularly-identical/", null, "https://python.engineering/python-check-whether-two-lists-circularly-identical/", null, "https://python.engineering/wp-content/uploads/2022/04/pye-best-laptop-for-sims-4.jpg", null, "https://python.engineering/wp-content/uploads/2022/04/pye-best-laptop-for-zoom.jpg", null, "https://python.engineering/wp-content/uploads/2022/04/pye-best-laptop-for-minecraft.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5803006,"math_prob":0.8873737,"size":2662,"snap":"2022-27-2022-33","text_gpt3_token_len":889,"char_repetition_ratio":0.16365689,"word_repetition_ratio":0.2984293,"special_character_ratio":0.3658903,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99474204,"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,2,null,5,null,5,null,5,null,5,null,5,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T00:19:54Z\",\"WARC-Record-ID\":\"<urn:uuid:65ef2bd0-15f2-4d89-84f3-683daae85d97>\",\"Content-Length\":\"38370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:705aff63-a253-41cb-b383-155b3f9d5d7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8cb065b-840f-4f51-8b94-fb1ce790d564>\",\"WARC-IP-Address\":\"172.67.181.212\",\"WARC-Target-URI\":\"https://python.engineering/python-check-whether-two-lists-circularly-identical/\",\"WARC-Payload-Digest\":\"sha1:EFKMJDV4X25HT2SEGZKMAM6OHMF6ASBM\",\"WARC-Block-Digest\":\"sha1:6WYAGRTQBIYOXIMHWXVTH7AMZZDMU26P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103619185.32_warc_CC-MAIN-20220628233925-20220629023925-00338.warc.gz\"}"}
https://www.geeksforgeeks.org/p5-js-createvector-function/	[ "# p5.js \| createVector() function\n\nThe createVector() function in p5.js is used to create the new p5 vector which contains both magnitude and direction. This provides a two or three dimensional vector, specifically a geometric vector.\n\nSyntax:\n\n`createVector([x], [y], [z])`\n\nParameters: This function accepts three parameters as mentioned above and described below:\n\n• x: This parameter stores the x component of the vector.\n• y: This parameter stores the y component of the vector.\n• z: This parameter stores the z component of the vector.\n• Below programs illustrate the createVector() function in p5.js:\n\nExample 1: This example uses createVector() function to draw a line.\n\n `function` `setup() { ` ` ` ` ``// Create a Canvas ` ` ``createCanvas(500, 550); ` `} ` ` ` `function` `draw() { ` ` ` ` ``// Vector initislisation ` ` ``// using createVector ` ` ``t1 = createVector(10, 40); ` ` ``t2 = createVector(411, 500); ` ` ` ` ``// Set background color ` ` ``background(200); ` ` ` ` ``// Set stroke weight ` ` ``strokeWeight(2); ` ` ` ` ``// line using vector ` ` ``line(t1.x, t1.y, t2.x, t2.y); ` ` ` ` ``translate(12, 54); ` ` ` ` ``line(t1.x, t1.y, t2.x, t2.y); ` `} `\n\nOutput:", null, "Example 2: This example uses createVector() function to draw a circle.\n\n `function` `setup() { ` ` ` ` ``// Create a Canvas ` ` ``createCanvas(500, 550); ` `} ` ` ` `function` `draw() { ` ` ` ` ``// Vector initislisation ` ` ``// using createVector ` ` ``t1 = createVector(10, 40); ` ` ``t2 = createVector(41, 50); ` ` ` ` ``// Set background color ` ` ``background(200); ` ` ` ` ``// Set stroke weight ` ` ``strokeWeight(2); ` ` ` ` ``// Fill yellow ` ` ``fill(``'yellow'``); ` ` ` ` ``// ellispe using vector ` ` ``ellipse(t1.xmillis() / 1000 20, ` ` ``t1.y, t2.x+100, t2.y+100); ` `} `\n\nOutput:", null, "My Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\n\nBe the First to upvote.\n\nPlease write to us at contribute@geeksforgeeks.org to report any issue with the above content." ]	[ null, "https://media.geeksforgeeks.org/wp-content/uploads/20190420232304/Screenshot-2019-04-20-at-11.22.51-PM.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20190422183953/circle6.png", null, "https://media.geeksforgeeks.org/auth/profile/6z3x57km8koa88tq98ev", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5256069,"math_prob":0.84704137,"size":2233,"snap":"2019-35-2019-39","text_gpt3_token_len":607,"char_repetition_ratio":0.19380888,"word_repetition_ratio":0.18879056,"special_character_ratio":0.3036274,"punctuation_ratio":0.1970803,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98314106,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T12:19:03Z\",\"WARC-Record-ID\":\"<urn:uuid:261ba453-f1db-4326-ac61-31459dc1efaa>\",\"Content-Length\":\"97849\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85e28a74-be45-4f0e-a286-1f215652c68b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f60d7bfc-d8a6-4d9e-87c1-d3ac34915d5e>\",\"WARC-IP-Address\":\"104.96.221.51\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/p5-js-createvector-function/\",\"WARC-Payload-Digest\":\"sha1:V4SHRNCH3KQWVEVHLMBXNMJ4OMBRRNSO\",\"WARC-Block-Digest\":\"sha1:C23FAIZHUNZX3Q7ZXTPLYK4TC5NXMM4X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317113.27_warc_CC-MAIN-20190822110215-20190822132215-00240.warc.gz\"}"}
https://discuss.pytorch.org/t/second-derivative-passing-from-linear-to-non-linear-formula/130435	[ "# Second derivative - passing from linear to non-linear formula\n\nHi guys!\n\nI am interested about passing from this linear function created below, to a non-linear one, in order to estimate the second derivative. Is there exist some trick that I can use to have the second derivative different from zero? I am trying to estimate a greek called “Gamma” for a Call Option, that is priced using Montecarlo method\n\n``````import torch as np\nvalues = torch.tensor([1., 1.1, 1.2], requires_grad=True)\n\ndef delta_gamma(xi):\n\nk,T,j,sigma = 1.5,1.,10000,0.5\n\nmean = -.5 * sigma * sigma * T\n\nvolatility = sigma\n\nBM = torch.randn(1, j)volatility+mean\n\nproduct = Snp.exp(BM)\n\np = torch.maximum(product-k,torch.zeros_like(product))\n\nresult = torch.mean(p, 1)\n\nreturn result\n\nfor i in values:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7919438,"math_prob":0.9863236,"size":792,"snap":"2022-05-2022-21","text_gpt3_token_len":217,"char_repetition_ratio":0.095177665,"word_repetition_ratio":0.0,"special_character_ratio":0.28156567,"punctuation_ratio":0.22950819,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99931204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T23:35:53Z\",\"WARC-Record-ID\":\"<urn:uuid:99b11fd3-adc3-44a6-826c-33bbb92972e7>\",\"Content-Length\":\"13264\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d7eb513-140f-4003-9e8d-4830aa65f762>\",\"WARC-Concurrent-To\":\"<urn:uuid:abd91548-dd80-49ce-985d-10cde31a415c>\",\"WARC-IP-Address\":\"159.203.145.104\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/second-derivative-passing-from-linear-to-non-linear-formula/130435\",\"WARC-Payload-Digest\":\"sha1:LMLRNTOPSMK6B7ODO5IUN7ZFO6FOJIMG\",\"WARC-Block-Digest\":\"sha1:Y5LA7ZRFRELWEJ34X6QVEFMUTTXSWDCQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663021405.92_warc_CC-MAIN-20220528220030-20220529010030-00596.warc.gz\"}"}
https://www.smartzworld.com/notes/network-theory-pdf-notes-nt-pdf-notes/	[ "# Network Theory Pdf Notes – NT Pdf Notes\n\nHere you can download the free lecture Notes of Neheory Ptwork Tdf Notes – NT Pdf Notes materials with multiple file links to download. Network Theory Notes Pdf – NT Notes Pdf book starts with the topics Introduction,Advantages of Three Phase is preferred Over Single Phase,Frequency-selective or filter circuits pass to the output only those input signals that are in a desired range of frequencies (called pass band).\n\n## The Network Theory Pdf Notes – NT Pdf Notes\n\nNetwork Theory Notes pdf – NT pdf notes – NT notes pdf file to download are listed below please check it –\n\n#### Network Theory Book\n\nComplete notes\n\nUnit 1\n\nUnit 2\n\nUnit 3\n\nUnit 4\n\nUnit 5\n\nNote :- These notes are according to the R09 Syllabus book of JNTU.In R13 and R15,8-units of R09 syllabus are combined into 5-units in R13 and R15 syllabus. If you have any doubts please refer to the JNTU Syllabus Book.\n\n### Network Theory Notes Pdf – NT Notes Pdf\n\nUnit-1:\n\nIntroduction,Advantages of Three Phase is preferred Over Single Phase,STAR CONNECTION,THE STAR CONNECTION,Delta Connection,ANALYSIS OF BALANCED THREE PHASE CIRCUITS,ANALYSIS OF UNBALANCED LOADS,Power in three-phase circuits.\n\nUnit-2:\n\nIntroduction,What is meant by Transients,Series RL Circuit,Series RC circuit,Transient During Discharging a Capacitor,Response of a series R-L-C circuit due to a dc voltage source,Overdamped response,Under damped response,Critically damped response.\n\nUnit-3:\n\nThe concept of complex frequency,Terminal pairs or poles,Network Function,RESTRICTION ON POLE AND ZERO LOCATIONS OF NETWORK FUNCTION,A pair of terminals through which a current may enter or leave a network is known as a port. Two-terminal devices or elements (such as resistors, capacitors, and inductors) result in one-port networks.\n\n### Network Theory Notes pdf Details\n\nUnit-4:\n\nA pair of terminals through which a current may enter or leave a network is known as a port. A port is an access to the network and consists of a pair of terminals; the current entering one terminal leaves through the other terminal so that the net current entering the port equals zero.\n\nUnit-5:\n\nFrequency-selective or filter circuits pass to the output only those input signals that are in a desired range of frequencies (called pass band). The amplitude of signals outside this range of frequencies (called stop band) is reduced (ideally reduced to zero). Typically in these circuits, the input and output currents are kept to a small value and as such, the current transfer function is not an important parameter. The main parameter is the voltage transfer function in the frequency domain.\n\n## Network Theory Notes Reviews\n\n#### Avg. Score\n\nUsers rating & review for Network Theory Pdf Notes (NT Pdf Notes)\n\nUser Rating: Be the first one !\n\n## Basic Electronics PDF VSSUT \| BE Notes VSSUT\n\nBasic Electronics PDF VSSUT – BE Notes VSSUT of Total Complete Notes Please find the …" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8564314,"math_prob":0.8259887,"size":2839,"snap":"2019-51-2020-05","text_gpt3_token_len":650,"char_repetition_ratio":0.1276896,"word_repetition_ratio":0.18666667,"special_character_ratio":0.20183162,"punctuation_ratio":0.10128913,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9686088,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T03:07:05Z\",\"WARC-Record-ID\":\"<urn:uuid:33899b2e-5465-4a77-8d46-c3d829266ffe>\",\"Content-Length\":\"75086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34c3e325-8583-4b8a-90b4-e2202172bbb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:80cbab17-e611-4ffa-82a1-8da6da33a484>\",\"WARC-IP-Address\":\"209.182.192.62\",\"WARC-Target-URI\":\"https://www.smartzworld.com/notes/network-theory-pdf-notes-nt-pdf-notes/\",\"WARC-Payload-Digest\":\"sha1:WK4DLGGZWLDQDZP7GDQ5OBSIWABSMKAU\",\"WARC-Block-Digest\":\"sha1:ZTONVV7XIYVXJOJEUQO24DCQKJHNTBV2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540548537.21_warc_CC-MAIN-20191213020114-20191213044114-00288.warc.gz\"}"}
https://www.hindawi.com/journals/aaa/2013/432509/	[ "#### Abstract\n\nThis paper considers the existence of solutions for two boundary value problems for fractional -Laplacian equation. Under certain nonlinear growth conditions of the nonlinearity, two new existence results are obtained by using Schaefer's fixed point theorem. As an application, an example to illustrate our results is given.\n\n#### 1. Introduction\n\nFractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. As is known to all, the problem for fractional derivative was originally raised by Leibniz in a letter, dated September 30, 1695. A fractional derivative arises from many physical processes, such as a non-Markovian diffusion process with memory , charge transport in amorphous semiconductors , and propagations of mechanical waves in viscoelastic media , and so forth. Moreover, phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and material science are also described by differential equations of fractional order . For instance, Pereira et al. considered the following fractional Van der Pol equation: where is the fractional derivative of order and is a control parameter that reflects the degree of nonlinearity of the system. Equation (1) is obtained by substituting the capacitance by a fractance in the nonlinear RLC circuit model.\n\nRecently, fractional differential equations have been of great interest due to the intensive development of the theory of fractional calculus itself and its applications. For example, for fractional initial value problems, the existence and multiplicity of solutions (or positive solutions) were discussed in . On the other hand, for fractional boundary value problems (FBVPs), Agarwal et al. considered a two-point boundary value problem at nonresonance, and Bai considered a -point boundary value problem at resonance. For more papers on FBVPs, see and the references therein.\n\nThe turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problems, Leibenson introduced the -Laplacian equation as follows: where . Obviously, is invertible and its inverse operator is , where is a constant such that .\n\nIn the past few decades, many important results relative to (2) with certain boundary value conditions have been obtained. We refer the reader to and the references cited therein. For boundary value problems of fractional -Laplacian equations, Chen and Liu considered an antiperiodic boundary value problem with the following form: where , , , and is Caputo fractional derivative. Under certain nonlinear growth conditions of the nonlinearity, an existence result was obtained by using degree theory. In addition, Yao et al. studied a three-point boundary value problem given by where is the standard Riemann-Liouville derivative with , , , , and the constant is a positive number satisfying . The monotone iterative technique was applied to establish the existence results on multiple positive solutions in .\n\nMotivated by the works mentioned previously, in this paper, we investigate the existence of solutions for fractional -Laplacian equation of the form subject to either boundary value conditions or where , , , is a Caputo fractional derivative, and is continuous.\n\nNote that the nonlinear operator reduces to the linear operator when and the additive index law holds under some reasonable constraints on the function .\n\nThe rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, based on Schaefer’s fixed point theorem, we establish two theorems on existence of solutions for FBVP (5) and (6) (Theorem 7) and FBVP (5) and (7) (Theorem 8). Finally, in Section 4, an explicit example is given to illustrate the main results. Our results are different from those of bibliographies listed in the previous texts.\n\n#### 2. Preliminaries\n\nFor the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory, which can be found, for instance, in [31, 32].\n\nDefinition 1. The Riemann-Liouville fractional integral operator of order of a function is given by provided that the right side integral is pointwise defined on .\n\nDefinition 2. The Caputo fractional derivative of order of a continuous function is given by where is the smallest integer greater than or equal to , provided that the right side integral is pointwise defined on .\n\nLemma 3 (see ). Let . Assume that . Then the following equality holds: where , ; here is the smallest integer greater than or equal to .\n\nLemma 4 (see ). For fixed , let one define Then the equation has a unique solution .\n\nIn this paper, we take with the norm and with the norm . By means of the linear functional analysis theory, we can prove that is a Banach space.\n\n#### 3. Existence Results\n\nIn this section, two theorems on existence of solutions for FBVP (5) and (6) and FBVP (5) and (7) will be given under nonlinear growth restriction of .\n\nAs a consequence of Lemma 3, we have the following results that are useful in what follows.\n\nLemma 5. Given , the unique solution of is where\n\nProof. Assume that satisfies the equation of FBVP (13); then Lemma 3 implies that From the boundary value condition , one has Thus, we have By condition , we get . The proof is complete.\n\nDefine the operator by where is the Nemytskii operator defined by Clearly, the fixed points of the operator are solutions of FBVP (5) and (6).\n\nLemma 6. Given , the unique solution of is where here is a constant dependent on .\n\nProof. Assume that satisfies the equation of FBVP (21); then Lemma 3 implies that From condition , one has Based on Lemma 4, we know that (25) has a unique solution . Moreover, by the integral mean value theorem, there exists a constant such that , which implies that . Thus, we have Hence From condition , we get . The proof is complete.\n\nDefine the operator by where and is the Nemytskii operator defined by (20). Clearly, the fixed points of the operator are solutions of FBVP (5) and (7).\n\nOur first result, based on Schaefer’s fixed point theorem and Lemma 5, is stated as follows.\n\nTheorem 7. Let be continuous. Assume that there exist nonnegative functions such that Then FBVP (5) and (6) has at least one solution, provided that\n\nProof. We will use Schaefer’s fixed point theorem to prove that has a fixed point. The proof will be given in the following two steps.\nStep 1. is completely continuous.\nLet be an open bounded subset. By the continuity of , we can get that is continuous and is bounded. Moreover, there exists a constant such that , for all , . Thus, in view of the Arzelà-Ascoli theorem, we need only to prove that is equicontinuous.\nFor and , we have Since is uniformly continuous on , we can obtain that is equicontinuous. A similar proof can show that is equicontinuous. This, together with the uniform continuity of on , yields that is also equicontinuous.\nStep 2 (priori bounds). Set Now it remains to show that the set is bounded.\nFrom Lemma 3 and boundary value condition , one has Thus, we get That is,\nFor , we have So, from , we obtain that which, together with and (35), yields that In view of (30), from (38), we can see that there exists a constant such that Thus, from (35), we get Combining (39) with (40), we have\nAs a consequence of Schaefer’s fixed point theorem, we deduce that has a fixed point which is the solution of FBVP (5) and (6). The proof is complete.\n\nOur second result, based on Schaefer’s fixed point theorem and Lemma 6, is stated as follows.\n\nTheorem 8. Let be continuous. Suppose that holds; then FBVP (5) and (7) has at least one solution, provided that (30) is satisfied.\n\nProof. The proof work is similar to the proof of Theorem 7, so we omit the details.\n\n#### 4. An Example\n\nIn this section, we will give an example to illustrate our main results.\n\nExample 1. Consider the following fractional -Laplacian equation: Corresponding to (5), we get that , , , and Choose , , and . By a simple calculation, we can obtain that , and Obviously, (42) subject to boundary value conditions (6) (or (7)) satisfies all assumptions of Theorem 7 (or Theorem 8). Hence, FBVP (42) and (6) (or FBVP (42) and (7)) has at least one solution.\n\n#### Acknowledgments\n\nThis work was supported by the Fundamental Research Funds for the Central Universities (2012QNA50) and the National Natural Science Foundation of China (11271364)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8599623,"math_prob":0.9888475,"size":16918,"snap":"2022-40-2023-06","text_gpt3_token_len":4301,"char_repetition_ratio":0.16235071,"word_repetition_ratio":0.14124294,"special_character_ratio":0.25198013,"punctuation_ratio":0.21600714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990263,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T17:51:00Z\",\"WARC-Record-ID\":\"<urn:uuid:9a627864-f25b-4bf2-b474-10b3a73d091c>\",\"Content-Length\":\"1049027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9457a28f-4f6e-4bd7-b4b5-4b1a35a217ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:331c70aa-4895-436d-9cc1-2bfff82da139>\",\"WARC-IP-Address\":\"13.249.39.61\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/aaa/2013/432509/\",\"WARC-Payload-Digest\":\"sha1:QHP7QPAS2X2BJQ2G554E6UKQSKO2PFW2\",\"WARC-Block-Digest\":\"sha1:Y2JSTRFXDPYCR4ML5EAC7EEPHLKRGSSA\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337428.0_warc_CC-MAIN-20221003164901-20221003194901-00655.warc.gz\"}"}
https://www.java67.com/2016/10/	[ "HTML\n\n# How to transpose a matrix in Java? Example Tutorial\n\nHello guys, continuing the tradition of this week, where I have mostly published articles about coding exercises for Java beginners, today also I am going to share an interesting coding problem, many of you have solved in your college or school days. Yes, it's about writing a Java program to transpose a matrix. In the last couple of tutorials, we have learned to how to add and subtract two matrices in Java (see here) and how to multiply two matrices in Java (see here). In this tutorial, I'll show you how to transpose a matrix in Java. The transpose of a matrix is a new matrix whose rows are the columns of the original. This means when you transpose a matrix the columns of the new matrix becomes the rows of the original matrix and vice-versa. In short, to transpose a matrix, just swap the rows and columns of the matrix. For example, if you have a matrix with 2 rows and 3 columns then transpose of that matrix will contain 3 rows and two columns.\n\n# How to Add and Subtract Two Matrices in Java\n\nThis is the second program in the series of matrices related programming exercises in Java. In the last program, you have learned matrix multiplication and in this program, you will learn how to perform addition and subtraction of two matrices in Java. We'll create methods to calculate both sum and difference of two matrices in Java program. In Mathematics, a matrix is a rectangular array with two dimensions known as rows and columns. In Java, your can use a two-dimensional array to represent a matrix because it also has two dimensions rows and columns. Since a 2D array is nothing but an array of the array in Java, the length of the outer array is equal to the number of rows and length of sub-array is equal to the number of columns.\n\n# How to calculate sum and difference of two complex numbers in Java\n\nFrom the last couple of articles, I am writing about coding exercises for beginners e.g. yesterday you learned how to write a program from matrix multiplication in Java (see here) and a couple of days back, you have learned recursive binary search algorithm. To continue that tradition today I am going to show you how to write a program for calculating sum and difference of two complex numbers in Java. If you remember the complex number from you maths classes, it has two part real and imaginary and to add a complex number we add their real and imaginary part separately, similar to subtract complex number we minus their real and imaginary part separately. For example, if first complex number is A + iB and the second complex number is X + iY then the addition of these two complex number will be equal to (A +X ) + i(B + Y).\n\n# How to Multiply Two Matrices in Java\n\nI first learned about matrix in class 12th and I first wrote the program to multiply two matrices on my first semester of engineering, so, when I thought about this program, It brings a lot of memories from the past. It's actually a beginner exercise to develop coding logic, much like Fibonacci, prime, and palindrome check, but what make this program interesting is the use of the two-dimensional array to represent a matrix in Java. Since matrix has both rows and columns, two-dimensional array just naturally fits into the requirement. Another important thing to solve this problem is to remember the rule of matrix multiplication in mathematics. If you don't remember the rule, just forget about how to solve this problem, unless you have access to Google. So, first, we'll refresh the rules of multiplication and then we'll look into coding aspect.\n\n# java.lang.OutOfMemoryError: unable to create new native thread - Cause and Solution\n\nThere are several types of OutOfMemoryError in Java e.g. OutOfMemoryError related to Java heap space and permgen space, and a new one coming in Java 8, Java.lang.OutOfMemoryError: MetaSpace. Each and every OutOfMemoryError has their own unique reason and corresponding unique solution. For example, java.langOutOfMemoryError: Java Heap Space comes when the application has exhausted all heap memory and tries to create an object which requires further memory allocation, that time JVM throws this error to tell the application that it's not possible to create any object. Similarly java.lang.OutOfMemoryError: PermGen Space comes when there is no more memory in permgen space and application tries to load more classes (as class metadata is stored in this area) or tries to create new String (because prior to Java 7 String pool has also existed on permgen space).\n\n# 5 Difference between StringBuffer, StringBuilder and String in Java\n\nThough all three classes StringBuffer, StringBuilder and String are used for representing text data in Java there are some significant differences between them. One of the most notable differences between StringBuilder, StringBuffer, and String in Java is that both StringBuffer and StrinBuilder are Mutable class but String is Immutable in Java. What this means is, you can add, remove or replace characters from StringBuffer and StringBuilder object but any change on String object e.g. converting uppercase to lowercase or appending a new character using String concatenation will always result in a new String object. Another key difference between them is that both StringBuffer and String are thread-safe but StringBuilder is not thread-safe in Java. String achieves its thread-safety from Immutability but StringBuffer achieves it via synchronization, which is also the main difference between the StringBuffer and StringBuilder in Java.\n\n# JDBC - How to solve java.sql.BatchUpdateException: String or binary data would be truncated.\n\nRecently I was working in Java application which uses Microsoft SQL Server at its backend. The architecture of Java application was old i.e. even though there was heavy database communication back and forth there was no ORM used e.g. no Hibernate, JPA, or Apache iBatis. The Java application was using old DAO design pattern, where the DB related classes which are responsible for loading and storing data from database was calling stored procedure to do their Job. These stored procedure takes data from Java application and insert into SQL Server tables. One day, one of my collegue called me to troubleshoot \"java.sql.BatchUpdateException: String or binary data would be truncated\", which it's application was throwing and he has no clue whatsoever that what is wrong with the data he is getting from other system and trying to store.\n\n# How to calculate average of all numbers of array in Java\n\nIn the last article, I teach you how to calculate the sum of all numbers in a given array and in this article, we'll go one more step. This time, you need to write a program to calculate the average of all numbers from a given array, for example, you will be passed salaries of Java developers in different states in the USA and you need to calculate the average salary of Java developer in the USA. The example of average salaries of Java developer is more interesting because everybody wants to know how much Java developers make, isn't it? Anyway, coming back to the requirement of the program, The array will contain integers, which can be both positive and negative, so you must handle them. Your program should also be robust e.g. it should not break if you pass empty array or null. In these case either you can throw IllegalArgumentException as returning any other number will be ambiguous.\n\n# How to calculate sum of array elements in Java\n\nIn today's coding problem, we'll see how to write a program to calculate the sum of array elements in Java. You need to write a method which will accept an integer array and it should return total sum of all the elements. The array could contain both positive and negative numbers but only decimal numbers are allowed. The array can also be null or empty so make sure your solution handle those as well. In the case of a null or empty array, your program can throw IllegalArgumentException. The empty array means, an array whose length is zero or there is no element inside it. Well, that's enough for the requirement of this simple coding problem. The solution is really simple, just loop through the array and keep adding elements into sum until you process all the elements.\n\n# 10 Reasons of java.lang.NumberFormatException in Java - Solution\n\nThe NumberFormatException is one of the most common errors in Java application along with NullPointerException. This error comes when you try to convert a String into numeric data types e.g. int, float, double, long, short, char or byte. The data type conversion methods like Integer.parseInt(), Float.parseFloat(), Double.parseDoulbe(), and Long.parseLong() throws NumberFormatException to signal that input String is not valid numeric value. Even though the root cause is always something which cannot be converted into a number, there are many reasons and input due to which NumberFormatException occurs in Java application. Most of the time I have faced this error while converting a String to int or Integer in Java, but there are other scenarios as well when this error occurs. In this article, I am sharing 10 of the most common reasons of java.lang.NumberFormatException in Java programs.\n\n# How to calculate area of triangle in Java - Program\n\nWriting a Java program to calculate the area of a triangle is one of the basic programming exercises to develop coding sense on beginner programmers. Like many mathematical conceptual programs e.g. square root, factorial, or prime number this also serves a good exercise for beginners. Now, if you remember in maths you might have seen two main ways to calculate the area of a triangle, using vertices and using base and height. In this program, I have created two methods to calculate the area of a triangle using both ways. In the first method area(Point a, Point b, Point c) we expect coordinates of three vertices of triangle and then we calculate area of triangle using the formula (Ax(By -Cy) + Bx(Cy -Ay) + Cx(Ay - By))/2, while in second method, area(int base, int height) we expect value of base and height and then we calculate are of triangle using formula (base * height) / 2.\n\n# 3 ways to convert String to JSON object in Java?\n\nIt's very common nowadays to receive JSON String from a Java web service instead of XML, but unfortunately, JDK doesn't yet support conversion between JSON String to JSON object. Keeping JSON as String always is not a good option because you cannot operate on it easily, you need to convert it into JSON object before you do anything else e.g. retrieve any field or set different values. Fortunately, there are many open-source libraries which allows you to create JSON object from JSON formatted String e.g. Gson from Google, Jackson, and json-simple. In this tutorial, you will learn how to use these 3 main libraries to do this conversion with step by step examples.\n\n# 3 Ways to Reverse an Array in Java - Coding Interview Question\n\nOne of the common coding questions is, how do you reverse an array in Java? Well, there are multiple ways to solve this problem. You can reverse array by writing your own function, which loops through the array and swaps elements until the array is sorted. That's actually should you be your first approach on interviews. Later you can impress the interviewer by a couple of other tricks, which is specific to Java development world. For example, you can reverse an array by converting array to ArrayList and then use this code to reverse the ArrayList. You can also use Apache Commons ArrayUtils.reverse() method to reverse any array in Java. This method is overloaded to reverse byte, short, long, int, float, double and String array. You can use any of the method depending upon your array type." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90767497,"math_prob":0.7706526,"size":11833,"snap":"2019-51-2020-05","text_gpt3_token_len":2365,"char_repetition_ratio":0.1323865,"word_repetition_ratio":0.026170107,"special_character_ratio":0.1963154,"punctuation_ratio":0.10128832,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9757091,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T22:41:49Z\",\"WARC-Record-ID\":\"<urn:uuid:091bb468-7013-4e73-bf41-8a86de6762c5>\",\"Content-Length\":\"276211\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a26dd1d-b656-41f0-b6c3-bb62e45be956>\",\"WARC-Concurrent-To\":\"<urn:uuid:194cdec1-1f60-474b-b4fc-d3f9e9ede080>\",\"WARC-IP-Address\":\"172.217.5.243\",\"WARC-Target-URI\":\"https://www.java67.com/2016/10/\",\"WARC-Payload-Digest\":\"sha1:NJHIERCFZDSBN66CNEVBU3M7IMTT67HI\",\"WARC-Block-Digest\":\"sha1:YTR645SBYQ2M2WMY42E4IVYM7HYKN466\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540491491.18_warc_CC-MAIN-20191206222837-20191207010837-00198.warc.gz\"}"}
https://www.math-for-all-grades.com/Worksheet-on-weighted-averages.html	[ "## WORKSHEET ON WEIGHTED AVERAGES:\n\nHi welcome visitors to our website:\n\nin this worksheet, I am going to discuss in detail the concept of weighted average and one very interesting short-cut to solve questions on weighted average.\n\nTo enrich your learning, you can also visit our you tube on the concept of weighted average.\n\nSo, let’s get on with the formula of weighted average, the property of weighted average and finally the short cut to find the ratio of the number of terms in two groups whose individual averages are given.\n\nLet us take up the following problem on weighted average:\n\nSOLVED PROBLEM 1:\n\nIn a group of 80 members, the number of males is 45 and the number of females is 35. The average age of the males is m and that of the females is f. compare the average age of the two groups compared to the average of the ages of the two groups i.e. (f + m)/2. f > m.\n\nSolution:\n\nFirst all, let us set up the data for weighted average as we do normally as follows:\n\nIf you have watched our webpage on weighted average and its property, then you will understand that weighted average will be a number nearer to the average of the group of males i.e. ‘m’, which is the average of the group having more number of terms.\n\n(here, terms represent ages of the males and females)\n\nNow, tell me if a number is closer to the average ‘m’ than to the other average ‘f’, then what will it be compared to (f + m)/2 which is the midpoint of the two average numbers m and f?\n\nAs per the data written above, you may conclude that since the weighted average age of the two groups males and females lies closer to the average age ‘m’ from the property of weighted average, therefore from the diagram drawn above in the data representation, weighted average age of the two groups will be less than the midpoint (f + m)/2.\n\nHere lies the mistake often committed by students while solving this question on weighted average age of the two groups of males and female in the above question.\n\nYour conclusion that weighted average age of the two groups is nearer to the average age ‘m’ than to the other average ‘f’ is true; but compared to (f + m)/2 it is less or greater cannot be still determined without knowing the relationship between ‘f’ and ‘m’.\n\nBecause there can exist three relationships between ‘f’ and ‘m’:\nf < m or f = m or f > m\n\ni.e. the average age of the females group may be less than the average age of the males or\n\nthe average age of the group of females may be same as that of the average age of the males or even\n\nthe average age of the group of females could be greater than the average of the males.\n\nThe exploration of the above three possibilities of relationship between the average age of the females and the average age of the males is triggered by the information given by the question : f > m\n\nSuppose the average age of the two groups of females and the males is same i.e. f = m, then how can you conclude from the property of weighted average that the weighted average age of the two groups will be less than the average of the average ages: (f + m)/2.\n\nWhat I am trying to drive home the point is that the weighted average property will enable you to tell that weighted average age of the two groups will definitely be closer to the average of the males, but what the property cannot tell you is what it will be compared to the average (f + m)/2.\n\nTo be able to determine the second point above i.e. what is weighted average compared to the average (f + m)/2, we need to know the relationship between the average ages ‘m’ and ‘f’, i.e. which of the three relationships between them exists as explored above.\n\nThat’s why the question has supplied the additional detail f > m.\nWe must realize the importance of the information f > m and its purpose is to determine if weighted average age, which is closer to the average age ‘m’ of the males, is less than or equal to or greater than the average of the average ages : f and m i.e. (f + m)/2\n\nNow, after having ascertained the purpose of the relationship between f and m i.e. f > m, we can now give the answer that weighted average age of the two groups will be less than the average (f + m)/2.\n\nIf it is not still clear, then we can make the explanation more clear by assuming some numbers for f and m based on the relationship f > m.\n\nLet us take average age of the males’ group, m as 20 and the average age of the females’ group as some 40\n\nSo by substituting the above assumed values for f and m, the average (f + m)/2 will be (40 + 20)/2 = 60/2 = 30\n\nNow, you can clearly see that the weighted average age of the two groups males and females will be less than 30.\n\nHow? Its not clear still. Ok, we will go as below:\n\nFrom the weighted average property, what we can say is the weighted average age of the two males and females whose individual average ages have been taken as 20 and 40 will be a number closer to the average age of the males which is 20.\n\nA number which is closer to the average age 20 than to the other average age 40 must by implication be less than the midpoint of 20 and 40 i.e. 30\n\nFor example, 28 is closer to 20 than to 40 as 28 is only 8 numbers away from 20 but slightly more away from 40, i.e. 12 away from 40.\n\nNow figure out in your mind what 28 is compared to 30.\nObviously less.\n\nThat’s why weighted average age of the two groups by being nearer to the average age of the males i.e. ‘m’ will be also less than the average of the average ages i.e. (f + m)/2 by taking into consideration the information f > m given in the question.\n\nNow I will tell you a short cut to solve a question on weighted average:\n\nConsider the following problem:\n\nThe average marks of a group having ‘p’ students in a class is 75 and the average marks of another group having ‘n’ students is 92. Find what value the fraction p/n will be, if the weighted average marks of the two groups is 84\n\nSolution:\n\nFirst of all, present the data for the weighted average as we do usually:\n\nTo find the value of the fraction p/n, move in the direction of the arrows as drawn above in the presentation:\n\np / n = (92 – 84)/(84 – 75) = 8/9.\n\nThat’s it.\n\nIsn’t that a wonderful short-cut.\n\nWell, the above fraction p/n signifies another important property of weighted average, which will discuss in another worksheet." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9541041,"math_prob":0.9913733,"size":6229,"snap":"2022-40-2023-06","text_gpt3_token_len":1452,"char_repetition_ratio":0.26473895,"word_repetition_ratio":0.13289036,"special_character_ratio":0.24418044,"punctuation_ratio":0.08175624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99558747,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T11:01:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c0df443f-db45-4c0c-a954-9aa59e19ba56>\",\"Content-Length\":\"38480\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e56960d-d928-40f9-8f6f-5ac940cc88d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b67a823-b36b-482f-a4f3-baaaf448de93>\",\"WARC-IP-Address\":\"104.21.77.198\",\"WARC-Target-URI\":\"https://www.math-for-all-grades.com/Worksheet-on-weighted-averages.html\",\"WARC-Payload-Digest\":\"sha1:7QEBB75TPYK4JAT2DIZNAYQYOGEACC2E\",\"WARC-Block-Digest\":\"sha1:XYDL5MLPEWFC6TU7RA2CQLXWFUD262PY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335350.36_warc_CC-MAIN-20220929100506-20220929130506-00778.warc.gz\"}"}
https://gmplib.org/list-archives/gmp-discuss/2004-October/001378.html	[ "# Computing A^x (mod n)\n\nPhilip Lee rocketman768 at gmail.com\nSat Oct 2 18:57:14 CEST 2004\n\n```I'm just wondering the algorithm behind mpz_powm(). It can't be doing\nthe calculation (mod n) inside whatever loop it's running, because if\nI take a a 572-bit number and raise it to a 576-bit number modulo a\n576-bit number, the program will terminate. I'm guessing it is doing\nthe exponential junk first and then just taking the modulus right?\nBecause this would cause an extreme overflow in such a case as mine.\n(As you can tell, I'm doing some RSA stuff ).\n\nSo my question is, \"if the mpz_powm() function isn't doing a running\ncalculation (mod n ), why is it not?\" As I said, this would cause\noverflows which would be prevented if it was instead doing a running\ncalculation (mod n). Anyways, just wondering.\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9055826,"math_prob":0.75676066,"size":848,"snap":"2023-14-2023-23","text_gpt3_token_len":219,"char_repetition_ratio":0.109004736,"word_repetition_ratio":0.013888889,"special_character_ratio":0.25589624,"punctuation_ratio":0.10465116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9805292,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T00:08:01Z\",\"WARC-Record-ID\":\"<urn:uuid:255c7c8d-cdc7-4cdd-9255-16d5ef5e8402>\",\"Content-Length\":\"3000\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f8751048-b19b-4471-bb6e-01492e6b84a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f28b0db6-fd58-4ea3-bafd-5502fe85d261>\",\"WARC-IP-Address\":\"130.242.124.102\",\"WARC-Target-URI\":\"https://gmplib.org/list-archives/gmp-discuss/2004-October/001378.html\",\"WARC-Payload-Digest\":\"sha1:4NUY3LL2LX2EBV76BUEDMZBZ4WAE2KBJ\",\"WARC-Block-Digest\":\"sha1:H46CKIUNE47YC7NTNOQIKWVSGPV54VMB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652184.68_warc_CC-MAIN-20230605221713-20230606011713-00395.warc.gz\"}"}
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/patricia4.html	[ "Subject: Problem Solving Inequalities Name: Patricia Who are you: Student Your quiz grades are 73, 75, 89 and 91. What is the lowest grade you can obtain on the last quiz and still achieve an average at least 85? Thank you. Hi Patricia, If you had all 5 grades and the average was 85 then the sum of the 5 grades divided by 5 would be 85. Thus the sum of the 5 grades would be 5", null, "85. Penny Hi Patricia. I'll show you how to do this with different numbers: Let's say your quiz grades are 75, 85, 95 and want to achieve an 80 average overall. What minimum score do you need on the fourth and final quiz? Let x be the score of the final quiz. Then the average for the four quizzes is: (75 + 85 + 95 + x ) / 4 and we want a minimum average of 80, so: (75 + 85 + 95 + x ) / 4 ≥ 80 Solve for x: (75 + 85 + 95 + x ) ≥ 80", null, "4 x ≥ 80", null, "4 - 75 - 85 - 95 x ≥ 65 So for this question, you would need at least a 64 on the final quiz to achieve an overall average of 80. Use this method to solve your own question. Stephen La Rocque." ]	[ null, "http://mathcentral.uregina.ca/images/multiply.gif", null, "http://mathcentral.uregina.ca/images/multiply.gif", null, "http://mathcentral.uregina.ca/images/multiply.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90940756,"math_prob":0.97308207,"size":792,"snap":"2020-24-2020-29","text_gpt3_token_len":253,"char_repetition_ratio":0.12182741,"word_repetition_ratio":0.10869565,"special_character_ratio":0.3699495,"punctuation_ratio":0.097826086,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928106,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T20:20:27Z\",\"WARC-Record-ID\":\"<urn:uuid:bb5e3c46-c833-4eba-9a28-ffc8f5f2acba>\",\"Content-Length\":\"6763\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:52366b8b-24ec-452a-a8a8-3e68ddd0fe12>\",\"WARC-Concurrent-To\":\"<urn:uuid:68d36697-384c-499c-8beb-359bce4ade07>\",\"WARC-IP-Address\":\"142.3.156.43\",\"WARC-Target-URI\":\"http://mathcentral.uregina.ca/QQ/database/QQ.09.06/patricia4.html\",\"WARC-Payload-Digest\":\"sha1:PCTKN6RAO2HUIOSUKJL4UBRSPLR7BTH7\",\"WARC-Block-Digest\":\"sha1:F7BPY6A35J4SJLTPIL2KDFFCRHX6GSAZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882934.6_warc_CC-MAIN-20200703184459-20200703214459-00285.warc.gz\"}"}
https://education.blurtit.com/181877/2x59	[ "# 2x+5=9?\n\n2x+5=9 2x=9-5=4 so, x=2\nthanked the writer.\nRULES TO SOLVE SUCH TYPE OF PROBLEMS IS :-\n\nfirst takes all the \"x\" variable on one side and all the single digit{constants} on\nthe other, remember when shifting the numbers , their signs should get change...\n\n2x + 5 = 9\n2x = 9 - 5\n2x = 4\nx = 4/2", null, "" ]	[ null, "https://cf.blurtitcdn.com/var/avatar/thumb_default_avatar.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7845951,"math_prob":0.99552155,"size":584,"snap":"2021-43-2021-49","text_gpt3_token_len":226,"char_repetition_ratio":0.10344828,"word_repetition_ratio":0.752,"special_character_ratio":0.38869864,"punctuation_ratio":0.120567374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98896503,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T23:25:58Z\",\"WARC-Record-ID\":\"<urn:uuid:a41a90e9-059b-4df7-b5d6-562aa3badb61>\",\"Content-Length\":\"44458\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56778836-ba06-4a23-8cef-373a59f3eb42>\",\"WARC-Concurrent-To\":\"<urn:uuid:dde1aafe-98ac-4cb6-8f5f-176079886811>\",\"WARC-IP-Address\":\"172.67.65.240\",\"WARC-Target-URI\":\"https://education.blurtit.com/181877/2x59\",\"WARC-Payload-Digest\":\"sha1:3M55ALL7KY3JEL6BQX2CDPBQHALFE4QV\",\"WARC-Block-Digest\":\"sha1:BEJ6AVEPXVVL57LELQC55A7RAK6RAZK4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585449.31_warc_CC-MAIN-20211021230549-20211022020549-00513.warc.gz\"}"}
https://philoid.com/question/34703-four-equal-circles-each-of-radius-5-cm-touch-each-other-as-shown-in-the-figure-find-the-area-included-between-them-take-314	[ "", null, "## Book: RS Aggarwal - Mathematics\n\n### Chapter: 18. Area of Circle, Sector and Segment\n\n#### Subject: Maths - Class 10th\n\n##### Q. No. 37 of Exercise 18B\n\nListen NCERT Audio Books - Kitabein Ab Bolengi\n\n37\n##### Four equal circles, each of radius 5 cm, touch each other, as shown in the figure. Find the area included between them. [Take π = 3.14]", null, "Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of the square formed by joining the center of adjacent circles. Therefore, we can say that the side of the square equal to the twice of the radius of circle. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.\n\nGiven radius of each circle = r = 5 cm\n\nCentral angle of each sector formed at corner = θ = 90°\n\nSide of square ABCD = a = 2×r = 2×5 = 10 cm", null, "where R = radius of circle", null, "(putting value of r and θ)", null, "", null, "Area of all 4 quadrants = 25π eqn2\n\nArea of square = side×side = a×a = a2\n\nArea of square = 102 (putting value of side of square)\n\nArea of square = 100 cm2 eqn3\n\nArea of shaded region = Area of square – Area of all 4 quadrants\n\nArea of shaded region = 100 – 25π (from eqn3 and eqn2)\n\n= 100 – (25×3.14) (put π = 3.14)\n\n= 100 – 78.5\n\n= 21.5 cm2\n\nThe area of shaded region is 21.5 cm2.\n\n37\n\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60" ]	[ null, "https://static.philoid.co/philoid-com/assets/images/cc-optimised/jrsa1cc.jpg", null, "https://static.insightsonindia.in/ncertusercontent/solutions/", null, "https://static.insightsonindia.in/ncertusercontent/solutions/", null, "https://static.insightsonindia.in/ncertusercontent/solutions/", null, "https://static.insightsonindia.in/ncertusercontent/solutions/", null, "https://static.insightsonindia.in/ncertusercontent/solutions/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89612645,"math_prob":0.9988864,"size":1948,"snap":"2023-40-2023-50","text_gpt3_token_len":605,"char_repetition_ratio":0.16100822,"word_repetition_ratio":0.031325303,"special_character_ratio":0.31570843,"punctuation_ratio":0.10046729,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99787086,"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\":\"2023-09-23T00:54:30Z\",\"WARC-Record-ID\":\"<urn:uuid:fb3e1ab1-e1b0-446c-8a42-160f3a85708f>\",\"Content-Length\":\"107611\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93fb0b1b-1a3c-473e-af19-66817b343232>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa268999-9c6a-4384-ac18-cd60ac3e2222>\",\"WARC-IP-Address\":\"172.105.252.124\",\"WARC-Target-URI\":\"https://philoid.com/question/34703-four-equal-circles-each-of-radius-5-cm-touch-each-other-as-shown-in-the-figure-find-the-area-included-between-them-take-314\",\"WARC-Payload-Digest\":\"sha1:SFGV3A2UNEEVQXAFPSWKXJB2Y7AMKOCI\",\"WARC-Block-Digest\":\"sha1:NJNQNNJ4S7HIHU2C5Q3UT4KM6FCAVW7Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506429.78_warc_CC-MAIN-20230922234442-20230923024442-00544.warc.gz\"}"}
https://portal.research.lu.se/portal/en/publications/twoview-orthographic-epipolar-geometry(0b2d14df-4a11-4639-887a-de2ed8101e60).html	[ "Two-View Orthographic Epipolar Geometry: Minimal and Optimal Solvers\n\nResearch output: Contribution to journalArticle\n\nAbstract\n\nWe will in this paper present methods and algorithms for estimating two-view geometry based on an orthographic camera model. We use a previously neglected nonlinear criterion on rigidity to estimate the calibrated essential matrix. We give efficient algorithms for estimating it minimally (using only three point correspondences), in a least squares sense (using four or more point correspondences), and optimally with respect to the number of inliers. The inlier-optimal algorithm is based on a three-point solver and gives a fourth-order polynomial time algorithm. These methods can be used as building blocks to robustly find inlier correspondences in the presence of high degrees of outliers. We show experimentally that our methods can be used in many instances, where the orthographic camera model isn’t generally used. A case of special interest is situations with repetitive structures, which give high amounts of outliers in the initial feature point matching.\n\nDetails\n\nAuthors\nOrganisations\nResearch areas and keywords\n\nSubject classification (UKÄ) – MANDATORY\n\n• Computational Mathematics\nOriginal language English 163-173 Journal of Mathematical Imaging and Vision 60 2 Published - 2018 Feb Research Yes" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8232522,"math_prob":0.6390784,"size":1432,"snap":"2019-43-2019-47","text_gpt3_token_len":283,"char_repetition_ratio":0.083333336,"word_repetition_ratio":0.0,"special_character_ratio":0.18784916,"punctuation_ratio":0.058558557,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9519454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T17:18:39Z\",\"WARC-Record-ID\":\"<urn:uuid:92d1b310-b594-4623-99b7-849cfeb25d01>\",\"Content-Length\":\"18292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0adb3ed-33e4-4a6f-818d-70682f79cbaa>\",\"WARC-Concurrent-To\":\"<urn:uuid:c029df59-685a-406d-b20b-c667584cb82c>\",\"WARC-IP-Address\":\"130.235.140.134\",\"WARC-Target-URI\":\"https://portal.research.lu.se/portal/en/publications/twoview-orthographic-epipolar-geometry(0b2d14df-4a11-4639-887a-de2ed8101e60).html\",\"WARC-Payload-Digest\":\"sha1:ZNPNQ2J6BMJSQS26K6F6UOFM7JAWHDP2\",\"WARC-Block-Digest\":\"sha1:XA2DSULUV7C4BAZ66LTIBZCFCIKYKJUX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660067.26_warc_CC-MAIN-20191015155056-20191015182556-00221.warc.gz\"}"}
https://lovethedax.com/qa/question-what-is-difference-between-speed-and-velocity.html	[ "", null, "# Question: What Is Difference Between Speed And Velocity?\n\n## How can speed be calculated?\n\nThe formula for speed is speed = distance ÷ time.\n\nTo work out what the units are for speed, you need to know the units for distance and time.\n\nIn this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s)..\n\n## What are the three differences between speed and velocity?\n\nVelocity: Velocity is a physical vector quantity. It has a magnitude as well as direction….Speed & Velocity.SpeedVelocitySpeed is a scalar quantityVelocity is a vector quantity.Speed ascertains how fast a body moves.Velocity ascertains the object’s speed and the direction it takes while moving.2 more rows\n\n## What is the SI unit of speed and velocity?\n\nmeter per secondThe SI unit of speed and velocity is the ratio of two — the meter per second .\n\n## How do you find speed and velocity?\n\nVelocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.\n\n## What is the relationship between velocity and acceleration?\n\nAcceleration is the rate of change of velocity. (when velocity changes -> acceleration exists) If an object is changing its velocity, i.e. changing its speed or changing its direction, then it is said to be accelerating. Acceleration = Velocity / Time (Acceleration)\n\n## What is an example of speed and velocity?\n\nSpeed is how fast something moves. Velocity is speed with a direction. Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed….Speed and Velocity.SpeedVelocityHas:magnitudemagnitude and directionExample:60 km/h60 km/h NorthExample:5 m/s5 m/s upwards\n\n## Can velocity be negative?\n\nVelocity is a vector quantity. If we’re moving along a line, positive velocity means we’re moving in one direction, and negative velocity means we’re moving in the other direction. Speed is the magnitude of the velocity vector, and hence is always positive.\n\n## What is velocity in physics class 9?\n\nVelocity: Velocity is the speed of an object moving in a definite direction. The SI unit of velocity is also metre per second. Velocity is a vector quantity; it has both magnitude and direction.\n\n## What is the definition of speed velocity and acceleration?\n\nAverage speed is distance divided by time. Velocity is speed in a given direction. Acceleration is change in velocity divided by time. Movement can be shown in distance-time and velocity-time graphs.\n\n## Can speed be negative in physics?\n\nThe ratio of distance travelled and the time taken by a body can be zero but not negative. Since distance and time are positive quantities and speed is obtained by the ratio of these two quantities, speed cannot be negative.\n\n## What is the correct SI unit of velocity?\n\nmeter per secondThe SI unit of velocity is meter per second (m/s).\n\n## What is the difference between speed and velocity Class 9?\n\nSpeed is the rate at which an object covers distance in a particular time. … Velocity is the displacement traveled by a body per unit time.\n\n## Which is faster speed or velocity?\n\nSpeed is the magnitude of velocity. Velocity is the speed of an object plus its direction. Speed is called a scalar quantity and velocity is a vector quantity. The fastest possible speed in the universe is the speed of light.\n\n## What is the main difference between speed and velocity quizlet?\n\nWhat is the difference between speed and velocity? Speed is just how fast you travel but velocity is the speed in a given direction.\n\n## What are the similarities and differences between speed and velocity?\n\nSpeed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a vector quantity; it is direction-aware.\n\n## What is SI unit of time?\n\nThe second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency Cs. , the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, to be 9 192 631 770 when expressed in the unit Hz, which is equal to s–1.\n\n## What is definition of velocity?\n\nThe velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. … Velocity is a physical vector quantity; both magnitude and direction are needed to define it.\n\n## Which is an example of speed?\n\nAn example of speed is a car being driven 45 miles per hour. An example of speed is someone cleaning a room in 10 minutes. An example of speed is how quickly a jaguar runs. Speed is defined as to help someone or something along, or move too quickly.\n\n## What is the difference between speed and velocity Brainly?\n\nSpeed measures an object’s change in position, while velocity measures an object’s change in position per unit time.\n\n## What is acceleration and its SI unit?\n\nIn physics or physical science, acceleration (symbol: a) is defined as the rate of change (or derivative with respect to time) of velocity. It is thus a vector quantity with dimension length/time². In SI units, acceleration is measured in meters/second² using an accelerometer." ]	[ null, "https://mc.yandex.ru/watch/67749877", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92949843,"math_prob":0.9749104,"size":6152,"snap":"2021-04-2021-17","text_gpt3_token_len":1314,"char_repetition_ratio":0.21877033,"word_repetition_ratio":0.20207743,"special_character_ratio":0.21407673,"punctuation_ratio":0.111853085,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982268,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T10:20:21Z\",\"WARC-Record-ID\":\"<urn:uuid:9f9bf61f-b116-4f52-bbc4-bb5138dff906>\",\"Content-Length\":\"41981\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b05630e-753e-4bd4-8304-cc0fbc8f42d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:b023a7c5-d41a-4f1e-9435-230434d8887f>\",\"WARC-IP-Address\":\"193.176.77.119\",\"WARC-Target-URI\":\"https://lovethedax.com/qa/question-what-is-difference-between-speed-and-velocity.html\",\"WARC-Payload-Digest\":\"sha1:UIUBDZARKIPJTYX67QX7WGXYVK3JD5JI\",\"WARC-Block-Digest\":\"sha1:PLYADEXY47I3YSCPIYIUYNBL7IJ54MKF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038061820.19_warc_CC-MAIN-20210411085610-20210411115610-00253.warc.gz\"}"}
https://www.shaalaa.com/question-bank-solutions/distinguish-between-percentage-method-counting-elasticity-demand-geometric-method-counting-elasticity-demand-supply-curve-schedule_51762	[ "Share\n\n# Distinguish Between :Percentage Method of Counting Elasticity of Demand and Geometric Method of Counting Elasticity of Demand. - Economics\n\nConceptSupply Curve and Schedule\n\n#### Question\n\nDistinguish between :\n\nPercentage method of counting elasticity of demand and Geometric method of counting elasticity of demand.\n\n#### Solution\n\n Basis of Difference Percentage method of counting elasticity of demand Geometric method of counting elasticity of demand Method $E_d = \\frac{\\text { Percentage change in demand}}{\\text { Percentage change in price}}$ $E_d = \\frac{\\text { Lower segment of the demand curve}}{\\text { Upper segment of the demand curve}}$ Also known as Arithmetic method Point elasticity method\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\n2012-2013 (March) (with solutions)\nQuestion 3.1.3 \| 2.00 marks\n\n#### Video TutorialsVIEW ALL \n\nSolution Distinguish Between :Percentage Method of Counting Elasticity of Demand and Geometric Method of Counting Elasticity of Demand. Concept: Supply Curve and Schedule.\nS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7880363,"math_prob":0.45660216,"size":766,"snap":"2020-10-2020-16","text_gpt3_token_len":163,"char_repetition_ratio":0.17585301,"word_repetition_ratio":0.11111111,"special_character_ratio":0.21018277,"punctuation_ratio":0.06140351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99783206,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T18:28:12Z\",\"WARC-Record-ID\":\"<urn:uuid:4f594cda-d936-4299-806f-d96bc3fdd9e9>\",\"Content-Length\":\"59136\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a34d7620-2292-4217-8bcd-c6736411a82e>\",\"WARC-Concurrent-To\":\"<urn:uuid:fdccff1e-728c-42fb-b946-0d8cad418b59>\",\"WARC-IP-Address\":\"172.105.37.75\",\"WARC-Target-URI\":\"https://www.shaalaa.com/question-bank-solutions/distinguish-between-percentage-method-counting-elasticity-demand-geometric-method-counting-elasticity-demand-supply-curve-schedule_51762\",\"WARC-Payload-Digest\":\"sha1:YFRA7S2ZDN23MQK6M54CMMGXF3PMUJWN\",\"WARC-Block-Digest\":\"sha1:HB6WN5EFVSBLVFDEH5DNHCULLJ7PWQKF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370503664.38_warc_CC-MAIN-20200331181930-20200331211930-00236.warc.gz\"}"}
https://logic.pdmi.ras.ru/seminars/complexity-seminar/2016-10-17	[ "## Понедельник 17.10. Krzysztof Sornat: \"Approximation and Parameterized Complexity of Minimax Approval Voting\"\n\nПонедельник, 17 октября, ауд. 106. Начало в 14:00.\n\nДокладчик: Krzysztof Sornat (University of Wroclaw).\n\nТема: Approximation and Parameterized Complexity of Minimax Approval Voting.\n\n### Abstract\n\nWe present three results on the complexity of Minimax Approval Voting that is a voting rule for choosing committee of fixed size k. Here, we see the votes and the choice as", null, "strings of length", null, "(characteristic vectors of the subsets). The goal is to minimize the maximum Hamming distance to a vote. First, we study Minimax Approval Voting parameterized by the Hamming distance", null, "from the solution to the votes. We show Minimax Approval Voting admits no algorithm running in time", null, ", unless the Exponential Time Hypothesis (ETH) fails. This means that the", null, "algorithm of Misra et al. [AAMAS 2015] is essentially optimal. Motivated by this, we then show a parameterized approximation scheme, running in time", null, ", which is essentially tight assuming ETH. Finally, we get a new polynomial-time randomized approximation scheme for Minimax Approval Voting, which runs in time", null, ", almost matching the running time of the fastest known PTAS for Closest String due to Ma and Sun [SIAM J. Comp. 2009].\n\nAuthors: Marek Cygan, Łukasz Kowalik, Arkadiusz Socała, Krzysztof Sornat\nhttp://arxiv.org/abs/1607.07906" ]	[ null, "https://logic.pdmi.ras.ru/sites/default/files/tex/3579696f06fe2e58f62d997256e0c9d669e94672.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/05b806fb3a39e75606f12e544fec1ae578c32b0b.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/f5efd7e164b692a1c06faa4659d1c840d8761de8.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/030cff1e4bd7c8031b93bcea4553dd5b04893ab2.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/460256232f10d4e18a7a2d96d1a24886ee755bbe.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/390a3b919015d40c7099e8debeafa0c16f637f11.png", null, "https://logic.pdmi.ras.ru/sites/default/files/tex/def26e76d55c16542a2900a14ec52d944e377953.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80214745,"math_prob":0.8655567,"size":1272,"snap":"2019-13-2019-22","text_gpt3_token_len":331,"char_repetition_ratio":0.11750789,"word_repetition_ratio":0.0,"special_character_ratio":0.21462265,"punctuation_ratio":0.15611814,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95883036,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,6,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-21T15:39:20Z\",\"WARC-Record-ID\":\"<urn:uuid:1cf3d6ce-17da-4807-8e14-4a0394a7b2dd>\",\"Content-Length\":\"10316\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c10be76-3775-4bd1-a061-2a3196d2092a>\",\"WARC-Concurrent-To\":\"<urn:uuid:dae84b12-64e6-42c3-a506-4ee30042fec3>\",\"WARC-IP-Address\":\"83.149.197.121\",\"WARC-Target-URI\":\"https://logic.pdmi.ras.ru/seminars/complexity-seminar/2016-10-17\",\"WARC-Payload-Digest\":\"sha1:LHGWFCYEXBCMMNZDT2SXXSQGZRIFL66K\",\"WARC-Block-Digest\":\"sha1:SPY7D6FZNSXLBOPBFHGX5CG4MBR2M27O\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202526.24_warc_CC-MAIN-20190321152638-20190321174638-00206.warc.gz\"}"}
https://www.geogebra.org/m/ex29ay9z	[ "# Baby-Step Modeling in 3D GC with AR: Part 1\n\n## MATH Ts and Ss:\n\nIf you're studying classes of functions and function transformations in 2D, YOU CAN build and model in 3D! In 2D, consider the equation . It's graph is a circle with center (0,0) and radius = 2. To see why, check out this resource here. Yet in this equation, if we replace y with z, this equation becomes . If we solve explicitly for z, we get (upper semicircle) and (lower semicircle). In 3D, think of z as the new DEPENDENT VARIABLE. If we graph these 2 surfaces in 3D, the value of y doesn't matter. Thus, these semicircles become infinitely long half-cylinders (see screencast below). Here, if we study the relationship of z with respect to x (only) -- while ignoring y --, we have cross sections of planes (that are parallel to the xAxis) that are circles (formed from the upper and lower semicircles put together). Now if we add \"0.3y\" to both surfaces, what happens? Why do we get a slide?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8950579,"math_prob":0.94124997,"size":1045,"snap":"2021-31-2021-39","text_gpt3_token_len":280,"char_repetition_ratio":0.10951009,"word_repetition_ratio":0.0,"special_character_ratio":0.25358853,"punctuation_ratio":0.1559633,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99699974,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T18:40:26Z\",\"WARC-Record-ID\":\"<urn:uuid:605d6b51-66e5-46ce-b45e-f6997ce6e211>\",\"Content-Length\":\"40803\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dfa0d9cc-17af-4649-b32e-bd07081b5794>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b818f1a-6f42-4e2c-a6d0-2fb0d3d4b484>\",\"WARC-IP-Address\":\"52.85.144.70\",\"WARC-Target-URI\":\"https://www.geogebra.org/m/ex29ay9z\",\"WARC-Payload-Digest\":\"sha1:BHN42NE2DZO535CKVNUGZE5QTS543Z2W\",\"WARC-Block-Digest\":\"sha1:63YVNC7BFYBAB2ET22OQJQ6EXK6LC6E3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.63_warc_CC-MAIN-20210801154943-20210801184943-00389.warc.gz\"}"}
https://brainmass.com/math/functional-analysis/polar-coordinates-43494	[ "Explore BrainMass\n\n# Polar coordinates\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nX and y represents rectangular coordinates. What is the given equation using polar coordinates (r, theta). x^2 = 4y\n\nhttps://brainmass.com/math/functional-analysis/polar-coordinates-43494\n\n#### Solution Preview\n\nusing x = r cos (theta) y = r sin( theta)\n\nwe obtain r^2 cos^2(theta) = 4 r ...\n\n#### Solution Summary\n\nThis shows how to translate an equation into polar coordinates.\n\n\\$2.49" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7788177,"math_prob":0.95611024,"size":643,"snap":"2021-31-2021-39","text_gpt3_token_len":168,"char_repetition_ratio":0.14710484,"word_repetition_ratio":0.0,"special_character_ratio":0.25660965,"punctuation_ratio":0.16,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972738,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T07:21:24Z\",\"WARC-Record-ID\":\"<urn:uuid:e07f17f2-868e-4e45-ad87-5b196ec32926>\",\"Content-Length\":\"330377\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86770a59-c346-4d2f-8c2c-87c29509b7a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f280104a-2e9a-4b2b-adf9-d6c025992f73>\",\"WARC-IP-Address\":\"104.26.5.245\",\"WARC-Target-URI\":\"https://brainmass.com/math/functional-analysis/polar-coordinates-43494\",\"WARC-Payload-Digest\":\"sha1:OS2ATV442A3V47J7FUC4S73QTMCBOWTI\",\"WARC-Block-Digest\":\"sha1:A5HKCPGOGIHWODWVWSPEFISNSYWGTCDV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154796.71_warc_CC-MAIN-20210804045226-20210804075226-00193.warc.gz\"}"}
https://veronak.savingadvice.com/2013/08/	[ "User Real IP - 34.204.200.74\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 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39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n)\n```\nArchive for August, 2013: Amber's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: August, 2013", null, "", null, "", null, "# Archive for August, 2013\n\n## I'm back! Again ;)\n\nAugust 25th, 2013 at 10:24 pm\n\nHi all!\nI am officially back, if I'm not mistaken my last post was the start of my summer classes in May , which by the way was super tough.\n\nIn the midst of all that I manage to add a little funds to CC, I had previously paid off but it's under control and should be paid off again in September", null, "A lot has happened, first I graduated! Yay I'm so excited. I now have a masters degree, the first and only in my family. Was it tough? Absolutely. Am I glad I stuck it out? Yes\nThe day of graduation I received a \\$15,000 raise, sweet. Unfortunately the money now must be paid on student loans.\n\nAlso some more great news, I've started a Human Resource firm. Basically the company is geared toward small businesses. I'd like to the point of contact for these businesses as their human resource department but out sourced. I have my first company, a construction business started by the owner of the company I am currently employed. I'll be drafting up the contract Wednesday for her to sign. It's great for me because construction is a tad bit different with all the OSHA requirements so this will help with me gaining more experience and knowledge\n\nPlease like my page on Facebook if you can, and share a few of my posts. The goal is to hit 100 likes by the end of the month and get a little traffic on the page. Name of the company is KHRServices\n\nNow that school is out of the way and I've started the new business I can start focusing on paying down the rest of my debt.\n\nI have been keeping up with my 52 week challenge.", null, "One thing though I have about \\$7000 in my savings which is about 6 months worth of expenses. I wonder if I should stop socking away the \\$25 a week and start paying that on the CC? Any thoughts on this?\n\nGlad to be officially back and don't forget to like my new business ( and share) on Facebook\n\nKHRServices" ]	[ null, "https://www.savingadvice.com/blogs/images/search/top_left.php", null, "https://www.savingadvice.com/blogs/images/search/top_right.php", null, "https://www.savingadvice.com/blogs/images/search/bottom_left.php", null, "https://www.savingadvice.com/forums/core/images/smilies/smile.png", null, "https://www.savingadvice.com/forums/core/images/smilies/smile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9783472,"math_prob":0.9955059,"size":1869,"snap":"2019-51-2020-05","text_gpt3_token_len":433,"char_repetition_ratio":0.09383378,"word_repetition_ratio":0.0,"special_character_ratio":0.23434992,"punctuation_ratio":0.09113924,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996884,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T13:47:21Z\",\"WARC-Record-ID\":\"<urn:uuid:5a3774bd-3212-46b9-b454-9739349f21e0>\",\"Content-Length\":\"70445\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6a50c7b-24cb-4adf-85c5-d038a5f36747>\",\"WARC-Concurrent-To\":\"<urn:uuid:f71d183e-a09b-4d98-8075-8886926bb151>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://veronak.savingadvice.com/2013/08/\",\"WARC-Payload-Digest\":\"sha1:BPN5YPZHRFAFPFZYPGDGXITOVSLJCDYA\",\"WARC-Block-Digest\":\"sha1:J4HRCWNMPBAD6KFVRF7AYN5G2C7CKGML\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250620381.59_warc_CC-MAIN-20200124130719-20200124155719-00480.warc.gz\"}"}
http://alameencollege.org/Al-ameen_departments_backend.php?id=9	[ "# Department of Mathematics\n\n## Department Of Mathematics\n\nThe Department of Mathematics was established in the year 1995, as a complementary subject for B.Sc Petrochemicals and B.Sc. Physics courses. In 2013 M.Sc. Mathematics has been started.\n\nOur vision is to develop a love and appreciation for the discipline as an intellectual endeavor. To foster a Solid competency in Mathematics. To create an awareness of the intrinsic relation of mathematics with other technical fields. To set a foundation for further studies and a career in mathematically related fields.\n\nOur Mission is to familiarize students with basic mathematical tools. To generate mathematical concepts and ways of thinking.To highlight the importance of mathematical thinking. To develop communication skills appropriate to the science of mathematics.\n\nGraduates Of The Mathematics Program Will Be Able To:\n\nApply Knowledge Of Mathematics, In All The Fields Of Learning Including Higher Research And Its Extensions.\n\nInnovate, Invent And Solve Complex Mathematical Problems Using The Knowledge Of Pure And Applied Mathematics\n\nTo Solve One Dimensional Wave And Heat Equations Employing The Methods In Partial Differential Equations.\n\nUtilize Number Theory In The Field Of Cryptography That Helps In Hiding Information And Maintaining Secrecy In Military Information Transmission, Computer Password And Electronic Commerce.\n\nFacilitate In The Study Of Crystallographic Groups In Chemistry And Lie Symmetry Groups In Physics.\n\nDemonstrate Risk Assessment In Financial Markets, Disease Spread In Biology And Punnett Squares In Ecology.\n\nLearn To Solve Improper Integrals.\n\nMake Use Of Linear Equations For Solving Any Differential Equations Understand Various Problems Related With Planar Graphs.\n\nUnderstand The Concepts Of Matrices And Linear Equations\n\nLearn Properties Of Inverse Laplace Transforms\n\nDemonstrate A Competence In Formulating, Analysing, And Solving Problems In Several Core Areas Of Mathematics At A Detailed Level, Including Analysis, Linear And Abstract Algebra, Statistics, Probability And Applications Of Mathematics\n\nDemonstrate An Advanced Knowledge And Fundamental Understanding Of A Number Of Specialist Mathematical Topics, Including The Ability To Solve Problems Related To Those Topics Using Appropriate Tools And Techniques\n\nCommunicate Clearly In Writing And Orally Knowledge, Ideas And Conclusions About Mathematics, Including Formulating Complex Mathematical Arguments, Using Abstract Mathematical Thinking, Synthesising Intuition About Mathematical Ideas And Their Applications\n\nDemonstrate That They Can Advance Their Own Knowledge And Understanding Of Mathematics And Its Applications With A High Degree Of Autonomy.\n\nMSc Mathematics Course Mainly Focus On The Thrust Areas Of Mathematics Like Linear Algebra, Discrete Mathematics, Combinatorics, Number Theory, Analysis, Differential Equations, Linear Programming And OR, Topology Etc. And Upon Successful Completion Of This Courses Students Will Be Able To:\n\nSolve Systems Of Linear Equations,\n\nRecognize The Concepts Of The Terms Span, Linear Independence, Basis, And Dimension, And Apply These Concepts To Various Vector Spaces And Subspaces,\n\nUse Matrix Algebra And The Related Matrices To Linear Transformations,\n\nUse Technological Tools Such As Computer Algebra Systems Or Graphing Calculators For Visualization And Calculation Of Linear Algebra Concepts.\n\nFormulate And Interpret Statements Presented In Boolean Logic. Reformulate Statements From Common Language To Formal Logic. Apply Truth Tables And The Rules Of Propositional And Predicate Calculus,\n\nDemonstrate A Working Knowledge Of Set Notation And Elementary Set Theory, Recognize The Connection Between Set Operations And Logic, Prove Elementary Results Involving Sets, And Explain Russells Paradox,\n\nGain An Historical Perspective Of The Development Of Modern Discrete Mathematics\n\nApply Diverse Counting Strategies To Solve Varied Problems Involving Strings, Combinations, Distributions, And Partitions,\n\nWrite And Analyze Combinatorial, Algebraic, Inductive, And Formal Proofs Of Combinatoric Identities, And\n\nRecognize Properties Of Graphs Such As Distinctive Circuits Or Trees.\n\nDefine And Interpret The Concepts Of Divisibility, Congruence, Greatest Common Divisor, Prime, And Prime-Factorization,\n\nApply The Law Of Quadratic Reciprocity And Other Methods To Classify Numbers As Primitive Roots, Quadratic Residues, And Quadratic Non-Residues,\n\nDetermine The Continuity, Differentiability, And Integrability Of Functions Defined On Subsets Of The Real Line,\n\nApply The Mean Value Theorem And The Fundamental Theorem Of Calculus To Problems In The Context Of Real Analysis, And\n\nWrite Solutions To Problems And Proofs Of Theorems That Meet Rigorous Standards Based On Content, Organization And Coherence, Argument And Support, And Style And Mechanics.\n\nSolve Differential Equations Of First Order Using Graphical, Numerical, And Analytical Methods,\n\nSolve And Apply Linear Differential Equations Of Second Order (And Higher),\n\nSolve Linear Differential Equations Using The Laplace Transform Technique,\n\nDevelop The Ability To Apply Differential Equations To Significant Applied And/Or Theoretical Problems.\n\nAssess Properties Implied By The Definitions Of Groups And Rings,\n\nUse Various Canonical Types Of Groups (Including Cyclic Groups And Groups Of Permutations) And Canonical Types Of Rings (Including Polynomial Rings And Modular Rings),\n\nProduce Rigorous Proofs Of Propositions Arising In The Context Of Abstract Algebra\n\nFormulate And Model A Linear Programming Problem From A Word Problem And Solve Them Graphically In 2 And 3 Dimensions, While Employing Some Convex Analysis,\n\nBe Able To Modify A Primal Problem, And Use The Fundamental Insight Of Linear Programming To Identify The New Solution, Or Use The Dual Simplex Method To Restore Feasibility.\n\nDefine And Illustrate The Concept Of Topological Spaces And Continuous Functions,\n\nProve A Selection Of Theorems Concerning Topological Spaces, Continuous Functions, Product Topologies, And Quotient Topologies\n\nRepresent Complex Numbers Algebraically And Geometrically,\n\nApply The Concept And Consequences Of Analyticity And The Cauchy-Riemann Equations And Of Results On Harmonic And Entire Functions Including The Fundamental Theorem Of Algebra,\n\nAnalyze Sequences And Series Of Analytic Functions And Types Of Convergence,\n\n• M.Sc Mathematics\n\n• Complementary Courses\n\n• Latex\n\n•Well Occupied Library\n\n•Provided With The Journal Of Ramanujan Mathematical Society\n\n•Csir Net/Jrf Coaching For Final Year M.Sc Students\n\nName Designation\nDr. Vinitha T Assistant Professor & HOD\nRilga K.O Assistant Professor on Contract\nSulfith Salim Assistant Professor on Contract\nShanthy Jose Assistant Professor on Contract\nSonia George Assistant Professor on Contract\nAnju Jose Assistant Professor on Contract\nSONI SEBASTIAN T Assistant Professor on Contract\n\nView More >>" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66842115,"math_prob":0.7091562,"size":6838,"snap":"2019-51-2020-05","text_gpt3_token_len":1338,"char_repetition_ratio":0.13227978,"word_repetition_ratio":0.002118644,"special_character_ratio":0.16145071,"punctuation_ratio":0.10570627,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9716541,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T03:18:23Z\",\"WARC-Record-ID\":\"<urn:uuid:711d7fb7-74bc-4c17-b457-c3e699e55118>\",\"Content-Length\":\"83991\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34631b17-8ee8-4f19-b7b4-48e2c46bbc26>\",\"WARC-Concurrent-To\":\"<urn:uuid:68d85a61-1903-480c-a729-e2fad02f53c1>\",\"WARC-IP-Address\":\"166.62.27.186\",\"WARC-Target-URI\":\"http://alameencollege.org/Al-ameen_departments_backend.php?id=9\",\"WARC-Payload-Digest\":\"sha1:YMR776FZN5JWHPIWDS5DXONOGYOFV2YV\",\"WARC-Block-Digest\":\"sha1:WRZFKR2WARG6VA23YHHLFFWHONCQPUX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251694176.67_warc_CC-MAIN-20200127020458-20200127050458-00365.warc.gz\"}"}
https://discourse.dhall-lang.org/t/are-there-type-variables-generic-types/348	[ "", null, "# Are there type variables / generic types?\n\n#1\n\nHi, just starting with Dhall. I will like to do something like this:\n\n``````let (Thing a) : Type =\n{ output : a\n, a : Text\n, b : Text\n}\n\nlet stringThings : List (Thing Text) =\n[ {output = \"out\", a= \"a\", b = \"b\"}\n, {output = \"out\", a= \"a\", b = \"b\"}\n]\n\nlet intThings : List (Thing Integer) =\n[ {output = 1, a= \"a\", b = \"b\"}\n, {output = 1, a= \"a\", b = \"b\"}\n]\n``````\n\nSo I would like to have a list of types where one or two attributes are generic.\nIs there such a thing in Dhall?\n\nOne way I can think of is to create different concrete types:\n\n``````let TextThing : Type\nlet IntThing: Type\nlet DoubleThing : Type\nlet OptionalTextThing : Type\n...\n``````\n\nBut this will create a lot of repetition.\n\nAnother way is to use a union in the `output`, But then elements in the list could have mixed outputs. I want all elements in a list to have the same output.\n\nAny other way to achieve something like this?\n\n#2\n\nPerhaps you can achieve that with a function to create the type, for example:\n\n``````let Thing = λ(a : Type) → { output : a, a : Text, b : Text }\n\nlet stringThings\n: List (Thing Text)\n= [ { output = \"out\", a = \"a\", b = \"b\" }\n, { output = \"out\", a = \"a\", b = \"b\" }\n]\n\nlet intThings\n: List (Thing Integer)\n= [ { output = +1, a = \"a\", b = \"b\" }, { output = +1, a = \"a\", b = \"b\" } ]\n\nlet Thing/getA = λ(a : Type) → λ(thing : Thing a) → thing.a\n\nin Thing/getA Text { output = \"out\", a = \"a\", b = \"b\" }\n``````\n\n#3\n\nNice, this works really well, thanks." ]	[ null, "https://dhall-discourse-uploads.s3.dualstack.us-west-1.amazonaws.com/original/1X/87f17b6b61493e8404c449e71ccbeb0b44db5fb8.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8294706,"math_prob":0.9234208,"size":862,"snap":"2020-34-2020-40","text_gpt3_token_len":251,"char_repetition_ratio":0.13986014,"word_repetition_ratio":0.13259669,"special_character_ratio":0.32134572,"punctuation_ratio":0.1904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984708,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T02:47:24Z\",\"WARC-Record-ID\":\"<urn:uuid:f68b8f05-8a90-42ba-86c8-3b5622ba1269>\",\"Content-Length\":\"10792\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fa52bab9-d67b-4f45-8f33-44649b55839c>\",\"WARC-Concurrent-To\":\"<urn:uuid:af0b5364-ece9-464b-bf48-ba704ae2b9b4>\",\"WARC-IP-Address\":\"173.255.251.40\",\"WARC-Target-URI\":\"https://discourse.dhall-lang.org/t/are-there-type-variables-generic-types/348\",\"WARC-Payload-Digest\":\"sha1:D6HNX4LWXJ5II2KOMGVM2XVCX4TX7XBX\",\"WARC-Block-Digest\":\"sha1:NUSNJ4LQVH7DMHVHBLGLXGSTOP3MM5H2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198887.3_warc_CC-MAIN-20200921014923-20200921044923-00293.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/1602.07370/	[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Exact solutions for logistic reaction-diffusion in biology\n\n1. Dept. of Mathematics and Statistics, La Trobe University, Victoria, Australia.\n2. Phenomics and Bioinformatics Research Centre, School of Information Technology and Mathematical Sciences, University of South Australia. Bronwyn.H\n###### Abstract\n\nReaction-diffusion equations with a nonlinear source have been widely used to model various systems, with particular application to biology. Here, we provide a solution technique for these types of equations in -dimensions. The nonclassical symmetry method leads to a single relationship between the nonlinear diffusion coefficient and the nonlinear reaction term; the subsequent solutions for the Kirchhoff variable are exponential in time (either growth or decay) and satisfy the linear Helmholtz equation in space. Example solutions are given in two dimensions for particular parameter sets for both quadratic and cubic reaction terms.\n\nKeywords: Nonclassical symmetries, Reaction-diffusion equations, Fisher equation, Fitzhugh-Nagumo equation, KPP equation, Exact solutions.\n\n## 1 Introduction\n\nLogistic reaction-diffusion equations with a nonlinear source are widely used to model many different systems, particularly in biology. One of the earliest appearances of such a model was in the seminal paper by Fisher in 1937 , where he introduced the equation\n\n ∂θ∂t=D∂2θ∂x2+sθ(1−θ). (1)\n\nFisher’s equation originally modelled the frequency in a diploid population, of a new advantageous recessive gene, labelled ‘’. In fact, for a sexually reproducing species, Fisher’s equation follows only if there are three different phenotypes , and whose relative fitness coefficients are in linear progression; otherwise Fisher’s assumptions lead to a cubic rather than a quadratic source term [2, 3]. Fisher’s equation remains the model of choice for many biological problems such as those in population dynamics (where is the population density divided by the carrying capacity of the environment ), and for biological cellular tissue growth (where is the cell population density divided by the steady-state tissue density, see for example ).\n\nSince cell mobility depends on cell density, such models naturally generalise to nonlinear reaction-diffusion with logistic reaction and nonlinear diffusion. Extending the model to three dimensions and including a broader range of source terms, the general form of a logistic reaction-diffusion equation for a density is\n\n θt=∇⋅[D(θ)∇θ]+R(θ) (2)\n\nwhere is the density dependent diffusion coefficient, is a logistic type source term which may be quadratic or cubic, and is the usual gradient operator in three dimensions. Some properties of these equations are shared more generally by equations of Kolmogorov-Piscounov-Petrov type but here, we consider to be either quadratic (as in the traditional Fisher-type model) or cubic (as in the Huxley and Fitzhugh-Nagumo models). The Fitzhugh-Nagumo equation has been used to model a nerve axon potential, the intermediate unstable steady state being the threshold electrical potential that separates the basins of attraction for the stable steady activated state and the quiescent state .\n\nDespite the wide use of equation (2), few exact analytic solutions are known even in the one-dimensional case. We have not been able to find previously published exact solutions for logistic reaction-diffusion when is non-constant. When (i.e. the one-dimensional case), is constant and is quadratic, equation (2) is known as the Fisher, Fisher-Kolmogorov or KPP equation due to the classic papers by these authors [1, 8]. An exact travelling wave solution was first presented by Ablowitz and Zeppetella in 1979 . In two dimensions, explicit and approximate travelling wave solutions have been presented by Brazhnik and Tyson .\n\nWhen , is constant, and is cubic or of higher order Huxley type, Kametaka found a travelling wave solution, while other travelling wave solutions have been found by McKean and Rinzel . Periodic solutions were found by Carpenter and Hastings . Arrigo, et al. , and Clarkson and Mansfield found some exact solutions using the method of nonclassical symmetry analysis and these solutions have been applied to the problem of a new advancing recessive gene [3, 18]. The solutions found using the nonclassical symmetry approach can also be found using the Painlevé approach [19, 20]. Other solutions have been presented by Kawahara and Tanaka , Kudryashov , Chen and Gu and Nikitin and Barannyk . The existence of solutions in the cubic case was investigated by Nagylaki and Conley .\n\nIn this paper, we use the nonclassical symmetry method to present exact analytical solutions to equation (2) when is quadratic with two real roots, cubic with one doubly repeated root, and cubic with three distinct roots. In each case, the nonlinear diffusion coefficient takes on a particular form.\n\nIn Section 2, we describe the solution technique; the solutions are presented in Sections 3, 4 and 5. The application of the solutions to population genetics is discussed in Section 6 and some final remarks are presented in Section 7.\n\n## 2 Nonclassical reduction to the Helmholtz equation\n\nA full Lie point symmetry classification of equation (2) was made by Dorodnitsyn et al. . In classical Lie point symmetry analysis, one seeks transformations that leave the governing equation invariant. In some cases, these transformations may be used to simplify the governing equation, leading to a possible analytic solution. Classically invariant solutions include travelling waves and scale-invariant solutions that may exhibit extinction, single-peak or multi-peak blow-up in finite or infinite time, with unbounded or compact spatial support on , only when the free functions for diffusivity and/or reaction are power-laws, exponentials or of form.\n\nFollowing Ovsiannikov’s general formulation of partial symmetries , the idea of nonclassical symmetries was pioneered by Bluman and Cole . This requires a transformation to leave the system consisting of the governing equation and the invariant surface condition, invariant. This extra requirement can sometimes give rise to transformations that cannot be found by the Lie point method. Nonclassical symmetry methods, also known as Q-conditional symmetries, have been applied to equations belonging to the class (2), and a number of forms of and have been found to admit strictly nonclassical symmetries [16, 17, 30, 31]. The complete nonclassical symmetry classification of equation (2) in two dimensions was given by Goard and Broadbridge . Some of the same nonclassical symmetries readily extend to -dimensions .\n\nBy writing equation (2) in terms of the Kirchhoff variable (see for example )\n\n u=u0+∫θθ0D(θ′) dθ′, (3)\n\nso that a boundary condition corresponds to , we obtain\n\n F(u)∂u∂t=∇2u+Q(u) (4)\n\nwhere and . This equation admits the nonclassical reduction operator \n\n Γ=∂∂t+Au∂∂u (5)\n\nwhenever and are related by\n\n Q(u)=AuF(u)+κu (6)\n\nfor constant. Given this gives by direct integration:\n\nHowever, given , is obtained by solving a differential equation:\n\n D(θ)=dudθ=AuR−κu,\n\nequivalently\n\n D(θ)=A∫θθ0DdθR(θ)−κ∫θθ0D dθ. (7)\n\nEquation (5) is a genuine nonclassical symmetry because it leaves equation (2) invariant only if one also makes use of the invariant surface condition, . However this conditional invariance allows a consistent reduction of the original PDE, to a differential equation among the invariants of the symmetry. Making use of this nonclassical reduction, equation (4) can be transformed to the linear Helmholtz equation\n\n ∇2Φ+κΦ=0withu=eAtΦ(x). (8)\n\nIn this manner, an arbitrary solution of the linear Helmholtz equation in or 3 dimensions may be used to construct a solution of the nonlinear reaction-diffusion equation. For example, one may use any of the solutions that have previously been constructed for the amplitude of a scattered acoustic wave . However, such solutions that represent scattering by a finite body, must approach the isotropic solution at large distances. Therefore the radial solutions are considered to be canonical, and indicative of features of fields scattered from aspherical boundaries. In the current context, there are three possible types of solution to be considered, namely those with , and .\n\nWith , is a radial solution of the Laplace equation, which can only be a linear combination of the trivial constant solution and the unit point source solution. Any such non-constant solution must be singular at the origin, where there is a steady flux from or to a point source or sink. The differential equation (7) is now linear, with general solution\n\n u=c1exp∫AR(θ)dθ. (9)\n\nWith , in two dimensions must be a linear combination of modified Bessel functions and which must either be infinite at the origin or be unbounded at large . A similar situation pertains in three dimensions when the modified Bessel functions are replaced by spherical modified Bessel functions.\n\nFinally, in the case , there are positive bounded solutions that are Bessel functions in two dimensions and spherical Bessel functions in three dimensions. They satisfy boundary conditions\n\n ur(0,t)=0, (10) u(r1,t)=0, (11)\n\nwhere , being the first zero of the Bessel function. This outer boundary condition may represent extreme total culling of a species at a boundary, for example extreme harvesting of a prey species at the boundary of a protection zone, selective culling of some genotype, or removal of outer tumour cells by radiation or chemo-therapy or diathermy. Alternatively, since varies from 0 at to at , the outer boundary may be relocated to some location where it satisfies a Robin condition for some pre-chosen parameter . Since is simply the radial flux density of the population, this may approximately represent individuals responding to an external chemoattractant with a fixed probability proportional to , except that when the diffusivity is not constant, is not exactly proportional to the population .\n\nFor convenience, from here on we set .\n\nA solution of (7) must be a fixed point of the map\n\n Dn+1(θ)=A∫θ0Dn dθR(θ)−κ∫θ0Dn dθ. (12)\n\nFor some values of the system parameters, this is a contraction map that converges to a unique solution . The modelling constraint may restrict the values of the temporal exponential growth parameter that can occur. These details will depend on the form of the reaction function . We now present some exact analytic solutions in the case where is quadratic or cubic.\n\n## 3 Fisher-type logistic equations, R(θ)=sθ(1−θ)\n\nThe standard generalisation of Fisher’s equation to -dimensions is\n\n ∂θ∂t=D∇2u+sθ(1−θ).\n\nSince cell mobility depends on cell density, Fisher’s equation naturally generalises to nonlinear reaction-diffusion (2) with nonlinear diffusion as well as logistic reaction. Even in one spatial dimension, there are very few known exact solutions, apart from the one-dimensional travelling wave solution with a special non-minimal group velocity . The logistic source changes sign at the carrying capacity, quite different from the positive definite source term of combustion. Despite that major difference, the construction that was previously applied to combustion modelling still can be applied to population dynamics after some restrictions on the system parameters.\n\nFirst consider the case . The diffusivity is given explicitly by\n\n D=−Asθ−2(θ−1−1)A/s.\n\nSince must be positive for , it must be true that . Then as . Since in biological applications the diffusivity must be bounded, the case is inadmissible.\n\nSecondly, consider the case . Since , equation (7) then implies . This can represent only a growing population (by equation (8)). If we presume that has an upper bound, then exponential growth of implies unbounded growth of the population density . However, this is problematic for population modelling since the source term is negative when . The radial solution is a linear combination of modified Bessel functions that have a point source or a singularity at the origin. Alternatively, may diverge at some finite value of so that remains bounded as diverges. However, an unbounded diffusivity is untenable in population modelling.\n\nFinally, we consider the applicable case . It follows from (7) that . Therefore the only valid solutions remaining are those with , so that approaches zero exponentially in time (by equation (8)). Since is now analytic at with leading order , compared to the case of Arrhenius combustion , is no longer negligible compared to at small . From (7) as ,\n\nimplying\n\n A=s−κD(0). (13)\n\nFor example if is the radius of a circular aquatic reserve, outside of which the population of a mobile species is practically zero, one may assume at from which it follows that ( being the first zero of Bessel function ), and (13) gives a condition that guarantees the non-existence of the undesirable solution with , that approaches extinction. This condition may be expressed\n\n r1>λ1√D(0)/s. (14)\n\nFor example, for a mobile species in which an individual’s range expands to km per year and the time scale for uninhibited exponential growth is years, the conservative safe diameter of a marine park would be km.\n\nBy choosing as the first estimate for , the constant value , it follows that all subsequent iterates of the map (12), must have the correct values of at and . The first few iterates are:\n\n D0 = D(0), (15) D1 = \|A\|D0/sθ+K2D0−1, (16) D2 = −A2D0log([sθ+K2D0−s]/[K2D0−s])s2θ(1−θ)+K2AD0log([sθ+K2D0−s]/[K2D0−s]). (17)\n\nThese are shown in Figure 1 for the example with , and . The iterated approximation closely agrees with the numerical approximation obtained by solving (7) with the third/fourth-order Runge-Kutta routine ode45 of Matlab. From to which is well over double the carrying capacity (), the diffusivity is positive-valued and decreasing.", null, "Figure 1: D(θ) for the Fisher source term constructed by Matlab routine ode45 (solid), as well as approximations D0 (solid), D1(θ) (dashed), and D2(θ) (dash-dot) (K=1, s=1 and A=−1.5).\n\nThe exact solution for the doomed population is shown in Figure 2.", null, "Figure 2: Solution for the decreasing population density versus radial coordinate, at times \|A\|t=−1.5,0.0,1.5,2.5, according to the Fisher-type equation with K=1 and s=1.\n\nThe shape of the population curve approaches that of the Kirchhoff variable but from above in this case, since is now a decreasing function of population density, a reasonable model for some populations and for cells whose mobility is decreased by close packing.\n\n## 4 Huxley equation, with R(θ)=sθ2(1−θ)\n\nIn this case, the equation in dimensions with nonlinear diffusivity may be written\n\n θt=∇⋅[D(θ)∇θ]+sθ2(1−θ).\n\nThis better models the frequency of a new advantageous recessive gene in the important case considered by Fisher, that of a Mendelian diploid sexually reproducing population. It also occurs in the Huxley approach to modelling the electrical potential in a nervous system.\n\nWe first consider the case . Equation (7) has general solution\n\n D(θ)=c1θ−1(1−θ)−2(1−θ−1)−(A/s)+1exp(−Asθ)\n\nThis expression is divergent as for , and also divergent as when . As such, the case when is inadmissible for modelling a population.\n\nSecondly, we consider the case when . Since , equation (7) implies that . As described in the previous section, the case when is not of interest in population modelling.\n\nFinally, consider the case when . Since , we find that , so that any valid solutions have , and by equation (8), approaches zero exponentially in time. When is small, , so that by using a Taylor series approximation for , in the limit as ,\n\nimplying .\n\nFor a problem in 2-dimensions, the solution to equation (8) is the Bessel function . Again, if we assume, for example, that is the radius of an aquatic reserve and that outside the reserve the population is practically zero, at , then , where is the first zero of the Bessel function . Within the reserve, and so .\n\nIt would be desirable if we could calculate a minimum radius for the aquatic reserve such that the population inside could be sustained and not become extinct. However, this can only happen if , and since and , this is not possible. As a result, the extinguishing solution will always exist for a population that can be appropriately modelled using the Huxley source term.\n\nThe nonlinear diffusivity can be calculated in the same way as for the Fisher case. By choosing for the first estimate for we find that\n\n D0 = D(0), (18) D1 = −A2K21sθ(1−θ)+A , (19) D2 = (20)\n\nwhere . These iterates for the nonlinear diffusion are shown in Figure 3, together with the solution to equation (7) found using Matlab’s ode45 routine.", null, "Figure 3: D(θ) for the Huxley source term, constructed by Matlabroutine ode45 (solid), as well as approximations D0 (solid), D1(θ) (dashed), and D2(θ) (dash-dot) (K=1, s=1 and A=−1.5). The second iterate, D2(θ), is almost indistinguishable from the numerical solution.\n\nThe exact solution for the population is shown in Figure 4, which shows the population density decreasing over time. In this case, in general the diffusivity is not a monotonic function of density but it has a local maximum. At late times, is small, is increasing for small , and the shape of the solution approaches the Bessel function from below. At early times, the range of densities may allow to be non-monotonic, and the scaled density may lie above the limiting Bessel function.", null, "Figure 4: Solution for the decreasing population density versus radial coordinate, for the Huxley source term, at times \|A\|t=0.0,0.1,0.2,0.3, according to the Huxley-type equation with K=1 and s=1.\n\nIn the population genetics application, this means that a new recessive engineered gene can be completely removed from an isolated population simply by extreme culling at the boundary. In the application to the nerve axon potential, it means that the potential can be set to a uniform value internally by applying that potential at the boundary.\n\n## 5 Fitzhugh-Nagumo equation, with R(θ)=sθ(1−θ)(θ−θ1)\n\nThe cubic reaction term as an appropriate model in biological applications was first introduced in the late 1960’s and early 1970s (see for example ). In this case, the generalisation to higher dimensions and to include nonlinear diffusion is\n\n θt=∇⋅[D(θ)∇θ]+sθ(1−θ)(θ−θ1).\n\nWhereas the Fisher and Huxley equations have a single stable steady state at , the Fitzhugh Nagumo equation is a bi-stable model, with an intermediate unstable steady state between the stable activated state and stable quiescent state . The model is best known in one spatial dimension for propagation of the axon potential. It could be considered in higher dimensions for bi-stable activation of a specific neurological function associated with one region of the nervous system. The model arises also in diploid population genetics in the generic case that three possible genotypes have fitness coefficients that are neither Mendelian nor in arithmetic progression [3, 18]. It could also represent a population that is artificially introduced to a new environment where it must have a threshold density of in order to survive.\n\nFirst, consider the case when . The solution to equation (7) is\n\n D(θ)=c1θ−(1+α/θ1)(θ−1)−1+α/(θ1−1)(θ−θ1)−1−α/(θ1(θ1−1)),\n\nwhere . When , the expression for is divergent as and when , it is divergent at both and . When , we see that is divergent at all three zeros of . As such, the case when is inadmissible when modelling a population.\n\nWhen , we can use either the fact that or to show that we must have in order to have positive. We may then use the same arguments presented in section 3 to conclude that this case is also inadmissible.\n\nFinally, we consider the case . From (7), we see that\n\nexcept in the singular case of the the denominator being zero, . Also . Since , we can deduce that and will approach zero exponentially in time from (8). Also, in the non-singular case,\n\n A=−(sθ1+K2D(0)).\n\nThis may represent an axon potential or a population decaying towards extinction, because of culling at the boundary even if it started above the activation threshold. For a 2-dimensional problem with rotational invariance, the solution to equation (8) is the Bessel function, . Once again, we can deduce the minimum size of an aquatic reserve such that the undesirable extinguishing solution does not exist. If is the radius of the aquatic reserve, outside of which the population is practically zero, then , where is the first zero of the Bessel function. For the population to be maintained, we require . For a population with a carrying capacity at , it must be true that so that the local growth rate is negative at higher populations. Since and , this means that . In that case, the quiescent state (or extinction state) is now the intermediate unstable steady state. It then follows that\n\n r1>λ1√D(0)s\|θ1\|\n\nis the minimum radius for a marine park in order for the extinction point not to be stable. For example and using the same parameters as before, if km per year, years, and , the reserve should be at least km in diameter.\n\nOnce again, an estimate for can be calculated using the iterative map (12). Since the carrying capacity better represents the ecological potential, we use the value as a starting value for the iterative procedure, guaranteeing that all subsequent iterates have the correct value at . The first two iterates are:\n\n D0 = −AK2 D1 = A2K2s1P(θ)\n\nwhere . The form for will depend on the values of the parameters , and . We have the necessary condition that since we must have . If has two real roots, then we also have that either or . In this case, the next iterate, may be written in terms of functions and, as such, it becomes singular for certain values of in the domain of interest.\n\nIf has a repeated root, the next iterate will depend inversely upon a quartic in . This iterate for the diffusivity then also has singularities for certain values of in the domain of interest and as such, we proceed no further.\n\nIf has two complex roots, then we require as well as . In this case, the integral of is an inverse tan function, and the next iterate, , is given by\n\n D2=A3K2 tan−1(1β(θ−\\large12(θ1+1)))+tan−1(12β(θ1+1))βs2θ(1−θ)(θ−θ1)−A2[tan−1(1β(θ−\\large12(θ1+1)))+tan−1(12β(θ1+1))],\n\nwhere . Some values of the parameters produce reasonable functions for , for example, if , , and , the first three iterates are shown in Figure 5, together with the numerical solution to equation (7) found using Matlab’s ode45 routine.", null, "Figure 5: D(θ) for the Fitzhugh-Nagumo source term, constructed by Matlabroutine ode45 (solid), as well as approximations D0 (solid), D1(θ) (dashed), D2(θ) (dash-dot) (K=1, s=1 and A=−1.5).\n\nThe exact solution for the population is plotted in Figure 6, showing the population is doomed to extinction.", null, "Figure 6: Solution for the decreasing population density versus radial coordinate, for the Fitzhugh-Nagumo source term, at times \|A\|t=0.0,0.1,0.2,0.3, according to the Fitzhugh-Nagumo type equation with K=1 and s=1.\n\n## 6 Application to population genetics\n\nLet be the frequency of new advantageous allele , and be the frequency of the original allele . Let be the fitness coefficients of genotypes with zero, one and two copies of respectively. Allele will be said to be partially recessive if . That is, addition of the second copy of gene gives a greater advantage than does the first copy. For example, in the reaction diffusion equations with linear diffusion, calculations in show that the generic source term is that of Fitzhugh-Nagumo-type, with\n\n s=γ11−2γ12+γ22;θ1=12−ν;ν=γ22−γ11γ22−γ12\n\nFor a partially recessive gene , and . For the nonlinear diffusion model with Fitzhugh-Nagumo-type source, the extinguishing solution developed in Section 5 exists only if the radius of the boundary where the new gene is selectively removed, is less than the critical value .\n\nThe case is a special case that gives rise to the Fisher equation with quadratic logistic source term . This arises in the context of diploid sexual population genetics only in the special case that the fitness coefficients are in arithmetic progression, ie . In that case, allele is neither partially recessive nor partially dominant. The Fisher equation arises also as the equation for the frequency of a new advantageous gene of an asexual species . The critical radius for existence of the extinguishing solution in the case of the quadratic logistic source term is .\n\nAnother special case arises when , the case of a totally recessive gene . The phenotype whose expression requires two copies of a new gene, is particularly vulnerable. Now , the value at which the critical radius diverges. The extinguishing solution of the Huxley reaction-diffusion equation (Section 4) always exists, no matter how large is the diameter of the boundary where individuals with the totally recessive new gene are removed.\n\nIf a new, perhaps genetically engineered, advantageous gene is totally recessive or if it is partially recessive within a population that is contained within a circle smaller than that with critical radius, then the frequency of the new gene can be reduced to zero uniformly just by selective culling at the boundary.\n\nIf the new gene is partially recessive and the circular domain of the species has a radius larger than the critical value, then its removal cannot be achieved by actions taken at the boundary alone.\n\n## 7 Discussion and final remarks\n\nReaction-diffusion equations, with a source term that is either quadratic or cubic, are commonly used to model various physical and biological systems. Here we have shown that equations of this type with nonlinear diffusion, are exactly solvable provided the nonlinear diffusion and source terms satisfy a particular relationship. This relationship can be found by first writing the equation in terms of the Kirchhoff variable, and then using the nonclassical symmetry reduction method. Provided that the relationship is satisfied, the equation may be separated into spatial and temporal parts. The solutions for the Kirchhoff variable are then exponential in time (either growth or decay), and satisfy either a Helmholtz, modified Helmholtz or Laplace equation in space.\n\nIn general, one may either specify the nonlinear diffusivity and then calculate the corresponding reaction term which will allow exact solution, or alternatively, the reaction term may be specified and the diffusivity calculated using the aforementioned relationship. In this paper, we are particularly interested in logistic Fisher-type models and cubic Huxley- or Fitzhugh Nagumo-type models, and so we calculate the nonlinear diffusivity after specifying the reaction term. In this case, the diffusivity can be constructed by solving a differential equation that is equivalent to an Abel equation if the Kirchhoff variable satisfies a Helmholtz (or modified Helmholtz) equation, or a separable equation if the Kirchhoff variable satisfies the Laplace equation. In cases where the nonlinear diffusivity cannot be written in closed form, it may be written in terms of a series expansion. It is important to note that, although the diffusivity is required to be a nonlinear function of the dependent variable, for reasonable parameters sets, it is well behaved and slowly varying over the relevant range of the population density.\n\nInterpretation of the obtained solutions, in the application of population genetics, is discussed in Section 6. The extinguishing solution of the Huxley reaction-diffusion equation (Section 4) always exists, no matter how large is the diameter of the boundary where individuals with the totally recessive new gene are removed. In practice, it is more likely that only individuals with the distinctive phenotype expressed from two copies of the recessive gene would be identified and removed at the boundary. This would locally remove a fraction of the new genes from the gene pool. This suggests a nonlinear Robin boundary condition , at , representing a selective removal from the domain of the new gene, in proportion to its frequency. If only the proportion of the new genes are detected, then the radiation constant is proportional to , , which is approximately with constant. The nonlinear radiation boundary condition therefore is\n\n −ur=Hu2; r=r1,\n\nwith (constant). This boundary condition would be of interest for the application to population genetics." ]	[ null, "https://media.arxiv-vanity.com/render-output/3658870/x1.png", null, "https://media.arxiv-vanity.com/render-output/3658870/x2.png", null, "https://media.arxiv-vanity.com/render-output/3658870/x3.png", null, "https://media.arxiv-vanity.com/render-output/3658870/x4.png", null, "https://media.arxiv-vanity.com/render-output/3658870/x5.png", null, "https://media.arxiv-vanity.com/render-output/3658870/x6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88791287,"math_prob":0.9886248,"size":31153,"snap":"2020-34-2020-40","text_gpt3_token_len":6949,"char_repetition_ratio":0.16315131,"word_repetition_ratio":0.040265754,"special_character_ratio":0.22395916,"punctuation_ratio":0.14457418,"nsfw_num_words":3,"has_unicode_error":false,"math_prob_llama3":0.9956901,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T12:56:38Z\",\"WARC-Record-ID\":\"<urn:uuid:c2e46a89-9daa-4fd0-b807-563f4b8e93cf>\",\"Content-Length\":\"616130\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec178dd5-d3cc-48ba-971f-f98b26e2c099>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9b7266a-6fa1-403d-8f83-38a24d6f1701>\",\"WARC-IP-Address\":\"104.28.21.249\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1602.07370/\",\"WARC-Payload-Digest\":\"sha1:O6NISDJADWMIKZKQCYC7J7NIXQEQDG6X\",\"WARC-Block-Digest\":\"sha1:PDPXJRGUVUVQNUYG65D2ZYF3JZHBNKA6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400279782.77_warc_CC-MAIN-20200927121105-20200927151105-00442.warc.gz\"}"}
https://vi-magento.com/library-function-in-qbasic-quick-answer/	[ "Chuyển tới nội dung\nTrang chủ » Library Function In Qbasic? Quick Answer\n\n# Library Function In Qbasic? Quick Answer\n\nAre you looking for an answer to the topic “library function in qbasic“? We answer all your questions at the website https://vi-magento.com in category: Top 794 tips update new. You will find the answer right below.\n\nLibrary function is also known as built in function or routine function. A library function in QBASIC may be a string function for numeric function. All string library functions can manipulate either string or numeric data and return a string value.\n\n## Functions in QBasic \| Library Functions in QBasic \| How to use Library Functions\n\nFunctions in QBasic \| Library Functions in QBasic \| How to use Library Functions\nFunctions in QBasic \| Library Functions in QBasic \| How to use Library Functions\n\nLibrary functions are built-in functions that are grouped together and placed in a common location called library. Each function here performs a specific operation. We can use this library functions to get the pre-defined output. All C standard library functions are declared by using many header files.\n\nLibrary Function in QBASIC\n• User-defined function.\n• Built-in function.\nFunctions in QBASIC are readymade programs, which take some data, manipulate them and return a value, which may be a string or a numeric type.\n\nREVIEW OF LIBRARY FUNCTION.\nFunction name Syntax Purpose\nSTR () STR\\$ (x) To convert a numeric expression to its string expression.\nSQR () SQR (x) It returns square root of any positive number.\nStandard C Library Functions Table, By Name\nFunction System Include File Function Prototype\nabort stdlib.h void abort(void);\nabs stdlib.h int abs(int n);\nacos math.h double acos(double x);\nasctime time.h char asctime(const struct tm time);\n\n## What are the two types of library functions in Qbasic?\n\nLibrary Function in QBASIC\n• User-defined function.\n• Built-in function.\n\nWhat are library functions in QBASIC?\n\nB) BUILD-IN OR LIBRARY FUNCTIONS: –\n\nThe functions which are provided by the QBASIC system and also allow the programmer to use them according to the requirement are known as BUILT-IN or Library Functions. These functions are required to be called by the programmer to use them in a program.\n\nWhat are library functions name two types of library functions in QBASIC?\n\nAnswer: It is a built-in or ready-made program which helps us to perform a certain task such as mathematical, financial, etc. Explanation: A function manipulates data passes to it and returns either a string or a numeric.\n\nHow many types of functions are there in QBASIC?\n\nIn QBasic, there are two types of functions: standard or “built-in” and user-defined. Standard or “built-in” functions are provided by the QBasic system or compiler environment and basically allow the programmer to use them so he/she doesn’t have to write code to handle certain situations.\n\nWhat are the two types of statements in QBASIC?\n\nThere are several different types statements in QBASIC Programming language. For example, CLS statement clears the screen, PRINT statement displays output and INPUT statement takes the input from the users.\n\n## What are the standard library functions in C?\n\nStandard C Library Functions Table, By Name\nFunction System Include File Function Prototype\nabort stdlib.h void abort(void);\nabs stdlib.h int abs(int n);\nacos math.h double acos(double x);\nasctime time.h char asctime(const struct tm time);\n\nWhat are the standard library functions?\n\nLibrary functions include standard input/output (stdio. h), string manipulation (string. h), math functions (math. h), and date and time functions (time.\n\nWhat are standard functions in C language?\n\nThe standard functions are also called as library functions or pre-defined functions. In C when we use standard functions, we must include the respective header file using #include statement. For example, the function printf() is defined in header file stdio. h (Standard Input Output header file).\n\nHow many standard library are there in C?\n\nThe ANSI C standard library consists of 24 C header files which can be included into a programmer’s project with a single directive. Each header file contains one or more function declarations, data type definitions and macros.\n\nWhat is standard library in C language?\n\nThe C standard library or libc is the standard library for the C programming language, as specified in the ISO C standard. Starting from the original ANSI C standard, it was developed at the same time as the C library POSIX specification, which is a superset of it.\n\n## Which is the library function?\n\nLibrary functions are built-in functions that are grouped together and placed in a common location called library. Each function here performs a specific operation. We can use this library functions to get the pre-defined output. All C standard library functions are declared by using many header files.\n\nWhat is library function example?\n\nExample: Square root using sqrt() function\n\nSuppose, you want to find the square root of a number. To compute the square root of a number, you can use the sqrt() library function. The function is defined in the math. h header file.\n\nLibrary functions are pre defined functions. Inorder to ease the work of the user, the developer provides some built in functions. For some operations we need not write our own code. We can simply use one of those library functions to get our work done.\n\nIs Main () a library function?\n\nmain is not a library function. However, the standard also imposes some requirements on its signature, and that it must not be overloaded or declared static or inline. In particular, it must not be used, meaning that you cannot call it.\n\nWhat are the two library functions?\n\nThe library functions to allocate memory is malloc() and calloc() methods. The answer to the question: Specify the 2 library functions to dynamically allocate memory is option C which is malloc() and calloc().\n\n## What are the 3 types of functions available in QBasic?\n\nCOS, SIN, and TAN Function\n\nThe CINT function gives results by rounding up the fractional portion. It supports only -32768 to +32767.\n\nWhat are the types of functions in QBasic?\n\nIn QBasic, there are two types of functions: standard or “built-in” and user-defined. Standard or “built-in” functions are provided by the QBasic system or compiler environment and basically allow the programmer to use them so he/she doesn’t have to write code to handle certain situations.\n\nWhat is QBasic and its function?\n\nQBasic is an integrated development environment (IDE) and interpreter for a variety of dialects of BASIC which are based on QuickBASIC. Code entered into the IDE is compiled to an intermediate representation (IR), and this IR is immediately executed on demand within the IDE.\n\nHow many types of QBasic are there?\n\nIn QBASIC, variables are also of two types: Numeric variable. String variable.\n\nWhat are the two types of library functions in QBasic?\n\nc. Functions provided by QBASIC system are called built-in or Standard or Libraryfunction. d. Val function is used to convert a number in string form into numeric value.\n\nReferences:\n\nReview Of Library Function In QBASIC Class 10 Computer Science …\n\nWorld Class Library Function in Qbasic – Schoolans.com\n\nLIBRARY FUNCTION IN QBASIC.pptx – Course Hero\n\nLibrary Functions in QBASIC-Introduction – PRAJWAL RAI\n\n## Information related to the topic library function in qbasic\n\nHere are the search results of the thread library function in qbasic from Bing. You can read more if you want.\n\nval function in qbasic\n\ncint function in qbasic\n\nqbasic functions list\n\nqbasic subroutines\n\nlibrary function example\n\nleft in qbasic\n\nstring function in qbasic\n\nlearn qbasic\n\nlibrary function in qbasic\n\nYou have just come across an article on the topic library function in qbasic. If you found this article useful, please share it. Thank you very much." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87876123,"math_prob":0.7494675,"size":7404,"snap":"2023-14-2023-23","text_gpt3_token_len":1597,"char_repetition_ratio":0.21851352,"word_repetition_ratio":0.24916667,"special_character_ratio":0.20813075,"punctuation_ratio":0.10885341,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97415036,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T09:40:15Z\",\"WARC-Record-ID\":\"<urn:uuid:17b123b4-720f-4f39-b9f3-d5667e69b4f6>\",\"Content-Length\":\"76146\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee70f637-d5fb-425d-b4ed-d1acaeae0f30>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c50c77b-ea64-4c05-93e1-669b9e772fc1>\",\"WARC-IP-Address\":\"45.252.250.21\",\"WARC-Target-URI\":\"https://vi-magento.com/library-function-in-qbasic-quick-answer/\",\"WARC-Payload-Digest\":\"sha1:ZVYLGASSHYKX44NLOJOFL337TTV6IZUU\",\"WARC-Block-Digest\":\"sha1:U37UMG3HD3KQ2DIWUCLUKNFL6NZHW5TM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943471.24_warc_CC-MAIN-20230320083513-20230320113513-00565.warc.gz\"}"}
https://www.vutbr.cz/en/students/courses/detail/209371?apid=209371	[ "Course detail\n\n# Signals and Systems\n\nThis module provides an introduction to the linear time-invariant continuous- and discrete-time signals and systems. Students are introduced with the various methods of description and analysis of the continuous- and discrete-time signals and systems: time domain, frequency domain, spectrum, Fourier series, sampling, transforms (Laplace, Fourier, Z) and difference or differential equations. These methods are used to analyse signals and systems properties and to determine basic characteristics: linearity, time-invariance, causality, stability, etc.\n\nLearning outcomes of the course unit\n\nAn absolvent is able to:\n- describe continuous and discrete time signals in time and frequency domain,\n- perform continuous and discrete time signal transform using the Fourier series, the Fourier transform, the Laplace transform and the Z-transform,\n- discuss practical interpretations of these transforms and their properties,\n- describe fundamental properties of LTI continuous-time systems,\n- describe fundamental properties of LTI discrete-time systems,\n- use the different methods to describe LTI systems,\n- determine system response of an LTI system to standard and general signals,\n- determine from the description of the LTI system its characteristics such as linearity, time-invariance, causality and stability.\n\nPrerequisites\n\nStudent should have sufficient knowledge of the mathematical analysis for bachelors, mainly: differential and integral calculus, series, basic transforms and complex numbers.\n\n• compulsory prerequisite\n\nCo-requisites\n\nNot applicable.\n\nRecommended optional programme components\n\nNot applicable.\n\nChi-Tsong Chen:System and Signal Analysis,Saunders College publishing, 1994. (EN)\nOPENHEIM, Alan, WILSKY, Alan. Signals and Systems. Second edition. New Jersey: Prentice Hall 1997, 957 s. ISBN 0-13-814757-4. (EN)\nCHI-TSONG CHEN. System and Signal Analysis. Oxford University press 2004. 424 s. ISBN 0-19-515661-7. (EN)\n\nPlanned learning activities and teaching methods\n\nTeaching methods include lectures with demonstrations of practical computations, and pc labs. Students have to finish individual projects during semester.\n\nAssesment methods and criteria linked to learning outcomes\n\n30 points for individual projects\n70 points for final exam\n\nLanguage of instruction\n\nEnglish\n\nWork placements\n\nNot applicable.\n\nCourse curriculum\n\n1. General introduction and motivation; continuous and discrete world, signals classification.\n2. Continuous-time and discrete-time signals - basic operations and manipulations, discretization of continuous-time signals.\n3. Frequency domain of the continuous-time signal, the Fourier series.\n4. The Fourier transform, examples.\n5. Time and frequency domain of the discrete-time signal, the discrete Fourier series, the discrete Fourier transform (DFT).\n6. Systems – definition, classification, the examples of real systems.\n7. Continuous-time LTI system – description using the differential equations, the Laplace transform.\n8. Continuous-time LTI system – transfer function, poles and zeros, stability of LTI systems.\n9. Continuous-time LTI system – response on the standard input signals, the relation to the BIBO stability.\n10. Discrete-time LTI system - description using the difference equations, the Z-transform.\n11. Discrete-time LTI system – transfer function, poles and zeros, stability of LTI systems.\n12. Discrete-time LTI system – response on the standard input signals, the relation to the BIBO stability.\n13. Summary.\n\nAims\n\nTo acquaint with the fundamentals of signals and systems with the continuous and discrete time. To learn to apply the fundamentals to real signals and systems.\n\nSpecification of controlled education, way of implementation and compensation for absences\n\nThe content and forms of instruction in the evaluated course are specified by a regulation issued by the lecturer responsible for the course and updated for every academic year.\n\nClassification of course in study plans\n\n• Programme BPA-ELE Bachelor's\n\nspecialization BPA-PSA , 2. year of study, winter semester, 6 credits, compulsory\n\n#### Type of course unit\n\nLecture\n\n52 hours, optionally\n\nTeacher / Lecturer\n\nExercise in computer lab\n\n13 hours, compulsory\n\nTeacher / Lecturer" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83098996,"math_prob":0.85611725,"size":4148,"snap":"2021-31-2021-39","text_gpt3_token_len":844,"char_repetition_ratio":0.15781853,"word_repetition_ratio":0.07760141,"special_character_ratio":0.19623916,"punctuation_ratio":0.15804598,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9589677,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-29T08:27:19Z\",\"WARC-Record-ID\":\"<urn:uuid:8a93fb85-c2e1-450a-b013-e38c41d0e817>\",\"Content-Length\":\"59292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:148a7aab-07e5-4b28-873d-2f2b3af56d1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc73af39-449d-43a2-a60f-f1a0e0815220>\",\"WARC-IP-Address\":\"147.229.2.90\",\"WARC-Target-URI\":\"https://www.vutbr.cz/en/students/courses/detail/209371?apid=209371\",\"WARC-Payload-Digest\":\"sha1:W4Z3I4P6FMWMLFA67UEDFLMCIF7UYNTM\",\"WARC-Block-Digest\":\"sha1:4ISKVC4CK4LH34AZKYJ2737DPO3TQLAQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153854.42_warc_CC-MAIN-20210729074313-20210729104313-00454.warc.gz\"}"}
https://gamedev.stackexchange.com/questions/53419/animating-a-background-pulse-with-easing-equation	[ "# Animating a background 'pulse' with easing equation\n\nI'm trying to find a suitable easing equation (or other method) to animate an object so that it 'pulses' (imagine a 'spike' on a music visualiser, or see the image I drew badly below)", null, "'v' is the value I'm using to scale. 't1' and 't2' mark the end of a single pulse.\n\nThe game I'm working on is a puzzler, but I'd like to have objects in the background pulse in time with user-selected music.\n\nI'm sure it's just a case of knowing what to search for, but any advice would be appreciated on how best to achieve this.\n\nMy answer is similar to Miro's, but I think the Math ought to be a lot simpler. Of course, the details of your curve make all the difference. If you don't care precisely what the curve looks like, then all you need is the basic sawtooth.\n\nvar clock = function(x) {\nreturn (1-x) - floor(1-x); //I like this method, though not the simplest.\n}\n\nvar clockVal = clock(time);\n\n\nThat yields:", null, "If you want to make it more curvy, raise it to some power:\n\nvar clockVal = Math.pow(clock(time), 2);\nvar clockVal = Math.pow(clock(time), 3);", null, "", null, "• Interim results say this looks good - though in my original drawing there was a small amount of time over which the pulse 'spikes' (rather than right on t = 1). Any ideas? – codinghands Apr 7 '13 at 4:56\n• Another way to give credit to @Miro; he already included that. Change the call to clock to input a longer period like so: var clockVal = clock(time / period); – Seth Battin Apr 7 '13 at 5:30\n\nI assume that you want to create periodic function so you need to periodize x:\n\np(x) = x/T - floor(x/T)\n\n\nThen you'll create rational function from two linear functions.\n\nf(x) = ( ax + b ) / ( cx + d )\n\n\nYou've got 2 points [0,V], [1,0] and together you have:\n\nf(x) = (V - V * p(x)) / (1 + p(x) * shape)", null, "• p(x) = x/T - floor(x/T) : ten times faster with : (x/T) % 1 . How to be sure (cx+ d) !== 0 ? And how do you use the rationnal function ? – GameAlchemist Apr 6 '13 at 20:40\n• @VincentPiel Modulo is usually associated with integers. Some languages may allow n % 1 to retrieve the decimal portion of the number, but it's not necessarily a good idea. And the claim that using the operator is \"ten times faster\" that some other operation is probably untrue. – Seth Battin Apr 6 '13 at 21:34\n• jsperf.com/trunc-or-1/2 floor 92% slower (% 1) is faster, and valid in Javascript. The fact that it doesn't work in other languages doesn't seem like a good reason not to use it in Javascript. How to be sure (cx+ d) !== 0 ? And how do you use the rationnal function ? – GameAlchemist Apr 6 '13 at 21:40\n• Wow. Well, congratulations on writing proving yourself correct. – Seth Battin Apr 6 '13 at 21:51\n• Hi Miro - thanks for your answer. As Vincent said: \"And how do you use the rational function?\" – codinghands Apr 7 '13 at 4:55" ]	[ null, "https://i.stack.imgur.com/D2l1N.png", null, "https://i.stack.imgur.com/KuTGh.png", null, "https://i.stack.imgur.com/COWKf.png", null, "https://i.stack.imgur.com/B4Qst.png", null, "https://i.stack.imgur.com/fAb9r.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9600001,"math_prob":0.9240426,"size":513,"snap":"2021-04-2021-17","text_gpt3_token_len":134,"char_repetition_ratio":0.08840864,"word_repetition_ratio":0.0,"special_character_ratio":0.24951267,"punctuation_ratio":0.06542056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96444654,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T06:07:12Z\",\"WARC-Record-ID\":\"<urn:uuid:1b5a471c-32fc-487f-aeeb-f1ab240fd642>\",\"Content-Length\":\"164148\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49171afb-2411-47da-bfdc-10a9738c11ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3815b6f-0310-4801-aa38-f3e5865f6ecf>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://gamedev.stackexchange.com/questions/53419/animating-a-background-pulse-with-easing-equation\",\"WARC-Payload-Digest\":\"sha1:42XE5JJWQHMTRGCZMZ4VB7A64IJJG5MU\",\"WARC-Block-Digest\":\"sha1:EUQOXIY5QRTG2YMQQEHCXX2SMKVOELEV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517966.39_warc_CC-MAIN-20210119042046-20210119072046-00320.warc.gz\"}"}
https://www.vatsalsaglani.dev/leetcode/leetcode-167	[ "# Leetcode - 167: Two Sum II\n\n## The Two sum II problem is the extension of the Two Sum problem which can be solved using the pointers technique\n\nMEDIUM\n–––\n\nAs seen in the question Two Sum in Python using dictionary given a list of `integers` and a `target` integer the algorithm needs to return the indices of two numbers which can be added to get the target. Here the only change is the given list is sorted in ascending order which can be used as an advantage.\n\n## Two Sum II\n\n### Question\n\nGiven an array of integers `numbers` that is already sorted in ascending order, find two numbers such that they add up to a specific `target` number.\n\nReturn the indices of the two numbers (1-indexed) as an integer array `answer` of size `2`, where `1 <= answer < answer <= numbers.length`.\n\nThe bruteforce solution will be same as the one in the previous blog.\n\n### Solution and Explanation\n\ntowsumtwo_onepass.py\n``````def twoSum(numbers, target):\n\nleft_ptr = 0\nright_ptr = len(numbers) - 1\n\nwhile left_ptr < right_ptr:\n\ncurrent_sum = numbers[right_ptr] + numbers[left_ptr]\n\nif current_sum > target:\n\nright_ptr -= 1\n\nelif current_sum < target:\n\nleft_ptr += 1\n\nelse:\n\nreturn [left_ptr + 1, right_ptr + 1]\n``````\n\n#### Explanation\n\nAs the numbers list is sorted we can use the one pass two pointer method wherein we can start with two different pointers one from the start of the list and anoter from the end.\n\nThe logic becomes more simpler. We just need to move the pointers based on the sum of the values of their indices.\n\nIf the sum is higher than the given target than move the right pointer towards the left i.e. `right_ptr -= 1`.\n\nIf the sum is lesser than the target then we need to move the left pointer towards the rigth i.e. `left_ptr += 1`.\n\nThe only condition left is when the sum is equal to the given target and as mentioned in the problem we need to return 1-indexed of the two indices. Hence, we will add 1 to both the `left_ptr` and `right_ptr`.\n\nThat's it for this blog, hope you found this helpful. You can connect with me on Twitter\n\nBuilt with ❤️ using\n\nI am available on\n\nSitemap" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8410859,"math_prob":0.9853875,"size":1940,"snap":"2023-14-2023-23","text_gpt3_token_len":464,"char_repetition_ratio":0.14359504,"word_repetition_ratio":0.0,"special_character_ratio":0.2386598,"punctuation_ratio":0.079155676,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993444,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-26T05:36:33Z\",\"WARC-Record-ID\":\"<urn:uuid:637d069a-c287-40d8-ad17-00f558b9edc4>\",\"Content-Length\":\"28213\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d171ff6-1706-44d2-851b-6d08d2dcf95a>\",\"WARC-Concurrent-To\":\"<urn:uuid:01bbe976-57ae-49ef-8ba9-167646fd0d32>\",\"WARC-IP-Address\":\"76.76.21.241\",\"WARC-Target-URI\":\"https://www.vatsalsaglani.dev/leetcode/leetcode-167\",\"WARC-Payload-Digest\":\"sha1:Y6XFAGV2RZYJZH6G4OFDTWCXMECDKPLV\",\"WARC-Block-Digest\":\"sha1:UNS5YYWI2ZHHGCO45OJ2AL3PYG2S43LS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945433.92_warc_CC-MAIN-20230326044821-20230326074821-00569.warc.gz\"}"}
https://spsstools.net/en/syntax/syntax-index/ranking-largest-values-sorting-grouping/calculate-mode/	[ "``` 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``` ```(Q)In my data set I have 4 categories of 5 questions each I would like to calculate the mode for each category for each subject, i.e., compute modecat1, modecat2, modecat3 and modeca4 for each subject. (A) Posted by rlevesque@videotron.ca to SPSSX-L on 2001/08/28 Note that when the distribution is multi-modal, the syntax selects the mode having the highest value. * This assumes you need the mode in the data editor. If it is sufficient to have the mode in the Output window, use FREQ with the /STATISTICS=mode subcommand. INPUT PROGRAM. SET SEED=987651423. VECTOR q(20F8.0). LOOP id=1 TO 25. + LEAVE id. + LOOP #=1 TO 20. + COMPUTE q(#)=TRUNC(UNIFORM(5)+1). + END LOOP. + END CASE. END LOOP. END FILE. END INPUT PROGRAM. EXECUTE. **** Start job. VECTOR resp=q1 TO q20. LOOP qnb=1 TO 20. COMPUTE resp=resp(qnb). COMPUTE cat=1+TRUNC(qnb/5.1). XSAVE OUTFILE='c:\\\\temp\\\\temp.sav' /KEEP=id cat qnb resp. END LOOP. EXECUTE. GET FILE='c:\\\\temp\\\\temp.sav'. * aggregate cases by response values within id and category. AGGREGATE /OUTFILE=* /BREAK=id cat resp /nbresp = N(qnb). * identify then keep the line with the largest number of cases. SORT CASES BY id cat nbresp. ADD FILES FILE=* /BY=id cat /LAST=last. SELECT IF last=1. VARIABLE LABELS nbresp 'Number of response' resp 'Response'. EXECUTE. ```\nRelated pages\n\n..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5314019,"math_prob":0.9612735,"size":1455,"snap":"2023-40-2023-50","text_gpt3_token_len":485,"char_repetition_ratio":0.1026878,"word_repetition_ratio":0.0,"special_character_ratio":0.35670102,"punctuation_ratio":0.15522388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834556,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T04:03:08Z\",\"WARC-Record-ID\":\"<urn:uuid:3c612cac-ed3d-41f4-a5c3-7257c4493cea>\",\"Content-Length\":\"15402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d668c57-b3a0-4941-9847-a65d34d92044>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a7905ba-e146-4f83-bcdc-999debd039ab>\",\"WARC-IP-Address\":\"78.108.88.70\",\"WARC-Target-URI\":\"https://spsstools.net/en/syntax/syntax-index/ranking-largest-values-sorting-grouping/calculate-mode/\",\"WARC-Payload-Digest\":\"sha1:XOXV5GMEWUVNQ3H45ZSKGFDJUDZXJQJO\",\"WARC-Block-Digest\":\"sha1:RQ63WHOQZZTJBEJYLZ6VXGEBTBCVWV5U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506559.11_warc_CC-MAIN-20230924023050-20230924053050-00106.warc.gz\"}"}