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https://www.keyword-suggest-tool.com/search/vertex+to+standard+quadratic+form+calculator/	[ "# Vertex to standard quadratic form calculator\n\nVertex to standard quadratic form calculator keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website\n\n## Keyword Suggestions\n\nStandard to vertex quadratic form calculator\n\n( Please select at least 2 keywords )\n\n#### Most Searched Keywords\n\n## Websites Listing\n\nWe found at least 10 Websites Listing below when search with vertex to standard quadratic form calculator on Search Engine\n\n### Quadratic Equation Calculator - Symbolab\n\nDA: 16 PA: 37 MOZ Rank: 53\n\n### Functions Vertex Calculator - Symbolab\n\nFree functions vertex calculator - find function's vertex step-by-step. This website uses cookies to ensure you get the best experience. ... Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile ... just click the link in the email we ...\n\nhttps://www.symbolab.com/solver/function-vertex-calculator\n\nDA: 16 PA: 34 MOZ Rank: 50\n\nOur quadratic equation formula solver is designed to solve all types of quadratic equations. Be it standard form, factored form or vertex form, you will get your answer within a few minutes. Works on all browsers. The best part of the online quadratic equation calculator is it is compatible with all type of window versions as well as browsers.\n\nDA: 20 PA: 31 MOZ Rank: 51\n\n### Vertex form calculator - Cannabisser\n\nWhy not try this out? Thousands of users are using our software to conquer their algebra homework. Instead of x², you can also write x^2. Contact - They are pretty smart and first class . Now that's customer service! From vertex form to standard form. However, I asked my son to give it a try. If the coefficient of the term x2 is positive, the vertex will be in the bottom of the U- shaped curve.\n\nhttps://www.cannabisser.com/site/0aff5d-vertex-form-calculator\n\nDA: 19 PA: 35 MOZ Rank: 54\n\n### Quadratic equations vertex and standard form online ...\n\nSte. 105-181 19179 Blanco Rd #181 San Antonio, TX 78258 USA\n\nDA: 12 PA: 50 MOZ Rank: 91\n\n### Braingenie \| Changing from Vertex to Standard Form\n\nIdentifying the Vertex as Max or Min Given Intercept Form Finding the Vertex Given Intercept Form Graphing y = a(x + p)(x + q) Using the Vertex and X-Intercepts Solving Word Problems Using Intercept Form Changing Between Vertex, Intercept and Standard Form\n\nhttps://braingenie.ck12.org/skills/106803\n\nDA: 19 PA: 14 MOZ Rank: 33\n\n### Relations and Functions Calculators\n\nEvaluate Functions. This calculator will: (1.) Evaluate a function for a specified value. (2.) Return the answer in the simplest form. (3.) Graph the function and indicate the specified value.\n\nhttps://relations-functions.appspot.com/calculators.html\n\nDA: 31 PA: 17 MOZ Rank: 48\n\n### Solver Convert to Vertex Form and Graph\n\nThis Solver (Convert to Vertex Form and Graph) was created by by ccs2011(207) : View Source, Show, Put on YOUR site About ccs2011: Convert to Vertex Form and Graph. Enter quadratic equation in standard form:--> x 2 + x + This solver has been accessed 2388320 times. ...\n\nDA: 15 PA: 50 MOZ Rank: 68\n\n### Mod 2 notes.pdf - MAT 110 Module 2 Notes Module 2 Notes ...\n\nMAT 110 - Module 2 Notes Standard to Vertex Form 1. When a quadratic function is in the form y = ax 2 + bx + c, to change is to vertex form: (a) Calculate the axis of symmetry using the formula x =-b 2 a (b) When you find this value, plug it back into the function. (c) the a value does not change (d) Write the equation in vertex for y = a (x-h ...\n\nhttps://www.coursehero.com/file/77791204/mod-2-notespdf/\n\nDA: 18 PA: 30 MOZ Rank: 48\n\n### Solver Completing the Square to Get a Quadratic into ...\n\nThis Solver (Completing the Square to Get a Quadratic into Vertex Form) was created by by jim_thompson5910(35256) : View Source, Show, Put on YOUR site About jim_thompson5910: If you need more math help, then you can email me. I charge \\$2 for steps, or \\$1 for answers only.\n\nhttps://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/completing-the-square.solver\n\nDA: 15 PA: 50 MOZ Rank: 98" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8761568,"math_prob":0.90105236,"size":3200,"snap":"2021-04-2021-17","text_gpt3_token_len":690,"char_repetition_ratio":0.1301627,"word_repetition_ratio":0.055045873,"special_character_ratio":0.225625,"punctuation_ratio":0.11092715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96956927,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-16T05:17:39Z\",\"WARC-Record-ID\":\"<urn:uuid:3947e6f7-8586-413c-acce-f334c79d5719>\",\"Content-Length\":\"41170\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e79dd8e1-66e4-4dde-985c-eae1c27137d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:1497187a-d4c4-406a-8060-a1ffe1f29e1e>\",\"WARC-IP-Address\":\"172.67.188.207\",\"WARC-Target-URI\":\"https://www.keyword-suggest-tool.com/search/vertex+to+standard+quadratic+form+calculator/\",\"WARC-Payload-Digest\":\"sha1:PFXNHVEFXUZPUNYILNMYZBWTPDROOWGD\",\"WARC-Block-Digest\":\"sha1:ZFL6SCJL2QP7LZBJAISM5AIDT7XG4LZZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703500028.5_warc_CC-MAIN-20210116044418-20210116074418-00295.warc.gz\"}"}
https://rdrr.io/cran/tipmap/f/inst/doc/introduction.Rmd	[ "# Introduction to the R package 'tipmap'\" In tipmap: Tipping Point Analysis for Bayesian Dynamic Borrowing\n\n```knitr::opts_chunk\\$set(\necho = T, collapse = T, warning = F, message = F,\nprompt = T, comment = \"#\",\nout.width = \"100%\"\n)\n```\n\n## Purpose of the package\n\nThe R package tipmap implements a tipping point analysis for clinical trials that employ Bayesian dynamic borrowing of a treatment effect from external evidence via robust meta-analytic predictive (MAP) priors. A tipping point analysis allows to assess how much weight on the informative component of a robust MAP prior is needed to conclude that the investigated treatment is efficacious, based on the total evidence. The package mainly provides an implementation of a graphical approach proposed by (Best et al., 2021) for different one-sided evidence levels (80%, 90%, 95%, 97.5%).\n\nTipping point analyses can be useful both at the planning and the analysis stage of a trial that uses external information. At the planning stage, they can help determining (pre-specifying) a weight of the informative component of the MAP prior for the primary analysis. Various possible results of the planned trial in the target population and the consequences for the treatment effect estimate and statistical inferences based on the total evidence may be explored for a range of weights. Through this exercise, in addition to other criteria, decision-makers can develop a sense under which circumstances they would still feel comfortable to establish efficacy with a specific level of certainty. A preferred weight can then rather straightforwardly be derived. The tipping point analysis can thus help to elicit sensible weights. At the analysis stage, tipping point analyses can be used as sensitivity analysis to assess the dependency of the treatment effect estimate and statistical inferences on weight of the informative component of the MAP prior. This may also be understood in the sense of a reverse-Bayes analysis (Held et al., 2022).\n\nThis vignette shows an exemplary application of the tipping point analysis with hypothetical data.\n\nFurther functions in this package (not illustrated in this vignette) facilitate the specification of a robust MAP prior via expert elicitation, specifically the choice of a primary weight, using the roulette method, as well as the computation of the posterior distribution of the treatment effect with stochastic weights elicited from a panel of experts.\n\nIntended use of the tipmap-package is the planning, analysis and interpretation of (small) clinical trials in pediatric drug development, where extrapolation of efficacy, often through Bayesian methods, has become increasingly common (Gamalo et al., 2022; ICH, 2022). However, the applicability of the package is generally wider.\n\nFor the implementation of the MAP prior approach, including computation of the posterior distribution, the RBesT-package is used (Weber et al., 2021).\n\n## Data generation for an exemplary tipping point analysis\n\nIn this vignette, we assume that results from three clinical trials conducted in adult patients (the source population) are available, which share key features with a new trial among pediatric patients (the target population). For example, they had been conducted in the same indication, studied the same drug and provided results on an endpoint of interest for the target population. This means a certain degree of exchangeability between the trials in the source and target population can be assumed. The similarity in disease and response to treatment between source and target population always need to be carefully considered in any setting, usually by clinical experts in the disease area.\n\nWe assume that it is supported by medical evidence and now planned to consider the trials in adult patients in a Bayesian dynamic borrowing approach, and we would like to create a robust MAP prior (Schmidli et al., 2014). The treatment effect measure of interest is assumed to be a mean difference between a treated group and a control group with respect to a continuous endpoint.\n\n### Derivation of MAP prior based on trials in the source population\n\nWe start by specifying an object that contains the prior data.\n\nThe function `create_prior_data()` takes vectors of total sample sizes, treatment effect estimates and their standard errors as arguments and generates a data frame. A study label is optional.\n\n```library(tipmap)\nprior_data <- create_prior_data(\nn_total = c(160, 240, 320),\nest = c(1.16, 1.43, 1.59),\nse = c(0.46, 0.35, 0.28)\n)\n```\n```# compute standard deviation of change in each arm (assumed equal);\n# for two-sample data:\nsd <- prior_data\\$se / sqrt(1/(prior_data\\$n_total/2) + 1/(prior_data\\$n_total/2))\nsigma1 <- mean(sd)\nsigma1 # = 2.70826\nnt <- 15; nc <- 15\nf <- (nt+nc)^2 / (ntnc)\nsigma2 <- sqrt(f)sigma1\nsigma2 # = 5.41652\n```\n```print(prior_data)\n```\n\nWe then generate a MAP prior from our prior data using the RBesT-package (Weber et al., 2021).\n\n```set.seed(123)\nuisd <- 5.42\nmap_mcmc <- RBesT::gMAP(\nformula = cbind(est, se) ~ 1 \| study_label,\ndata = prior_data,\nfamily = gaussian,\nweights = n_total,\ntau.dist = \"HalfNormal\",\ntau.prior = cbind(0, uisd / 16),\nbeta.prior = cbind(0, uisd)\n)\n```\n\nA few additional specifications are needed to be made to fit the MAP prior model; for details see Neuenschwander and Schmidli (2020) or Weber et al. (2021). The variable `uisd` here represents an assumed unit-information standard deviation and the specification of the prior on between-trial heterogeneity parameter tau follows recommendations to consider moderate heterogeneity for a two-group parameter, such as the mean difference (Neuenschwander and Schmidli, 2020).\n\nThis is a summary of the fitted model based on samples from the posterior distribution:\n\n```summary(map_mcmc)\n```\n\nA forest plot of the Bayesian meta-analysis is shown in Figure 1. It is augmented with meta-analytic shrinkage estimates per trial. The figure shows the per-trial point estimates (light point) and the 95% frequentist confidence intervals (dashed line) and the model derived median (dark point) and the 95% credible interval of the meta-analytic model.\n\n```plot(map_mcmc)\\$forest_model\n```\n\nSubsequently, the MAP prior is approximated by a mixture of conjugate normal distributions. The parametric form facilitates the computation of posteriors when the MAP prior is combined with results from the trial in the target population.\n\n```map_prior <- RBesT::automixfit(\nsample = map_mcmc,\nNc = seq(1, 4),\nk = 6,\nthresh = -Inf\n)\n```\n\nThe approximation yields a mixture of two normals:\n\n```print(map_prior)\n```\n\nThe density of the parametric mixture together with a histogram of MCMC samples from the `map_mcmc` object is shown in Figure 2.\n\n```plot(map_prior)\\$mix\n```\n\nThe derivation of the MAP prior is now complete. For normal likelihoods the parametric representation by a mixture of normals can be used to calculate posterior distributions analytically.\n\n### Trial results in the target population\n\nWe now create a numeric vector with data on pediatric trial (the total sample size, the treatment effect estimate and its standard error). In the planning stage, this may be an expected result.\n\n```pediatric_trial <- create_new_trial_data(n_total = 30, est = 1.02, se = 1.4)\n```\n```print(pediatric_trial)\n```\n\nThe function `create_new_trial_data()` computes quantiles, assuming normally distributed errors. This is merely used to plot a confidence interval for the treatment effect estimate obtained in the target trial in the tipping point plot.\n\n## Performing the tipping point analysis\n\n### Computation of posteriors for a range of weights\n\nWe can now compute posterior distributions for a range of weights using the function `create_posterior_data()`.\n\n```posterior <- create_posterior_data(\nmap_prior = map_prior,\nnew_trial_data = pediatric_trial,\nsigma = uisd)\n```\n```head(posterior, 4)\n```\n```tail(posterior, 4)\n```\n\nThe resulting data frame has `r dim(posterior)` rows and `r dim(posterior)` columns. The weights increase incrementally in steps of 0.005 from 0 to 1, i.e.\\ posterior quantiles for `r length(posterior\\$weight)` weights are computed. For each weight the data frame contains the following `r dim(posterior)-1` posterior quantiles.\n\n```length(posterior\\$weight)\ndim(posterior)\nclass(posterior)\ncolnames(posterior)[-1]\n```\n```colnames(posterior)[-1]\n```\n\nThese posterior quantiles can be directly used for inferences based on the total evidence (new data and prior combined). They reflect one-sided 99%, 97.5%, 95%, 90%, 80%, and 50% evidence levels for a given weight, respectively.\n\n### Creating the tipping point plot\n\nThe function to produce the tipping point plot is called `tipmap_plot()`, it requires a dataframe with data on all components generated by the function `create_tipmap_data()`.\n\n```tipmap_data <- create_tipmap_data(\nnew_trial_data = pediatric_trial,\nposterior = posterior,\nmap_prior = map_prior)\n```\n```(p1 <- tipmap_plot(tipmap_data = tipmap_data))\n```\n\nIn the center of the plot, a funnel-shaped display of quantiles of the posterior distribution (reflecting one-sided evidence-levels) is shown for given weights of the informative component of the MAP prior. The intersections between the lines connecting the respective quantiles and the horizontal line at 0 (the null effect) are referred to as tipping points (indicated by vertical lines in red color). They indictae the minimum weight that is required to conclude that the treatment is efficacious for a given one-sided evidence level (Best et al., 2021). On the left and right side of the plot, the treatment effect estimate obtained in the trial in the (pediatric) target population (with 95% confidence interval) and the MAP prior (with 95% credible interval) are shown, respectively.\n\nThe plot is a `ggplot`-object that can be modified accordingly. For example, if we had chosen a primary weight of 0.38, we could add a vertical reference line at this position. There are additional features to customize the plot in the `tipmap_plot()` function, see `help(tipmap_plot)`.\n\n```primary_weight <- 0.38\n(p2 <- p1 + ggplot2::geom_vline(xintercept = primary_weight, col=\"green4\"))\n```\n\nWe see from Figure 4 that, for a weight of 0.38, there is a probability of larger than 90% but less than 95% based on the posterior distribution that the treatment effect is larger than 0, i.e. the treatment is efficacious.\n\n### Extracting quantities of interest\n\nThe data frame with posteriors for all weights can be filtered to obtain posterior quantiles for weights of specific interest by the function `get_posterior_by_weight()`:\n\n```get_posterior_by_weight(\nposterior = posterior,\nweight = c(primary_weight)\n)\n```\n\nThe function `get_tipping_points()` extracts tipping points for one-sided 80%, 90%, 95% and 97.5% evidence levels, respectively.\n\n```tipp_points <- get_tipping_points(\ntipmap_data = tipmap_data,\nquantile = c(0.2, 0.1, 0.05, 0.025)\n)\ntipp_points\n```\n\nCalculating the precise posterior probability that treatment effect exceeds a threshold value is possible via functions in the RBesT-package.\n\n```prior_primary <- RBesT::robustify(\npriormix = map_prior,\nweight = (1 - primary_weight),\nm = 0,\nn = 1,\nsigma = uisd\n)\n```\n```posterior_primary <- RBesT::postmix(\npriormix = prior_primary,\nm = pediatric_trial[\"mean\"],\nse = pediatric_trial[\"se\"]\n)\n```\n```summary(posterior_primary)\n```\n\nThe posterior probability that the treatment effect is larger than 0, 0.5 and 1, respectively, can be assessed through the cumulative distribution function of the posterior.\n\n```round(1 - RBesT::pmix(posterior_primary, q = 0), 3)\nround(1 - RBesT::pmix(posterior_primary, q = 0.5), 3)\nround(1 - RBesT::pmix(posterior_primary, q = 1), 3)\n```\n\nThis is illustrated by a cumulative density curve of the posterior.\n\n```library(ggplot2)\nplot(posterior_primary, fun = RBesT::pmix) +\nscale_x_continuous(breaks = seq(-1, 2, 0.5)) +\nscale_y_continuous(breaks = 1-c(1, 0.927, 0.879, 0.782, 0.5, 0),\nlimits = c(0,1),\nexpand = c(0,0)\n) +\nylab(\"Cumulative density of posterior with w=0.38\") +\nxlab(\"Quantile\") +\ngeom_segment(aes(x = 0,\ny = RBesT::pmix(mix = posterior_primary, q = 0),\nxend = 0,\nyend = 1),\ncol=\"red\") +\ngeom_segment(aes(x = 0.5,\ny = RBesT::pmix(mix = posterior_primary, q = 0.5),\nxend = 0.5,\nyend = 1),\ncol=\"red\") +\ngeom_segment(aes(x = 1,\ny = RBesT::pmix(mix = posterior_primary, q = 1),\nxend = 1,\nyend = 1),\ncol=\"red\") +\ntheme_bw()\n```\n\nAs a further example, for the weight corresponding to the tipping point of the one-sided evidence-level of 95% (=0.51), we would obtain a posterior probability of 95% that the treatment effect is larger than 0.\n\n```tipp_points\n```\n```prior_95p <- RBesT::robustify(\npriormix = map_prior,\nweight = (1 - tipp_points),\nm = 0,\nn = 1,\nsigma = uisd\n)\n```\n```posterior_95p <- RBesT::postmix(\npriormix = prior_95p,\nm = pediatric_trial[\"mean\"],\nse = pediatric_trial[\"se\"]\n)\n```\n```round(1 - RBesT::pmix(posterior_95p, q = 0), 3)\n```\n\n## References\n\nBest, N., Price, R. G., Pouliquen, I. J., et al. (2021) Assessing efficacy in important subgroups in confirmatory trials: An example using Bayesian dynamic borrowing. Pharm Stat, 20, 551–562. DOI: 10.1002/pst.2093\n\nGamalo, M., Bucci-Rechtweg, C., Nelson, R. M., et al. (2022) Extrapolation as a Default Strategy in Pediatric Drug Development. Ther Innov Regul Sci, 56, 883–894. DOI: 10.1007/s43441-021-00367-9\n\nHeld, L., Matthews, R., Ott, M., et al. (2022) Reverse-Bayes methods for evidence assessment and research synthesis. Res Synth Methods, 13, 295–314. DOI: 10.1002/jrsm.1538\n\nICH (2022) International Council for Harmonisation of Technical Requirements for Pharmaceuticals for Human Use (ICH) guideline E11A on pediatric extrapolation. EMA/CHMP/ICH/205218/2022.\n\nNeuenschwander, B. and Schmidli, H. (2020) Use of Historical Data. In Bayesian Methods in Pharmaceutical Research (eds E. Lesaffre, G. Baio, and B. Boulanger), pp. 111–129. 1st ed. Chapman & Hall/ CRC Biostatistics Series. DOI: 10.1201/9781315180212\n\nSchmidli, H., Gsteiger, S., Roychoudhury, S., et al. (2014) Robust meta-analytic-predictive priors in clinical trials with historical control information. Biometrics, 70, 1023–1032. DOI: 10.1111/biom.12242\n\nWeber, S., Li, Y., Seaman III, J. W., et al. (2021) Applying Meta-Analytic-Predictive Priors with the R Bayesian Evidence Synthesis Tools. J Stat Softw, 100, 1–32. DOI: 10.18637/jss.v100.i19\n\n## Try the tipmap package in your browser\n\nAny scripts or data that you put into this service are public.\n\ntipmap documentation built on Dec. 8, 2022, 1:13 a.m." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7995667,"math_prob":0.98164,"size":13775,"snap":"2022-40-2023-06","text_gpt3_token_len":3483,"char_repetition_ratio":0.13027376,"word_repetition_ratio":0.04612365,"special_character_ratio":0.26098004,"punctuation_ratio":0.1639732,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9854581,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T14:53:53Z\",\"WARC-Record-ID\":\"<urn:uuid:456fd465-5878-4500-aa06-19d2225bab6d>\",\"Content-Length\":\"56584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90023f95-5f2e-4253-be3d-799f8d37e788>\",\"WARC-Concurrent-To\":\"<urn:uuid:4be495b6-b793-47b4-8cd3-1b3d99a9d434>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/tipmap/f/inst/doc/introduction.Rmd\",\"WARC-Payload-Digest\":\"sha1:7UBR4Y3P7XVX5IK4C6N6A2UB5DBWVMGQ\",\"WARC-Block-Digest\":\"sha1:ZGS6HHNT3UXKB62XA2VRSAUH6MSV6KQL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499871.68_warc_CC-MAIN-20230131122916-20230131152916-00788.warc.gz\"}"}
https://biology.reachingfordreams.com/chemistry-cheat-sheet/chemical-equilibrium/40-changes-in-equilibrium-systems	[ "Le Châtelier’s principle - a generalization that states that:\n\nchemical systems at equilibrium shift to restore equilibrium when a change occurs that disturbs the equilibrium.\n\nAn adjustment by a system at equilibrium that results in a change in the concentrations of reactants and products is called an equilibrium shift.\n\nLe Châtelier’s principle allows chemists to predict the qualitative effects of changes in concentration, pressure, and temperature on a chemical reaction system at equilibrium.\n\nWe can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (K).\n\nIf a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes.\n\nSince this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K.\n\nTo re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value as K.\n\n## The Relationship Between the Equilibrium Constant and the Reaction Quotient", null, "If reaction quotient (Qc) is greater than equilibrium constant (Kc), the numerator of the reaction quotient expression must be very large.\n\nThe concentrations of the chemicals on the right side of the equation\n\n(mA + nB + ⇌ xC + yD)\n\nmust be more than their concentrations at equilibrium.\n\nIn this situation, the system attains equilibrium by moving to the left.\n\nConversely, if Qc is less than Kc , the system attains equilibrium by moving to the right.\n\n## Effect of Change in Concentration on Equilibrium", null, "(a) The test tube contains 0.1 M Fe3+.\n\n(b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)2+ ion.\n\nFe3+ (aq) + SCN (aq) ⇌ Fe(SCN)2+ (aq).\n\n(c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN − as the white solid AgSCN.\n\nAg+ (aq) + SCN (aq) ⇌ AgSCN(s).\n\nThe decrease in the SCN concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)2+.\n\nThe stress on the system in the image above is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause Q to be larger than K).\n\nAs a consequence, Le Châtelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.\n\n### Collision Theory and Concentration Changes in an Equilibrium System\n\nAccording to collision theory, entities in a chemical system must collide to react.\n\nWhen the concentration of an entity in a chemical reaction system is increased, it is more likely that that entity will collide with other entities. There are simply more of them present. However, only collisions between reactant entities can potentially contribute to a chemical reaction. Even so, the more frequently collisions occur overall, the more likely it is that a chemical reaction will take place.\n\nCollision theory explains the response of a chemical reaction system at equilibrium to a change in concentration as the result of random collisions and probability.\n\nWhen we add more reactant entities in an equilibrium system, the equilibrium shifts to the right because the number of successful collisions for the forward reaction increases.\n\nIf, instead, we add more product entities, then the number of successive collisions for the reverse reaction will increase and the equilibrium will shift to the left.\n\n## Le Châtelier’s Principle and Changes in Energy\n\nFor a given system at equilibrium, the value of the equilibrium constant depends only on temperature.\n\nChanging the temperature of a reacting mixture changes the rate of the forward and reverse reactions by different amounts, because the forward and reverse reactions have different activation energies.\n\nA reacting mixture at one temperature has an equilibrium constant whose value changes if the mixture is allowed to reach equilibrium at a different temperature.\n\nTo apply Le Châtelier’s principle and predict how a change in energy will affect a chemical system at equilibrium, we can think of energy as a reactant or a product.\n\nEnergy is absorbed in an endothermic reaction. If we consider energy to be a reactant, we can write the word equation:\n\nreactants + energy ⇌ products\n\nSince energy is released during an exothermic reaction, we can consider energy to be a reactant and write\n\nreactants ⇌ products + energy\n\nTo use Le Châtelier’s principle to predict how an equilibrium will shift in response to a change in energy, consider how the system can counteract this shift.\n\n### Endothermic Reactions\n\nIf an endothermic reaction is cooled (thermal energy removed), we can consider that the quantity of one of the reactants has been decreased.\n\nWe can therefore predict that the equilibrium will shift to the left (toward the reactants), and energy will be released.\n\nIf thermal energy were added to this equilibrium system by heating, the equilibrium would likely shift to the right.\n\n### Exothermic Reactions\n\nIf thermal energy is removed from an exothermic reaction—where energy is a product—then the equilibrium will shift to the right (toward the products), and energy will be released to counteract the change.\n\nIf energy is added to an exothermic reaction, the equilibrium will shift to the left to compensate for the change, and the energy will be used as products are converted to reactants.\n\n## Le Châtelier’s Principle and Changes in Gas Volume\n\nSometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system.\n\nAn easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium.\n\nWhile evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for Kp).\n\nSome changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression.\n\nThus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.\n\nAs we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components.\n\nIn accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress.\n\nThe reverse reaction would be favored by a decrease in pressure.\n\nConsider what happens when we increase the pressure on a system in which N2, H2, and NH3 are at equilibrium:\n\nN(g) + 3H2 (g) ⇌ 2NH3 (g)\n\nThe formation of additional amounts of NH3 decreases the total number of molecules in the system because each time two molecules of NH3 form, a total of four molecules of N2 and H2 are consumed.\n\nThis reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure.\n\nOn the other hand, a decrease in the pressure on the system favors decomposition of NH3 into N2 and H2 , which tends to restore the pressure.\n\nLe Châtelier’s Principle and Changes in Gas Volume", null, "(a) The container holds the equilibrium reaction N(g) + 3H2 (g) ⇌ 2NH3 (g) .\n\n(b) The volume of the container is decreased by half, which doubles the total pressure of the system.\n\n(c) The reaction shifts to the right, which decreases the total number of particles in the container and the total pressure by producing more ammonia, NH3 (g).\n\n## The Effect of a Catalyst on Equilibrium\n\nA catalyst speeds up the rate of a reaction, either by allowing a different reaction mechanism or by providing additional mechanisms.\n\nThe overall effect is to lower the activation energy, which increases the rate of reaction.\n\nThe activation energy is lowered the same amount for the forward and reverse reactions, however. There is the same increase in reaction rates for both reactions.\n\nAs a result, a catalyst does not affect the position of equilibrium. It only affects the time that is taken to achieve equilibrium.\n\nThe Effects of Changing Conditions on a System at Equilibrium\n\n Type of reaction Change to system Effect on Kc Direction of change all reactions increasing any reactant concentration, or decreasing any product concentration no effect toward products decreasing any reactant concentration, or increasing any product concentration no effect toward reactants using a catalyst no effect no change exothermic increasing temperature decreases toward reactants decreasing temperature increases toward products endothermic increasing temperature increases toward products decreasing temperature decreases toward reactants equal number of reactant and product gas molecules changing the volume of the container, or adding a non-reacting gas no effect no change more gaseous product molecules than reactant gaseous molecules decreasing the volume of the container at constant temperature no effect toward reactants increasing the volume of the container at constant temperature, or adding a non-reacting gas at contstant pressure no effect toward products fewer gaseous product molecules than reactant gaseous molecules decreasing the volume of the container at constant temperature no effect toward products increasing the volume of the container at constant temperature no effect toward reactants" ]	[ null, "https://biology.reachingfordreams.com/plugins/system/lazyloadforjoomla/src/assets/images/blank.gif", null, "https://biology.reachingfordreams.com/plugins/system/lazyloadforjoomla/src/assets/images/blank.gif", null, "https://biology.reachingfordreams.com/plugins/system/lazyloadforjoomla/src/assets/images/blank.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8755222,"math_prob":0.9781787,"size":9983,"snap":"2022-40-2023-06","text_gpt3_token_len":2134,"char_repetition_ratio":0.19320573,"word_repetition_ratio":0.062227752,"special_character_ratio":0.19362918,"punctuation_ratio":0.07204611,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9718802,"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\":\"2023-01-27T23:35:03Z\",\"WARC-Record-ID\":\"<urn:uuid:d04efa31-2963-4eb2-a50d-22e2e5c99e63>\",\"Content-Length\":\"49213\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8073de29-5f58-4c41-b5c7-1d613267a5c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:676f274b-fc6f-4225-8b7c-5fc371d769bd>\",\"WARC-IP-Address\":\"65.39.193.37\",\"WARC-Target-URI\":\"https://biology.reachingfordreams.com/chemistry-cheat-sheet/chemical-equilibrium/40-changes-in-equilibrium-systems\",\"WARC-Payload-Digest\":\"sha1:LR7A4SKF5WPUVATVJMUTQEZWHKRVI44A\",\"WARC-Block-Digest\":\"sha1:TM4LXX63NWBKXCVZTLBXNYFPPMIV4BLK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499468.22_warc_CC-MAIN-20230127231443-20230128021443-00830.warc.gz\"}"}
https://ux.stackexchange.com/questions/140997/when-to-use-white-text-on-top-of-a-color-for-readability/140998	[ "# When to use white text on top of a color for readability?\n\nI am setting the background of a textbox to some variable color.\n\nI know the hex/rgb of the color I'm setting the background to.\n\nNormally the text is black (or very very dark gray).\n\nOn some colors, it would be far easier to read the text if it were white instead of black.\n\nHow can it be programmatically decided, given a hex/rgb color, whether white or black text would be more readable if placed on top of it?\n\nThis is standardised by W3C's Web Content Accessibility Guidelines (WCAG) and decided based on the contrast ratio between colors, which you can calculate programmatically.\n\nIn your case, you would want to use whichever of white or black has a higher contrast ratio with the background color.\n\nHere is just one of many articles on how to calculate contrast ratio in JavaScript: dev.to: Building your own color contrast checker.\n\nFrom that article:\n\nThe contrast ratio between two colors is calculated using a mathematical function that we will see below. The resulting value of that calculation will go from 1 to 21 (normally represented as 1:1 or 21:1 respectively). If both are the exact same color, the contrast ratio will be 1; The highest contrast is between white and black, and exact opposites in the color wheel (complementary colors) will have high values too.\n\nThe mathematical formula is documented by WCAG.\n\ncontrast ratio\n(L1 + 0.05) / (L2 + 0.05), where\n\n• L1 is the relative luminance of the lighter of the colors, and\n• L2 is the relative luminance of the darker of the colors.\n\nIt uses relative luminance:\n\nFor the sRGB colorspace, the relative luminance of a color is defined as L = 0.2126 * R + 0.7152 * G + 0.0722 * B where R, G and B are defined as:\n\n• if RsRGB <= 0.03928 then R = RsRGB/12.92 else R = ((RsRGB+0.055)/1.055) ^ 2.4\n• if GsRGB <= 0.03928 then G = GsRGB/12.92 else G = ((GsRGB+0.055)/1.055) ^ 2.4\n• if BsRGB <= 0.03928 then B = BsRGB/12.92 else B = ((BsRGB+0.055)/1.055) ^ 2.4\n\nand RsRGB, GsRGB, and BsRGB are defined as:\n\n• RsRGB = R8bit/255\n• GsRGB = G8bit/255\n• BsRGB = B8bit/255\n\nThere exist free online contrast calculators like this one from WebAIM. In your case, add #fff as one value and your other color as the other. In most cases, you only need WCAG AA passing to be considered accessible. WCAG AA considers a 4.5:1 ratio to be passing. Usually it's good to go just a little darker than a minimum passing value, and the contrast checker can help you find a value for that.\n\n• 4.5:1 is passing, but 7:1 is the recommendation for main body text (for example, the text of the answers here on SE), and speaking from experience as someone who actually has vision issues affected by this, ‘passing’ should not be considered ‘good enough’ unless you are dealing with large font sizes or exceptionally good fonts. Sep 14, 2021 at 11:52" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8898681,"math_prob":0.9620837,"size":2004,"snap":"2022-27-2022-33","text_gpt3_token_len":562,"char_repetition_ratio":0.128,"word_repetition_ratio":0.011204482,"special_character_ratio":0.2994012,"punctuation_ratio":0.13679245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9820744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T16:19:31Z\",\"WARC-Record-ID\":\"<urn:uuid:fa640734-43d3-4b47-9cce-349eacf3f5a1>\",\"Content-Length\":\"226262\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de92e75b-1c75-4bc4-bb58-a352dd22deab>\",\"WARC-Concurrent-To\":\"<urn:uuid:485019cd-d0e7-4d93-a3f9-552920317ff7>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://ux.stackexchange.com/questions/140997/when-to-use-white-text-on-top-of-a-color-for-readability/140998\",\"WARC-Payload-Digest\":\"sha1:C5X2D2GPV3UEUTQGAC6J5S2JU56QSYMP\",\"WARC-Block-Digest\":\"sha1:4XP3LLNDR274D7D5ODWBA4UAKJVEZFYO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103640328.37_warc_CC-MAIN-20220629150145-20220629180145-00048.warc.gz\"}"}
https://se.mathworks.com/help/control/ref/dlyapchol.html	[ "# dlyapchol\n\nSquare-root solver for discrete-time Lyapunov equations\n\n## Syntax\n\n```R = dlyapchol(A,B) X = dlyapchol(A,B,E) ```\n\n## Description\n\n`R = dlyapchol(A,B)` computes a Cholesky factorization `X = R'R` of the solution `X` to the Lyapunov matrix equation:\n\n```AXA'- X + BB' = 0 ```\n\nAll eigenvalues of `A` matrix must lie in the open unit disk for `R` to exist.\n\n`X = dlyapchol(A,B,E)` computes a Cholesky factorization `X = R'R` of `X` solving the Sylvester equation\n\n```AXA' - EXE' + BB' = 0 ```\n\nAll generalized eigenvalues of (`A`,`E`) must lie in the open unit disk for `R` to exist.\n\n## Algorithms\n\n`dlyapchol` uses SLICOT routines SB03OD and SG03BD.\n\n## References\n\n Bartels, R.H. and G.W. Stewart, \"Solution of the Matrix Equation AX + XB = C,\" Comm. of the ACM, Vol. 15, No. 9, 1972.\n\n Hammarling, S.J., “Numerical solution of the stable, non-negative definite Lyapunov equation,” IMA J. Num. Anal., Vol. 2, pp. 303-325, 1982.\n\n Penzl, T., ”Numerical solution of generalized Lyapunov equations,” Advances in Comp. Math., Vol. 8, pp. 33-48, 1998." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73798996,"math_prob":0.9952847,"size":953,"snap":"2020-24-2020-29","text_gpt3_token_len":343,"char_repetition_ratio":0.12539515,"word_repetition_ratio":0.15189873,"special_character_ratio":0.3221406,"punctuation_ratio":0.22608696,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997141,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T18:45:56Z\",\"WARC-Record-ID\":\"<urn:uuid:ec0ef821-108b-46df-b13e-10640abccbdc>\",\"Content-Length\":\"65381\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b957db1-a6eb-49c5-8397-74fac074f9d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b8f8f52-6eff-4ac9-9615-891b8cd7392d>\",\"WARC-IP-Address\":\"23.50.112.17\",\"WARC-Target-URI\":\"https://se.mathworks.com/help/control/ref/dlyapchol.html\",\"WARC-Payload-Digest\":\"sha1:DL4IXZ22E3K2BDYEI66C5OPM7IVSOUBV\",\"WARC-Block-Digest\":\"sha1:I5KCQ3VYEPJW2G5J26GAVXQ4QHO3BJUQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347391277.13_warc_CC-MAIN-20200526160400-20200526190400-00101.warc.gz\"}"}
https://forum.inductiveautomation.com/t/best-way-to-construct-timestamp-from-elements/11919	[ "# Best way to construct timestamp from elements\n\nI want to construct a timestamp from individual elements (year, month, day, hour, minute, second).\nThe elements are initially coming from a PLC and then I’m writing into a database.\n\nThis script (admittedly with some hard-coded values) appears to work:\n\n[code]from datetime import datetime\ndt_y = 2016\ndt_m = 1\ndt_d = 20\ndt_h = 12\ndt_mn = 34\ndt_s = 56\ntstamp=datetime(dt_y, dt_m, dt_d, dt_h, dt_mn, dt_s)\n\nsql = “UPDATE mytable SET t_stamp=? WHERE ID=1”\nsystem.db.runPrepUpdate(sql,[tstamp])[/code]\n\nBut is this the best way of doing it?\n\nThere’s probably not a “right” answer to this. What you have is certainly acceptable. Since Ignition is built with java and jython has to convert to java objects to move data through the system, I tend to use the corresponding java objects directly. There are various differences, but the same end effect. Somewhat more efficient if needed in big loops.[code]from java.util import Date\ndt_y = 2016\ndt_m = 1\ndt_d = 20\ndt_h = 12\ndt_mn = 34\ndt_s = 56\ntstamp=Date(dt_y-1900, dt_m-1, dt_d, dt_h, dt_mn, dt_s)\n\nsql = “UPDATE mytable SET t_stamp=? WHERE ID=1”\nsystem.db.runPrepUpdate(sql,[tstamp])[/code]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8116069,"math_prob":0.9444778,"size":596,"snap":"2023-40-2023-50","text_gpt3_token_len":176,"char_repetition_ratio":0.091216214,"word_repetition_ratio":0.0,"special_character_ratio":0.2885906,"punctuation_ratio":0.14166667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9900034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T12:29:35Z\",\"WARC-Record-ID\":\"<urn:uuid:c7ce1f37-825b-4a5f-aac7-d99f82f09b9a>\",\"Content-Length\":\"21208\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34311495-4cfa-4b1c-b0f4-e72d3e085358>\",\"WARC-Concurrent-To\":\"<urn:uuid:96aa0503-3e45-4b1b-8e76-56144e61034c>\",\"WARC-IP-Address\":\"184.105.99.43\",\"WARC-Target-URI\":\"https://forum.inductiveautomation.com/t/best-way-to-construct-timestamp-from-elements/11919\",\"WARC-Payload-Digest\":\"sha1:RML6BTDZRNP4CIG5ZFLN236QIPVUAP72\",\"WARC-Block-Digest\":\"sha1:HNS2JYTPBSWQ2VEE6ZRG7JRSQY5SZ5OD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00336.warc.gz\"}"}
https://lapsedgeographer.london/2020-04/case_when/	[ "# When's the case for case_when() ?", null, "Last week I was taking some colleagues through the code for my COVID19 PDF scraping and afterwards one sent me a message asking about a chunk of code that used the `dplyr::case_when()` function. In particular they wanted to know why `case_when()` uses the tilde (`~`)1, which led to a bit more of a generalised conversation about `case_when()` and how it works.\n\nIn your script, when you assign the values to the entity and position columns inside your `case_when()` call why do you use `~` rather than `=` or `<-`? It felt too specific and potentially obvious to others to ask during your talk. I understand why it’s used in model building for regression etc… but not really in this context.\n\nThis is a great question and formulating the answer actually helped crystallise something I knew about R but hadn’t really processed before then.\n\n``````subnational_datapoints <- subnational_data %>%\nfilter(y == 36 \| y == 104 \| y == 242 \|\ny == 363 \| y == 431 \| y == 568) %>%\nmutate(\nentity = case_when(\ny == 36 ~ \"location\",\ny == 104 & x == 36 ~ \"retail_recr\",\ny == 104 & x == 210 ~ \"grocery_pharm\",\ny == 104 & x == 384 ~ \"parks\",\ny == 242 & x == 36 ~ \"transit\",\ny == 242 & x == 210 ~ \"workplace\",\ny == 242 & x == 384 ~ \"residential\",\ny == 363 ~ \"location\",\ny == 431 & x == 36 ~ \"retail_recr\",\ny == 431 & x == 210 ~ \"grocery_pharm\",\ny == 431 & x == 384 ~ \"parks\",\ny == 568 & x == 36 ~ \"transit\",\ny == 568 & x == 210 ~ \"workplace\",\ny == 568 & x == 384 ~ \"residential\"),\nposition = case_when(\ny == 36 ~ \"first\",\ny == 104 ~ \"first\",\ny == 242 ~ \"first\",\ny == 363 ~ \"second\",\ny == 431 ~ \"second\",\ny == 568 ~ \"second\")\n)\n``````\n\n## The basics\n\nThe documentation for `dplyr::case_when()` calls it a “A general vectorised if”, which is a good explanation if you’re well-versed in R. But a lot of people aren’t. The best way to describe it is that helps you to avoid nasty nested `if` conditions. You remember when you used Excel2 and you’d want a cell to display something based on an entry in another cell and so you combined `IF()` functions together, well `case_when()` is basically a much nicer and easier to process approach to that. Let’s take our example from the code chunk, we have a set of x and y coordinates and we want to generate the entity given the values of x and y. Let’s also pretend we’re back using Excel (or Sheets, or you spreadsheeting application of choice), so we’ve got a table with our x and y values in columns A and B and we’ve written a formula to provide the entity in column C.\n\n``````\| \| A \| B \| C \|\n\|:--\|----:\|----:\|:--------------\|\n\| 1 \| y \| x \| entity \|\n\| 2 \| 104 \| 36 \| retail_recr \|\n\| 3 \| 104 \| 210 \| grocery_pharm \|\n\| 4 \| 104 \| 384 \| parks \|\n\| 5 \| 242 \| 36 \| transit \|\n\| 6 \| 242 \| 210 \| workplace \|\n\| 7 \| 242 \| 284 \| residential \|\n``````\n\nThe simplest way to write the formula is as follows:\n\n=IF(A2=104,IF(B2=36,“retail_recr”,IF(B2=210,“grocery_pharm”,“parks”)), IF(B2=36,“transit”,IF(B2=210,“workplace”,“residential)))\n\nThere’s a fair potential for error here. Firstly, it only does one test of y, so if the value in A6 was 268 it would still continue to process the second-half of these nested IF statements, the same is also the case for the final set of x values, no matter the value in B4 you’ll get parks if A4 is 104, and in B7 no matter what values you have in either A7 or B7 you’ll get residential.\n\nWe could write this set of IF statements that is more explicit, and will return `FALSE` in any cell where there are incorrect values:\n\n=IF(AND(A2=104,B2=36),“retail_recr”,IF(AND(A2=104,B2=210),“grocery_pharm”, IF(AND(A2=104,B2=384),“parks”,IF(AND(A2=242,B2=36),“transit”, IF(AND(A2=242,B2=210),“workplace”,IF(AND(A2=242,B2=384),“residential”))))))\n\nThis does the job and is error-proof in the sense that it only returns valid results for valid inputs. However, if you’re anything like me then your Excel formulae are lazy and so you’d have written the former. Plus you’re not protected against copy/paste, transposition or overwriting errors.\n\n## case_when() to the rescue\n\nBase R has the `ifelse()` function, and `{dplyr}` has a stricter interpretation `if_else()`, we could nest a set of calls to either of these just as we do in Excel (`ifelse(x==1,\"a\", ifelse(x==2, \"b\", \"c\"))`). But we can avoid messy code by using `dplyr::case_when()`.\n\nIt turns out I use `case_when()` a lot in my code. Some 92 of the 211 `.R` and `.Rmd` files in my live R project folders contain at least one call to `case_when`, a whopping 43.6% of my current scripts! Before today I knew that I used it fairly regularly, but if you’d have asked me to guess I’d probably have said that maybe a quarter or a third of my code used it, not almost half!!\n\nSo what makes `case_when()` so great? Well it’s a very handy function for handling multiple conditions. Let’s take a subset of the code from above, we’ve got a tibble that contains the variables x and y, we’re filtering it to three specific values of y and then creating a new variable called entity using `case_when()`.\n\n``````subnational_datapoints <- subnational_data %>%\nfilter(y == 36 \| y == 104 \| y == 242) %>%\nmutate(\nentity = case_when(\ny == 36 ~ \"location\",\ny == 104 & x == 36 ~ \"retail_recr\",\ny == 104 & x == 210 ~ \"grocery_pharm\",\ny == 104 & x == 384 ~ \"parks\",\ny == 242 & x == 36 ~ \"transit\",\ny == 242 & x == 210 ~ \"workplace\",\ny == 242 & x == 384 ~ \"residential\")\n``````\n\nWhile you can use `case_when()` working outside of data.frame operations that’s not something I’ve done and I think more conventional nested `base::if()` statements are better, `case_when()` is designed for working with its `{dplyr}` siblings, especially `dplyr::mutate()`. The arguments to `case_when()` are a set of formula pairs, which is where our friend the tilde (`~`) from my colleague’s question comes in. R’s formula notation is something you’ll mainly come across when constructing a linear model, e.g. `y ~ x + z`, but it’s not restricted to its use in model building. A formula is a type of object that R explicitly understands to be a pair of items, a left-hand side (LHS) and a right-hand side (RHS) split by a tilde, which we can generalise this as `LHS ~ RHS`. This two-part nature of a formula pair is what `case_when()` is exploiting.\n\n`case_when()` uses the LHS part as the condition to test for and the RHS as the value to return, it works in sequence through the conditions and when it obtains a `TRUE` result it will return the value associated with that condition. Our first pair, `y == 36 ~ \"location`, tests if y is equal to 36 and if so then it returns “location”. The next pair, `y == 104 & x == 36 ~ \"retail_recr\"`, tests if y is equal to 104 AND x is equal to 36. This code replicates the second Excel formula written above, but I certainly find it easier to read and follow.\n\nThere are a couple of important things to remember when using `case_when()`. Firstly, it’s crucial that the all of the items in your RHS are of the same type, i.e. they’re all character strings, or they’re all numerics, if they’re not the same type then the call will fail. Secondly, if a row in your data doesn’t match any of the conditions then it’ll get an `NA` value, but what if you wanted to have a catch-all value for all other cases, well that’s pretty simple. `case_when()` is testing conditions and looking for a TRUE result, so all you need to do is add a final line of the form `TRUE ~ value`\n\n``````df <- tibble(val = c(1, 2, 1, 3, 4, 5)) %>%\nmutate(new_val = case_when(\nval == 1 ~ \"One\",\nval > 3 ~ \"More than three\",\nTRUE ~ \"Other\"\n))\n``````\n\nIn this code we look at val and if it’s equal to 1 or greater than 3 then we’ll get the response we’ve specified, but in all other cases we’ll get the value “Other”.\n\n``````> df\n\n\| val\|new_val \|\n\|---:\|:----------------\|\n\| 1 \| One \|\n\| 2 \| Other \|\n\| 1 \| One \|\n\| 3 \| Other \|\n\| 4 \| More than three \|\n\| 5 \| More than three \|\n\n``````\n\nSo when’s the case for `case_when()`? Almost half the time if my code is anything to go by.\n\n1. This opens up a segue to the best R prank ever, a tweet that has since been deleted: Today we were learning R Coding and I made a function so that whenever someone typed a ~ it automatically added “Swinton” after it. Three hours it took my colleagues to get it, three hours. - Joe (@raptorbaitjoe) May 29, 2019. ↩︎\n\n2. But not like Kelly Rowland:\n\n↩︎" ]	[ null, "https://lapsedgeographer.london/img/post/2020-04-25-tilda-swinton.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85073173,"math_prob":0.97695607,"size":8102,"snap":"2023-40-2023-50","text_gpt3_token_len":2297,"char_repetition_ratio":0.12200543,"word_repetition_ratio":0.122715406,"special_character_ratio":0.33399162,"punctuation_ratio":0.11131498,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98800886,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T00:38:11Z\",\"WARC-Record-ID\":\"<urn:uuid:964e1419-2ec6-42eb-943b-d762eca89a67>\",\"Content-Length\":\"24563\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0de7a5d7-05f5-4d22-8450-6b940144bf2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1ed892b-1080-4316-bc2a-9ad0f9959650>\",\"WARC-IP-Address\":\"18.213.222.111\",\"WARC-Target-URI\":\"https://lapsedgeographer.london/2020-04/case_when/\",\"WARC-Payload-Digest\":\"sha1:U4EXMLL2Q67ILOZ6P4PUATGVRS6CK7RS\",\"WARC-Block-Digest\":\"sha1:PDKG2B24NAK65PE6OU3E6POWGGBKDIII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100476.94_warc_CC-MAIN-20231202235258-20231203025258-00174.warc.gz\"}"}
http://www.expertsmind.com/questions/portion-and-variation-301141728.aspx	[ "## PORTION AND VARIATION, Algebra\n\nAssignment Help:\nSUPPOSE Y IS DIRECTLY PROPORTIONAL TO X AND THAT Y = 35 WHEN X = 5\nFIND THE CONSTANT OF PROPORTIONALITY K\nK=\n\n#### Elimination with 2 problems., {5x-4y=-21} {-2x+4y=18}\n\n{5x-4y=-21} {-2x+4y=18}\n\n(-5,-2) y=5/2x+3\n\n#### Word problems, Five star coffee charges a 40 dollar start up fee and then 8...\n\nFive star coffee charges a 40 dollar start up fee and then 8 dollars per month. Custom coffee company charges a 25 dollar start up fee and 12.50 dollars per month. In which time fr\n\n#### Eponents, How do you write 14x14x14x14x14x14x14= in exponential form?\n\nHow do you write 14x14x14x14x14x14x14= in exponential form?\n\n#### Algebra 2, has a y-intercept of 5 and a slope of 2/3. solve for the standa...\n\nhas a y-intercept of 5 and a slope of 2/3. solve for the standard equation\n\n#### College algebra, how to use the factor theorem\n\nhow to use the factor theorem\n\n#### Areas, does total surface area mean total exposed area\n\ndoes total surface area mean total exposed area\n\n#### Identies, (x+a) (x+b) (x+c) =\n\n(x+a) (x+b) (x+c) =\n\n#### Solving quadratic functions, Sum of two numbers is 10 and their multipicati...\n\nSum of two numbers is 10 and their multipication is 21,find the two numbers. x^2 -10x +21=0\n\n#### Determinant, Power cofactor theorem\n\nPower cofactor theorem", null, "", null, "" ]	[ null, "http://www.expertsmind.com/questions/CaptchaImage.axd", null, "http://www.expertsmind.com/prostyles/images/3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84038883,"math_prob":0.97568846,"size":1166,"snap":"2020-45-2020-50","text_gpt3_token_len":323,"char_repetition_ratio":0.09208262,"word_repetition_ratio":0.010928961,"special_character_ratio":0.2890223,"punctuation_ratio":0.105932206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986599,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T05:51:55Z\",\"WARC-Record-ID\":\"<urn:uuid:1be9c61a-492b-40c5-85ec-2d782097eda9>\",\"Content-Length\":\"61241\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02757bfa-41e9-4bf6-8a87-f910bc4e257a>\",\"WARC-Concurrent-To\":\"<urn:uuid:fcdcf795-cfb3-4bee-b2cb-6243e95b1632>\",\"WARC-IP-Address\":\"198.38.85.49\",\"WARC-Target-URI\":\"http://www.expertsmind.com/questions/portion-and-variation-301141728.aspx\",\"WARC-Payload-Digest\":\"sha1:TVV6FILMZUYEBBMGDDUJK55RFRD2LLKS\",\"WARC-Block-Digest\":\"sha1:42HC5Y7VJ2VJHQII5IDIDTHRSMROE63L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107882102.31_warc_CC-MAIN-20201024051926-20201024081926-00342.warc.gz\"}"}
https://www.bankersadda.com/ibps-rrb-mains-reasoning-quiz-21-2	[ "IBPS RRB PO/Clerk Mains Reasoning Quiz: 21st September 2019\n\nDo you follow a proper plan or strategy for IBPS RRB Mains 2019? Are you aiming IBPS RRB 2019 this time? If yes, then this is the section which can help you to do wonders if practiced well. A good attempt with a mix of accuracy can help you fetch good marks. The reasoning is a game of wits and mind. It is all about logics that a question may have. Speed and accuracy are what that matters the most in this section. The only way to achieve an ambitious goal is by practicing only. So, attempt the quiz of Reasoning ability that inculcates the important questions from the important topics. Do not miss out to practice the Reasoning Ability Quiz that is being provided on Bankersadda.\n\nDirections (1-5): Study the given information and answer the questions:\nWhen a number arrangement machine is given an input line of numbers, it arranges them following a particular rule. The following is an illustration of an input and its rearrangement.", null, "Step IV, is the last step of the above arrangement as the intended arrangement is obtained.\nAs per the rules followed in the given steps find out the appropriate steps for the given input:\n\nInput: 61 22 15 52 71 36 18 11\n\nQ1. What is the sum of the numbers at both the ends in step III of the given arrangement?\n(a) 36\n(b) 33\n(c) 60\n(d) 123\n(e) None of the above\n\nQ2. Which element is 2nd to the right of the one which is 4th from the left end in step I?\n(a) 22\n(b) 44\n(c) 53\n(d) 72\n(e) None of these\n\nQ3. What is the products of the numbers which is 2nd from the right end and 2nd from the left end in final step of the given arrangement?\n(a) 180\n(b) 72\n(c) 13.5\n(d) 180\n(e) None of these\n\nQ4. Which element is 2nd to the left of the one which is 4th from the left end in step II?\n(a) 22\n(b) 44\n(c) 89\n(d) 72\n(e) None of these\n\nQ5. What is the sum of the numbers at both the ends in step IV of the given arrangement?\n(a) 36\n(b) 63\n(c) 60\n(d) 16.5\n(e) None of the above\n\nSolutions (1-5):\nSol. In this input output question numbers are arranged following different rules in each step. Let us understand the logic behind it- In each step the numbers are arranged\nIn step 1: the numbers are multiplied with 3 and then 2 alternatively.\nStep 2: the first two numbers are subtracted, then 2nd and 3rd number are added and alternatively process is repeated till the last number.\nStep 3: Multiplication of the digits of the resultant in the previous step.\nStep4: The numbers in the previous step is divided by 2.", null, "S1. Ans(b)\nS2. Ans(d)\nS3. Ans(e)\nS4. Ans (c)\nS5. Ans (d)\n\nDirections (6-10): In each of the questions given below, a group of digits is given followed by four combinations of letters/ symbols numbered (a), (b), (c) and (d). You have to find out which of the four combinations correctly represents the group of digits based on the letter/ symbol codes and the conditions given below. If none of the four combinations represents the group of digits correctly, give (e) ‘None of these’ as the answer.", null, "Condition for coding the group digits:\n(i) If the first digit is odd and last digit is even, the codes for the first and the last digits are to be interchanged.\n(ii) If the first as well as the last digit is even, both are to be coded by the code for the last digit.\n(iii) If the first as well as the last digit is odd, both are to be coded as ‘X’.\n\nQ6. 285961\n(a) @PD=S\n(b) @∆D=S\n(c) @PV=S\n(d) @PD=SV\n(e) None of these\n\nS6. (a)\nSol. None of the conditions are applied.\n\nQ7. 972486\n(a) =∆@VPS\n(b) S∆@VP=\n(c) SD@VP=\n(d) SA@P=\n(e) None of these\n\nS7. (b)\nSol. Condition (i) Applies.\n\nQ8. 834762\n(a) PMV∆SP\n(b) PMV∆S@\n(c) @MVA∆SP\n(d) @MV∆S@\n(e) None of these\n\nS8. (d)\nSol. Condition (ii) applies\n\nQ9. 785291\n(a) ∆PD@X\n(b) ∆PD@=X\n(c) XPD@=X\n(d) XPD@=\n(e) None of these\n\nS9. (c)\nSol. Condition (iii) applies.\n\nQ10. 748956\n(a) AVP=DS\n(b) SVP=DS\n(c) ∆VP=D∆\n(d) SP=D∆\n(e) None of these\n\nS10. (e)\nSol. Condition (i) applies.\n\nDirections (11-15): Study the following information carefully and answer the given questions.\nA, B, C, D, W, X, Y and Z are eight friends sitting around a square table, two on each side. All of them are facing away from the centre and each is opposite to another. They all live on different floors of a two storey building with topmost floor being numbered as second and the bottommost floor being numbered as first. There are three female members and no two females are seated next to one another. Four of them live on each floor. X sits between D and Z. One of the male members whose immediate neighbours are also males lives on the topmost floor. Y is a female member who sits second to the left of X. Z is not a female member but sits opposite A, who is a female. C sits third to the left of W and is not male member. Only one female lives on the floor below the floor on which D lives.The person who sit opposite to B lives on second floor. A does not lives on topmost floor.C and D are immediate neighbours.\n\nQ11. Who among the following is not a male member?\n(a) W.\n(b) X\n(c) Y\n(d) B\n(e) None of these\n\nQ12. Which of the following statements is true about W and X?\n(a) Both are opposite to each other\n(b) They do not live on the same floor.\n(c) W is female, but X is a male\n(d) Both are females\n(e) None of these\n\nQ13. Which of the following groups includes only females?\n(a) YAW\n(b) ACB\n(c) XYZ\n(d) ACY.\n(e) None of these\n\nQ14. Who among the following is sitting between B and the female whose immediate neighbours live below her, when counted from the right of B?\n(a) A\n(b) C and D\n(c) C\n(d) A and W.\n(e) None of these\n\nQ15. Who amongst the following pair of females lives on the same floor?\n(a) DY\n(b) AY\n(c) AC\n(d) AB\n(e) None of these.\n\nSolutions (11-15):", null, "S11. Ans.(c)\nS12. Ans.(b)\nS13. Ans.(d)\nS14. Ans.(d)\nS15. Ans.(e)\n\nIf you are preparing for Bank exams, then you can also check out a video for Reasoning below:\n\nYou may also like to Read:" ]	[ null, "https://www.bankersadda.com/wp-content/uploads/2019/09/20150723/1.2.png", null, "https://www.bankersadda.com/wp-content/uploads/2019/09/20150725/ans01.png", null, "https://www.bankersadda.com/wp-content/uploads/2019/09/20151214/2221.png", null, "https://www.bankersadda.com/wp-content/uploads/2019/09/20151435/1-to5-300x235.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90158916,"math_prob":0.9649394,"size":5785,"snap":"2019-43-2019-47","text_gpt3_token_len":1682,"char_repetition_ratio":0.1539526,"word_repetition_ratio":0.064631954,"special_character_ratio":0.293172,"punctuation_ratio":0.10741886,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9779347,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T11:08:03Z\",\"WARC-Record-ID\":\"<urn:uuid:2a1db3de-c0f9-4d0c-9772-3724e3b6f570>\",\"Content-Length\":\"132624\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3248148-ffc3-4a87-9952-44073e52f61c>\",\"WARC-Concurrent-To\":\"<urn:uuid:4411c460-638f-4126-9802-1a94cf70eaad>\",\"WARC-IP-Address\":\"99.84.104.65\",\"WARC-Target-URI\":\"https://www.bankersadda.com/ibps-rrb-mains-reasoning-quiz-21-2\",\"WARC-Payload-Digest\":\"sha1:Z6BDPT6YMCJYH6XWOARSFPGRWMDHWA6W\",\"WARC-Block-Digest\":\"sha1:4JPAGMH5CD4YPY3BWDMELBCXNBVUYYOP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986658566.9_warc_CC-MAIN-20191015104838-20191015132338-00519.warc.gz\"}"}
https://linearalgebras.com/solution-abstract-algebra-exercise-5-1-9.html	[ "If you find any mistakes, please make a comment! Thank you.\n\n## Exhibit symmetric group as a subgroup of a general linear group\n\nSolution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.9\n\nSolution: We know that if $F$ is a finite field then $\\mathsf{Aut}(F^n) \\cong GL_n(F)$. This isomorphism $\\zeta$ can be defined as follows: given $\\theta \\in \\mathsf{Aut}(F^n)$, $\\zeta(\\theta)$ is the matrix in $GL_n(F)$ whose $i$-th row is precisely $\\theta(e_i)$. (In particular, $\\zeta$ is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In Exercises 5.1.7 and 5.1.8, we found an injective group homomorphism $\\psi : S_n \\rightarrow \\mathsf{Aut}(F^n)$. Combining these results we have an injective group homomorphism $\\Psi : S_n \\rightarrow GL_n(F)$, computed as follows: $$\\Psi(\\pi) = [e_{\\pi(1)}\\ \\cdots\\ e_{\\pi(n)}]^T.$$ We can see that each $\\Psi(\\pi)$ is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus $S_n$ is identified with a subgroup of $GL_n(F)$ consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.\n\nThus we have proven the result for a finite field $F$; we began with this case because it was shown previously that $\\mathsf{Aut}(F^n) \\cong GL_n(F)$. If this is true for arbitrary fields then the same proof carries over to arbitrary fields $F$; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields $F$ by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not $k = 0$ in $F$ for some integer $k$ never arises, so that all computations hold for arbitrary fields (in which $1 \\neq 0$) and in fact the set of permutation matrices over any $F$ is closed under matrix multiplication." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88668376,"math_prob":0.9994154,"size":1888,"snap":"2022-27-2022-33","text_gpt3_token_len":511,"char_repetition_ratio":0.10244162,"word_repetition_ratio":0.0,"special_character_ratio":0.2706568,"punctuation_ratio":0.09840426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999567,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T05:47:17Z\",\"WARC-Record-ID\":\"<urn:uuid:49c55833-6406-41d3-b1cf-ab2e5c147926>\",\"Content-Length\":\"35845\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8db292f8-dd5f-440e-b2b9-ed8b019c8a37>\",\"WARC-Concurrent-To\":\"<urn:uuid:2766fe33-d30a-47ba-8ee0-7d61aa1a3cf5>\",\"WARC-IP-Address\":\"104.21.5.104\",\"WARC-Target-URI\":\"https://linearalgebras.com/solution-abstract-algebra-exercise-5-1-9.html\",\"WARC-Payload-Digest\":\"sha1:UMHODIKZHYOP76ZV5NI7O22BYI2YC7Q6\",\"WARC-Block-Digest\":\"sha1:VWZEA6BIUV4PR5U7F3WRJK2357RBHGKI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104354651.73_warc_CC-MAIN-20220704050055-20220704080055-00561.warc.gz\"}"}
https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-021-02017-y	[ "# A four-quadrant mobility model-based routing protocol for post-earthquake emergency communication network\n\n## Abstract\n\nEmergency communication network (ECN) is the important infrastructure to acquire the real-time information after disaster, which is essential for rescue task. However, the existed routing protocols seldom consider the uneven distribution of rescue area so that cannot satisfy the ECN’s requirement and the quality of rescue needs improvement. In this paper, we avoided the traditional linear propulsion from rim to core and proposed a novel four-quadrant mobility model (FQMM), which makes rescuers arrive the most intensity core area first, which can improve the rescue quality. Then, a FQMM-based protocol (FQMMBP) for ECN is designed, which aims to improve the performance of ECN in terms of package delivery rate (PDR) and end-to-end delay. Finally, we set up a virtual earthquake scenario to simulate our proposed protocol in NS2. Results show that the proposed protocol outperforms to the three compared routing protocols, i.e., AODV, DSDV and DSR. The average FDR is improved by 16.31%, and the average reduction of delay is 64.45%, which shows our proposed scheme’s advantages in quality of rescue.\n\n## Introduction\n\n### Background\n\nAfter the devastating earthquake, emergency rescue crews are needed to rush to the scene for rescue as soon as possible. Post-earthquake emergency rescue includes material dispatching, team dispatching, personnel evacuation, post-disaster reconstruction, etc. Due to the different catastrophic degrees caused by the earthquake in different affected areas, the rescue urgency degrees (RUDs) are different for those areas. Seismic intensity, which refers to the catastrophic degree of earthquake impact on the ground or artificial buildings in a certain area , can be used to express the rescue urgencies of disaster areas. The seismic intensity can be obtained in the following ways : In the area where seismic observation equipment is intensively deployed, we can directly obtain the intensity distribution map of the instrument; In the area where seismic observation equipment is sparsely deployed, we can obtain the intensity distribution map of the instrument by grid interpolation; In the area where seismic observation equipment is rarely deployed, we can obtain the intensity distribution map either by historical seismic data statistics or earthquake simulations. Therefore, it is feasible to carry out emergency rescue according to the emergency degree of the disaster area reflected by seismic intensity. When seismic intensity is obtained, the degree of emergency rescue can be determined by the value of seismic intensity, e.g., after an earthquake, the degree of emergency rescue in high intensity areas is higher than that in low intensity areas, while in the region with equal seismic intensity, the degrees of emergency rescue are equal.\n\nAfter the devastating earthquake, the communication infrastructure and power system in the disaster area will suffer from different degrees of damage, resulting in communication system paralysis. Rescue crews have to quickly establish ECN in the disaster area to meet the communication demands both inside and outside the affected area. Unlike traditional self-organized network, the nodes in ECN have the characteristics of energy limitation, high energy consumption in communication and low energy consumption in data calculation. In recent years, the research on ECN routing protocols has been widely concerned by scholars.\n\n### Relate works\n\nIn the existing research on the mobility model of emergency rescue crews, the traditional task assignment model, e.g., dynamic multi-stage allocation model , optimal scheduling strategy and multi-objective optimization model , usually focuses on processing of rescue time. Song Ye et al. established an optimization model for the assignment of earthquake emergency rescue teams, aiming at improving rescue efficiency and maximizing the satisfaction of rescue time. Yuan Jinsha et al. considered demand urgency and satisfaction and constructed a dispatch model of multi-reserve point and multi-disaster point in two-ant group algorithm, which provides a reference to the disaster relief strategy for decision makers. Md. Ibrahim Talukdar et al. investigated shortest path map based (SPMB), random way point (RWP) and random walk (RW) models to measure the efficiency of mobility models. Cao Cejun et al. formulated a bi-level integer programming model (BIPM) to minimize total weighted travel time at the upper level and to maximize total weighted survivors’ perceived satisfaction at the lower level, and then, a case study from Wenchuan earthquake was presented to illustrate the proposed model and solution strategies. Song Yinghua et al. proposed a satisfaction function to measure the psychology of the victims. The emergency material distribution plan optimization model (EMDPOM) was established based on the satisfaction of the victims and the minimum total system cost, combined with the location problem of the emergency distribution center in the two-level distribution network. The model was solved by a real-coded genetic algorithm, and the data in the Wenchuan earthquake were taken as an example to verify the validity and feasibility of the model and algorithm. However, the above researches studied the mobility models from the perspective of rescue capacity and relief materials repertory, without considering seismic intensity in disaster areas. Furthermore, the above researches did not reasonably allocate the communication and rescue nodes based on the characteristics of ECN. Comparisons of traditional mobility models are listed in Table 1.\n\nIn this paper, we first propose a seismic intensity-based four-quadrant rescue mobility model (FQMM), and then, we simulate earthquake scene and propose a FQMM-based protocol for ECN. The main contributions of this paper are as follows.\n\nFirstly, this paper gives the topology of ECN for the scene of post-earthquake emergency rescue. In the proposed post-earthquake ECN, both fixed nodes (e.g., emergency communication vehicles, base stations) and mobile nodes (e.g., Unmanned Aerial Vehicles, portable communication devices) are included. The characteristics of ECN are frequent link breakage, inconsistencies in data rates, incompatibility of resources, the temporary unavailability of needed resources and communications links.\n\nSecondly, the FQMM mobility model for post-earthquake emergency rescue crews is proposed, which based on the topological structure of ECN. With FQMM, first responders not only can carry out grid search in disaster areas to ensure the effectiveness of rescue without missing any victims, but also can carry out radiation diffusion rescue from epicenter to ensure the worst affected areas are on top of the rescue list.\n\nFinally, this paper proposed a FQMM-based protocol (FQMMBP) for ECN. According to the location of the mobile node, FQMMBP predicts hop counts from source to destination, so that it can update the routing table in real time by selecting the optimal next-hop node, thus improving package delivery rate, reducing end-to-end delay.\n\nAlthough the routing protocol proposed in this paper only considers the factor of seismic intensity, the designed mobility model is universal and can be applied to other natural disasters, such as floods, fires, typhoons and so on.\n\n### Problem description\n\nThis paper aims to solve the problems of mobility, communication and rescue task arrangement in the earthquake rescue scenario. The detailed descriptions are as follows:\n\n1. 1.\n\nMobility problem: Based on the common sense of earthquake, the core, rather than the rim of earthquake, has more important emergency to rescue. Therefore, we intend to solve the mobility problem that realizes more rescuers reach the core of earthquake with the minimal traveling time.\n\n2. 2.\n\nRescue task arrangement problem: When the rescuers’ mobility is finished, the paper will solve the problem of rescue task arrangement problem. There is more than one rescuers in each divided zone. And what is the optimal number of arranged rescuers for each zones should be modeled and solved via algorithms.\n\n3. 3.\n\nCommunication problem: Due to the mobility of rescuers, the paper should consider the real-time connectivity communication among the rescue vehicle and rescuers. The proposed routing protocols should guarantee the communication link from rescuers to rescue vehicles during the process of mobility and rescue task, which based on relay connective probability.\n\nOverall, this paper mainly solves the problems defined as above.\n\n## Method\n\n### Scenario and assumption\n\nIn the post-earthquake ECN, the emergency communication vehicle (ECV) and base station (BS) are regarded as fixed nodes, while rescue crews equipped with individual communication equipment are regarded as mobile nodes. Unmanned aerial vehicle (UAV) can be used not only as a fixed node for relaying communication, but also as a mobile node for collecting disaster information. A typical post-earthquake ECN is shown in Fig. 1.\n\nAfter the earthquake, rescuers equipped with individual communication equipment were scattered to various disaster areas for rescue. Each communication node in ECN, such as ECV, UAV and portable individual equipment (PIE), has its own communication coverage. Since ECV cannot reach the center of the disaster area due to traffic control, it starts at the midpoint of length, where rescue crews start and stop rescue missions, as shown in Fig. 1. In the rescue process, ECV can provide real-time communication for rescue crews. Due to its mobility and air superiority, UAV can go deep into the disaster area and play the role of communication relay or disaster information collector. However, due to the limited battery capacity and high costs of UAV, we are unable to deploy a large number of UAVs in the disaster area. Rescuers can enter the disaster area with PIE, however, due to the limited coverage, high cost and large energy consumption, PIE cannot play the role of hot spot or relay for a long time. Therefore, how to cooperate with the communication equipment in ECN to improve the communication efficiency is one of the key issues that need to be paid attention to in post-earthquake emergency rescue. Therefore, heterogeneous communication devices in ECN need to cooperate with each other through appropriate routing protocols to improve communication efficiency.\n\nAfter the earthquake, we use rectangle to represent the earthquake influence range. The length and width of the rectangle are ai and bi (i = 1…n), respectively. Before studying the mobile model, we propose the following assumptions for simulation calculation:\n\n### Assumption 1\n\nIn regions with equal seismic intensity, the values of RUD are equal.\n\n### Assumption 2\n\nEach rescuer has got equal rescue efficiency.\n\n### Assumption 3\n\nRescue tasks in low RUD areas are at high priority. Rescue crews cannot start rescue tasks in higher RUD areas until rescue tasks in low RUD areas are completed.\n\n### Assumption 4\n\nRandom waypoint mobility model for rescuers is used in each disaster area. In order to ensure the fairness and rationality of the rescue sequence, the rescuers are approximately evenly distributed in disaster areas with equal RUD values.\n\n### Assumption 5\n\nLocations and number of rescue tasks are known by rescue crews in advance, since the Global Positioning System (GPS) is used.\n\nBased on seismic intensity and number of rescuers in disaster areas, this paper proposes a four-quadrant mobility model for rescue crews, which improves the rescue efficiency and ensures the safety of the victims' lives and properties. The proposed mobility model mainly includes two aspects: Firstly, the most seriously affected areas are rescued as soon as possible; secondly, the rescue crews are allocated to each appropriate quadrant, which improves the rescue efficiency and shortens the time of rescue.\n\nOur proposed protocols and algorithms are deployed into the rescuers, ECV and UAVs, which use the WSN technologies as platform and infrastructures. Considering the damage of base station and internet on the earthquake scenario, we use ad hoc networks to realize our proposed communication. The ECV is regarded as a tiny server, which is stopping at the rim of the damage area and calculating the RUD and mobility results for each rescuers. And all the rescuers’ mobile information and rescue task status will be send back to the ECV. We assume that every rescuer is implemented an on-broad unit (OBU), which uses WSN technologies. The rescuers can share their basic information, including the real-time position, the task procedure, the remaining energy and the mobile speed, etc. The OBU can realize the rescuers’ communication to other rescuers, ECV and UAVs. As the temporal relay, UAV will improve the communication performance when one of rescuer on the communication link is broken or exhaust energy. UAV will relay the broken communication link and re-arrange the task that exhausted rescuer to other working rescuers.\n\nBased on the overall idea, our proposed algorithms can be implemented by current technologies and can realize effective and real-time communication among rescuers, ECV, and UAVs.\n\n### Disaster area division\n\nDisaster area is divided into n layers according to seismic intensity. The epicenter is located in the first layer (RUD = 1) where length is a1 and width is b1. We define RUD of each region as 1, 2, …, n. Epicenter located at the center of rectangle where RUD = 1. All areas are rectangular rings, except for the one with RUD = 1, which is a rectangle. ECV parks at the midpoint of the length outside the rectangular ring with RUD = n, as shown in Fig. 2.\n\nThe length and width of each disaster area are marked as {RUD\|ai, bi}; then, each region can be expressed as follows:\n\n\\begin{aligned} {\\text{Region 1:}} & \\;\\left\\{ {{\\text{RUD}} = 1\|a_{1} ,b_{1} } \\right\\}; \\\\ {\\text{Region 2:}} & \\;\\left\\{ {{\\text{RUD}} = 2\|a_{2} = a_{1} + \\frac{{a_{1} }}{2} + \\frac{{a_{1} }}{2} = 2a_{1} ,b_{2} = b_{1} + \\frac{{b_{1} }}{2} + \\frac{{b_{1} }}{2} = 2b_{1} } \\right\\}; \\\\ {\\text{Region 3:}} & \\;\\left\\{ {{\\text{RUD}} = 3\|a_{3} = a_{2} + \\frac{{a_{2} }}{2} + \\frac{{a_{2} }}{2} = 2a_{2} = 4a_{1} ,b_{3} = b_{2} + \\frac{{b_{2} }}{2} + \\frac{{b_{2} }}{2} = 2b_{2} = 4b_{1} } \\right\\}; \\\\ \\end{aligned}\n\nAccording to mathematical induction, we can get:\n\nRegion i: {RUD = 3\|$$a_{i} = 2a_{{i - 1}} = 2^{{i - 1}} a_{1}$$,$${\\text{~}}b_{i} = 2b_{{i - 1}} = 2^{{i - 1}} b_{1}$$}.\n\n### FQMM model\n\nIn this section, a four-quadrant rescuers allocation scheme is proposed. Based on this scheme, a four-quadrant mobility model (FQMM) for rescuers is proposed. The objectives of the allocation scheme are: (1) carrying out grid search in disaster areas to ensure the effectiveness of rescue without missing any victims; (2) carrying out radiation diffusion rescue from epicenter to ensure the worst affected areas are on top of the rescue list. From the disaster area division discussed in Sect. 3.2, it can be seen that when RUD = 1, the region is a rectangle, and when RUD > 1, the regions are rectangular rings. Therefore, the discussion of rescuers allocation scheme includes two aspects: RUD = 1 and RUD > 1. The core ideas of rescuers allocation scheme for regions with RUD > 1 are similar, though the lengths of the regions are different.\n\n#### Mobility model for rescuers in regions with RUD = 1\n\nFive steps (Step 1 to Step 5) are included in the mobility model of rescuers for regions with RUD = 1.\n\nStep 1 As shown in Fig. 3, we establish a coordinate system which takes the epicenter as the origin. X and Y axes are the length and width of the rectangle. Four quadrants, QID = 1, QID = 2, QID = 3 and QID = 4 are counter-clockwise defined according to the definition of coordinate system. Each quadrant is described as follows:\n\n\\begin{aligned} & {\\text{Quadrant 1:}}\\;\\left\\{ {{\\text{RUD}} = 1\|{\\text{QID}} = 1} \\right\\};\\;{\\text{ Quadrant 2:}}\\;\\left\\{ {{\\text{RUD}} = 1\|{\\text{QID}} = 2} \\right\\}; \\\\ & {\\text{Quadrant 3:}}\\;\\left\\{ {{\\text{RUD}} = 1\|{\\text{QID}} = 3} \\right\\};\\;{\\text{Quadrant 4:}}\\;\\left\\{ {{\\text{RUD}} = 1\|{\\text{QID}} = 4} \\right\\}. \\\\ \\end{aligned}\n\nSuppose that the total number of rescuers is M, and the total number of rescue tasks in the area with RUD = 1 is Q, which is, $${\\text{RT}}_{{{\\text{QL}} = 0}}^{{{\\text{RUD}} = 1}} = Q$$, where $${\\text{RT}}_{{{\\text{QL}} = 0}}^{{{\\text{RUD}} = 1}}$$ stands for total number of rescue tasks to be completed in disaster areas with RUD = 1 and QL = 0. The number of tasks to be performed in each quadrant is described as: $${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}} = Q/4,\\quad \\left( {k = 1,2,3,4} \\right)$$, where $${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}}$$ stands for the number of tasks to be performed in disaster areas with RUD = 1 and QID = k (k = 1, 2, 3, 4). Rescuers are evenly allocated to four quadrants with QL = 1, and the number of rescuers allocated (NRA) in each quadrant can be discussed from four cases (Case 1 to Case 4) as follows:\n\nCase 1: Mod[M/4] = 0, ere function Mod[x/y] is used to return the remainder of the division of two numbers x, y. In this case, a quarter of rescuers are evenly allocated to four quadrants with QL = 1, which is:\n\n$${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}} = \\frac{M}{4},\\quad \\left( {k = 1,2,3,4} \\right)$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in each quadrant with RUD = 1 and QID = k.\n\nCase 2: Mod[M/4] = 1. In this case, one more rescuer is allocated to Quadrant 1, which are:\n\n$${\\text{~NRA}}_{{{\\text{QID}} = 1}}^{{{\\text{RUD}} = 1}} = \\frac{{M - 1}}{4} + 1$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = 1}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadrant 1 with RUD = 1, and $${\\text{NRA}}_{{{\\text{QID}} \\ne 1}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadrant 2, Quadrant 3 and Quadrant 4 with RUD = 1.\n\nCase 3: Mod[M/4] = 2. In this case, one more rescuer is allocated to Quadrant 1 and Quadrant 2, respectively, which are:\n\n$${\\text{NRA}}_{{{\\text{QID}} = 1\\;{\\text{or}}\\;2}}^{{{\\text{RUD}} = 1}} = \\frac{{M - 2}}{4} + 1$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = 1\\;{\\text{or}}\\;2}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadrant 1 and Quadrant 2 with RUD = 1, and $${\\text{NRA}}_{{{\\text{QID}} = 3\\;{\\text{or}}\\;4}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadrant and Quadrant 4 with RUD = 1.\n\nCase 4: Mod[M/4] = 3. In this case, one more rescuer is allocated to Quadrant 1, Quadrant 2, Quadrant 4 which are:\n\n$${\\text{NRA}}_{{{\\text{QID}} \\ne 3}}^{{{\\text{RUD}} = 1}} = \\frac{{M - 3}}{4} + 1$$\n\nwhere $${\\text{NDN}}_{{{\\text{QID}} \\ne 3}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadrant 1, Quadrant 2 and Quadrant 4 with RUD = 1, and $${\\text{NRA}}_{{{\\text{QID}} = 3}}^{{{\\text{RUD}} = 1}}$$ stands for the number of rescuers allocated in Quadran3 with RUD = 1.\n\nStep 2 According to the procedure of step 1, each quadrant with RUD = 1 is further divided into four secondary quadrants as shown in Fig. 4. Two elements are included in QID set, which record the quadrant ID from the first level to the last level. For example, {RUD = 1\|QID = {1, 3}} in Fig. 4 means region with QL = 1 is located in Quadrant 1 and region with QL = 2 is located in Quadrant 3.\n\nStep 3 For QL = x, replace the value of M with $${\\text{RT}}_{{{\\text{QL}} = x}}^{{{\\text{RUD}} = 1}}$$, repeat Step 1 and calculate the number of rescuers allocated for each quadrant with QL = x.\n\nStep 4 Repeat Step 2 and Step 3, until any of the following Abort conditions is satisfied.\n\nAbort condition 1 There is only one rescuer in each area of quadrant with QL = x, which is: $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}} = 1,\\quad \\left( {k = 1,2,3,4} \\right)$$, where $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}}$$ is the number of rescuers allocated in Quadrant k (k = 1, 2, 3, 4) with RUD = 1 and QL = x. Under this condition, further quadrant division is no longer meaningful, so the single rescuer will complete all rescue tasks in the region of quadrant with QL = x.\n\nAbort condition 2 The number of rescue tasks is less than that of rescuers assigned in the quadrant with QL = x, which is:$$~{\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}} \\ge {\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}} ,\\quad \\left( {k = 1,2,3,4} \\right)$$, where $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}}$$ is the number of rescuers allocated in Quadrant k (k = 1, 2, 3, 4) with RUD = 1 and QL = x, and $${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = 1}}$$ is number of rescue tasks in Quadrant k (k = 1, 2, 3, 4) with RUD = 1 and QL = x. Under this condition, each rescuer is assigned to at most one rescue task, and there are unoccupied rescuers in the area.\n\nStep 5 Quadrant division is suspended and rescuers begin to perform their own rescue tasks.\n\nFigure 5 shows the algorithm flow of rescue node allocation in rectangular disaster area when RUD = 1.\n\n#### Mobility model for rescuers in regions with RUD > 1\n\nAccording to the mobility model, rescue crews cannot start rescue tasks in higher RUD areas until rescue tasks in low RUD areas are completed. As shown in Fig. 6, in the gray area with RUD = 1, rescuers are carrying out rescue tasks. When the task is over, rescuers will be assigned to the white rectangular ring area with RUD = 2.\n\nFive steps (Step 6 to Step 10) are included in the mobility model of rescuers for regions with RUD > 1.\n\nStep 6 As shown in Fig. 7, we establish four coordinate systems which takes O1, O2, O3 and O4 as the origins, respectively. Then, the rectangular ring is equally divided into twelve quadrants (the white parts in Fig. 7) by four coordinate systems. In each coordinate system, four quadrants are equal with each other in area, as shown in Fig. 7 where four triangle marks.\n\nStep 7 It can be seen from Step 6 that in each coordinate system, there are three quadrants waiting for allocation of rescuers, and the left quadrant is allocated with rescue crews. As shown in Fig. 7, in the coordinate system with O3 as origin, there are four quadrants where Triangle (1), Triangle (2), Triangle (3) and Triangle (4) marks. Area where Triangle (1) marks is allocated with rescue crews, while areas where Triangle (2), Triangle (3) and Triangle (4) marks are waiting for rescuers allocations. We take coordinate system O3 as an example. Suppose there are M rescuers in Quadrant 1 with RUD = i − 1, which marked with Triangle (1). There are Q rescue tasks in areas with RUD = i, which is: $${\\text{RT}}_{{{\\text{QL}} = 0}}^{{{\\text{RUD}} = i}} = Q$$. Rescuers are evenly allocated to three quadrants [Triangle (2), Triangle (3) and Triangle (4)] with QL = 1, and the number of rescuers allocated (NRA) in each quadrant can be discussed from three cases (Case 5 to Case 7) as follows:\n\nCase 5 Mod[M/3] = 0, where function Mod[x/y] is used to return the remainder of the division of two numbers x, y. In this case, one-third rescuers are evenly allocated to three quadrants, which is:\n\n$${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}} = \\frac{M}{3},\\quad \\left( {k = 2,3,4} \\right)$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}}$$ stands for the number of rescuers allocated in each quadrant with RUD = i and QID = k.\n\nCase 6 Mod[M/3] = 1. In this case, one more rescuer is allocated to Quadrant 2, which are:\n\n$${\\text{NRA}}_{{{\\text{QID}} = 2}}^{{{\\text{RUD}} = i}} = \\frac{{M - 1}}{3} + 1$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = 2}}^{{{\\text{RUD}} = i}}$$ stands for the number of rescuers allocated in Quadrant 2 with RUD = i, and $${\\text{NRA}}_{{{\\text{QID}} = 3\\;{\\text{or}}\\;4}}^{{{\\text{RUD}} = i}}$$ stands for the number of rescuers allocated in Quadrant 3 and Quadrant 4 with R = i.\n\nCase 7 Mod[M/3] = 2. In this case, one more rescuer is allocated to Quadrant 2 and Quadrant 4, respectively, which are:\n\n$${\\text{NRA}}_{{{\\text{QID}} = 2\\;~{\\text{or}}\\;4}}^{{{\\text{RUD}} = i}} = \\frac{{M - 2}}{3} + 1$$\n\nwhere $${\\text{NRA}}_{{{\\text{QID}} = 2~\\;{\\text{or}}\\;4}}^{{{\\text{RUD}} = i}}$$ stands for number of rescuers allocated in Quadrant 2 and Quadrant 4 with RUD = i, and $${\\text{NRA}}_{{{\\text{QID}} = 3}}^{{{\\text{RUD}} = i}}$$ stands for the number of rescuers allocated in Quadrant 3 with RUD = i.\n\nThe number of rescue tasks in each quadrant with RUD = i is:$${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}} = Q/12,\\quad \\left( {k = 2,3,4} \\right)$$, where $${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}}$$ stands for the number of rescue tasks in Quadrant k (k = 2, 3, 4) wh RUD = i.\n\nStep 8 Quadrants (QL ≥ 2) division methods are the same as illustrated in Step 6, which take the middle points of quadrants with QL = 1 as the origins and establish the coordinate systems. Repeat Step 7, until any of the following Abort conditions is satisfied.\n\nAbort condition 3 There is only one rescuer in each area of quadrant with QL = x, which is:$${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}} = 1,\\quad \\left( {k = 1,2,3,4} \\right)$$, where $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}}$$ is the number of rescuers allocated in Quadrant k (k = 1, 2, 3, 4) with RUD = i and QL = x. Under this condition, further quadrant division is no longer meaningful, so the single rescuer will complete all rescue tasks in the region of quadrant with QL = x.\n\nAbort condition 4 The number of rescue tasks is less than that of rescuers assigned in the quadrant with QL = x, which is:$${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}} \\ge {\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}} ,\\quad \\left( {k = 1,2,3,4} \\right)$$, where $${\\text{NRA}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}}$$ is the number of rescuers allocated in Quadrant k (k = 1, 2, 3, 4) with RUD = i and QL = x, and $${\\text{RT}}_{{{\\text{QID}} = k}}^{{{\\text{RUD}} = i}}$$ is number of rescue tasks in Quadrant k (k = 1, 2, 3, 4) with RUD = i and QL = x. Under this condition, each rescuer is assigned to at most one rescue task, and there are unoccupied rescuers in the area.\n\nStep 9 Quadrant division is suspended and rescuers begin to perform their own rescue tasks.\n\nFigure 8 shows the algorithm flow of rescue node allocation in rectangular disaster area when RUD > 1.\n\n### FDMM-based routing protocol\n\nIn this section, a proposal of FQMM-based protocol (FQMMBP) for ECN is presented. First, we discuss route discovery and maintenance of FQMMBP, and then we propose FQMMBP protocol by analyzing the communication mechanism of nodes in ECN.\n\nPost-earthquake ECN, which is a hybrid self-organized network, mainly includes ECV, UAV, portable devices carried by rescue crews, base station, satellite, etc. Those communication devices can be classified as fixed communication nodes (such as ECV, base station, UAV used for relay.) and mobile communication nodes (such as portable devices carried by rescue crews, satellite.). As shown in Fig. 1, each portable device is equipped with a wireless communication module. Rescuers can use portable devices to build self-organized networks, through which they can communicate with each other and complete the transmission process of data packets, such as receiving, transmitting and relaying. Rescuers can also communicate with ECV, UAV and other fixed nodes within the coverages of portable devices. As a fixed communication node, ECV can communicate with rescuers and UAVs in the disaster area through downlink, or with the base station, satellite and other communication facilities through uplink. Through the backbone network, commanders can get the information about rescue crews, rescue process and suffering condition in time, so as to make scientific rescue decision and emergency response quickly.\n\nWe define the communication radius of mobile node as R. The mobile node sends data to the emergency communication vehicle in the way of multi-hop. In such communication mode, the source is the mobile node in the disaster area, the destination is ECV at the edge of the disaster area, and the relay nodes are mobile nodes or fixed nodes (UAVs, etc.) in the disaster area. On the premise of ensuring the stability of communication links, we choose relay nodes or links that can minimize the delay for data transmission. Based on the mobility model we discussed in Sect. 3.3, it can be seen that rescue crews are nearly evenly distributed in disaster areas with the same RUD. In order to ensure the stability of the communication link in the rescue process, we place fixed UAVs at the edges of disaster areas with different RUD values; however, they do not undertake the rescue tasks. When all rescue tasks in the disaster area with a specific RUD value are completed, the relay UAV in the area returns to the location of ECV.\n\n#### Routing setup\n\nIn order to reduce the data transmission delay, we select the next hop relay node according to RUD value, QL and QID of each region where mobile nodes are located in. RUD and QL are used to define the level of the region where mobile nodes are located in. QID is a set which includes 1, 2, 3 and 4, and there are QL elements in the set. According to the set, the mobile nodes learn the division of the regions and the quadrants position, so as to select the appropriate next hop mobile node. Therefore, in the process of routing discovery, it is necessary to extend the routing request (RREQ) message, so as to meet the requirements of our proposed communication mechanism of FQMMBP. We improved the traditional RREQ message by adding four new fields: RUD, QL, QID and QW. RUD describes the area where nodes are located and judge whether the area is a rectangle or a rectangular ring as shown in Fig. 2. QL describes the level of the quadrant where nodes are located, and it determines the relative position of the mobile rescue node. QID describes the division of the quadrant where nodes are located and judge its relative position to ECV. When the QL of quadrants are the same, QID helps pick up the mobile rescue node which is closer to ECV. QW calculates weights of nodes according to QL and QID, so as to select the most appropriate next-hop relay node.\n\nIn the process of route discovery, nodes broadcast RREQ messages to send request information to their neighbors. The improved RREQ data format is shown in Table 2.\n\nEach field of RREQ data format is described as follows:\n\n• Type: The length of this field is 8bit, and the type value of RREQ is 1.\n\n• Type: The length of this field is 8bit, and the type value of RREQ is 1.\n\n• Flags: The length of this field is 5 bits, and five identities (J, R, G, D, and U) are included in this field. \"J\" is a joint flag and is generally used for multicast. “R” is for route repairing and is used for multicast transmission. “G” represents the list of nodes around ECV which can be communicated. Flag “G” determines whether the RREQ message can be directly sent to the destination or not. “D” is the reply flag of the destination node which determines whether the destination node is allowed to reply to the received RREQ message or not. “U” is the flag of unknown serial number. U = 1 means that the serial number of the node is unknown.\n\n• Reserved: Reserved field for further improvement of RREQ message.\n\n• Hop Count: This field registers the hop counts that RREQ passes from the source node to the current node.\n\n• RREQ ID: This field is the unique identity of RREQ message.\n\n• RUD: Rescue Urgency Degree, which is inversely proportional to the seismic intensity value in this area.\n\n• QL: Quadrant Level, which is the number of times the quadrant divided.\n\n• QW: Weights of nodes in quadrant.\n\n#### Communication mechanism\n\n$$D_{{{\\text{QID}}}} = \\left\\{ {\\begin{array}{{20}l} c \\hfill & {{\\text{QID}} = 1\\;{\\text{or}}\\;2} \\hfill \\\\ d \\hfill & {{\\text{QID}} = 3\\;{\\text{or}}\\;4} \\hfill \\\\ \\end{array} } \\right.$$\n(1)\n$$c = \\sqrt {\\left( {\\frac{a}{4}} \\right)^{2} + \\left( {\\frac{b}{4}} \\right)^{2} }$$\n(2)\n$$d = \\sqrt {\\left( {\\frac{a}{4}} \\right)^{2} + \\left( {\\frac{{3b}}{4}} \\right)^{2} }$$\n(3)\n\nwhere a, b are length and width of disaster area, respectively.$$~D_{{PH}}$$ stands for the distance between the center of each quadrant and ECV. c is the distance between the center of Quadrant 1 and ECV, which equals with the distance between the center of Quadrant 2 and ECV. d is the distance between the center of Quadrant 3 and ECV, which equals with the distance between the center of Quadrant 4 and ECV. Take the weight of Quadrant 1 (or Quadrant 2) as the benchmark, which is: $$W_{{{\\text{QID}} = 1\\;{\\text{or}}\\;2}} = 1$$, where $$W_{{{\\text{QID}} = 1\\;{\\text{or}}\\;2}}$$ is the weight of Quadrant 1 or Quadrant 2. The weight of Quadrant 3 (or Quadrant 4) can be expressed as:\n\n$$W_{{{\\text{QID}} = 3\\;{\\text{or}}\\;4}} = \\frac{d}{c} = \\frac{{\\sqrt {a^{2} + 9b^{2} } }}{{\\sqrt {a^{2} + b^{2} } }}$$\n(4)\n\nwhere $$W_{{{\\text{QID}} = 3\\;{\\text{or}}\\;4}}$$ is the weight of Quadrant 3 or Quadrant 4.\n\nDefinition $${\\text{QID}}_{i}$$ is the Quadrant ID with QL = i. For example, $${\\text{QID}}_{3} = 4$$ means Quadrant 4 is with QL = 3. From Eqs. (1) and (4), we can get:\n\n$$D_{{{\\text{QID}}_{{\\text{i}}} }} = \\left\\{ {\\begin{array}{{20}l} {c_{i} } \\hfill & {{\\text{QID}}_{i} = 1\\;{\\text{or}}\\;2} \\hfill \\\\ {d_{i} } \\hfill & {{\\text{QID}}_{i} = 3\\;{\\text{or}}\\;4} \\hfill \\\\ \\end{array} } \\right.$$\n(5)\n\nwhere $$D_{{{\\text{QID}}_{i} }}$$ stands for the distance between the center of quadrant with QIDi and ECV. ci is distance between the center of Quadrant 1 with QL = i and ECV, which equals with the distance between the center of Quadrant 2 with QL = i and ECV. di is distance between the center of Quadrant 3 with QL = i and ECV, which equals with the distance between the center of Quadrant 4 with QL = i and ECV. From Eq. (4), we can get:\n\n$$W_{{{\\text{QID}}_{i} }} = \\frac{{\\sqrt {a_{i} ^{2} + 9b_{i} ^{2} } }}{{\\sqrt {a_{i} ^{2} + b_{i} ^{2} } }}$$\n(6)\n\nwhere $$W_{{{\\text{QID}}_{i} }}$$ is the weight of node located in Quadrant $${\\text{QID}}_{i}$$. ai, bi are length and width of disaster area with QL = i, respectively. According to the QID and QL, we can get the total weight of each node in quadrants with different levels by:\n\n$${\\text{QW}} = \\mathop \\sum \\limits_{{i = 1}}^{{{\\text{QL}}}} W_{{{\\text{QID}}_{i} }}$$\n(7)\n\nwhere QW is the total weight of one node in quadrants with different levels.$${\\text{~}}W_{{{\\text{QID}}_{i} }}$$ is the weight of node located in Quadrant $${\\text{QID}}_{i}.$$\n\n#### Routing discover and maintain\n\nIn this section, we propose FQMMBP protocol based on FQMM. Routing discovery and routing maintenance process are illustrated in Figs. 10 and 11, respectively.\n\nAs shown in Fig. 10, source node (denoted as NS) checks whether there is a route to ECV. If there is no route to ECV, the node broadcasts RREQ message to its neighbor node (denoted as NN). After receiving the message, NN first checks whether NS is a valid node. If NS is invalid, NN discards the received RREQ message; otherwise, NN updates the routing table and finds out whether there is routing information to ECV. If there is no route to ECV, NN forwards the received RREQ message. Repeat the broadcasting and checking process until one or more valid routes to the ECV are found.\n\nAs shown in Fig. 11, source node (denoted as NS) sends hello packets to the candidate nodes within its coverage by broadcasting and receives RREQ replies from them. NS determines whether there is a RREQ reply sent by ECV by checking whether there is a candidate node with RUD = n and both QL and QID are empty in RREQ reply. If there is an ECV within the coverage of NS, data packets are directly transmitted to ECV which is the destination of the communication, and communication ends. If there is no ECV within the coverage of NS, then NS determines whether there is a RREQ reply sent by fixed UAV by checking whether there is a candidate node with RUD ≠ n and both QL and QID are empty in RREQ reply. If there is an UAV within the coverage of NS, data packets are forwarded to UAV, and UAV forwarded data packets to its neighbor. Repeat this process until ECV is founded, finally communication ends. If there are neither ECVs nor UAVs in the candidate nodes, we compare QW values. If there is only one candidate node corresponding to the maximum QW value, then this node is the optimal next hop node (denoted as NOPT). NOPT becomes a new source node NS, and then it sends the hello packets to repeat the process. If the maximum QW value corresponds to more than one candidate nodes, we should update candidate nodes set by removing nodes whose QW is not the maximum. Moreover, the protocols flowchart adds a judgment module to take the energy consumption into consideration. The model will test the candidate relays’ remaining energy and selects the suitable candidate as best relay to achieve the communication so that the ECNs which remain the fewer energy will be weeded out from the candidate set to guarantee the quality of rescue. After that, we select NOPT from the new set according to the following Cases:\n\nCase 1: There is only one candidate node with the maximum number of Quadrant 1. This node is selected as NOPT.\n\nCase 2: The numbers of Quadrant 1 are equal, while there is only one candidate node with the maximum number of Quadrant 2. This node is selected as NOPT.\n\nCase 3: The numbers of Quadrant 2 are equal, while there is only one candidate node with the maximum number of Quadrant 4. This node is selected as NOPT.\n\nCase 4: The numbers of Quadrant 4 are equal, while there is only one candidate node with the maximum number of Quadrant 3. This node is selected as NOPT.\n\nCase 5: Recalculate the weights of each candidate node according to MCDM-ECP algorithm and then select NOPT.\n\nThe description of FQMMBP routing maintenance process is completed.\n\n### An example\n\nIn order to understand the entire processes of our propose FQMMBP protocol clearly, we give an example to illustrate the main processes including the disaster area division, the main mobility in different RUD, and the relay selection in communication link based on FQMM.\n\nAs shown as Fig. 12a, we give an example of a disaster area of 6 km * 4 km. According to the degree of emergency importance, the 1 km * 2 km core of earthquake will be divided as RUD = 1, which means the mobile rescuers will do the task in this area firstly. Based on the equation forehead Sect. 2.2, we can determine and calculate the length and width of each RUD. Besides RUD = 1 area is rectangular, all the areas of RUD > 1 are rectangular rings. Based on the results of disaster area division, the smaller RUD id will be rescued earlier. For this instance, we divide 3 disaster areas, RUD = 1 will be rescued first. RUD = 2 area is rescued once the tasks in RUD = 1 area are finished. So are the tasks in area of RUD = 3.\n\nAs shown as Fig. 12b, we suppose that 50 rescuers are set into the disaster area. When RUD = 1, the shape of area is rectangular, we first divide the area into four quadrants. According to the formulae in Sect. 2.3.1 50 mod 4 = 2. Therefore, the remaining 2 rescuers is arranged into first and second quadrants. As a result, 50 rescuers are arranged into four quadrants as 13, 13, 12 and 12 rescuers.\n\nAs shown as Fig. 12c, RUD = 2 and RUD = 3 are the same rectangular rings, which the rescuers are focused in one quadrant of the whole sub-quadrants. For the upper 2 sub-quadrants, 13 mod 3 = 1. Therefore, 13 rescuers are arranged to the other 3 quadrants as 5, 4 and 4 rescuers. For the lower 2 sub-quadrants, 12 mod 3 = 0. Therefore, the 12 rescuers mobile into the other 3 quadrants evenly. Attentionally, the rescuers must mobile and begin the next RUD’s tasks after the last RUD’s tasks are finished. And the additional one rescuer is often arranged to the quadrants that are nearly to the central line.\n\nAs shown as Fig. 12d, the communication link is setup from the communication vehicle that located at the rim of disaster area to the rescuers in four quadrants. The core communication link goes throughout the central line of disaster area so that the conditions of all quadrants can be acquired. Based on the formulae in Sect. 2.4, every rescuers as a sequence to identify its own position in the quadrants. The routing protocol will select the suitable relay nodes, as shown as red triangular in the figure to realize the communication to rescue vehicle. As cluster head, these relay nodes will also collect the real-time traffic and rescue information in their own sub-quadrants.\n\n## Results and discussion\n\nIn this section, we present the network level performance evaluation and simulations results of the proposed routing protocol based on NS-2 platform. For comparison, with the respect to each relevant class of routing protocols, we selected the pertinent protocols which are the most widely used (e.g., AODV, DSR, DSDV).\n\n### Experiment setup\n\nThe simulation environment is a 10 km × 10 km earthquake disaster area, in which the epicenter of the most severely affected area is 6 km × 4 km. There are 48 mobile rescue nodes, two fixed UAVs and one ECV. The simulation setup and respective parameters are detailed in Table 3.\n\n### Experiment results\n\nIn this section, four routing protocols, e.g., FQMMBP, AODV, DSDV and DSR, are analyzed from the perspective of package delivery rate (PDR), end-to-end delay (delay) and overhead.\n\n#### Package delivery rate\n\nFigure 13 shows the performance of Package Delivery Rate (PDR) for the different studied protocols. Our proposed FQMMBP achieves the best performance with nearly 91.24% of average PDR while the other three protocols with about 80% of average PDR. This proves that FQMMBP improves communication efficiency and stabilizes communication link status in a long period of time. DSDV achieves the worst performance with 79.23% of average PDR mainly due to the mobility of communication nodes. DSDV is a proactive protocol, and each node maintains a route table. In the emergency rescue scenario, routing tables are dynamic due to the mobility of rescuers, thus causing the cost of maintaining routing tables. The superiority of FQMMBP over other three routing protocols is mainly attributed to FQMM mobility model. In each disaster area under this model, the distribution of rescue mobile nodes tends to be uniform, which increases the probability that each node can find the appropriate next-hop node to forward data and ensures the integrity and reliability of communication link. Packets are likely to be delivered successfully in such distributions of nodes.\n\nCompared to other protocols, our proposed FQMMBP has the best performance in FDR, which is increased to AODV, DSDV and DSR as 15.81%, 17.59% and 15.54%, respectively. Therefore, our proposed routing protocol can improve average FDR as 16.31%. And when the simulation time denotes 450 s, our FQMMBP has the best increments in FDR compared to DSDV, which rises 32.87%. However, the FDR performance of FQMMBP at the beginning of simulation has the fewer advantage, because the rescuers should move into the core area of earthquake rather than the rim of scenario, which the FDR will be weak.\n\nPDR performances of the four compared protocols are almost the same before 150 s, that is because all nodes are concentrated in the central area with RUD = 1 at the beginning of simulation. At this point, the distributions of nodes in the four protocols are similar, and each node has not started to disperse or move to the next area according to the mobility model. As time goes on, the PDR of each protocol decreases gradually. This is because when the nodes complete the rescue tasks in the central area, they will spread to the next peripheral rectangular ring, resulting in the scattered distribution of nodes and the increase in distance between nodes, so the stability of communication link becomes poor. On the contrary, mobile nodes are evenly distributed and close to ECV under our proposed FQMMBP, which improves PDR.\n\n#### End-to-end delay\n\nFigure 14 shows the performance of End-to-end delay (delay) for the different studied protocols. In terms of delay, FQMMBP is significantly lower than other classical routing protocols. During the simulation, the average delay decreased from 1.44 s of DSR to 0.55 s of FQMMBP, which proves the significant improvement of communication performance. This is because in each rescue area, the distribution of mobile nodes tends to be uniform under FQMMBP, which increases the probability that each node can find the next hop node to communicate with. Within the coverage of each relay node, the existence probability of candidate next hop node increases, which shortens the time to find the candidate next hop node as well as the time to transmit data to ECV.\n\nCompared to other protocols, our proposed FQMMBP has the best performance in delay, which is reduced to AODV, DSDV and DSR as 60.20%, 66.56% and 66.58%, respectively. Therefore, our proposed routing protocol can decrease average delay as 64.45%. And when the simulation time denotes 400 s, our FQMMBP has the most decrement in delay compared to DSR, which is 1.459393 s. The reason of the trend of FQMMBP at the beginning is also above the rescuers’ mobility.\n\nPerformances of four compared protocols are almost the same before 150 s. This is because at the beginning of the simulation, all nodes are concentrated in the central rectangular area with the most serious seismic intensity. At this time, the distributions of nodes are relatively centralized and similar, and the nodes have not started to disperse and move according to their respective mobility models, so the performance differences of the four protocols are not significant. As time goes on, the delay of each protocol increases gradually, because all nodes will spread to the next peripheral rectangular ring when the rescue tasks in the central area are completed. The dispersion of nodes leads to the increase in distance between nodes; therefore, it takes longer for relay nodes to find the appropriate next hop nodes as the decreasing numbers of candidate next hop nodes. The distribution of mobile nodes is more uniform under the proposed FQMMBP, which shortens the delay.\n\nIn the period of 200–300 s, the delay increases obviously as the nodes spread to the next area around when they finish the rescue tasks in the central area. The node distribution changes from the original rectangular centralized to the rectangular ring decentralized, so the node distribution density becomes low, which increases the delay. After 300 s, the delay tends to be stable. This is because there is no significant change in the distributions of nodes when they move from the area with RUD = 2 to the areas with RUD > 2. In that case, the delay tends to be stable because of the little impact on FQMMBP.\n\nFigure 15 shows the performance of overhead for the different studied protocols.\n\nIn terms of overhead, FQMMBP is significantly larger than other classical routing protocols. That is because four new fields (RUD, QL, QID and QW) are added to RREQ message. Furthermore, the dispersion of nodes increases the number of forwarding packets, which also contributes to the increase in overhead. The data transmission overhead of AODV, DSDV and DSR protocols increases with time, while FQMMBP meets the same rule from 100 to 300 s. However, the overhead of FQMMBP is decreasing between 300 and 450 s, that is because before 300 s, the rescue nodes are distributed centrally, and they are far away from the destination nodes. Most of the data transmissions rely on fixed relay nodes for forwarding, so the overhead is increasing. In the period of 300–450 s, the distribution of rescue nodes is scattered, and the average distance between rescue nodes and destination nodes is close. In that case, the number of data forwarding is reduced, so the overhead is reduced. After 450 s, the distribution of rescue nodes is more scattered than before, and the data communication between nodes relies on multi-hop forwarding, which makes the data transmission overhead increasing. After 600 s, the overhead of FQMMBP can be controlled within 2.5%.\n\nAs a negative impact of lower delay and higher FDR, our FQMMBP lost some extents of overhead, which is a metric relating to the energy and rescuer’s cost. Compared to other protocols, our proposed FQMMBP has increased overhead to AODV, DSDV and DSR as 49.89%, 46.83% and 43.92%. Our proposed routing protocol has average 46.88% increased overhead. We trade 46.88% overhead loss to gain the benefits of 64.45% delay and 16.31% FDR. It is a remarkable and acceptable results for the specific earthquake scenario, which the victims’ life is much more important than rescuers’ battery. Although the performance of FQMMBP in overhead is not as good as the other three protocols, it is worthwhile for the emergency rescue.\n\n### Discussions\n\n#### Energy consumption\n\nThere is no doubt that energy consumption is an important metric in mobile rescue issue. In our previous work, we have analyzed the energy consumption during the communication and mobility .\n\nOverall in two sides, the energy consumption is taken into consideration in our previous work, and the energy exhaustion problem can be solved in the realistic emergency rescue scenario.\n\n#### Performance metrics selection\n\nWe have two anticipations for our proposed protocols: one is that the ECNs can be set up as soon as possible, which is related to the metric of delay. This metric (delay) can guarantee the rescuers acquire and collect the real-time information about the disaster and victim with short time. The other is the communication link can maintain as well as possible, which means the ECNs can acquire the sufficient information and related to the metric of FDR. For this work, all selected performance metrics (delay and FDR) are useful for the victims in the earthquake. For general scenario, energy consumption is a vital metric for communication modules. The lower energy consumption, the longer modules can remain. However, in this specific earthquake scenario, the lives of victims (human) are much more important than the modules’ (electric rescuers). We should guarantee the human being’s life with the fast (lower delay) and successful (high FDR) information, not one communication module’s duration in the communication link. The life of human is more vital and indispensable than the life of communication module. Actually, in order to save victims lives, we scarify ECN’s energy as the addition cost. This is our primary reason to select the delay and FDR as metrics, rather than energy consumption. Moreover, based on our revision, which took the energy consumption into consideration, the protocol’s selection results are not changed.\n\n#### Routing protocols selection\n\nIn our previous work’s simulation results, the selection policy has the lowest energy consumption. Therefore, we improve the selection policy by considering the rescuers’ mobility model and update it as FQMMBP. The idea of disaster area’s division, and the four-quadrant mobility model are our main contribution about the proposed routing protocol. This protocol has two advantages: one is that the protocol inherits the previous lower energy consumption advantage. The other is that the protocol can guarantee the quality of rescue in earthquake because of the disaster area division and rescuers’ mobility model.\n\nThe other routing protocols cannot work well in the earthquake emergency scenario. DSDV is adapted from the conventional Routing Information Protocol (RIP) to ad hoc networks routing. It adds a new attribute, sequence number, to each route table entry of the conventional RIP. Using the newly added sequence number, the mobile nodes can distinguish stale route information from the new and thus prevent the formation of routing loops. DSR is set up with the route discovery and route maintenance mechanism to make sure that the packages can be set to the destination node when the source lack the information about the destination node’s grouping ID. AODV is a table-driven routing protocol, which will set up the routing table first and then search the effective relay node in the table to transmit the package. The routing protocols above have the relative good performance in MANET and VANET. However, they cannot fit the earthquake emergency scenario. Because these routing protocols do not consider the disaster areas’ distribution character and mobility model’s influence on the rescue quality.\n\n## Conclusion\n\nRouting protocols for post-earthquake emergency communication network are different with traditional ones because of the heterogeneity and dynamicity of the network. After the earthquake, rescue urgency degree is related to the affected areas. In this paper, we first divide the whole disaster area into several regions with different RUD values according to catastrophic intensity obtained. A four-quadrant mobility model (FQMM) for rescuers based on RUD is proposed. With FQMM, the most seriously affected areas are rescued as soon as possible, furthermore, the rescue crews are allocated to each appropriate quadrant, which improves the rescue efficiency and shortens the time of rescue. Under this mobility model, we propose the FQMMBP protocol for emergency communication network, which improves the RREQ message by adding four new fields: RUD, QL, QID and QW.\n\nA virtual earthquake scenario is simulated, and results show that FQMMBP is superior to traditional routing protocols (AODV, DSDV and DSR) in performances of PDR and Delay. We trade 46.88% overhead loss to gain the benefits of 64.45% delay and 16.31% FDR. It is a remarkable and acceptable results for the specific earthquake scenario, which the victims’ life is much more important than rescuers’ battery.\n\nIn our further work, the model will focus on the global rescue, which means that one task may be finished by several rescuers. Once one rescuer exhausts out, the nearby rescuers can receive this information based on the ECNs in this paper and continues the remaining rescue task, which depends on the optimization model.\n\n## Availability of data and materials\n\nData sharing not applicable to this article as no datasets were generated or analyzed during the current study.\n\n## Abbreviations\n\nECV:\n\nThe emergency communication vehicle\n\nBS:\n\nBase station\n\nUAV:\n\nUnnamed aerial vehicle\n\nRUD:\n\nRescue urgency degree, which is inversely proportional to the seismic intensity value in this area. The closer the area is to the epicenter, the lower the value of RUD is, which means the disaster is serious and the priority of rescue is high; Otherwise, the opposite. When RUD = 1, the epicenter disaster area is a rectangle; When RUD > 1, the disaster areas are rectangle rings that expands outwards in turn\n\na i :\n\nLength of area where RUD = i\n\nb i :\n\nWidth of area where RUD = i\n\nQL:\n\nQW:\n\nTotal weight of the node located in Qid\n\nQID:\n\nRT:\n\nNRA:\n\nNumber of rescuers allocated\n\nw QID :\n\nWeight of QID\n\nD QID :\n\nDistance between the center of each quadrant and ECV\n\nM :\n\nNumber of mobile nodes\n\n## References\n\n1. Y. Hu. Earthquake engineering in China. Earthq. Engin. Engin. Vib. 1, 1–9 (2002). https://doi.org/10.1007/s11803-002-0001-5.\n\n2. W. Decai, N. Sidao, L. Jun, Research status of rapid assessment on seismic intensity. Prog. Geophys. 28(4), 1772–1784 (2013). https://doi.org/10.6038/pg20130418\n\n3. W. Yanyan, S. 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Sensors 19(5), 1126 (2019)\n\n## Acknowledgements\n\nWe would like to thank Ping Li who guided and gave us constructive suggestions at the beginning of the research. This research was funded by China Earthquake Administration under Grant No.XH21009. The preparation of this paper has also received support from Shanghai Sheshan National Geophysical Observatory under Grant No.2020Z04, Technology Commission of Shanghai Municipality under Grant No. 18DZ1200500, and Shanghai Sailing Program under Grant No.21YF1432800.\n\n### Authors' information\n\nXiaoming Wang: received his M.S. degree from Lanzhou University, Lanzhou, China, in 2008. He is currently working as a senior engineer in Shanghai Earthquake Administration; meanwhile, he is working towards his Ph.D. degree in Intelligence Communication in Donghua University, Shanghai, China His research interest includes vehicular networks, sensor networks and emergency communications.. Chang Guo: received the B.S. degree in Communication engineering from Donghua University in 2014 and Ph.D. in Information and intelligent communication system from Donghua University in 2020. Her research interests include the data delivery delay in VANETs, the traffic information acquisition, short-term traffic flow prediction, real-time path planning and load balance optimization in urban area.\n\n## Funding\n\nThis research was funded by China Earthquake Administration under Grant No.XH21009. The preparation of this paper has also received support from Shanghai Sheshan National Geophysical Observatory under Grant No.2020Z04, Technology Commission of Shanghai Municipality under Grant No. 18DZ1200500 and Shanghai Sailing Program under Grant No.21YF1432800.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nXW contributed on the work design, acquisition and analysis of data, and drafted the manuscript. CG did the works on the design of model and algorithms, and revision the manuscript based on the comments. All authors read and approved the final manuscript.\n\n### Corresponding author\n\nCorrespondence to Chang Guo.\n\n## Ethics declarations\n\n### Competing interest\n\nThe project is partially funded by China Earthquake Administration, other parts are paid directly by projects partners. One of the main aims of the activities is the generation of UAV’s FQBMMBP routing protocol. So, it is given to publish all the main funding in the project to force standardization in this area.", null, "" ]	[ null, "https://jwcn-eurasipjournals.springeropen.com/track/article/10.1186/s13638-021-02017-y", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9144809,"math_prob":0.96592474,"size":68259,"snap":"2022-27-2022-33","text_gpt3_token_len":16532,"char_repetition_ratio":0.17689547,"word_repetition_ratio":0.15005054,"special_character_ratio":0.23788804,"punctuation_ratio":0.12845516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9695041,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T05:22:09Z\",\"WARC-Record-ID\":\"<urn:uuid:7a87f154-f124-4d87-929f-5134f711d880>\",\"Content-Length\":\"335424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88b0da66-1102-4824-8587-8a94045d4680>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d84d83d-6327-49dd-96dd-f75612880ff9>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-021-02017-y\",\"WARC-Payload-Digest\":\"sha1:R2336PEJGFSPLCS3LBTRLYORJ7UYD7CN\",\"WARC-Block-Digest\":\"sha1:IUXC5S2RGWPQW2VPOKRSXWMTRLSZVAL4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571147.84_warc_CC-MAIN-20220810040253-20220810070253-00243.warc.gz\"}"}
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-5-section-5-5-factoring-special-forms-exercise-set-page-371/34	[ "## Intermediate Algebra for College Students (7th Edition)\n\n$2y(1-x^3y)(1+x^3y)$\nRECALL: A difference of two squares can be factored using the formula: $m^2-n^2=(m-n)(m+n)$ Factor out the greatest common factor, $2y$, to obtain: $=2y(1-x^6y^2) \\\\=2y[1^2-(x^3y)^2]$ The binomial is a difference of two squares. Factor the given difference of two squares using the formula above with $m=1$ and $n=x^3y$ to obtain: $=2y(1-x^3y)(1+x^3y)$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7565937,"math_prob":0.9999862,"size":398,"snap":"2023-40-2023-50","text_gpt3_token_len":152,"char_repetition_ratio":0.13959391,"word_repetition_ratio":0.0,"special_character_ratio":0.36934674,"punctuation_ratio":0.07446808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000068,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T02:09:19Z\",\"WARC-Record-ID\":\"<urn:uuid:800c86f7-d9f9-4577-985c-2c5a1a06b4fb>\",\"Content-Length\":\"81248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc35bd65-baa7-4486-bc1f-155b21509b7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:6540d34c-8d9d-4b28-bdcc-97b0af968317>\",\"WARC-IP-Address\":\"162.159.140.98\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-5-section-5-5-factoring-special-forms-exercise-set-page-371/34\",\"WARC-Payload-Digest\":\"sha1:IXW352NVLXRS6TTXHRJK5W3N7J44AT3C\",\"WARC-Block-Digest\":\"sha1:RKDD2ULBICHXOIPEUNRTQ7FJWXLG5MIK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510734.55_warc_CC-MAIN-20231001005750-20231001035750-00243.warc.gz\"}"}
https://www.premiumessays.net/sample-paper-on-difference-between-p-value-of-one-tailed-and-two-tailed-test/	[ "# Sample Paper on Difference between P-Value of One-tailed and Two-tailed Test\n\nWhen conducting a statistical significance test, p-values are generally provided in the output of the population selected. Therefore, ANOVA, regression, and correlation majorly apply the formula during hypothesis testing to assist in understanding the data transmitted. A one-tailed test is one-sided on the critical distribution area, and it is greater than or less than a particular figure, but not both (Hick 316). However, the two-tailed tests is a technique where the critical region is two-sided and confirms if the samples are within the range of values identified in a null hypothesis.\n\nFirst, alpha levels assist in providing relatively accurate results. Hence, it is the probability of arriving an incorrect decision when the null hypothesis is correct. Therefore, one-tailed uses the aggregate value, and it can only appear on either the left or right side. However, a two-tailed splits the alpha into half. The distinction between and two-tailed P-value is more straightforward in context. In the interpretation of a one-tail P value, it is possible to interpret the group that will have extensive data (Mourougan Sendi and Sethurama 37). Besides, a two-tailed selection involves a random range of benefits from the sample in case the null hypothesis is correct.\n\nIn differentiating the two tests, power must also be considered because it provides a relevant distinction. One-tailed tests subject the researcher into focusing on one side of the distribution table. Moreover, there is more power to one-tiled tests because people are always sure of the decisions made. The two-tailed test considers the probable outcomes making it difficult to rely on one provision. Therefore, it is easier to make conclusions of the sample with the one-tail null hypothesis being correct.\n\nWork Cited\n\nHick, W. E. “A note on one-tailed and two-tailed tests.” (1952): 316.\n\nMourougan, Sendil, and K. Sethuraman. “Hypothesis Development and Testing.” J. Bus. Manag 19 (2017): 34-40." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93551666,"math_prob":0.8269672,"size":1971,"snap":"2022-27-2022-33","text_gpt3_token_len":422,"char_repetition_ratio":0.13065583,"word_repetition_ratio":0.0066225166,"special_character_ratio":0.20547946,"punctuation_ratio":0.12903225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95639277,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T14:28:59Z\",\"WARC-Record-ID\":\"<urn:uuid:b7f73aa5-69f3-4908-92b3-7ecf628e4e28>\",\"Content-Length\":\"80959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5422487-ed02-43c2-a073-1cbfbb456477>\",\"WARC-Concurrent-To\":\"<urn:uuid:d511a0ae-7802-4794-b7e8-47df6cad7548>\",\"WARC-IP-Address\":\"162.0.210.8\",\"WARC-Target-URI\":\"https://www.premiumessays.net/sample-paper-on-difference-between-p-value-of-one-tailed-and-two-tailed-test/\",\"WARC-Payload-Digest\":\"sha1:XEVYTTO4CYME3QCAGQDE5OEMJIXB6FEZ\",\"WARC-Block-Digest\":\"sha1:USI732ACI6JMOOP6FTUILR2YGTDOPOBX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103821173.44_warc_CC-MAIN-20220630122857-20220630152857-00257.warc.gz\"}"}
https://codereview.stackexchange.com/questions/57943/determining-maximum-profit-to-be-made-from-selling-shares	[ "# Determining maximum profit to be made from selling shares\n\nQuestion\n\nYour algorithms have become so good at predicting the market that you now know what the share price of Wooden Orange Toothpicks Inc. (WOT) will be for the next $N$ days.\n\nEach day, you can either buy one share of WOT, sell any number of shares of WOT that you own, or not make any transaction at all. What is the maximum profit you can obtain with an optimum trading strategy?\n\nInput\n\nThe first line contains the number of test cases $T$. $T$ test cases follow:\n\nThe first line of each test case contains a number $N$. The next line contains $N$ integers, denoting the predicted price of WOT shares for the next $N$ days.\n\nOutput\n\nOutput $T$ lines, containing the maximum profit which can be obtained for the corresponding test case.\n\nConstraints\n\n$1 <= T <= 10$\n$1 <= N <= 50000$\n\nAll share prices are between 1 and 100000\n\nCode\n\nimport java.io.;\nimport java.util.;\nimport java.text.;\nimport java.math.;\nimport java.util.regex.;\nclass Solution{\npublic static void main(String[] args){\nScanner stdin = new Scanner(System.in);\nint t = stdin.nextInt();\nfor(int m=0;m<t;m++){\nlong n = stdin.nextLong();\nlong[] ar = new long[(int)n];\nfor(int i=0;i<n;i++){\nar[i] = stdin.nextLong();\n}\nlong max = maximum(ar);\nSystem.out.println(max);\n}\n}\npublic static long maximum(long[] ar){\nint j=0;\nlong costPrice=0;\nlong sellingPrice=0;\nint k=0;\nlong max = maximumKey(j,ar);\nwhile(j<ar.length){\nif(ar[j]<max){\ncostPrice += ar[j];\nj++;\n}\nelse if(ar[j]==max){\nsellingPrice += ((j-k)ar[j]);\nj++;\nk=j;\nmax = maximumKey(j,ar);\n}\n}\nreturn sellingPrice-costPrice;\n}\npublic static long maximumKey(int j, long[] ar){\nlong max = 0;\nfor(int i=j;i<ar.length;i++){\nif(max<ar[i])\nmax = ar[i];\n}\nreturn max;\n}\n}\n\n\nIn the above code I didn't use DP. But this is supposed to be solved using DP with time complexity $O(n)$.\n\n• There is a well known way to do this O(N) time with dynamic if you could only hold on to 1 share at a time. However, I do not think that it is possible in this case (with as many shares as can be purchased), since the most profit would stem from buying all the way until the last global maximum (if it is duplicated), waiting until prices have fallen to the level of the next peak, and recursing. Repeatedly finding the global maximum will cause this to be a O(N^2) problem in the worst case (with constantly rising and falling prices, and diminishing maximums). Jul 24 '14 at 23:57\n• Yes my program is of n^2 level whearas as i saw people saying they were able to solve in O(n) time by using DP. Now how can this question be solved using DP hackerrank.com/challenges/stockmax Jul 25 '14 at 1:15\n• Ah, there is indeed a special trick for determining next global max in O(1), see my answer below. Jul 25 '14 at 5:06\n\nThe trick to getting this problem in O(n) is some clever pre-processing.\n\n# Examples\n\nLet's take 2 trials:\n\nTrial 1: 1, 2, 3, 4, 5\n\nTrail 2: 5, 4, 3, 2, 1\n\nObviously best case in Trial 1 is to buy for 4 days and sell on the 5th, for profit of 10. In Trial 2, there is no profit to be gained, because the price never increases, because there is never a higher maximum down the line.\n\n# Trick\n\nThis points towards the trick: starting from the end make a note of the maximum encountered thus far.\n\nFor Trial 1, the maximums are 5, 5, 5, 5, 5\n\nFor Trial 2, the maximums are the same as the prices\n\nLet's try this for a more complex scenario: 10, 1, 9, 2, 8, 3, 7\n\nThe corresponding maximums would be: 10, 9, 9, 8, 8, 7, 7\n\nThen, for every day that price < maximum, you should buy. If price = maximum, you should sell, and if price > maximum, wait it out.\n\n# Code\n\nThis yields the following code, assuming you parse the prices into a non-empty long[] named prices:\n\nlong[] maximums = new long[prices.length];\nmaximums[prices.length - 1] = prices[prices.length - 1];\n// fill maximums array, from the end\nfor (int i = prices.length - 2; i >= 0; i--) {\nif (prices[i] > maximums[i + 1])\nmaximums[i] = prices[i];\nelse\nmaximums[i] = maximums[i + 1];\n}\n\nlong profit = 0;\nfor (int i = 0; i < prices.length; i++) {\nif (prices[i] < maximums[i]) {\n// BUY now and SELL at max\nprofit += maximums[i] - prices[i];\n}\n}\n\n• Got it.I was calculating the maximum everytime inside the for loop,instead i should have made an array storing the maximum value. Jul 25 '14 at 16:36\n• @mleyfman Thanks your solution helped me.But i guess there is a slight mistake.if(price[i]>maximums[i-1]) should be if price[i]>maximums[i+1],and maximums[i]=maximums[i-1] should be maximums[i]=maximums[i+1]. Correct me if i am wrong Jun 19 '15 at 7:43\n• Thanks for the trick and code. The idea is correct, however, some small typos are in your code which needed to be corrected when you fill out the maximums from the end: if (prices[i] > maximums[i + 1]) maximums[i] = prices[i]; else maximums[i] = maximums[i + 1];\n– user103564\nApr 22 '16 at 6:50\n• @naz20z Thanks for pointing that out, updated answer to reflect it. Apr 30 '16 at 22:38\n• This can be optimized to be single-pass constant space: long max = Long.MIN_VALUE, profit = 0; for(int i = prices.length - 1; i >= 0; --i) if(prices[i] >= max) max = prices[i]; else profit += max - prices[i]; Feb 14 '17 at 22:30" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65104973,"math_prob":0.990254,"size":1810,"snap":"2021-43-2021-49","text_gpt3_token_len":483,"char_repetition_ratio":0.1013289,"word_repetition_ratio":0.0071174377,"special_character_ratio":0.3121547,"punctuation_ratio":0.17974684,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969968,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T03:00:51Z\",\"WARC-Record-ID\":\"<urn:uuid:d7750e7f-7fcc-4b25-b277-69cee3d1fd82>\",\"Content-Length\":\"184578\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aef338db-c353-43de-9354-71b73fb73426>\",\"WARC-Concurrent-To\":\"<urn:uuid:11416925-f853-4a80-8d85-e4f0fe0e0fa9>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/57943/determining-maximum-profit-to-be-made-from-selling-shares\",\"WARC-Payload-Digest\":\"sha1:XGSPCSSBIQAQGA7E65NLNUQQC4TLETBW\",\"WARC-Block-Digest\":\"sha1:JCO65PP4NCUUMMSI6LA6HPJG6KJNX3KQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587794.19_warc_CC-MAIN-20211026011138-20211026041138-00606.warc.gz\"}"}
https://proofwiki.org/wiki/Sum_of_Sines_of_Multiples_of_Angle	[ "# Sum of Sines of Multiples of Angle\n\n## Theorem\n\n $\\displaystyle \\sum_{k \\mathop = 1}^n \\sin k x$ $=$ $\\displaystyle \\sin x + \\sin 2 x + \\sin 3 x + \\cdots + \\sin n x$ $\\displaystyle$ $=$ $\\displaystyle \\frac {\\sin \\frac {\\paren {n + 1} x} 2 \\sin \\frac {n x} 2} {\\sin \\frac x 2}$\n\nwhere $x$ is not an integer multiple of $2 \\pi$.\n\n## Proof 1\n\n$2 \\sin \\alpha \\sin \\beta = \\map \\cos {\\alpha - \\beta} - \\map \\cos {\\alpha + \\beta}$\n\nThus we establish the following sequence of identities:\n\n $\\displaystyle 2 \\sin x \\sin \\frac x 2$ $=$ $\\displaystyle \\cos \\frac x 2 - \\cos \\frac {3 x} 2$ $\\displaystyle 2 \\sin 2 x \\sin \\frac x 2$ $=$ $\\displaystyle \\cos \\frac {3 x} 2 - \\cos \\frac {5 x} 2$ $\\displaystyle$ $\\cdots$ $\\displaystyle$ $\\displaystyle 2 \\sin n x \\sin \\frac x 2$ $=$ $\\displaystyle \\cos \\frac {\\paren {2 n - 1} x} 2 - \\cos \\frac {\\paren {2 n + 1} x} 2$\n\nSumming the above:\n\n $\\displaystyle 2 \\sin \\frac x 2 \\paren {\\sum_{k \\mathop = 1}^n \\sin k x}$ $=$ $\\displaystyle \\cos \\frac x 2 - \\cos \\frac {\\paren {2 n + 1} x} 2$ Sums on right hand side form Telescoping Series $\\displaystyle$ $=$ $\\displaystyle -2 \\map \\sin {\\dfrac {\\frac x 2 + \\frac {\\paren {2 n + 1} x} 2} 2} \\map \\sin {\\dfrac {\\frac x 2 - \\frac {\\paren {2 n + 1} x} 2} 2}$ Prosthaphaeresis Formula for Cosine minus Cosine $\\displaystyle$ $=$ $\\displaystyle -2 \\sin \\dfrac {\\paren {n + 1} x} 2 \\sin \\dfrac {-n x} 2$ $\\displaystyle$ $=$ $\\displaystyle 2 \\sin \\dfrac {\\paren {n + 1} x} 2 \\sin \\dfrac {n x} 2$ Sine Function is Odd\n\nThe result follows by dividing both sides by $2 \\sin \\dfrac x 2$.\n\nIt is noted that when $x$ is a multiple of $2 \\pi$ then:\n\n$\\sin \\dfrac x 2 = 0$\n\nleaving the right hand side undefined.\n\n$\\blacksquare$\n\n## Proof 2\n\nLet $x$ be a real number that is not a integer multiple of $2 \\pi$.\n\nLet $k$ be a non-negative integer.\n\nWe have, from Euler's Formula:\n\n$\\map \\exp {i k x} = i \\sin k x + \\cos k x$\n\nSumming from $k = 0$ to $k = n$, we have:\n\n$\\displaystyle \\sum_{k \\mathop = 0}^n \\map \\exp {i k x} = i \\sum_{k \\mathop = 0}^n \\sin k x + \\sum_{k \\mathop = 0}^n \\cos k x$\n\nAs $\\sin k x$ and $\\cos k x$ are both real for real $k, x$, we have:\n\n $\\displaystyle \\sum_{k \\mathop = 0}^n \\sin k x$ $=$ $\\displaystyle \\map \\Im {\\sum_{k \\mathop = 0}^n \\map \\exp {i k x} }$ $\\displaystyle$ $=$ $\\displaystyle \\map \\Im {\\paren {i \\sin \\frac {n x} 2 + \\cos \\frac {n x} 2} \\frac {\\map \\sin {\\frac {\\paren {n + 1} x} 2} } {\\sin \\frac x 2} }$ Sum of $\\map \\exp {i k x}$ $\\displaystyle$ $=$ $\\displaystyle \\frac {\\sin \\frac {\\paren {n + 1} x} 2 \\sin \\frac {n x} 2} {\\sin \\frac x 2}$\n\n$\\blacksquare$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51919204,"math_prob":1.0000098,"size":2909,"snap":"2020-34-2020-40","text_gpt3_token_len":1162,"char_repetition_ratio":0.2726334,"word_repetition_ratio":0.22924188,"special_character_ratio":0.44070128,"punctuation_ratio":0.054307114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T08:31:03Z\",\"WARC-Record-ID\":\"<urn:uuid:0873ee88-5634-4d05-b73d-6315e3b64da4>\",\"Content-Length\":\"44620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1a0dee00-c48b-4f72-ad23-a1a28fafdce2>\",\"WARC-Concurrent-To\":\"<urn:uuid:e589dd93-8626-4edd-bef1-aceafd53195d>\",\"WARC-IP-Address\":\"172.67.198.93\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Sum_of_Sines_of_Multiples_of_Angle\",\"WARC-Payload-Digest\":\"sha1:U6D5TSMXTYUTKCUDH2NWQNIZKJ6TUNS6\",\"WARC-Block-Digest\":\"sha1:KYAYHYCTHBEJX45H4HEN4OVF7BKQUFLU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738964.20_warc_CC-MAIN-20200813073451-20200813103451-00004.warc.gz\"}"}
https://codecrucks.com/bubble-sort/	[ "# Bubble Sort\n\nBubble sort is comparison based sorting method, and also known as sinking sort. It is perhaps most simple sorting algorithm.\n\nBubble Sort is a simple method for sorting a given set of n elements provided in the form of an array with n elements. It analyzes each element individually and sorts them based on their values.\n\nIt compares the first and second elements of the array; if the first element is greater than the second element, it will swap both elements, and then compare the second and third elements, and so on.\n\nIn general, sorting is accomplished by comparing neighboring elements A[j] and A[j+1]. If the components being compared are out of sequence, they are switched. The biggest element bubbles up at the final place in the unsorted array at the end of each iteration. For array of size n, this process is repeated n – 1 times.\n\nSorting an array starts from the end. The largest element is moved to the last place in the first iteration, the second largest element is moved to the second last position in the second iteration, and so on.\n\nIt is called as bubble sort because, with each complete iteration, the largest element in the given array bubbles up to the last position or the highest index, similar to how a water bubble rises to the water’s surface.\n\nThis basic method performs badly in practice and is mostly used as an educational tool. Sorting libraries integrated into popular computer languages like as Python and Java utilize more efficient algorithms such as quicksort, timsort, or merge sort.\n\n## Example of Bubble Sort\n\nBubble sort compares adjacent elements and swaps them if they are out of order. In every iteration, the largest element from unsorted array moves to the last location. Following figures show the step by step simulation of it.\n\nlet us sort the letters of word “DESIGN” in alphabetical order using bubble sort.\n\nStep by step comparison and output of every pass is depicted in following figures. Adjacent cells in light gray colors are under comparison and dark gray cells indicate sorted elements.\n\n## Algorithm for Bubble Sort\n\nAlgorithm BUBBLE_SORT(A)\n// A is an array of size n\n\nfor i ← 1 to n do\nfor j ← 1 to n – i do\nif A[j] > A[j+1] do\nswap(A[j], A[j+1])\nend\nend\nend\n\nAlthough the above logic would sort an unsorted array, the technique is inefficient since the outer for loop will continue to execute for n iterations even if the array is sorted.\n\nOptimized version of the algorithm is mentioned below:\n\nAlgorithm OPTIMIZED_BUBBLE_SORT(A)\n// A is an array of size n\n\nfor i ← 1 to n do\nflag ← 1\nfor j ← 1 to n – i do\nif A[j] > A[j+1] do\ntemp ← A[j]\nA[i] ← A[j+1]\nA[j+1] ← temp\nflag ← 0\nend\nif (flag) do\nbreak\nend\nend\nend\n\n## Complexity Analysis\n\nFor first iteration of outer loop, inner loop does n comparisons. For second iteration of outer loop, inner loop dose n – 1 comparison and so on. In last iteration of outer loop, inner loop does only one comparison. Outer loop itself iterates n times. Running time of algorithm is defined as total number of comparisons. Thus,\n\nDespite the fact that statements within the inner loop do not execute, the complexity of bubble sort is O(n2). The quadratic complexity is caused by the two nested for loops, which iterate regardless of the input data pattern. As a result, the complexity of bubble sort is the same whether the situation is best, worst, or average.\n\nIt’s self-explanatory, that if no swapping occurs in the first run, the list is already sorted. By setting an appropriate flag, we can break the cycle. In this scenario, the best-case bubble sort running time would be linear, i.e. O(n).\n\nDuring the first iteration of the outer loop, the inner loop executes n times without changing the flag, indicating that the data is sorted.\n\nFollowing table shows the time complexity of bubble sort for both versions, normal (without flag) and optimized (with flag):\n\nOther O(n2) sorting algorithms, such as insertion sort, are typically quicker than bubble sort and are not any more complex. As a result, bubble sort is not a useful sorting algorithm.\n\nThe single important benefit it has over most other algorithms, including quicksort but not insertion sort, is that the ability to discover if list is sorted. The complexity of this algorithm is only O(n) when the list is already sorted (best-case).\n\nMost other algorithms, even ones with lower average-case complexity, execute their whole sorting operation on the set, making them more complicated.\n\n## Simulation\n\nWorking of bubble sort is nicely simulated in following diagram\n\n## Rabbits and Turtles\n\nBecause elements flow in different directions at different speeds, the distance and direction that elements must move during the sort influence the performance of bubble sort.\n\nLarger element may participate in consecutive swaps and moves faster toward end. smaller elements moves quite slowly towards beginning. Consider following data patter, where the largest element is at beginning and the smallest is at end.\n\nA = <8, 7, 6, 5, 4, 3, 2, 1>\n\nAfter first iteration, array A would be,\n\nA = <7, 6, 5, 4, 3, 2, 1, 8>\n\nAs can be seen, in one iteration, 8 moved from first to last position (moved seven positions.) Whereas, element 1 moved from last to second last position (moved one position only).\n\nSo larger element, which moved quickly towards the end are known as hare, and smaller elements which moves slowly towards beginning are known as turtle.\n\n## Discussion and Comparison\n\nSeveral attempts have been made to eliminate turtles in order to increase the speed of bubble sorting. Cocktail sort is a bidirectional bubble sort that moves from beginning to end and then reverses direction, moving from end to beginning. It can move turtles rather effectively, but its worst-case complexity is O(n2).\n\nIn his famous book “The Art of Computer Programming”, Donald Knuth concludes that “the bubble sort appears to have little to recommend it, except a catchy name and the fact that it leads to some interesting theoretical issues\n\nComb sort compares elements separated by large gaps and can move turtles fast before moving on to smaller and smaller gaps to smooth down the list. Its average performance is comparable to that of quicker algorithms such as quicksort.\n\nIn the worst situation, bubble sort is asymptotically identical to insertion sort in terms of running time, but the two algorithms differ significantly in terms of the number of swaps required. Astrachan’s experimental results also demonstrate that insertion sort works significantly better even on random lists. For these reasons, many current algorithm textbooks prefer the insertion sort method over the bubble sort technique.\n\nBubble sort also has a bad interaction with current CPU technology. It generates at least twice the number of writes as insertion sort, double the number of cache misses, and asymptotically more branch mispredictions. Astrachan’s experiments sorting strings in Java demonstrate that bubble sort is about one-fifth as quick as an insertion sort and 70% faster than a selection sort.\n\n## Code for Bubble Sort\n\nFollowing are the few popular sorting algorithms used in computer science:\n\n### 2 Responses\n\n1.", null, "paresh says:\n\n•", null, "codecrucks says:" ]	[ null, "https://secure.gravatar.com/avatar/6be4caaf700cc7a2b66503633b54a0ca", null, "https://secure.gravatar.com/avatar/390e4c6d856849f1ace10209aca7584b", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90881497,"math_prob":0.8794561,"size":7498,"snap":"2023-40-2023-50","text_gpt3_token_len":1626,"char_repetition_ratio":0.14985321,"word_repetition_ratio":0.05401112,"special_character_ratio":0.21125634,"punctuation_ratio":0.09909281,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.971633,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T00:34:24Z\",\"WARC-Record-ID\":\"<urn:uuid:8bd1ddfe-1e9d-4704-b9d9-de4eafe776fb>\",\"Content-Length\":\"183282\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77f1c755-4017-4ef3-8129-8b2215569d2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc2de8a9-062d-4ca3-a5e0-56de4e1b1b05>\",\"WARC-IP-Address\":\"104.21.12.82\",\"WARC-Target-URI\":\"https://codecrucks.com/bubble-sort/\",\"WARC-Payload-Digest\":\"sha1:QWB5WOXBEG76W6WRADHYML57KINY3YYC\",\"WARC-Block-Digest\":\"sha1:UJX6RSNMHMCNKENNMIGTTT34AT4IRV3Y\",\"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-20230922234442-20230923024442-00222.warc.gz\"}"}
https://www.isrgrajan.com/equations-you-should-know-that-changed-the-future-of-physics.html	[ "Thursday, September 28, 2023\nEducationEquations You Should Know: That Changed The Future Of Physics\n\n# Equations You Should Know: That Changed The Future Of Physics\n\nIt is well known that science without physics is nothing, but lame. Even physics can be considered the backbone of science. Every great invention is interlinked to physics. Physics can be divided majorly in two sections that is classical and modern physics. However, classical is very old one and deals with mainly person’s like Galileo and Archimedes, but however we are going to sketch only modern physics and the theories that developed and changed the whole physics. There are many laws dealing the same, but we are going to know about the basic ones which is of importance in the view of instruments and devices which we use in our daily life or the theories which developed and helped in broadening our minds thinking about outside of the earth.\n\n### 1. Newton’s Law Of Universal Gravitation:\n\nThe law and theory were developed by the great scientist Issac Newton. It got published in Printipia in July 1687 for the first time. The law is very special because it deals with the facts of revolving planet’s around the sun in the orbit and the factors affecting the revolution. Secondly, the law deals with the gravity of the revolving planets. This is the main section which is the basis for the Aerospace education and helps in mainly finding out all the concepts required to eject the satellites outside the space.\n\n### 2. Theory Of Relativity:\n\nThis law was proposed by Albert Einstein in 1905. This is the basic equation taught to every high school student of physics, the equation which is taught to students is E=mc^2. Why it is so important? This is the most accepted theory on the relationship between space and earth, it was derived after around two hundred years of Newton’s law of gravitation. Most of the Knowledge which we have about space and the universe is known with this Einstein’s great relation. Well, after he proposed this theory many other theories got developed with respect to relativity accordingly, but thus the basic equation which helps everyone who is presenting developed one.\n\n### 3. Maxwell’s Equations:\n\nThis law was proposed by James Clark Maxwell. Maxwell proposed this law in around 1860’s. This law deals with how charged particles give rise to electric and magnetic field and by this time you would have understood why it is so important to study this topic? This is basic theory which is helpful in running the great trains, running of different motors and magnetic instruments. Maxwell’s equations are the primary syllabus of the electrical and electronics graduates they basically have to study this equation in their entire curriculum. The basic application is observed in our induction cooking, ever wondered on which principle it runs, obviously it is a Maxwells law of electromagnetic induction which is monitoring it.\n\n### 4. Law’s Of Thermodynamics:\n\nThere are three law’s of Thermodynamics but second one plays a key role. This law was developed by Rudolf Clausius. This law deals with how energy flows from higher concentration to lower concentration, it was developed in 1865. This law is responsible for the development of the instruments and appliances which are used in industries, household and everywhere we look at. For example the refrigerator Which we use in our house is developed according to this law, motors, compressors everything runs on this principle. When dealing with studies this law is useful for the students of mechanical engineering.\n\n### 5. Schrödinger’s Equation:\n\nThis equation was proposed by Austrian physicist Erwin Schrödinger in 1926. This law is the basic equation of the area quantum physics. This law, mainly deals with how the quantum state of a quantum system with respect to time. The law helped in developing various instruments including electron microscope, microchips and lot more. This law helped in widening the facts and myths related to atoms and it’s constituents present inside it.\n\n### 6. Newton’s law of motion:\n\nEvery one of us are aware about the incident took place with Newton which led to the discovery of gravity. Yes, you are right falling of an apple. Most of them include this topic in classical physics but it is almost helpful in modern physics equally. These laws are about three in number, first one speaks about inertia, the second deals with momentum and the last one deals with the reaction force. The laws are fundamental for all the Engineers.\n\n### 7. Logaritms:\n\nWhen it comes to logarithms it is the most accepted practice when a bigger calculation has to be done. It was developed by Scottish Laird, John Napier of Merchiston. When it comes to solving multiplication of large numbers, for example 3456×56789 is really time consuming and needs good skills. To overcome this problem Napier came up with logarithms which us boon for calculations. Now, due to the advent of computers and calculators is of less use because it is inbuilt whereas, scientists always use logarithms in their day to day life. The equations are mainly used in finding exponential equations, compound interest and radioactive decay by a Scientist, Astronimer and an Engineer.\n\n### 8. Coloumb’s Law Of Electrostatics:\n\nWhen it comes to electricity electrostatics is the main branch dealing it and in turn coulombs law is the basic law of electrostatics. This law explains how force is related to charges and the distance between them. Interestingly, it helped in developing many electrical instruments which we use nowadays from fan to its capacitors, generators and what not.\n\nLast but not least there are many laws are proposed by scientist’s day by day, each law has got its own importance, it’s own dominancy, its own applications and every law is of great importance. But the major laws which led to great transformation in our society are the one discussed above. If you are a science student it is obvious that you may know all these laws, but it is exactly not like that even being general everyone should know about this law at least basic terminology because every appliance which we use has a principle due to which it is working and as we are using it we have to know the basic law which is governing that appliance or instrument." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.97213185,"math_prob":0.8786967,"size":6091,"snap":"2023-40-2023-50","text_gpt3_token_len":1216,"char_repetition_ratio":0.12354197,"word_repetition_ratio":0.003933137,"special_character_ratio":0.19323592,"punctuation_ratio":0.0877193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9640769,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T14:55:41Z\",\"WARC-Record-ID\":\"<urn:uuid:8c16b0e6-1470-482c-beee-eab338a24a83>\",\"Content-Length\":\"102014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db263baa-c436-42f9-ba49-d7d720e42c3f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5efaaab-4a5b-425e-8160-2b824031df80>\",\"WARC-IP-Address\":\"104.21.34.129\",\"WARC-Target-URI\":\"https://www.isrgrajan.com/equations-you-should-know-that-changed-the-future-of-physics.html\",\"WARC-Payload-Digest\":\"sha1:MYYSR2MAYYC4AY72HVQQZUJPG2QSSE7I\",\"WARC-Block-Digest\":\"sha1:F3LU7WKHVMGSDMRUGYP556TYV45FQIMF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510412.43_warc_CC-MAIN-20230928130936-20230928160936-00715.warc.gz\"}"}
http://docs.juliadiffeq.org/latest/tutorials/ode_example.html	[ "Ordinary Differential Equations\n\n# Ordinary Differential Equations\n\nThis tutorial will introduce you to the functionality for solving ODEs. Other introductions can be found by checking out DiffEqTutorials.jl. Additionally, a video tutorial walks through this material.\n\n## Example 1 : Solving Scalar Equations\n\nIn this example we will solve the equation\n\n$\\frac{du}{dt} = f(u,p,t)$\n\non the time interval $t\\in[0,1]$ where $f(u,p,t)=αu$. We know by Calculus that the solution to this equation is $u(t)=u₀\\exp(αt)$.\n\nThe general workflow is to define a problem, solve the problem, and then analyze the solution. The full code for solving this problem is:\n\nusing DifferentialEquations\nf(u,p,t) = 1.01u\nu0=1/2\ntspan = (0.0,1.0)\nprob = ODEProblem(f,u0,tspan)\nsol = solve(prob,Tsit5(),reltol=1e-8,abstol=1e-8)\nusing Plots\nplot(sol,linewidth=5,title=\"Solution to the linear ODE with a thick line\",\nxaxis=\"Time (t)\",yaxis=\"u(t) (in μm)\",label=\"My Thick Line!\") # legend=false\nplot!(sol.t, t->0.5exp(1.01t),lw=3,ls=:dash,label=\"True Solution!\")\n\nwhere the pieces are described below.\n\n### Step 1: Defining a Problem\n\nTo solve this numerically, we define a problem type by giving it the equation, the initial condition, and the timespan to solve over:\n\nusing DifferentialEquations\nf(u,p,t) = 1.01u\nu0=1/2\ntspan = (0.0,1.0)\nprob = ODEProblem(f,u0,tspan)\n\nNote that DifferentialEquations.jl will choose the types for the problem based on the types used to define the problem type. For our example, notice that u0 is a Float64, and therefore this will solve with the dependent variables being Float64. Since tspan = (0.0,1.0) is a tuple of Float64's, the independent variables will be solved using Float64's (note that the start time and end time must match types). You can use this to choose to solve with arbitrary precision numbers, unitful numbers, etc. Please see the notebook tutorials for more examples.\n\nThe problem types include many other features, including the ability to define mass matrices and hold callbacks for events. Each problem type has a page which details its constructor and the available fields. For ODEs, the appropriate page is here. In addition, a user can specify additional functions to be associated with the function in order to speed up the solvers. These are detailed at the performance overloads page.\n\n### Step 2: Solving a Problem\n\n#### Controlling the Solvers\n\nAfter defining a problem, you solve it using solve.\n\nsol = solve(prob)\n\nThe solvers can be controlled using the available options are described on the Common Solver Options manual page. For example, we can lower the relative tolerance (in order to get a more correct result, at the cost of more timesteps) by using the command reltol:\n\nsol = solve(prob,reltol=1e-6)\n\nThere are many controls for handling outputs. For example, we can choose to have the solver save every 0.1 time points by setting saveat=0.1. Chaining this with the tolerance choice looks like:\n\nsol = solve(prob,reltol=1e-6,saveat=0.1)\n\nMore generally, saveat can be any collection of time points to save at. Note that this uses interpolations to keep the timestep unconstrained to speed up the solution. In addition, if we only care about the endpoint, we can turn off intermediate saving in general:\n\nsol = solve(prob,reltol=1e-6,save_everystep=false)\n\nwhich will only save the final time point.\n\n#### Choosing a Solver Algorithm\n\nDifferentialEquations.jl has a method for choosing the default solver algorithm which will find an efficient method to solve your problem. To help users receive the right algorithm, DifferentialEquations.jl offers a method for choosing algorithms through hints. This default chooser utilizes the precisions of the number types and the keyword arguments (such as the tolerances) to select an algorithm. Additionally one can provide alg_hints to help choose good defaults using properties of the problem and necessary features for the solution. For example, if we have a stiff problem where we need high accuracy, but don't know the best stiff algorithm for this problem, we can use:\n\nsol = solve(prob,alg_hints=[:stiff],reltol=1e-8,abstol=1e-8)\n\nYou can also explicitly choose the algorithm to use. DifferentialEquations.jl offers a much wider variety of solver algorithms than traditional differential equations libraries. Many of these algorithms are from recent research and have been shown to be more efficient than the \"standard\" algorithms. For example, we can choose a 5th order Tsitouras method:\n\nsol = solve(prob,Tsit5())\n\nNote that the solver controls can be combined with the algorithm choice. Thus we can for example solve the problem using Tsit5() with a lower tolerance via:\n\nsol = solve(prob,Tsit5(),reltol=1e-8,abstol=1e-8)\n\nIn DifferentialEquations.jl, some good \"go-to\" choices for ODEs are:\n\n• AutoTsit5(Rosenbrock23()) handles both stiff and non-stiff equations. This is a good algorithm to use if you know nothing about the equation.\n• BS3() for fast low accuracy non-stiff.\n• Tsit5() for standard non-stiff. This is the first algorithm to try in most cases.\n• Vern7() for high accuracy non-stiff.\n• Rodas4() for stiff equations with Julia-defined types, events, etc.\n• radau() for really high accuracy stiff equations (requires installing ODEInterfaceDiffEq.jl)\n\nFor a comprehensive list of the available algorithms and detailed recommendations, Please see the solver documentation. Every problem type has an associated page detailing all of the solvers associated with the problem.\n\n### Step 3: Analyzing the Solution\n\n#### Handling the Solution Type\n\nThe result of solve is a solution object. We can access the 5th value of the solution with:\n\nsol #.637\n\nor get the time of the 8th timestep by:\n\nsol.t #.438\n\nConvenience features are also included. We can build an array using a comprehension over the solution tuples via:\n\n[t+u for (u,t) in tuples(sol)]\n\nor more generally\n\n[t+2u for (u,t) in zip(sol.u,sol.t)]\n\nallows one to use more parts of the solution type. The object that is returned by default acts as a continuous solution via an interpolation. We can access the interpolated values by treating sol as a function, for example:\n\nsol(0.45) # The value of the solution at t=0.45\n\nNote the difference between these: indexing with [i] is the value at the ith step, while (t) is an interpolation at time t!\n\nIf in the solver dense=true (this is the default unless saveat is used), then this interpolation is a high order interpolation and thus usually matches the error of the solution time points. The interpolations associated with each solver is detailed at the solver algorithm page. If dense=false (unless specifically set, this only occurs when save_everystep=false or saveat is used) then this defaults to giving a linear interpolation.\n\nFor details on more handling the output, see the solution handling page.\n\n#### Plotting Solutions\n\nWhile one can directly plot solution time points using the tools given above, convenience commands are defined by recipes for Plots.jl. To plot the solution object, simply call plot:\n\n#]add Plots # You need to install Plots.jl before your first time using it!\nusing Plots\n#plotly() # You can optionally choose a plotting backend\nplot(sol)", null, "If you are in Juno, this will plot to the plot pane. To open an interactive GUI (dependent on the backend), use the gui command:\n\ngui()\n\nThe plot function can be formatted using the attributes available in Plots.jl. Additional DiffEq-specific controls are documented at the plotting page.\n\nFor example, from the Plots.jl attribute page we see that the line width can be set via the argument linewidth. Additionally, a title can be set with title. Thus we add these to our plot command to get the correct output, fix up some axis labels, and change the legend (note we can disable the legend with legend=false) to get a nice looking plot:\n\nplot(sol,linewidth=5,title=\"Solution to the linear ODE with a thick line\",\nxaxis=\"Time (t)\",yaxis=\"u(t) (in μm)\",label=\"My Thick Line!\") # legend=false\n\nWe can then add to the plot using the plot! command:\n\nplot!(sol.t,t->0.5exp(1.01t),lw=3,ls=:dash,label=\"True Solution!\")", null, "## Example 2: Solving Systems of Equations\n\nIn this example we will solve the Lorenz equations:\n\n\\begin{align} \\frac{dx}{dt} &= σ(y-x) \\\\ \\frac{dy}{dt} &= x(ρ-z) - y \\\\ \\frac{dz}{dt} &= xy - βz \\\\ \\end{align}\n\nDefining your ODE function to be in-place updating can have performance benefits. What this means is that, instead of writing a function which outputs its solution, you write a function which updates a vector that is designated to hold the solution. By doing this, DifferentialEquations.jl's solver packages are able to reduce the amount of array allocations and achieve better performance.\n\nThe way we do this is we simply write the output to the 1st input of the function. For example, our Lorenz equation problem would be defined by the function:\n\nfunction lorenz(du,u,p,t)\ndu = 10.0(u-u)\ndu = u(28.0-u) - u\ndu = uu - (8/3)u\nend\n\nand then we can use this function in a problem:\n\nu0 = [1.0;0.0;0.0]\ntspan = (0.0,100.0)\nprob = ODEProblem(lorenz,u0,tspan)\nsol = solve(prob)\n\nUsing the plot recipe tools defined on the plotting page, we can choose to do a 3D phase space plot between the different variables:\n\nplot(sol,vars=(1,2,3))", null, "Note that the default plot for multi-dimensional systems is an overlay of each timeseries. We can plot the timeseries of just the second component using the variable choices interface once more:\n\nplot(sol,vars=(0,2))", null, "Note that here \"variable 0\" corresponds to the independent variable (\"time\").\n\n## Defining Parameterized Functions\n\nIn many cases you may want to explicitly have parameters associated with your differential equations. This can be used by things like parameter estimation routines. In this case, you use the p values via the syntax:\n\nfunction parameterized_lorenz(du,u,p,t)\ndu = p(u-u)\ndu = u(p-u) - u\ndu = uu - pu\nend\n\nand then we add the parameters to the ODEProblem:\n\nu0 = [1.0,0.0,0.0]\ntspan = (0.0,1.0)\np = [10.0,28.0,8/3]\nprob = ODEProblem(parameterized_lorenz,u0,tspan,p)\n\nWe can make our functions look nicer by doing a few tricks. For example:\n\nfunction parameterized_lorenz(du,u,p,t)\nx,y,z = u\nσ,ρ,β = p\ndu = dx = σ(y-x)\ndu = dy = x(ρ-z) - y\ndu = dz = xy - βz\nend\n\nNote that the type for the parameters p can be anything: you can use arrays, static arrays, named tuples, etc. to enclose your parameters in a way that is sensible for your problem.\n\nAdditionally, there exists a @ode_def macro allows for \"defining your ODE in pseudocode\" and getting a function which is efficient and runnable. To use the macro, you write out your system of equations with the left-hand side being d_ and those variables will be parsed as the dependent variables. The independent variable is t, and the other variables are parameters which you pass at the end. For example, we can write the Lorenz system as:\n\n#]add ParameterizedFunctions\nusing ParameterizedFunctions\ng = @ode_def begin\ndx = σ(y-x)\ndy = x(ρ-z) - y\ndz = xy - βz\nend σ ρ β\n\nDifferentialEquations.jl will automatically translate this to be exactly the same as f. The result is more legible code with no performance loss. For more information on the macro Domain Specific Language (DSL) and its limitations, please see the parameterized function page The result is that g is a function which you can now use to define the Lorenz problem.\n\nu0 = [1.0;0.0;0.0]\ntspan = (0.0,1.0)\np = [10.0,28.0,8/3]\nprob = ODEProblem(g,u0,tspan,p)\n\nThe macro does \"behind-the-scenes\" symbolic calculations to pre-compute things like the Jacobian, inverse Jacobian, etc. in order to speed up calculations. Thus not only will this lead to legible ODE definitions, but \"unfairly fast\" code! We can turn off some of the calculations by using a more specific macro, like @ode_def_bare. See ParameterizedFunctions.jl for more details.\n\nSince the parameters exist within the function, functions defined in this manner can also be used for sensitivity analysis, parameter estimation routines, and bifurcation plotting. This makes DifferentialEquations.jl a full-stop solution for differential equation analysis which also achieves high performance.\n\n## Example 3: Using Other Types for Systems of Equations\n\nDifferentialEquations.jl can handle many different dependent variable types (generally, anything with a linear index should work!). So instead of solving a vector equation, let's let u be a matrix! To do this, we simply need to have u0 be a matrix, and define f such that it takes in a matrix and outputs a matrix. We can define a matrix of linear ODEs as follows:\n\nA = [1. 0 0 -5\n4 -2 4 -3\n-4 0 0 1\n5 -2 2 3]\nu0 = rand(4,2)\ntspan = (0.0,1.0)\nf(u,p,t) = Au\nprob = ODEProblem(f,u0,tspan)\n\nHere our ODE is on a 4x2 matrix, and the ODE is the linear system defined by multiplication by A. To solve the ODE, we do the same steps as before.\n\nsol = solve(prob)\nplot(sol)", null, "We can instead use the in-place form by using Julia's in-place matrix multiplication function mul!:\n\nusing LinearAlgebra\nf(du,u,p,t) = mul!(du,A,u)\n\nAdditionally, we can use non-traditional array types as well. For example, StaticArrays.jl offers immutable arrays which are stack-allocated, meaning that their usage does not require any (slow) heap-allocations that arrays normally have. This means that they can be used to solve the same problem as above, with the only change being the type for the initial condition and constants:\n\nusing StaticArrays, DifferentialEquations\nA = @SMatrix [ 1.0 0.0 0.0 -5.0\n4.0 -2.0 4.0 -3.0\n-4.0 0.0 0.0 1.0\n5.0 -2.0 2.0 3.0]\nu0 = @SMatrix rand(4,2)\ntspan = (0.0,1.0)\nf(u,p,t) = Au\nprob = ODEProblem(f,u0,tspan)\nsol = solve(prob)\nusing Plots; plot(sol)\n\nNote that the analysis tools generalize over to systems of equations as well.\n\nsol\n\nstill returns the solution at the fourth timestep. It also indexes into the array as well. The last value is the timestep, and the beginning values are for the component. This means\n\nsol[5,3]\n\nis the value of the 5th component (by linear indexing) at the 3rd timepoint, or\n\nsol[2,1,:]\n\nis the timeseries for the component which is the 2nd row and 1 column.\n\n## Going Beyond ODEs: How to Use the Documentation\n\nNot everything can be covered in the tutorials. Instead, this tutorial will end by pointing you in the directions for the next steps.\n\n#### Common API for Defining, Solving, and Plotting\n\nOne feature of DifferentialEquations.jl is that this pattern for solving equations is conserved across the different types of differential equations. Every equation has a problem type, a solution type, and the same solution handling (+ plotting) setup. Thus the solver and plotting commands in the Basics section applies to all sorts of equations, like stochastic differential equations and delay differential equations. Each of these different problem types are defined in the Problem Types section of the docs. Every associated solver algorithm is detailed in the Solver Algorithms section, sorted by problem type. The same steps for ODEs can then be used for the analysis of the solution.\n\n#### Additional Features and Analysis Tools\n\nIn many cases, the common workflow only starts with solving the differential equation. Many common setups have built-in solutions in DifferentialEquations.jl. For example, check out the features for:\n\nMany more are defined in the relevant sections of the docs. Please explore the rest of the documentation, including tutorials for getting started with other types of equations. 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https://deeplearn.org/arxiv/96589/implicit-neural-solver-for-time-dependent-linear-pdes-with-convergence-guarantee	[ "### Implicit Neural Solver for Time-dependent Linear PDEs with Convergence Guarantee\n\n• 2019-10-08 15:20:01\n• Suprosanna Shit, Abinav Ravi, Ivan Ezhov, Jana Lipkova, Marie Piraud, Bjoern Menze\n• 1\n\n### Abstract\n\nFast and accurate solution of time-dependent partial differential equations(PDEs) is of key interest in many research fields including physics,engineering, and biology. Generally, implicit schemes are preferred over theexplicit ones for better stability and correctness. The existing implicitschemes are usually iterative and employ a general-purpose solver which may besub-optimal for a specific class of PDEs. In this paper, we propose a neuralsolver to learn an optimal iterative scheme for a class of PDEs, in adata-driven fashion. We achieve this goal by modifying an iteration of anexisting semi-implicit solver using a deep neural network. Further, we providetheoretical proof that our approach preserves the correctness and convergenceguarantees provided by the existing iterative-solvers. We also demonstrate thatour model generalizes to a different parameter setting than the one seen duringtraining and achieves faster convergence compared to the semi-implicit schemes.\n\n# Implicit Neural Solver for Time-dependent Linear PDEs with Convergence Guarantee\n\nSuprosanna Shit\nTechnical University Munich\nAbinav Ravi\nTechnical University Munich\nIvan Ezhov\nTechnical University Munich\nJana Lipkova\nTechnical University Munich\nMarie Piraud\nKonica Minolta Laboratory Europe\nBjoern Menze\nTechnical University Munich\n###### Abstract\n\nFast and accurate solution of time-dependent partial differential equations (PDEs) is of key interest in many research fields including physics, engineering, and biology. Generally, implicit schemes are preferred over the explicit ones for better stability and correctness. The existing implicit schemes are usually iterative and employ a general-purpose solver which may be sub-optimal for a specific class of PDEs. In this paper, we propose a neural solver to learn an optimal iterative scheme for a class of PDEs, in a data-driven fashion. We achieve this goal by modifying an iteration of an existing semi-implicit solver using a deep neural network. Further, we provide theoretical proof that our approach preserves the correctness and convergence guarantees provided by the existing iterative-solvers. We also demonstrate that our model generalizes to a different parameter setting than the one seen during training and achieves faster convergence compared to the semi-implicit schemes.\n\nImplicit Neural Solver for Time-dependent Linear PDEs with Convergence Guarantee\n\nSuprosanna Shit Technical University Munich Abinav Ravi Technical University Munich Ivan Ezhov Technical University Munich Jana Lipkova Technical University Munich Marie Piraud Konica Minolta Laboratory Europe Bjoern Menze Technical University Munich\n\n\\@float\n\nnoticebox[b]33rd Conference on Neural Information Processing Systems (NeurIPS 2019), Vancouver, Canada.\n\n## 1 Introduction\n\nTime-dependent partial differential equations (PDEs) are an essential mathematical tool to describe numerous physical processes such as heat transfer, wave propagation, quantum transport, and tumor growth. Solving the initial-value problem (IVP) and boundary-value problem (BVP) accurately and efficiently is of primary research interest in computational science. Numerical solution of time-dependant PDEs requires appropriate spatio-temporal discretization. Spatial discretization can be cast as either finite difference method (FDM) or finite element method (FEM) or finite volume method (FVM), whereas temporal discretization relies on either explicit, implicit or semi-implicit methods. Explicit temporal update rules are generally a single or few forward computation steps, while implicit or semi-implicit update rules, such as Crank-Nicolson’s scheme, resort to a fixed-point iterative scheme. Small time and spatial resolution facilitate a more accurate solution, however, increases computational burden at the same time. Moreover, maximally allowed spatio-temporal resolution is not only constrained by the desired accuracy but also limited to numerical stability criteria. Note that implicit and semi-implicit methods offer a relaxed stability constraints (sometimes unconditionally stable) at the expense of an increased computational cost caused by the iterative schemes.\n\nIn recent times, deep neural networks have gained significant attention by numerical computation community due to its superior performance in solving forward simulations and inverse-problems . Recent work by Tompson et al. shows a data-driven convolutional neural network (CNN) can accelerate fluid simulation compared to traditional numerical schemes. Long et al. shows that learned differential operator can outperform hand-designed discrete schemes. However, on the contrary to the well understood and theoretically grounded classical methods, the deep learning-based approaches rely mainly on empirical validity. Recently, Hsieh et al. develop a promising way to learn numerical solver while providing a theoretical convergence guarantee. They demonstrate that a feed-forward CNN, which is trained to mimic a single iteration of a linear solver, can deliver faster solution than the handcrafted solver. Astonishingly for time-dependent PDEs, the temporal update step of the previously proposed neural schemes relies on an explicit forward Euler method and none of them is capable of making use of the advanced implicit and semi-implicit methods. This limitation restricts the general use of neural architectures in solving time-dependent PDEs.\n\nIn this paper, we introduce a novel neural solver for time-dependant linear PDEs. Motivated by we construct a neural iterator from a semi-implicit update rule. We replace a single iteration of the semi-implicit scheme with a learnable parameterized function such that the fixed point of the algorithm is preserved. To leverage the theoretical guarantees we perform data-driven learning to enhance the convergence speed. As a result, our approach provides: (i) theoretical guarantees of convergence to the correct stationary solution, (ii) faster convergence than existing solvers, and (iii) generalizes to a resolutions and parameter settings very different from the ones seen at training time.\n\n## 2 Methodology\n\nIn this section, we first introduce the necessary background on the semi-implicit method for time-dependant linear PDEs and subsequently describe our proposed neural solver.\n\n### 2.1 Background\n\nIn the following, we consider the general IVP form of time-dependant linear PDE for the variable of interest $u$ within the computation domain $\\Omega$, w.r.t. the spatial variable $x$ and temporal variable $t$, subject to Dirichlet boundary condition $b(x,t)$ at the boundary $\\Gamma$\n\n $\\frac{\\partial u}{\\partial t}=\\mathcal{F}(u,x,t;\\Theta),\\leavevmode\\nobreak\\ % \\forall x\\in\\Omega,\\mbox{ s.t. }u(x,t)=b(x,t),\\leavevmode\\nobreak\\ \\forall x% \\in\\Gamma\\mbox{ and }u_{t_{0}}=u_{0};$ (1)\n\n$\\mathcal{F}(u,x,t;\\Theta)$ is a linear operator and can be discretized as $\\sum_{i=1}^{N}\\frac{\\Theta_{i}\\partial_{i}}{\\delta x^{p_{i}}}u$, with uniform spatial discretization step $\\delta x$ and PDE parameter set $\\Theta=\\{\\Theta_{i}\\}_{i=1:N}$, where $\\Theta_{i}$ is a diagonal matrix comprising the coefficients of the differential operator $\\partial_{i}$ of order $p_{i}$. Let’s call $u(x,t)$ as $u_{t}$ for simplicity. A first order semi-implicit update rule to get $u_{t+\\delta t}$ from $u_{t}$ (with time step $\\delta t$) is given by\n\n $\\frac{u_{t+\\delta t}-u_{t}}{\\delta t}=\\epsilon\\mathcal{F}(u,x,t+\\delta t;% \\Theta)+(1-\\epsilon)\\mathcal{F}(u,x,t;\\Theta);\\leavevmode\\nobreak\\ [0<\\epsilon% \\leq 1]$ (2)\n\nTo obtain $u_{t+\\delta t}$ one needs to solve the following linear system of equations\n\n $\\left(I-\\delta t\\epsilon\\sum_{i=1}^{N}\\frac{\\Theta_{i}d_{i}}{\\delta x^{p_{i}}}% \\right)u_{t+\\delta t}=\\delta t\\epsilon\\sum_{i=1}^{N}\\frac{\\Theta_{i}(\\partial_% {i}-d_{i}I)}{\\delta x^{p_{i}}}u_{t+\\delta t}+c(u_{t},\\Theta,\\delta x,\\delta t,% \\epsilon;\\partial)$ (3)\n\nwhere $c$ is independent of $u_{t+\\delta t}$ and $d_{i}$ is the central element of the central difference discretization of operator $\\partial_{i}$. Note that for central difference scheme, $\\partial_{i}-d_{i}I$ is real, zero-diagonal, and either circulant or skew-circulant matrix. One can use an iterative scheme to compute $u_{t+\\delta t}$ from an arbitrary initialization $u_{t+\\delta t}^{0}$ on the right-hand-side of Eq. 3. For simplicity of notation, we refer to $\\left(I-\\delta t\\epsilon\\sum_{j=1}^{N}\\frac{\\Theta_{j}d_{j}}{\\delta x^{p_{j}}}% \\right)^{-1}\\frac{\\delta t\\epsilon\\Theta_{i}}{\\delta x^{p_{i}}}$ as $\\Lambda_{i}$, and, we drop the subscript of $u$ and use a superscript to denote a single iteration at a time $t+\\delta t$. We enforce Dirichlet boundary condition using a projection step with a binary boundary mask $G$.\n\n $u^{m+1}=G\\left(\\sum_{i=1}^{N}\\Lambda_{i}(\\partial_{i}-d_{i}I)u^{m}+c\\right)+(I% -G)b$ (4)\n\nEq. 4 can be seen as a linear operator $u^{m+1}=\\Psi(u^{m})=Tu^{m}+k$. We can guarantee the spectral radius of the linear transformer $T$, i.e. $\\rho(T)<1$, by appropriately selecting $\\delta x,\\delta t,\\mbox{ and }\\epsilon$ [see Appendix A], leading to a fixed-point algorithm.", null, "Figure 1: Qualitative comparison of u (c.f. Eq 7) from the neural scheme (10 iterations) and a semi-implicit scheme (25 iterations) against the FEniCS solution for a test sequence of 10 time points. All methods use same initial- and boundary condition. The neural update shows consistently faster convergence than semi-implicit one.\n\n### 2.2 Neural Solver\n\nWe propose the following iterator\n\n $\\Phi_{H}(u)=\\Psi(u)+G\\left(\\sum_{i=1}^{N}\\Lambda_{i}H_{i}w\\right)$ (5)\n\nwhere $w=\\Psi(u)-u$, and $H_{i}$ is a learned linear operator which satisfies $H_{i}0=0$ for $i=1:N$. Substituting $w$ in Eq. 5 we get $\\Phi_{H}(u)=T^{\\prime}u+k^{\\prime}$, where $k^{\\prime}$ denotes the additive part which is independent of $u$, and $T^{\\prime}=T+G\\sum_{i=1}^{N}\\Lambda_{i}H_{i}(T-I)$.\n\n###### Lemma 1.\n\nFor a linear PDE problem $(\\{\\Theta_{i},\\partial_{i}\\}_{i=1:N},G,u_{0},b,\\delta x,\\delta t,\\epsilon)$ and choice of $\\{H_{i}\\}_{i=1:N}$ if $u^{}$ is a fixed point of $\\Psi,$ it is a fixed point of $\\Phi_{H}$ in Eq. 5.\n\n###### Proof.\n\nBased on the iterative rule in Eq.5 $,$ if $u^{}$ satisfies $\\Psi\\left(u^{}\\right)=u^{}$ then $w=\\Psi\\left(u^{}\\right)-u^{}=0$ . Therefore, $\\Phi_{H}\\left(u^{}\\right)=\\Psi\\left(u^{}\\right)+G\\sum_{i=1}^{N}\\Lambda_{i}H_% {i}0=u^{*}.$\n\nMoreover, the space of $\\Phi_{H}$ subsumes the standard solver $\\Psi.$ If $H_{i}=0,$ then $\\Phi_{H}=\\Psi.$ Furthermore, if $H_{i}=\\partial_{i}$ , then since $GT=T$\n\n $\\Phi_{H}(u)=\\Psi(u)+GT(\\Psi(u)-u)=T\\Psi(u)+c=\\Psi^{2}(u)$ (6)\n\nwhich is equal to two iterations of $\\Psi.$ Because computing $\\Phi$ requires two convolutions: ${H_{i}}$ and $T$ one iteration with $\\Phi_{H}$ has the same number of convolution operations as two iterations of $\\Psi.$ This shows that we can learn an ${H_{i}}$ such that our iterator $\\Phi_{H}$ is at least as the standard solver $\\Psi.$ In the following theorem, we show that there is a convex open set of ${H_{i}}$ that the learning algorithm can explore.\n\n###### Theorem 1.\n\nFor fixed $G,\\{\\Theta_{i}\\}_{i=1:N},u_{0},b,\\delta x,\\delta t$, and $\\epsilon$ the spectral norm of $\\Phi_{H}(u;G,\\{\\Theta_{i}\\}_{i=1:N},u_{0},b,\\delta x,\\delta t,\\epsilon)$ is a convex function of $\\{H_{i}\\}_{i=1:N}$ and the set of $\\{H_{i}\\}_{i=1:N}$ such that the spectral norm of $\\Phi_{H}(u)<1$ is a convex open set.\n\n###### Proof.\n\nSee Appendix B. ∎\n\nOn a stark contrast with previous work , we have several sets of parameters $\\{\\Theta_{i}\\}_{i=1:N},\\delta x,\\delta t$, and $\\epsilon$ attached to the PDEs governing equation. Although we train on a single domain, the model posses a generalization properties, which we show in the following.\n\n###### Proposition 1.\n\nFor fixed {$\\partial_{i}\\}_{i=1:N},G$, and $\\{H_{i}\\}_{i=1:N}$, and some $u^{\\prime}_{0},b^{\\prime},\\{\\Theta^{\\prime}_{i}\\}_{i=1:N},\\delta x^{\\prime},% \\delta t^{\\prime}$, and $\\epsilon^{\\prime}$, if $\\Phi_{H}(u)$ is valid iterator for the PDE problem $(\\{\\Theta^{\\prime}_{i},\\partial_{i}\\}_{i=1:N},G,u^{\\prime}_{0},b^{\\prime},% \\delta x^{\\prime},\\delta t^{\\prime},\\epsilon^{\\prime})$, then for all $u_{0}$ and $b,$ the iterator $\\Phi_{H}(u)$ is valid iterator for the PDE problem $(\\{\\Theta_{i},\\partial_{i}\\}_{i=1:N},G,u_{0},b,\\delta x,\\delta t,\\epsilon)$ if $\\left\\\|\\Lambda_{i}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|\\partial_{j}-d_{j}I% \\right\\\|},\\leavevmode\\nobreak\\ \\forall i=1:N$.\n\n###### Proof.\n\nSee Appendix B. ∎\n\n## 3 Experiments\n\nWe consider a 2-D advection-diffusion equation of the following form\n\n $\\displaystyle\\frac{\\partial u}{\\partial t}=\\mathbf{v}^{\\top}\\cdot\\begin{% bmatrix}\\partial_{x}u\\\\ \\partial_{y}u\\end{bmatrix}+\\mathbf{D}^{\\top}\\cdot\\begin{bmatrix}\\partial_{xx}u% \\\\ \\partial_{yy}u\\end{bmatrix}\\mbox{; subject to }u_{t_{0}}=u_{0}(x,y)$ (7)\n\nwhere $\\mathbf{v}=[v_{x},v_{y}]^{\\top}$ and $\\mathbf{D}=[D_{xx},D_{yy}]^{\\top}$ are advection velocity and diffusivity respectively. We minimize the following loss function\n\n $\\mathcal{L}=\\frac{1}{L}\\sum_{n=1}^{N}\\sum_{l=1}^{L}\\left\\\|\\Phi^{n,k}_{H}(u^{l}% _{t})-u^{l}_{t+n\\delta t}\\right\\\|;k\\sim\\mathcal{U}[1,20]$\n\nwhere $n$ is the number of time-step and $k$ iteration for a single time step is denoted as $\\Phi_{H}(\\Phi_{H}\\dots(\\Phi_{H}))=\\Phi^{k}_{H}$.\n\nData Generation: We consider a rectangular domain of $\\Omega=[0,2\\pi]\\times[0,2\\pi]$. Elements of $\\mathbf{v}$ and $\\mathbf{D}$ are drawn from a uniform distribution of $\\mathcal{U}[-2.0,2.0]$ and $\\mathcal{U}[0.2,0.8]$ respectively. The computational domain is discretized into 64 x 64 regular mesh. We assume zero Dirichlet boundary condition and the initial value is generated according to as $u_{0}=\\lambda cos(kx+ly)+\\gamma sin(kx+ly)$ where $\\gamma$ and $\\lambda$ are drawn from a normal distribution of $\\mathcal{N}(0,0.02)$, and, $k$ and $l$ are values drawn in random from a uniform distribution of $\\mathcal{U}[1,9]$. We generate 200 simulations each having 50 time steps, using FEniCS for $\\delta t=0.2$. Further, we take the train, test, and validation split of the simulated time series as $80\\%:10\\%:10\\%$. A time series of a test data is shown in Fig 1.\n\nImplementation Details: Following , we use a three-layer convolutional neural network to model each of the $H_{i}$. We use zero-padding to enforce zero Dirichlet condition at the boundary and use a kernel size of 3x3. During training, we fixed the following parameters as follows $\\delta x=0.098,\\delta t=0.2,\\epsilon=0.9$. We use PyTorch framework and train our network with Adam Optimizer .\n\n### 3.1 Results and Discussion", null, "Figure 2: (a), (b), and (c) shows the mean-squared error (between FEniCS solution and Semi-implicit scheme and neural scheme) vs time plot for different parameters during test time as specified. The banded curves indicate the 25% and 75% percentile of the normalized errors among 20 test samples.\n\nGeneralization for Different Parameters: We investigate the effect of different parameter settings than those used during, to validate the generalizability of the neural scheme. To study the effect of different $\\epsilon$ we use the original test set. We generate two additional test cases; varying one parameter at a time : a) $\\delta t=0.12$, and b) $\\delta x=0.049$. The elements of $\\mathbf{v}$ and $\\mathbf{D}$ are drawn from the same distribution as before. The average error propagation over ten time step between semi-implicit finite difference method and the proposed neural implicit solver is being compared in Figure 2.\n\nWe observe that error from the neural scheme (10 iterations per time step) is less compared to the error from the semi-implicit FDM (25 iterations per time step) scheme for all three different test sets. This affirms our hypothesis that the neural solver is more accurate compared to the semi-implicit FDM and generalizable to other parameter settings at the same time.\n\nRun-time Comparison: We compare the run time for the neural solver (10 iterations per time step) and semi-implicit scheme (25 iterations per time step), for one time-steps. The experiments are conducted on an Intel Xeon W-2123 CPU @ 3.60GHz, with code running on one of the four cores. We report that the trained neural solver takes circa 0.0148s compared to 0.0141s for the semi-implicit scheme, whereas the FEniCS solution takes 3.19s for machine precision convergence.\n\n## 4 Conclusion\n\nThis abstract introduces a novel implicit neural scheme to solve linear time-dependent PDEs. We leverage an existing semi-implicit update rule to design a learnable iterator which provides theoretical guarantees. The learned iterator achieves faster convergence compared to the existing semi-implicit solver and produces a more accurate solution for a fixed computation budget. More importantly, we empirically demonstrate that training on a single parameter setting is enough to generalize over other parameter settings which confirms our theoretical results. The learned neural solver can be a faster alternative for simulation of various physical processes.\n\n#### Acknowledgments\n\nSuprosanna Shit and Ivan Ezhov are supported by the Translational Brain Imaging Training Network (TRABIT) under the European Union’s ‘Horizon 2020’ research & innovation programme (Grant agreement ID: 765148).\n\n## References\n\n• M. Alnæs et al. (2015) The FEniCS project version 1.5. Archive of Numerical Software 3 (100). Cited by: §3.\n• J. Cai et al. (2012) Image restoration: total variation, wavelet frames, and beyond. Journal of the American Mathematical Society 25 (4), pp. 1033–1089. Cited by: Appendix C.\n• I. Ezhov et al. (2019) Neural parameters estimation for brain tumor growth modeling. In Proceedings of the International Conference on Medical Image Computing and Computer Assisted Intervention, Cited by: §1.\n• R. Horn and C. Johnson (1991) Topics in matrix analysis. Cambridge University Press. Cited by: Appendix A, Appendix A.\n• J. Hsieh et al. (2019) Learning neural PDE solvers with convergence guarantees. In Proceedings of the International Conference on Learning Representations, Cited by: Appendix B, §1, §1, §2.2, §3.\n• D. P. Kingma and J. Ba (2014) Adam: a method for stochastic optimization. In Proceedings of the International Conference on Learning Representations, Cited by: §3.\n• A. Logg et al. (2012) Automated solution of differential equations by the finite element method: the FEniCS book. Vol. 84, Springer Science & Business Media. Cited by: §3.\n• Z. Long et al. (2018) PDE-net: learning PDEs from data. In Proceedings of the 35th International Conference on Machine Learning, Vol. 80, pp. 3208–3216. Cited by: §1, §3.\n• M. Magill et al. (2018) Neural networks trained to solve differential equations learn general representations. In Advances in Neural Information Processing Systems, pp. 4071–4081. Cited by: §1.\n• M. Raissi et al. (2019) Physics-informed neural networks: a deep learning framework for solving forward and inverse problems involving nonlinear partial differential equations. Journal of Computational Physics 378, pp. 686 – 707. Cited by: §1.\n• J. Tompson et al. (2017) Accelerating Eulerian fluid simulation with convolutional networks. In Proceedings of the 34th International Conference on Machine Learning, Vol. 70, pp. 3424–3433. Cited by: §1.\n\n## Appendix A Convergence Criteria for Semi-implicit Update\n\nThe spectral radius of $T$ can be expresses as following\n\n $\\displaystyle\\rho(T)$ $\\displaystyle\\leq\\left\\\|T\\right\\\|$ $\\displaystyle=\\left\\\|G\\sum_{i=1}^{N}\\Lambda_{i}(\\partial_{i}-d_{i}I)\\right\\\|$ $\\displaystyle\\leq\\left\\\|G\\right\\\|\\sum_{i=1}^{N}\\left\\\|\\Lambda_{i}\\right\\\|\\left% \\\|(\\partial_{i}-d_{i}I)\\right\\\|;\\leavevmode\\nobreak\\ [\\mbox{ Invoking norm % inequalities\\@@cite[cite]{[\\@@bibref{}{horn1991topics}{}{}]}}]$ $\\displaystyle=\\sum_{i=1}^{N}\\left\\\|\\Lambda_{i}\\right\\\|\\left\\\|(\\partial_{i}-d_{% i}I)\\right\\\|;\\leavevmode\\nobreak\\ [\\\|G\\\|=1]$\n\nGiven $\\left\\\|\\Lambda_{i}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|(\\partial_{j}-d_{j}I)% \\right\\\|};\\leavevmode\\nobreak\\ \\forall i=1:N$, we have $\\rho(T)<1$.\n\n## Appendix B Proofs\n\nSee 1\n\n###### Proof.\n\nThe spectral norm $\\\|\\cdot\\\|$ is convex from the sub-additive property, and $T^{\\prime}$ is linear in $\\{H_{i}\\}_{i=1:N}$. To prove that it is open, observe that $\\\|\\cdot\\\|$ is a continuous function, so $(T+G\\sum_{i=1}^{N}\\Lambda_{i}H_{i}(T-I))$ is continuous in $\\{H_{i}\\}_{i=1:N}$. Given $\\rho(T^{\\prime})$, the set of $\\{H_{i}\\}_{i=1:N}$ is the preimage under this continuous function of $(0,1-\\zeta)$ for some $\\zeta>0$, and the inverse image of open set $(0,1-\\zeta)$ must be open. ∎\n\n###### Lemma 2.\n\nThe upper bound of the spectral norm of $\\Phi_{H}$ is independent of $\\{\\Theta_{i}\\}_{i=1:N},\\delta x,\\delta t$, and $\\epsilon$ Given $\\left\\\|\\Lambda_{i}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|\\partial_{j}-d_{j}I% \\right\\\|},\\leavevmode\\nobreak\\ \\forall i=1:N$.\n\n###### Proof.\n\nConsidering the spectral norm of $T^{\\prime}$ and invoking product and triangular inequality of norm, we obtain the following tight bound\n\n $\\displaystyle\\left\\\|T^{\\prime}\\right\\\|$ $\\displaystyle=\\left\\\|G\\sum_{i=1}^{N}\\Lambda_{i}(\\partial_{i}-d_{i}I-H_{i})+G% \\sum_{i=1}^{N}(\\Lambda_{i}H_{i})T\\right\\\|$ $\\displaystyle\\leq\\\|G\\\|\\sum_{i=1}^{N}\\left\\\|\\Lambda_{i}\\right\\\|\\left\\\|\\partial_% {i}-d_{i}I-H_{i}\\right\\\|+\\\|G\\\|\\sum_{i=1}^{N}\\left\\\|\\Lambda_{i}\\right\\\|\\left\\\|H% _{i}\\right\\\|\\left\\\|T\\right\\\|$ $\\displaystyle<\\sum_{i=1}^{N}\\left\\\|\\Lambda_{i}\\right\\\|\\left(\\left\\\|\\partial_{i% }-d_{i}I-H_{i}\\right\\\|+\\left\\\|H_{i}\\right\\\|\\right)\\leavevmode\\nobreak\\ [\\\|G\\\|=% 1,\\\|T\\\|<1]$\n\nGiven $\\left\\\|\\Lambda_{i}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|\\partial_{j}-d_{j}I% \\right\\\|},\\leavevmode\\nobreak\\ \\forall i=1:N$, we have\n\n $\\displaystyle\\left\\\|T^{\\prime}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|\\partial_% {j}-d_{j}I\\right\\\|}\\sum_{i=1}^{N}\\left(\\left\\\|\\partial_{i}-d_{i}I-H_{i}\\right% \\\|+\\left\\\|H_{i}\\right\\\|\\right)$\n\nSee 1\n\n###### Proof.\n\nFrom Theorem 1 of and Lemma 1, our iterator is valid if and only if $\\rho\\left(T^{\\prime}\\right)<1$. From Lemma 2 the upper bound of spectral norm of iterator only depends on {$\\partial_{i}\\}_{i=1:N}$ and $\\{H_{i}\\}_{i=1:N}$ given $\\left\\\|\\Lambda_{i}\\right\\\|<\\frac{1}{\\sum_{j=1}^{N}\\left\\\|\\partial_{j}-d_{j}I% \\right\\\|},\\leavevmode\\nobreak\\ \\forall i=1:N$. Nonetheless, for any matrix spectral radius is upper bounded by its spectral norm. Thus, if the iterator is valid for some $u^{\\prime}_{0},b^{\\prime},\\{\\Theta^{\\prime}_{i}\\}_{i=1:N},\\delta x^{\\prime},% \\delta t^{\\prime}$, and $\\epsilon^{\\prime}$, then it is valid for any feasible choice of $u_{0},b,\\{\\Theta_{i}\\}_{i=1:N},\\delta x,\\delta t$, and $\\epsilon$ satisfying the constraints. ∎\n\n## Appendix C Geometric Interpretation\n\nSurprisingly, we find that the form of $\\\|T^{\\prime}\\\|$ has a special structure. As the denominator is constant the objective is to minimize $\\left\\\|\\partial_{i}-d_{i}I-H_{i}\\right\\\|+\\left\\\|H_{i}\\right\\\|$ w.r.t. $\\left\\\|H_{i}\\right\\\|;\\forall i=1:N$. Invoking triangular inequality of norm we have the lower-bound\n\n $\\left\\\|\\partial_{i}-d_{i}I-H_{i}\\right\\\|+\\left\\\|H_{i}\\right\\\|\\geq\\left\\\|% \\partial_{i}-d_{i}I\\right\\\|$\n\nTaking square of both side, we have\n\n $\\mbox{Find }{H_{i}}\\mbox{ such that }(\\partial_{i}-d_{i}I)^{\\top}H_{i}\\leq% \\left\\\|H_{i}\\right\\\|$\n\nWhen the equality holds the local optima is the surface of the hyper-sphere with center at $\\frac{1}{2}(\\partial_{i}-d_{i}I)$ with radius $\\frac{1}{2}\\left\\\|\\partial_{i}-d_{i}I\\right\\\|$. We interpret that each $H_{i}$ explores an optimal discretization scheme near the manifold of $\\partial_{i}$ by learning the sum of order rules as described in ." ]	[ null, "https://deeplearn.org/arxiv_files/1910.03452v1/Figs/sequence.png", null, "https://deeplearn.org/arxiv_files/1910.03452v1/Figs/delta_0_75_var.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87706816,"math_prob":0.99876654,"size":16123,"snap":"2020-10-2020-16","text_gpt3_token_len":3398,"char_repetition_ratio":0.12767541,"word_repetition_ratio":0.084033616,"special_character_ratio":0.2064132,"punctuation_ratio":0.111267604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99993324,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-06T10:00:42Z\",\"WARC-Record-ID\":\"<urn:uuid:816effad-092d-42cb-9ea7-1591d696622b>\",\"Content-Length\":\"123183\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3441f924-4545-4425-abde-d1bc1682b4c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:436aee8d-290c-4206-ac13-0dc996669a36>\",\"WARC-IP-Address\":\"104.24.115.104\",\"WARC-Target-URI\":\"https://deeplearn.org/arxiv/96589/implicit-neural-solver-for-time-dependent-linear-pdes-with-convergence-guarantee\",\"WARC-Payload-Digest\":\"sha1:C2JIGPF6DLY7M3SZDSB2NR4FUWTCUD7K\",\"WARC-Block-Digest\":\"sha1:QLXFPWNVK3EQNWDLBEXWFQLCCSZIDB2O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371620338.63_warc_CC-MAIN-20200406070848-20200406101348-00464.warc.gz\"}"}
https://www.cuemath.com/8th-grade-math/	[ "By Grade 8 children are well versed with a bunch of topics. Higher-order concepts like rational numbers, irrational numbers, exponents are all introduced in Grade 8. Children in Grade 8 are expected to practice these concepts and get a hold over them. Concepts learnt in Grade 8 go a long way in higher grade concepts.\n\n## Exponents\n\n• Basics of Exponents\n• Laws of Exponents\n• Negative Exponents\n• ## Scientific Notation\n\n• Expressing Very Large and Very Small Numbers in the Standard Form\n• Operations on Numbers Expressed in Standard Form\n• ## Lines, Angles, and Triangles\n\n• Angles Formed by Intersection of Parallel Lines and Transversals\n• The Angle Sum Property and Exterior Angle Property of Triangles\n\n• Translations\n• Reflections\n• Rotations\n• ## Congruence\n\n• Congruent Figures\n• ## Similarity\n\n• Dilation\n• Similar Figures\n• ## Linear Equations in One Variable\n\n• Solving Simple Linear Equations in One Variable\n• Solving Equations Reducible to the Linear Form\n• Applications of Linear Equations in One Variable\n• ## Straight Lines\n\n• Graphs of Proportional Relationships\n• Comparing Proportional Relationships\n• The Slope-Intercept Form of a Line's Equation\n• ## Simultaneous Linear Equations\n\n• Solving a Pair of Linear Equations Graphically\n• Solving a Pair of Linear Equations Algebraically\n• Applications of Simultaneous Linear Equations\n• ## Functions\n\n• Relation and Functions\n• Representation of Functions\n• Linear and Nonlinear Functions\n• Analyzing and Sketching Graphs\n• ## Volume\n\n• Volume of Cylinders\n• Volume of Cones\n• Volume of Spheres\n• ## Data Analysis and Displays\n\n• Scatter Plots\n• Lines of Fit\n• Two-Way Tables\n• Choosing a Data Display\n• ## Real Numbers\n\n• Classification of Rational Numbers\n• Conversion of Recurring Decimals to Fractions\n• Representation of Recurring Decimals on the Number Line\n• Irrational Numbers\n• ## Square Roots and Cube Roots\n\n• Finding Square Roots\n• Finding Cube Roots\n• ## The Pythagoras Theorem\n\n• The Pythagoras Theorem and Its Application\n• The Distance Formula\n\n• All Topics\n• ## 8th Grade Math Curriculum\n\nScoring high marks in Grade 8 Mathematics just got easier! Your kid will now be familiar with Translations, Reflections, Rotations & will start performing operations on exponents with his learnings from Cuemath Online Tuitions. Your child will start asking the why behind every fact, instead of being satisfied being just the what of that fact..get ready to have a curious grade 8 kid, who just wants to be tutored more!\n\nBy the end of this grade, your kids will learn to solve simple linear equations in one variable. Cuemath tutors will encourage your kids to learn the concepts using objects, pictures, and visual models. In the online Math tuitions with Cuemath, your kid will get personalized attention right from the first class in Grade 8. Our tutors are handpicked and the best in the market. Your child will be tutored online for every concept in grade 8 by an expert teacher online. As each tuition class progresses, you will see the improvement in your kids’ mathematical abilities.\n\n## Best tutors for kids in Grade 8\n\nOur Grade 8 online tuition classes with Cuemath tutors will help your kid tackle any Math problem. Cuemath’s online tuitions are designed to teach, challenge, and boost the confidence of budding Mathematicians. One of the most important things to understand about your kid in Grade 8 is that the more he practices in online tuitions, the faster he will grasp the formulas and techniques to get ahead. Get your child enrolled in Cuemath online classes today, and see a difference right after a couple of tuitions!\n\n>> Visualize Math with Interactive 8th Grade Math Worksheets\n\n## Grade 8 kids can learn Math Online\n\nCuemath has over a dozen unique activities and alluring characters- for kids in grade 8 who are tutoring online with us. During these activities, the tutors observe Grade 8 students picking up on Relation and Functions, Analyzing and Sketching Graphs, and so many more things. Cuemath tutors observe the performance of your kid and create a personalized learning plan instead of robotic completion of courses." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88059336,"math_prob":0.626268,"size":2569,"snap":"2023-40-2023-50","text_gpt3_token_len":558,"char_repetition_ratio":0.15126705,"word_repetition_ratio":0.004784689,"special_character_ratio":0.18723239,"punctuation_ratio":0.0845987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97752243,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T15:30:46Z\",\"WARC-Record-ID\":\"<urn:uuid:cd47da7a-7123-476c-9c0e-b5ee6a069fae>\",\"Content-Length\":\"143033\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:468ed392-3361-434c-a485-e19c0d5acf15>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f6045ad-1f1f-4f27-b127-a70cf2ee3e5a>\",\"WARC-IP-Address\":\"104.18.19.168\",\"WARC-Target-URI\":\"https://www.cuemath.com/8th-grade-math/\",\"WARC-Payload-Digest\":\"sha1:T6MVVU34EEJLJT3F5WKTFDTE7HDYAEQF\",\"WARC-Block-Digest\":\"sha1:FY5ST6QCA6EPT7QYRWYIAWUK2BWPNAIC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506481.17_warc_CC-MAIN-20230923130827-20230923160827-00248.warc.gz\"}"}
https://web2.0calc.com/questions/math-homework-help-please_1	[ "+0\n\n0\n323\n3\n\n1) The three angle measures of a triangle are x/2, x/3 and x/6 degrees. What is the measure of the smallest angle?\n\n2)Each pair of opposite sides of an octagon are parallel. We know three interior angles of the octagon are 115,125 and,135. What is the degree measure of the largest interior angle of the octagon?\n\n3)What is the ratio of the number of degrees in the interior angle of a regular pentagon to the number of degrees in the interior angle of a regular hexagon?\n\nThanks!\n\nFeb 5, 2019\n\n#1\n+1\n\n1) If x/2, x/3, and x/6 are angle measures of a triangle, they must add up to 180.\n\nx/2 + x/3 + x/6 = 180\n\nFind the common denominator:\n\n3x/6+2x/6 + x/6 = 180\n\nMultiply both sides by 6:\n\n3x+2x+x=1080\n\n6x=1080,\n\nx = 180\n\nx/6, the smallest angle, is 30.\n\n2) Correct me if I'm wrong, but isn't it just 135?\n\n3) 108/120 = 9/10\n\nFeb 5, 2019\n#2\n0\n\nFor #2, 135 isn't correct but thank you for helping me solve 1 and 3. I did not look at #1 in the way that you showed.\n\nFeb 5, 2019\n#3\n+1\n\nI believe this might be the second one\n\nThe sum of the interior angles of an octagon = 135 * 8 = 1080°\n\nSo....the remaining two angles must sum to 1080 - 2(115 + 125 + 135) = 330°\n\nSo....since the opposite sides are parallel.....the largest remaining angles each measure 330/2 = 165°", null, "", null, "", null, "Feb 6, 2019" ]	[ null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92687136,"math_prob":0.99110496,"size":1230,"snap":"2020-34-2020-40","text_gpt3_token_len":379,"char_repetition_ratio":0.1411093,"word_repetition_ratio":0.05957447,"special_character_ratio":0.35447153,"punctuation_ratio":0.13356164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987884,"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\":\"2020-08-05T02:13:32Z\",\"WARC-Record-ID\":\"<urn:uuid:0a74e913-47db-4295-8398-49986fbf9067>\",\"Content-Length\":\"28694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c98714c-8b2e-4005-8e75-cc32f7df1aba>\",\"WARC-Concurrent-To\":\"<urn:uuid:629ccf17-b21b-464d-b3ed-0f9ca5d90866>\",\"WARC-IP-Address\":\"209.126.117.101\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/math-homework-help-please_1\",\"WARC-Payload-Digest\":\"sha1:5ABWMPUBPPM6NXRU46KX3NFXCJ2BACWL\",\"WARC-Block-Digest\":\"sha1:XHRJRVSMBK36CQVSLBNOK7AQXDFYTFJY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735906.77_warc_CC-MAIN-20200805010001-20200805040001-00503.warc.gz\"}"}
https://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real/54521	[ "# Is this Batman equation for real? [closed]\n\nHardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?", null, "• Why don't you just try it? Jul 29, 2011 at 21:19\n• @Jim: If you mouse over the downvote button, you see: \"This question does not show any research effort; it is unclear or not useful.\" I downvoted because the OP was too lazy to type in the equation himself to any plotting program or calculator, which would have immediately shown that the equation is \"for real\". If the OP were asking for an explanation of how such an equation might be derived, as ShreevatsaR has done, that would be an appropriate question. Jul 30, 2011 at 17:08\n• @Zev Chonoles: i do not know of any web-site that can plot that equation. i wouldn't know where to begin. Also i don't understand how any solver could plot such a diagram. (Hence the question). Jul 30, 2011 at 19:35\n• I don't understand why this question has so many upvotes. Aug 3, 2011 at 18:36\n• Aug 31, 2011 at 14:40\n\nAs Willie Wong observed, including an expression of the form $\\displaystyle \\frac{\|\\alpha\|}{\\alpha}$ is a way of ensuring that $\\alpha > 0$. (As $\\sqrt{\|\\alpha\|/\\alpha}$ is $1$ if $\\alpha > 0$ and non-real if $\\alpha < 0$.)\n\nThe ellipse $\\displaystyle \\left( \\frac{x}{7} \\right)^{2} + \\left( \\frac{y}{3} \\right)^{2} - 1 = 0$ looks like this:", null, "So the curve $\\left( \\frac{x}{7} \\right)^{2}\\sqrt{\\frac{\\left\| \\left\| x \\right\|-3 \\right\|}{\\left\| x \\right\|-3}} + \\left( \\frac{y}{3} \\right)^{2}\\sqrt{\\frac{\\left\| y+3\\frac{\\sqrt{33}}{7} \\right\|}{y+3\\frac{\\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $\|x\|>3$ and $y > -3\\sqrt{33}/7$:", null, "That's the first factor.\n\nThe second factor is quite ingeniously done. The curve $\\left\| \\frac{x}{2} \\right\|\\; -\\; \\frac{\\left( 3\\sqrt{33}-7 \\right)}{112}x^{2}\\; -\\; 3\\; +\\; \\sqrt{1-\\left( \\left\| \\left\| x \\right\|-2 \\right\|-1 \\right)^{2}}-y=0$ looks like:", null, "This is got by adding $y = \\left\| \\frac{x}{2} \\right\| - \\frac{\\left( 3\\sqrt{33}-7 \\right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:", null, "and $y = \\sqrt{1-\\left( \\left\| \\left\| x \\right\|-2 \\right\|-1 \\right)^{2}}$, the upper halves of the four circles $\\left( \\left\| \\left\| x \\right\|-2 \\right\|-1 \\right)^2 + y^2 = 1$:", null, "The third factor $9\\sqrt{\\frac{\\left( \\left\| \\left( 1-\\left\| x \\right\| \\right)\\left( \\left\| x \\right\|-.75 \\right) \\right\| \\right)}{\\left( 1-\\left\| x \\right\| \\right)\\left( \\left\| x \\right\|-.75 \\right)}}\\; -\\; 8\\left\| x \\right\|\\; -\\; y\\; =\\; 0$ is just the pair of lines y = 9 - 8\|x\|:", null, "truncated to the region $0.75 < \|x\| < 1$.\n\nSimilarly, the fourth factor $3\\left\| x \\right\|\\; +\\; .75\\sqrt{\\left( \\frac{\\left\| \\left( .75-\\left\| x \\right\| \\right)\\left( \\left\| x \\right\|-.5 \\right) \\right\|}{\\left( .75-\\left\| x \\right\| \\right)\\left( \\left\| x \\right\|-.5 \\right)} \\right)}\\; -\\; y\\; =\\; 0$ is the pair of lines $y = 3\|x\| + 0.75$:", null, "truncated to the region $0.5 < \|x\| < 0.75$.\n\nThe fifth factor $2.25\\sqrt{\\frac{\\left\| \\left( .5-x \\right)\\left( x+.5 \\right) \\right\|}{\\left( .5-x \\right)\\left( x+.5 \\right)}}\\; -\\; y\\; =\\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.\n\nFinally, $\\frac{6\\sqrt{10}}{7}\\; +\\; \\left( 1.5\\; -\\; .5\\left\| x \\right\| \\right)\\; -\\; \\frac{\\left( 6\\sqrt{10} \\right)}{14}\\sqrt{4-\\left( \\left\| x \\right\|-1 \\right)^{2}}\\; -\\; y\\; =\\; 0$ looks like:", null, "so the sixth factor $\\frac{6\\sqrt{10}}{7}\\; +\\; \\left( 1.5\\; -\\; .5\\left\| x \\right\| \\right)\\sqrt{\\frac{\\left\| \\left\| x \\right\|-1 \\right\|}{\\left\| x \\right\|-1}}\\; -\\; \\frac{\\left( 6\\sqrt{10} \\right)}{14}\\sqrt{4-\\left( \\left\| x \\right\|-1 \\right)^{2}}\\; -\\; y\\; =\\; 0$ looks like", null, "As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)", null, "• I tip my hat to you for this comprehensive dissection. Jul 30, 2011 at 14:06\n• If there were only no rep-cap, Shree would be swimming in rep now... ;P Jul 30, 2011 at 19:56\n• \"I don’t know how ShreevatsaR did it but he sure brought a large luggage when intelligence showered the Earth.\" yangkidudel.wordpress.com/2011/08/02/love-and-mathematics Aug 3, 2011 at 0:15\n• @Jonas Meyer: LOL, that's embarrassing! :P But then again, if Batman is what it takes for someone to appreciate mathematics a little, well good for Batman. :-) Aug 3, 2011 at 10:46\n• @Jack: Grapher, which comes by default on Mac OS X. I mentioned it in the answer actually, just before the last figure. Aug 3, 2011 at 17:34\n\nYou may be able to see more easily the correspondences between the equations and the graph through the following picture which is from the link I got after a curious search on Google(link broken now):", null, "• Geometer's Sketchpad? Aug 3, 2011 at 18:28\n• @Isaac: Probably right. Sep 6, 2011 at 12:27\n• to see the graph just google the equation: 2sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqrt(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9 Aug 20, 2012 at 18:47\n• @HelderVelez 2014: \"'abs' (and any subsequent words) was ignored because we limit queries to 32 words.\" - Google\n– Baby\nNov 28, 2014 at 8:11\n• x1(y), x2(y) should be 7 sqrt( 1 - y^2 / 9) and -7 * sqrt( 1 - y^2 / 9), respectively Jun 25, 2017 at 4:39\n\nHere's what I got from the equation using Maple...", null, "• What if Commissioner Gordon uses Mathematica ???? Jul 30, 2011 at 5:25\n• @The Chaz: Then Commissioner Gordon should support the Mathematica SE Site Proposal on Area 51. Jul 30, 2011 at 19:22\n• touché .... Jul 30, 2011 at 19:24\n• Here's Mathematica code. See Heike's post. I tried it on M8 and it works fine. groups.google.com/group/comp.soft-sys.math.mathematica/…\n– Sol\nAug 5, 2011 at 22:48\n• @Sol: You can do better; see my answer. Aug 9, 2011 at 7:56\n\nLooking at the equation, it looks like it contains terms of the form $$\\sqrt{\\frac{\| \|x\| - 1 \|}{\|x\| - 1}}$$ which evaluates to $$\\begin{cases} 1 & \|x\| > 1\\\\ i & \|x\| < 1\\end{cases}$$\n\nSince any non-zero real number $y$ cannot be equal to a purely imaginary non-zero number, the presence of that term is a way of writing a piece-wise defined function as a single expression. My guess is that if you try to plot this in $\\mathbb{C}^2$ instead of $\\mathbb{R}^2$ you will get all kinds of awful.\n\n• Yeah, the equation looks too contrived to me. :) A parametric form (it's just quadratic and linear arcs sewn together, it looks) would still be messy, but not as messy. (Probably a good job for splines...) Jul 30, 2011 at 2:43\n• +1 i was wondering how they split it up into sections. Jul 30, 2011 at 19:38\n• \" My guess is that if you try to plot this in C2 instead of R2 you will get all kinds of awful.\" What did you expect? The analytic continuation of the Batman symbol??\n– jwg\nAug 1, 2013 at 8:43\n\nSince people (not from this site, but still...) keep bugging me, and I am unable to edit my previous answer, here's Mathematica code for plotting this monster:\n\nPlot[{With[{w = 3 Sqrt[1 - (x/7)^2],\nl = 6/7 Sqrt + (3 + x)/2 - 3/7 Sqrt Sqrt[4 - (x + 1)^2],\nh = (3 (Abs[x - 1/2] + Abs[x + 1/2] + 6) -\n11 (Abs[x - 3/4] + Abs[x + 3/4]))/2,\nr = 6/7 Sqrt + (3 - x)/2 - 3/7 Sqrt Sqrt[4 - (x - 1)^2]},\nw + (l - w) UnitStep[x + 3] + (h - l) UnitStep[x + 1] +\n(r - h) UnitStep[x - 1] + (w - r) UnitStep[x - 3]],\n1/2 (3 Sqrt[1 - (x/7)^2] + Sqrt[1 - (Abs[Abs[x] - 2] - 1)^2] + Abs[x/2] -\n((3 Sqrt - 7)/112) x^2 - 3) (Sign[x + 4] - Sign[x - 4]) - 3Sqrt[1 - (x/7)^2]},\n{x, -7, 7}, AspectRatio -> Automatic, Axes -> None, Frame -> True,\nPlotStyle -> Black]", null, "This should work even for versions that do not have the Piecewise[] construct. Enjoy. :P\n\nIn fact, the five linear pieces that consist the \"head\" (corresponding to the third, fourth, and fifth pieces in Shreevatsa's answer) can be expressed in a less complicated manner, like so:\n\n$$y=\\frac{\\sqrt{\\mathrm{sign}(1-\|x\|)}}{2}\\left(3\\left(\\left\|x-\\frac12\\right\|+\\left\|x+\\frac12\\right\|+6\\right)-11\\left(\\left\|x-\\frac34\\right\|+\\left\|x+\\frac34\\right\|\\right)\\right)$$\n\nThis can be derived by noting that the functions\n\n$$\\begin{cases}f(x)&\\text{if }x<c\\\\g(x)&\\text{if }c<x\\end{cases}$$\n\nand $f(x)+(g(x)-f(x))U(x-c)$ (where $U(x)$ is the unit step function) are equivalent, and using the \"relation\"\n\n$$U(x)=\\frac{x+\|x\|}{2x}$$\n\nNote that the elliptic sections (both ends of the \"wings\", corresponding to the first piece in Shreevatsa's answer) were cut along the lines $y=-\\frac37\\left((2\\sqrt{10}+\\sqrt{33})\|x\|-8\\sqrt{10}-3\\sqrt{33}\\right)$, so the elliptic potion can alternatively be expressed as\n\n$$\\left(\\left(\\frac{x}{7}\\right)^2+\\left(\\frac{y}{3}\\right)^2-1\\right)\\sqrt{\\mathrm{sign}\\left(y+\\frac37\\left((2\\sqrt{10}+\\sqrt{33})\|x\|-8\\sqrt{10}-3\\sqrt{33}\\right)\\right)}=0$$\n\nTheoretically, since all you have are arcs of linear and quadratic curves, the chimera can be expressed parametrically using rational B-splines, but I'll leave that for someone else to explore...\n\n• Okay, make that an hour. Sheesh. facepalm* Jul 30, 2011 at 19:56\n• Great, this would be much clearer. BTW, I think it's safer to leave it at \"can be expressed using B-splines\" than to actually try it out: who knows how many hours that will waste, right? :-) Jul 31, 2011 at 3:20\n\nThe following is what I got from the equations using MATLAB:", null, "Here is the M-File (thanks to this link):\n\nclf; clc; clear all;\nsyms x y\n\neq1 = ((x/7)^2sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2sqrt(abs(y+3/7sqrt(33))/(y+3/7sqrt(33)))-1);\neq2 = (abs(x/2)-((3sqrt(33)-7)/112)x^2-3+sqrt(1-(abs(abs(x)-2)-1)^2)-y);\neq3 = (9sqrt(abs((abs(x)-1)(abs(x)-.75))/((1-abs(x))(abs(x)-.75)))-8abs(x)-y);\neq4 = (3abs(x)+.75sqrt(abs((abs(x)-.75)(abs(x)-.5))/((.75-abs(x))(abs(x)-.5)))-y);\neq5 = (2.25sqrt(abs((x-.5)(x+.5))/((.5-x)(.5+x)))-y);\neq6 = (6sqrt(10)/7+(1.5-.5abs(x))sqrt(abs(abs(x)-1)/(abs(x)-1))-(6sqrt(10)/14)sqrt(4-(abs(x)-1)^2)-y);\n\naxes('Xlim', [-7.25 7.25], 'Ylim', [-5 5]);\nhold on\n\nezplot(eq1,[-8 8 -3sqrt(33)/7 6-4sqrt(33)/7]);\nezplot(eq2,[-4 4]);\nezplot(eq3,[-1 -0.75 -5 5]);\nezplot(eq3,[0.75 1 -5 5]);\nezplot(eq4,[-0.75 0.75 2.25 5]);\nezplot(eq5,[-0.5 0.5 -5 5]);\nezplot(eq6,[-3 -1 -5 5]);\nezplot(eq6,[1 3 -5 5]);\ncolormap([0 0 1])\n\ntitle('Batman');\nxlabel('');\nylabel('');\nhold off\n\n\nThe 'Batman equation' above relies on an artifact of the plotting software used which blithely ignores the fact that the value $\\sqrt{\\frac{\|x\|}{x}}$ is undefined when $x=0$. Indeed, since we’re dealing with real numbers, this value is really only defined when $x>0$. It seems a little ‘sneaky’ to rely on the solver to ignore complex values and also to conveniently ignore undefined values.\n\nA nicer solution would be one that is unequivocally defined everywhere (in the real, as opposed to complex, world). Furthermore, a nice solution would be ‘robust’ in that small variations (such as those arising from, say, roundoff) would perturb the solution slightly (as opposed to eliminating large chunks).\n\nTry the following in Maxima (actually wxmaxima) which is free. The resulting plot is not quite as nice as the plot above (the lines around the head don’t have that nice ‘straight line’ look), but seems more ‘legitimate’ to me (in that any reasonable solver should plot a similar shape). Please excuse the code mess.\n\n/* [wxMaxima batch file version 1] [ DO NOT EDIT BY HAND! ]/\n/ [ Created with wxMaxima version 0.8.5 ] /\n\n/ [wxMaxima: input start ] /\n/ [wxMaxima: input end ] /\n\n/ [wxMaxima: input start ] /\nf(a,b,x,y):=ax^2+by^2;\n/ [wxMaxima: input end ] /\n\n/ [wxMaxima: input start ] /\nc1:sqrt(26);\n/ [wxMaxima: input end ] /\n\n/ [wxMaxima: input start ] /\ndraw2d(implicit(\nf(1/36,1/9,x,y)\n+max(0,2-f(1.5,1,x+3,y+2.7))\n+max(0,2-f(1.5,1,x-3,y+2.7))\n+max(0,2-f(1.9,1/1.7,(5(x+1)+(y+3.5))/c1,(-(x+1)+5(y+3.5))/c1))\n+max(0,2-f(1.9,1/1.7,(5(x-1)-(y+3.5))/c1,((x-1)+5(y+3.5))/c1))\n+max(0,2-((1.1(x-2))^4-(y-2.1)))\n+max(0,2-((1.1(x+2))^4-(y-2.1)))\n+max(0,2-((1.5x)^8-(y-3.5)))\n-1,\nx,-6,6,y,-4,4));\n/* [wxMaxima: input end ] /\n\n/ Maxima can't load/batch files which end with a comment! /\n\"Created with wxMaxima\"$ The resulting plot is:", null, "(Note that this is, more or less, a copy of the entry I made on http://blog.makezine.com.) • I really think that an indeterminate value multiplied by zero equals zero, so it seems to be legit. Is there any reason 0 0/0 should not be defined to be zero? Jun 15, 2015 at 19:45 • @dbanet: What are you referring to? The issue above is that the original equations rely on the plotting software ignoring undefined values, which is peculiar, to say the least. The expression$\\sqrt{\\frac{\|x\|}{x}}$(with$x$being replaced by some expression) is what I referred to and it appears without being multiplied by$x$. Jun 15, 2015 at 19:54 • @copper-hat: The function$f(x)=\\sqrt{\\frac{\|x\|}{x}}$appears only in boolean expressions$F:x\\to \\{\\text{True},\\text{False}\\}$of form$f(x)g(x)=0$, so if$g(x)$is defined as$g:x\\in\\mathbb{C}\\to{0}$, I would rather evaluate$F(0)$to$\\text{True}$than to$\\text{False}$, as$\\forall{x}:f(x)g(x)=0\\Longleftrightarrow \\Big(f(x)=0\\lor g(x)=0\\Big)$. Jun 15, 2015 at 21:57 • @dbanet: I'm really not sure what you are getting at. Look at the expressions in the question. They rely on the expression$\\sqrt{\\frac{\|x\|}{x}}$returning zero for$x \\le 0$, which is strange (look at Willie's answer math.stackexchange.com/a/54521/27978). My answer plots level sets, which avoids this whole issue. Jun 15, 2015 at 22:10 • @dbanet: I don't really get your point. In the formula in the question there are lots of expressions of the above form that are multiplied by quantities that do not evaluate to zero when$x<0\\$. If they did, there would be no need to have the strange expression in the first place. Jun 15, 2015 at 22:21\n\nHere's the equations typed out if you want save time with writing it yourself.\n\n(x/7)^2SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\\SQRT(ABS(y+3SQRT(33)/7)/(y+3SQRT(33)/7))-1=0\nABS(x/2)-((3SQRT(33)-7)/112)x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0\n9SQRT(ABS((ABS(x)-1)(ABS(x)-0.75))/((1-ABS(x))(ABS(x)-0.75)))-8ABS(x)-y=0\n3ABS(x)+0.75SQRT(ABS((ABS(x)-0.75)(ABS(x)-0.5))/((0.75-ABS(x))(ABS(x)-0.5)))-y=0\n2.25SQRT(ABS((x-0.5)(x+0.5))/((0.5-x)(0.5+x)))-y=0\n(6SQRT(10))/7+(1.5-0.5ABS(x))SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6SQRT(10))/14)SQRT(4-(ABS(x)-1)^2)-y=0\n\n• The multiple \"=0\" lines are different than the mulitiplications in the original, there are some backslashes in there that throw things off, and the formatting is hard to read. The pastebin is better. Jul 30, 2011 at 16:42\n• @copper, it is just because of the algorithm that draw uses for drawing implicit functions. You need to setup the variables ip_grid and ip_grid_in, that are the sampling values in your region. For example draw2d(ip_grid=[60,60], ip_grid_in=[20,20], implicit(y^2=x^3-2*x+1, x, -4,4, y, -4,4) ); Oct 16, 2014 at 19:21" ]	[ null, "https://i.stack.imgur.com/VYKfg.jpg", null, "https://i.stack.imgur.com/PXv4W.png", null, "https://i.stack.imgur.com/oeCdG.png", null, "https://i.stack.imgur.com/vAoFe.png", null, "https://i.stack.imgur.com/Vfyre.png", null, "https://i.stack.imgur.com/69Pdf.png", null, "https://i.stack.imgur.com/3CJGO.png", null, "https://i.stack.imgur.com/Sh0Bp.png", null, "https://i.stack.imgur.com/XKs3Z.png", null, "https://i.stack.imgur.com/OO3np.png", null, "https://i.stack.imgur.com/YmA4v.png", null, "https://i.stack.imgur.com/EHNR8.png", null, "https://i.stack.imgur.com/6pXe9.jpg", null, "https://i.stack.imgur.com/zYXKB.png", null, "https://i.stack.imgur.com/vHI8K.jpg", null, "https://i.stack.imgur.com/dMzTL.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57418066,"math_prob":0.9993474,"size":2786,"snap":"2022-40-2023-06","text_gpt3_token_len":1151,"char_repetition_ratio":0.2929547,"word_repetition_ratio":0.121212125,"special_character_ratio":0.4648241,"punctuation_ratio":0.15640599,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984694,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,10,null,10,null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T22:28:03Z\",\"WARC-Record-ID\":\"<urn:uuid:bdd20f3a-2149-4bbf-9eea-3adb923183ec>\",\"Content-Length\":\"258378\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc3498cf-3d1d-401b-9f05-5795282ac474>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e6ab01f-df22-4f4a-a1df-2eabc0366659>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real/54521\",\"WARC-Payload-Digest\":\"sha1:CCRK5RPXFRXNS3GTNEGCZOGBC5G2VNW2\",\"WARC-Block-Digest\":\"sha1:DIMP2TVTYTSAHZ3764XVKBWCL4YXEJVX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500365.52_warc_CC-MAIN-20230206212647-20230207002647-00249.warc.gz\"}"}
https://www.esaral.com/q/two-angles-of-a-triangle-are-equal-and-the-third-angle-is-greater-than-each-of-those-angles-by-30-84307	[ "# Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°.\n\nQuestion:\n\nTwo angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.\n\nSolution:\n\nGiven that,\n\nTwo angles of a triangle are equal and the third angle is greater than each of those angles by 30°.\n\nLet x, x, x + 30° be the angles of a triangle\n\nWe know that,\n\nSum of all angles in a triangle is 180°\n\nx + x + x + 30° = 180°\n\n3x + 30° = 180°\n\n3x = 180° − 30°\n\n3x = 150°\n\nx = 50°\n\nTherefore, the three angles are 50°, 50°, 80°." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73009676,"math_prob":0.9999119,"size":3598,"snap":"2022-40-2023-06","text_gpt3_token_len":946,"char_repetition_ratio":0.1460768,"word_repetition_ratio":0.86092716,"special_character_ratio":0.23457476,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999287,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T21:11:28Z\",\"WARC-Record-ID\":\"<urn:uuid:2f91f02e-4d8a-44bb-b886-6dcf947d562a>\",\"Content-Length\":\"470221\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77fa40fb-a285-4d35-9e2f-8278bca3fe74>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b3f6d19-0ce8-4ab0-a4b8-3f730ced9094>\",\"WARC-IP-Address\":\"104.21.61.187\",\"WARC-Target-URI\":\"https://www.esaral.com/q/two-angles-of-a-triangle-are-equal-and-the-third-angle-is-greater-than-each-of-those-angles-by-30-84307\",\"WARC-Payload-Digest\":\"sha1:FHG2RVUKQFCBLIEW6PLCRGJW4CW45JBH\",\"WARC-Block-Digest\":\"sha1:SI74KDESD6VOQ4VFQ5BJG3ZQ6Y6PBS3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500154.33_warc_CC-MAIN-20230204205328-20230204235328-00305.warc.gz\"}"}
https://www.programcreek.com/2014/03/create-java-string-by-double-quotes-vs-by-constructor/	[ "# Create Java String Using ” ” or Constructor?\n\nIn Java, a string can be created in two ways:\n\n```String x = \"abc\"; String y = new String(\"abc\");```\n\nWhat is the difference between using the double quotes and using the constructor?\n\n1. Double Quotes vs. Constructor\n\nThis question can be answered by using two simple examples.\n\nExample 1:\n\n```String a = \"abcd\"; String b = \"abcd\"; System.out.println(a == b); // True System.out.println(a.equals(b)); // True```\n\n`a==b` is true because `a` and `b` are referring to the same string literal in the method area. The memory references are the same.\n\nWhen the same string literal is created more than once, only one copy of each distinct string value is stored. This is called \"string interning\". All compile-time constant strings in Java are automatically interned.\n\nExample 2:\n\n```String c = new String(\"abcd\"); String d = new String(\"abcd\"); System.out.println(c == d); // False System.out.println(c.equals(d)); // True```\n\n`c==d` is false because `c` and `d` refer to two different objects in the heap. Different objects always have different memory references.\n\nThis diagram illustrate the two situations above:", null, "2. Run-Time String Interning\n\nThanks to LukasEder (his comment below):\n\nString interning can still be done at run-time, even if two strings are constructed with constructors:\n\n```String c = new String(\"abcd\").intern(); String d = new String(\"abcd\").intern(); System.out.println(c == d); // Now true System.out.println(c.equals(d)); // True```\n\n3. When to Use Which\n\nBecause the literal \"abcd\" is already of type String, using constructor will create an extra unnecessary object. Therefore, double quotes should be used if you just need to create a String.\n\nIf you do need to create a new object in the heap, constructor should be used. Here is a use case.\n\nCategory >> Basics >> Diagram >> Java\nIf you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:\n```<pre><code>\nString foo = \"bar\";\n</code></pre>\n```\n• sss\n\nin jdk 1.6 is also true\n\n• affan\n\nSir I am newly joined Java programming certificate course my preceptor give me an project & I am stack on it for instance sir kindly advice me that how could I will manage program where I using string,double,integer variable in simultaneously please give some example.\n\n• Walt Corey\n\nThis is how one writes code nobody can understand. Good way to be top of the list come next layoff. I knew a guy did the same crap wit C++, so ‘highly optimized” nobody could read it, and it also didn’t work but broke in mysterious ways. Rather than seeing how obtuse you can code, see how readable you can. Readability is FAR more cost effective than so-called highly optimized code.\n\n• Ritter Liu\n\nThank you for your share 😛\n\n• Charly\n\nYou are worng, In JDK1.6 and JDK1.7 return true. Test rourself\n\n• 智陶\n\nString c = new String(“abcd”).intern();\nString d = new String(“abcd”).intern();\nSystem.out.println(c == d); // True in JDK1.7 while false in JDK 1.6.\ncorrect me if i am wrong\n\n• Feng Sun\n\ngood to know, thanks\n\n• LukasEder\n\nWhen the same string literal is created more than once, JVM store only one copy of each distinct string value. This is called “string interning“.\n\nThat’s not entirely correct. In your example, the compiler will already make sure that the same string constant is effectively referenced. String interning can still be done at runtime, even if two strings are constructed with constructors:\n\nString c = new String(“abcd”).intern();\nString d = new String(“abcd”).intern();\nSystem.out.println(c == d); // Now true\nSystem.out.println(c.equals(d)); // True" ]	[ null, "https://www.programcreek.com/wp-content/uploads/2014/03/constructor-vs-double-quotes-Java-String-New-Page-650x324.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70389056,"math_prob":0.81539696,"size":1883,"snap":"2023-40-2023-50","text_gpt3_token_len":451,"char_repetition_ratio":0.14316125,"word_repetition_ratio":0.0,"special_character_ratio":0.25703666,"punctuation_ratio":0.1631579,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9643094,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T23:17:26Z\",\"WARC-Record-ID\":\"<urn:uuid:a30bc89a-e700-4367-a6f4-fcf1522d84db>\",\"Content-Length\":\"29097\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72345139-dc2c-4148-8bda-3c113cb367de>\",\"WARC-Concurrent-To\":\"<urn:uuid:8097ec61-8326-4b90-9bf2-d9bac3d51b2f>\",\"WARC-IP-Address\":\"172.67.71.73\",\"WARC-Target-URI\":\"https://www.programcreek.com/2014/03/create-java-string-by-double-quotes-vs-by-constructor/\",\"WARC-Payload-Digest\":\"sha1:CVIIKI67NJKODUCHBEKBUJ5N7UYS5NOI\",\"WARC-Block-Digest\":\"sha1:3ETDWL6SIOUWKN3ZQXTF5EVH7MUKD44C\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510462.75_warc_CC-MAIN-20230928230810-20230929020810-00780.warc.gz\"}"}
https://present5.com/ministry-of-education-and-science-of-ukraine-national/	[ "", null, "Скачать презентацию Ministry of Education and Science of Ukraine National\n\ned_-_3.ppt\n\n• Размер: 14.4 Mегабайта\n• Количество слайдов: 81\n\n## Описание презентации Ministry of Education and Science of Ukraine National по слайдам", null, "Ministry of Education and Science of Ukraine National Aerospace University n. a. N. Y. . Zhukovsky « « Kh. AI » » Department of Graphical and Computer Modeling DESCRIPTIVE GEOMETRY", null, "Surfaces.", null, "Surfaces. Classification. Determinant. Outline.", null, "Surfaces. Classification. Determinant. Outline. Fundamentals : Surface of geometric solid is multiple of boundary points of a given body. B oundary point is a center of a sphere which always contains inner and outer points for this body while its radius tend to zero. To specify a surface in a drawing means to indicate the conditions enabling us to construct each point of this surface. Representation of a surface : 1. Analytical method ; 2. Graphical method. a ) framing; b ) kinematic. Kinematic surface represents a locus of lines moving in space according to a certain law. These lines, which may be straight or curved, are called generating lines ( generatrices ). In general, there is a variety of laws for generating a certain surface. It is desirable to select those laws and shapes of the generating lines which are most simple and convenient both for representing the required surface and solving the problems associated with it. Determinant of kinematic surface is minimum number of parameters and constraints which univocally represent given surface. Determinant structure : Determinant = {(geometrical parameters )( algorithm of constructions ) }. Surface on orthographical view can be represented by projections of its determinant.", null, "Surfaces. Classification. Surfaces Curved Polyhedral (multifaceted) Kinematical Framing. R uled (linear) N onlinear Prismatic. Pyramidal. Continuous / Discrete D evelopable C onical C ylindrical Torse G eneral view C hannel surface N ondevelopable G eneral view C atalan C yclic R evolution General view generatrix Parallel shift with Plane Director (parallelism ) Plane parallel to axes. To systematize properties of surfaces we can classify them by the type of generatrices and character of their motion.", null, "Surfaces. Classification. Determinant. Outline of a Surface. For a surface to be specified, it is sufficient to have the projections of its directrix (guide line) and adequate information on the method for constructing the generatrix passing through any point of the directrix. But if it is desirable to make the representation more obvious and expressive, then it is advisable to draw also the outline of the surface, several positions of the generatrix, most important lines and points on the surface, etc. Representation of a surface on a drawing by projections of its determinant can’t provide proper obviousness. It’s reasonable to supplement drawing of a surface by its outline. Example: a Cone, represented by its determinant : КВ = {(ℓ‚ t ) (ℓ∩ t , ℓ = R ( t )}, Difficult to imagine such a figure without additional explanations. t 2 l 1 t", null, "Surfaces. Classification. Determinant. Outline of a Surface. П 2 S Contour line Outline. Projectors O 1 O 2 O 3 П 1 On parallel projection of a surface Ф on the plane П some projectors touch with the surface Ф and form tangent cylindrical projecting surface Q. Tangent curve of Q and Ф surfaces is called Contour Line (Contour Curve), its projection on the plane П – Outline of the surface Ф. This curve can be plane or space. Outline of a surface is projection of its contour line. Thus Outline is visibility border for points which belong to a surface. For instance, the sphere on the planes П 1 and П 2 has contour curves – circles q 2 and m 1. They project on the other planes as curves of visibility q 1 and m 2 – outline curves. m 2 m 1 q 1 q", null, "Surfaces. Classification. Determinant. Outline of a surface. Frontal Visible Contour. Frontal Outline Horizontal Visible Contour Horizontal Outline", null, "Surfaces of Revolution. Forming. Definitions. Parallel Meridian Generatrix Neck Axis of revolution l 1 i m 3 m 2 m 1 o 2 o 1 o 3 i Meridian plane. A surface of revolution is generated by the revolution of a curved-line or straight-line generatrix about a fixed straight line called the axis of the revolution. Determinant of the surface of revolution — generatrix and axis of revolution : SR = {( l , I )( l = R ( I ))} Each point of the generatrix describes a circle. Hence, a plane perpendicular to the axis of revolution cuts this surface in circles. Such circles are called parallels. The biggest parallel is called the equator , the smallest parallel is called the neck of a surface. Neck and equator project to outlines. A plane passing through the axis of a surface of revolution is termed a meridian plane. The line where a meridian plane intersects a surface of revolution is called the meridian of the surface. Equator Cylinder of revolution. Generatrix – l — straight line axis of revolution ί ; n — curve , equally spaced from axis of revolution. If we imagine the totality of rectilinear generatrices, revolving around the axes, and the totality of generating circles, moving along that axes, then each line of one totality (of one «family» of lines) will intersect all lines of the other totality (of the other «family» of lines). As a result, the mesh of the given surface is obtained. Other surfaces may be thought of in the same way.", null, "Surfaces of Revolution. General.", null, "Surfaces of Revolution. Forming. Surface of revolution. Generatrix – Straight Line. Relative position of Generatrix and Axes of revolution. CYLINDER CONE SHEET HYPERBOLOID OF REVOLUTION PARALLEL INTERSECTING SKEW HYPERBOL", null, "Surfaces of Revolution. Forming. Surface of revolution. Generatrix – Circle. R>r R<r R=0 i n Rr i TORUS CLOSED TORUS SPHER", null, "Surfaces of Revolution. Examples. C O B A 12′ I I=I ’10» 1′ 2′ 3′ 4′ 5′ 6’9 ‘ ‘ 11» 12» 1» 2» 3» I=I ‘ m 1 m 2 I ‘I 1’12’ 2’11’ 10′ 9′ 8′ 3′ 7′ 6′ 5′ 4’5» 4» 3» 2» 1»6» 10»9» 8» 7» 12»11»For the hyperboloid of revolution the meridian is a hyperbola. If the hyperbola is revolved about its imaginary axis, then we have a hyperboloid of revolution of one sheet. A hyperboloid of revolution of one sheet can also be generated by the revolution of a straight line if the generatrix and the axis of revolution are skew lines.", null, "Surfaces of Revolution. Examples. C O B I I ‘ A B 1 B 2 A 1 B 1 A 2 B 2 M 2 t 1 t 2 C 2 I=I ‘A 1 C 1 M 1 Neck", null, "Surfaces of Revolution. Sections by Planes. O 14 23 2 2 21 2 (1) 1 =3 1 (2) 1 =4 1 1 1 8 1 2 1 3 1 4 1 5 16 17 11 2 2 2 =8 2 5 2 4 2 =6 2 3 2 =7 21′ 2’8′ 3’7′ 4’6′ 5′ O 1 Minor axes Major axes Minor axes", null, "Surfaces of Revolution. Sections by Planes. S ELLIPSE – TWO OUTLINES FROM ONE SIDE OF AN APEX ARE INTERSECTED PARABOLA – PARALLEL TO THE OUTLINESTRAIGHT LINES – THROUGH APEX ALONG GENERATRIX HYPERBOLA – TWO OUTLINES FROM DIFFERENT SIDES OF AN APEX ARE INTERSECTED CIRCLE – PERPENDICULAR TO AN AXESCone of revolution. Conical Sections.", null, "Surfaces of Revolution. Sections by Planes. Point Ellipse Circle Double Parabola Two straight Lines Hyperbola", null, "Surfaces of Revolution. Sections by Planes. A C B D D ‘ B ‘ C ‘ O ‘ A ‘ B 2 =D 2 =O 2 C 2 C ‘A 2 C ‘D ‘ O ‘O ELLIPS", null, "Surfaces of Revolution. Sections by Planes. O ‘M 2 =N 2=O 2 M ‘K 2 J 2 =L 2 L ‘ J ‘ N ‘K ‘ L’R jl T 2 = P 2 T ‘σ 2 σ ‘ 2 T ‘ P ‘HYPERBOL", null, "Surfaces of Revolution. Sections by Planes. O ‘ M 2 =N 2=O 2 M ‘M ‘ K 2 J 2 =L 2 L ‘J ‘ N ‘K ‘ L’ R jl T 2 = P 2 T ‘ σ 2 σ ‘ 2 T ‘P ‘ PARABOL", null, "Sections of Solids. 1. Designate all character points of intersection between cutting plane and surfaces of figure. 2. Locate axes of section 1 – 7 parallel to a cutting line. Mark distances between points in question. 3. Construct true sizes of full sections from corresponding marked points ( for example, А-А section – ellipse from cylinder cutting between points 1 – 7 , rectangle from pyramid cutting between points 3 – 6 , etc. ). 4. Designate obtained section according to a standard (ISO, GOST, DSTU, etc. ). BA A B — B 2 2′ 71 2 3 4 5 6 7 7 6 5 4 1 22 1 4 3 6′ 3’1′ 3′ 6 ‘ 1 ‘ B 34 4’5 5′ A — A 3 3’6 6’ 1 Combined figure sectioned by a plane", null, "Intersection between a Solid and a Plane. Three ways how to construct conical section. 1. Using Major and Minor Axis of Ellipse 2. Using Parallels 3. Using Generatrices AA S 1 1 1 3 11 2 (2) 2 3 2 (4) 2 2 1 4 1 5 15 2 S 2 BB S 11 2 (3) 22 2 (5) 2 3 1 4 2 S 2 8 2 10 2 12 2 (11) 2 2 1 6 17 1 12 1 4 15 1 8 110 11 1 19 16 2 (7) 2 (9) 2 1 1 СС S 11 2 (3) 22 2 (5) 2 3 1 4 2 S 2 8 2 10 2 12 2 (11) 2 2 1 12 15 1 8 110 11 1 16 2 (7) 2 (9)", null, "Intersection between a Solid and a Plane. S 1(1) 2 3 2 2 2 5 2 3 1 4 2 S 2 2 1 5 1 6 2 1 1 6 14 1 1 3 3 3 2 3 (5) 34 3 S 3 (6)", null, "Surfaces. Ruled surfaces. A surface which can be generated by a straight line is called the ruled surface. Hence, the ruled surface represents a locus of straight lines. A surface which may be generated only by a curved line will be called the double- curved surface.", null, "Surfaces. Developable Ruled surfaces. Some curved surfaces can be developed so that they coincide completely (with all their points) with a plane without stretching or shrinking. Such surfaces will be called developable. They comprise only ruled surfaces in which adjacent rectilinear generatrices are parallel or intersect , or are tangent to sphere curve. All double-curved surfaces and the ruled surfaces which cannot be developed into a plane are called nondevelopable (or warped ) surfaces", null, "Surfaces. Ruled surfaces with one Diretrix. Conical Surface m 1 m 2 AS m M 2 M 1 S 2 l 1 A 1 S 1 mssmsm, , Con. S A conical surface is generated by a straight line passing through a fixed point and through all the points (in succession) of a curved directing line. This line is also called directrix or guide line. The fixed point S is called the vertex of a conical surface. S – vertex m – directrix ℓ – generatrix S 2 S", null, "Surfaces. Ruled surfaces with one Diretrix. Cylindrical surface A 2 m 1 A 1 ml 1 A ll 2 A 11 ll, , , Cyl. Siillm. Sm If vertex of conical surface is moved to infinity, then conical surface turns into cylindrical. A cylindrical surface is generated by a straight line which is parallel in all its positions to a given straight line and passes in succession through all points of a curved directing line. S – vertex → ∞ m – directrix ℓ — generatrix", null, "Surfaces. Ruled surfaces with one Diretrix. n , k – border linesmm, , TSl 1 m 1 B 1 m 2 l 2 B 2 A 1 A 2 l A m The surface called a surface with a cuspidal edge or torse surface is generated by a rectilinear generatrix performing continuous motion and touching a space curve at all its positions. This space curve serves as the directrix for the surface under consideration and is called the cuspidal edge. Obviously, by specifying the projections of the cuspidal edge, we can specify a surface in the drawing. Determinant of a torse surface : 1. Curve line m as a directrix. 2. Straight line ℓ as a generatrix. n k.", null, "Surfaces. Ruled surfaces with one Diretrix. Ruled surfaces in mechanical engineering Beveled Gear surface Spur Gear surface", null, "Surfaces. Ruled surfaces with two Diretrices. llnmnm. PS, , , where, l — generatrix , m , n — directrices ; — plane director. Ruled surface can be determined by two directrices ( guide lines ) and plane director ( plane of parallelism ). Determinant of a surface :", null, "Surfaces. Ruled surfaces with two Diretrices. CYLINDROID –– both directrices are curves. CONOID –– one directrix is a curve while the other is a straight line. WARPED PLANE –– both directrices are straight lines.", null, "Surfaces. Ruled surfaces with two Diretrices. σ 2 – plane director. l 1 l ‘ 1 m 1 M 1 l 2 M 2 m 2 l ‘ 2 σ 2 n 2 n 1 CYLINDROID The surface called a cylindroid is generated by a moving straight line ( generatrix ) which in all its positions remains parallel to a given plane (called «the plane director “ plane of parallelism ” ) and intersects two curved guide lines (two directrices). If the directrices are plane curves, then, of course, they must lie in different planes.", null, "Surfaces. Ruled surfaces with two Diretrices. σ 2 — plane director. х m 2 n 1 m 1 σ 2 n 2 M 2 M 1 CONOID The surface called a conoid is generated by a moving straight line ( generatrix ) which all the time remains parallel to a given plane (called the plane director or “ plane of parallelism ” ) and intersects two directrices one of which is a curve , the other being a straight line. If the curve is a plane one, then it must not lie in the same plane with the second directrix which is a straight line.", null, "Surfaces. Ruled surfaces with two Diretrices. σ 1 — plane director. σ х σ m 2 M 1 n 2 n 1 m 1 M 2 The surface of a warped plane ( hyperbolic paraboloid ) is determined by a plane director and two noncoplanar (skew) straight-line directrices. A straight-line generatrix moving along the directrices (and remaining parallel to the plane director) describes the surface of a hyperbolic paraboloid. It can also be obtained by planar parallel motion of one parabola as generatrix along the other parabola as directrix. WARPED PLANE (hyperbolic paraboloid)", null, "Surfaces. Ruled surfaces with two Diretrices. WARPED PLANE –– both directrices are straight lines. x m 2 n 1 m 1 m, n – directrices , П 1 – plane director. l 1 2 l 1 1 M 2 M", null, "Surfaces. Ruled surfaces with two Diretrices. Right Helicoid — ruled surface with a plane director perpendicular to one directrix — straight line (axis of helicoid) while the other directrix is a helix. Straight line directrix and axis of helix base cylinder are collinear (coinciding) lines. Thus Right helicoid surface is conoid. Algorithm: how to construct Right Helicoid. Source data : – directrix – straight line plane director ( P 1 ), m – other directrix – helix , ℓ — generatrix. • Construct two views of helix – sinusoid and circle. • Construct a few intermediate locations of generatrix ℓ which intersect both directrices and remain parallel to the plane director P 1. • Multiple locations of generatrix form the helicoid surface . Point A belongs to the helicoidal surface. Missing projection can be found by constructing proper projections of generatrix. i 1 i 2 m 2 A 2 l 2 m 1 A 1 l", null, "Surfaces. Ruled surfaces with two Diretrices. Skew Helicoid – ruled surface where generatrix ℓ moves along two directrices (one is m – helix , the other i – its axis ) and remains parallel to the base cone of revolution ( constant angle between axis and generatrix of a cone ). Curve of intersection between skew helicoid surface and a plane, perpendicular to its axis is called Archimedean spiral. Point A belongs to the skew helicoidal surface. Missing projection can be found by constructing proper projections of generatrix. i 1 i 2 m 2 A 2 l 2 m 1 A 1 l", null, "Surfaces. Ruled surfaces with two Diretrices. Right Helicoid Skew Helicoid", null, "Surfaces. Ruled surfaces with three Diretrices. 1 – general view (3 directrices – curved lines ) 2 – double-skew cylindroid (2 directrices – curved lines ) 3 – double-skew conoid (2 directrices – straight lines ) 4 – one-sheet hyperboloid (3 directrices – straight lines ) Generally Ruled Surface unambiguously represented by moving generatrix along three directrices.", null, "Surfaces. Ruled surfaces with three Diretrices. One-sheet Hyperboloid — can be formed by moving straight line generatrix along three skew lines (straight line directrices) which are not parallel to any plane. Generatrices can be directrices and vice versa. Outline — hyperbola. HYPERBOL", null, "Surfaces. Ruled surfaces with three Diretrices. Skew Wedge surface ( type of double-skew cylindroid surface ) — two directrices are smooth curves and one is a straight line. All directrices are lying in parallel planes. On construction Chords are divided proportionally. А 1 1 1 D 1 В 1 С 1 1 1 А 2 1 2 D 2 В 2 С 2 1 2 n 1 2 n 2 2 n 3 1 n 2 1 n", null, "Channel Surfaces.", null, "Channel Surfaces. O n L n. L 1 m", null, "Channel Surfaces. A 1 A m 1 O 1 lm 2 l 1 O 2 A 2 O m", null, "Channel Surfaces. 0 r=f()h 1=r h=F()", null, "Channel Surfaces. Screw Torus", null, "Surfaces. Positional problems. Intersection between a Line and a Surface. 1 1 1 21 2 2 2 T 11 1 =2 12 2 1 2 S 1 S 2 a 1 = B 1 K 1 B 2 T 2 S 1 K 2 a 2 S 2 3 Intersection between 5 — faced prismatic horizontally perpendicular surface and a straight line. Segment of a frontally perpendicular cylindrical surface intersects with a straight line Horizontally projecting straight line intersects: (а) — pyramidal surface ( b ) – conical surface", null, "Surfaces. Positional problems. Intersection between a Line and a Surface. M 1 M 2 A 1 b 2 A 2 R 1 K 1 B 1 T 11 1 4 23 2 2 1 4 1 N 1 3 1 N 2 K 21 2 2 2 T 2 R 2 B 2 x 5 b 1 Intersection between АВ -line and cylindrical surface. K – piercing point.", null, "Intersection between АВ -line and surface of revolution. Surfaces. Positional problems. Intersection between a Line and a Surface. 6 a 1 1 1 a 2 K ‘ 1 1 ‘ 2 b 1 b 2 5 ‘ 1 2 1 3 1 4 1 5 1 4 ‘ 1 3 ‘ 1 2 ‘ 1 1 ‘ 1 K ‘ 2 K 1 2 ‘ 23 ‘ 24 ‘ 25 ‘", null, "Surfaces. Positional problems. Intersection between a Line and a Surface. 81 2 N 2 1 1 x M 1 N 1 O 1 M 2 2 12 2 x’ 2 ‘ 1 M ‘ 11 ‘ 1 r O 2 N ‘ 1 7 N 114 113 2 11 2 9 1 l 1 5 17 1 1 1 M 1 M 2 3 25 2 11 1 7 29 2 l 2 1 2 6 1 2 14 18 110 112 2 12 114 2 8 2 6 24 2 2 2 1 3 1 N 2 10 2 O ‘ 1 Projecting surfaces Intersection between a sphere and oblique line", null, "Surfaces. Positional problems. Intersection between a Line and a Surface. 10 a 2 k 2 a 1 k 1 h 2 h 1 b 1 b 2 m 2 n 2 m 1 n 1 h", null, "Surfaces. Positional problems. Intersection between a Line and a Surface. n 1 12 m 1 1 1 a 1 2 1 k 1 b 1 h 1 S 1 h 2 b 2 k 2 a 2 n 2 m 2 S", null, "Surfaces. Positional problems. Intersection between surfaces. Positional problems determine relative position and mutual belongings of objects and can be solved on their orthographical views. Basic positional problem: construct curve of intersection between two surfaces. If 1– 2 surfaces are perpendicular to principal planes (projecting surfaces) , then positional problems can be solved without additional cutting surfaces.", null, "Intersection between Surfaces A 2 A 1 B 2 C 1 B 1 A 3 =C 3 B 3 S 2 S 3 S 11 2 3 2 (5) 2 (7) 2(8) 2 1 3 3 35 3 8 3 (7) 3 (6) 3 4 3 ( 2 ) 3 4 1 1 1 3 1 2 16 18 1 5 17 1 2 24 2 (6) 2 Surface perpendicular to a principal plane", null, "Intersection between Surfaces. Y 341 2 ( 2 ‘ ) 2 (3 ‘ ) 24 2 (5 ‘ ) 2 (4 ‘ ) 2 3 22 2(1 ‘ ) 2 5 2 1 1 4 1 (2) 1 (3) 1(3 ‘ ) 14 ‘ 1 5 1 (2 ‘ ) 15 ‘ 1 1 ‘ 1 Y 1 ‘Y 6 6 2 6 16 ‘ 1 ( 6 ‘ ) 2 Y 6 Y 1 ‘Y 34 Both Surfaces are perpendicular to principal planes Y 1 1 3 3 3 2 34 3 5 3 6 3 1 ‘ 33 ‘ 32 ‘ 36 ‘ 35 ‘ 3 4 ‘ 3 Circle Ellipse. Straight Line", null, "МТК Intersection between Surfaces. Polyhedrons. Both Surfaces are perpendicular to principal planes", null, "МТК Intersection between Surfaces. f 2 h 16 2 B 1 2 2 3 2 4 25 27 2 f 1 h", null, "Intersection between Surfaces. B 2 A 2 C 1 B 1 A 1 1 1 2 1 3 1 4 15 16 1 S 1 4 23 2(6) 21 2 ( 5 ) 22 2 S ‘ 2 П ‘ 2 6 ‘ 2 4 ‘ 2 5 ‘ 2 3 ‘ 2 1 ‘ 2 2 ‘ 2 K ‘ 2 L ‘ 2 M ‘ 2 N ‘ 2 O ‘ 2 P ‘ 2 B ‘ 2 A ‘ 2 =C ‘ 2 K 1 L 1 N 1 M 1 P 1 O 1 L 2 N 2 P 2 O 2 M 2 K 2 h 1 f 2 f", null, "Intersection between Surfaces. Cutting Surface Method. To construct curve of intersection between surfaces we can use Cutting Surface method. As cutting surfaces we can use Cutting Planes or Cutting Spheres. Projection of a Curve of intersection between surfaces lies in an area of intersection between projections of outlines of surfaces. To construct curve of intersection it’s desirable to indicate characteristic points : — on the outline ( upper , lower , left , right ). Ф 2 n. Ф 1 m MN", null, "Intersection between Surfaces. Cutting Plane Method. S 2 ( 2 ‘ ) 2 (3 ‘ ) 24 2 (5 ‘ ) 2 (4 ‘ ) 2 3 22 25 2 (4 ‘ ) 3 (4) 3 6 36 ‘ 3 (3) 3 2 32 ‘ 3 1 11 ‘ 1 2 1 3 13 ‘ 12 ‘ 1 4 14 ‘ 1 6 16 ‘ 1 S 3 S 1 Algorithm Determine area where curve of intersection can exist. 1. Determine which surfaces of a model are perpendicular to principal planes. 2. Define base points for curve of intersection. 3. Assign type of cutting surfaces. 4. Define maximum number of cutting surfaces and their range. 5. Construct points of a curve of intersection. 6. Define base points where visibility of a curve changes. 7. Join all obtained points by a curve line in a proper sequence. (3 ‘ ) 3 1 3 hyperbolaellipse circle 6 2 (6 ‘ ) 2 5 35 ‘ 3 1 2 (1 ‘ ) 2 5 ‘ 1 5 1 Y 3 Y", null, "Intersection between Surfaces.", null, "Intersection between Surfaces. Cutting Plane Method. R C 1 R C 2 R К 1 R К 2 R", null, "Intersection between Surfaces. Τ 1 Γ 2 2 11 1 4 23 2 Σ", null, "Intersection between Surfaces. Cutting Plane Method. а 16 1 8 1 4 14 2 9 2 3 1 Σ 1 Δ 1 = f 17 25 2 2 2 b 2 h 2 1 2 f 2 5 16 1 A 1 7 1 h ‘ 2 h » 2 а 2 B 1 D 1 S 1 2 1 h ‘ 2 b 13 28 1 A 2 B 2 C 2 1 1 Common symmetry plane S", null, "Intersection between Surfaces. Cutting Sphere Method. Coaxial surfaces of revolution (i. e. surfaces with a common axis) intersect along circles.", null, "Intersection between Surfaces. Cutting Sphere Method. Cutting surfaces — spheres Theorem: Two coaxial surfaces of revolution intersect at circles which lie in planes, perpendicular to the axis of revolution, and centers of the circles lie on that axis. If center of a sphere lies on the axes of revolution, then sphere intersects surface of revolution at circles. Application of the method : 1. Both surfaces are surfaces of revolution ; 2. Axes of these surfaces are intersecting lines ; 3. A plane, formed by intersecting axes, should be parallel to a principal plane. а. c. b. d. A B A =", null, "Intersection between Surfaces. Cutting Sphere Method. Algorithm how to construct curves of intersection between two surfaces of revolution (cones). 1. Determine point of intersection between axes of revolution as a center of cutting spheres (О 2 point ). 2. Find projections of base points for curves of intersection. 3. Define band of radii R for cutting spheres: R MAX and R MIN. R MAX – distance from the center of a sphere to the outermost point , which belongs to the curve of intersection. R MIN – radius of a sphere, which is tangent to the one surface of revolution and intersects the other one. 4. Construct sphere of a radius R: R MIN < R MAX and with a center О 2. The sphere intersect each surface at circles. These circles project on П 2 plane as straight lines. Point of intersection between these lines belongs to the curve of intersection between two surfaces. 5. Construct multiple of cutting spheres for different R. Join gained points and obtain curve of intersection between two surfaces. Cutting surfaces – concentric spheres O 2 R min R max", null, "Intersection between Surfaces. Cutting Sphere Method. Cutting surfaces – eccentric spheres Algorithm how to construct curves of intersection between two surfaces of revolution (a cone and a torus). Application of the method: both surfaces of revolution have a common plane of symmetry. 1. Determine base points 1 2 , 2 2. 2. Construct plane ∑ 2 ( sigma ), which includes axes of a torus t 2. It intersects the torus at a circle of R-radius (radius of tube). 3. On the axis of a cone find center of a sphere, which intersects torus at that circle — point О 2. 4. Draw a sphere of R’-radius and О 2 – center. 5. Find a circle of intersection between a sphere and a cone. (on the plane П 2 projects to a line ). 6. Find points of intersection between two circles – points 3(4). 7. Through the axis t 2 draw new cutting plane ∑ ‘ 2 and repeat constructions. 8. Join points 1, 3, 5, 2 by a smooth curve. R’’O ‘ 2 O 2 R ’ t 2 3 21 2 5 2 2 2Σ 2 Σ ‘ 2=(4) 2 =(6) 2 R", null, "Intersection between Surfaces. Cutting Sphere Method. Curve of intersection. Concentric spheres Eccentric spheres", null, "Intersection between Surfaces. Particular case. Monge’s theorem. In general two second-order surfaces of revolution intersect at a four-degree curve. But in some cases the curve of intersection decomposes into two planar second-degree curves. It happens when both intersecting surfaces of revolution (a cylinder and a cone, two cones, an ellipsoid and a cone, etc. ) are circumscribed about a common sphere.", null, "Intersection between Surfaces. Cutting sphere method. Particular case. Monge’s theorem. Two second-order surfaces circumscribed about a third second-order surface intersect each other along two second-order curves.", null, "Intersection between Surfaces. Cutting sphere method. Particular case. Monge’s theorem. Two second-order surfaces circumscribed about a third second-order surface intersect each other along two second-order curves.", null, "Intersection between Surfaces. Cutting sphere method. Particular case. Monge’s theorem. Curve of intersection — elliptical arcs", null, "Developments. Properties of developments. 1. Each point on the development corresponds to a single point of a surface. 2. Straight lines of a surface remain straight on a development. 3. Straight line segments preserve their lengths. 4. An angle formed by lines on a surface remains equal to an angle between the corresponding lines on the development. 5. The area of a closed domain on a surface retains its magnitude within the corresponding closed domain on the development. 6. Shortest distance (beeline) on a surface develops into a straight line on a development (developed distance). 7. Parallel lines develop to parallel lines. Some curved surfaces can be developed so that they coincide completely (with all their points) with a plane without stretching or shrinking. Obtained planar figure is called development of a surface while surfaces are called developable. They comprise polyhedrons and ruled surfaces where adjacent rectilinear generatrices are parallel or intersect, or are tangent to sphere curve. All double-curved surfaces and the ruled surfaces which cannot be developed into a plane are called nondevelopable (or warped) surfaces.", null, "Methods of exact development. 1. Triangulation method. 2. Radial-line method. 3. Stretch-out-line (right section) method. Developments. Types of developments: 1. Exact. 2. Approximate. 3. Conventional. Exact developments. Developments of polyhedrons, right circular cylinder and cone can be constructed theoretically exactly. Cylinder develops to a rectangle, cone – to a circular sector. D D HH π D L L", null, "Developments. Triangulation method. Development of a pyramid. Lateral faces of a pyramid are triangles. To find true sizes of triangles you can use transformation of a drawing or find true lengths of corresponding edges. Because base of a pyramid is parallel to П 1 -plane, it’s sufficient to find true lengths of lateral edges AS , BS , CS. After that construct, for example, В CS-face, and supplement construction adding ASB and А SC faces. S 2 C 2 B 2 A 2 B ‘ 2 C ‘ 2 A ‘ 2 S 1 C 1 B 1 A 1 A ‘ 1 B ‘ 1 C ‘ 1 S 0 C 0 B 0 A 0 B 0 R SA R SB R S С R AB R С B R BCR BA R", null, "Developments. Radial Line method. C 0 A 1 A 2 C 0 A 0 B 0 D 1 C 1 F 1 B 1 C 2 B 2 E 1 F 0 F 2 D 2 E 0 D 0 This method is used when a base of a figure is parallel to the one principal plane of projection and generatrices are parallel to the other principal plane. Find true size of А 2 D 2 F 0 C 0 –face revolving it around frontal А 2 D 2. Point F 2 moves perpendicularly to А 2 D 2 to location F 0 , which can be found by protracting А 1 В 1 – line segment (the true size of АВ ) from the point А 2. From the point B 2 draw perpendicular to А 2 D 2 and find B 0 , protracting it from the point C 0 by В 1 С 1 -line segment, etc. Thus we obtained row of points А 0 А 2 , В 0 , С 0 . . . , which define a margin for developed bottom base of a figure.", null, "Developments. Stretch out Line method. This method is used when generatrices are parallel to a principal plane of projection. Cut ABCDEF-prism by the plane , perpendicular to the lateral edges of prism. Construct section of the prism by the plane in question – 123. Define true lengths of sides of 123 -triangle. On a free space of a drawing draw horizontal straight line a. From the arbitrary point 1 0 on this straight line protract line segments [1 0 2 0 ], [2 0 3 0 ], [3 0 1 0 ], equal to the sides of 123 -triangle. Through the points 1 0 , 2 0 , 3 0 , 1 0 draw straight lines , perpendicular to the a -line , and protract from the points 1 0 , 2 0 , 3 0 , 1 0 line segments , equal to lengths of corresponding lateral edges ([1 2 A 2 ], [1 2 D 2 ], [2 2 B 2 ], [2 2 E 2 ], [3 2 С 2 ], [3 2 F 2 ]), Then join obtained points A 0 B 0 C 0 A 0 and D 0 E 0 F 0 D 0 line segments. Obtained planar figure A 0 B 0 C 0 A 0 D 0 E 0 F 0 D 0 is a development of a lateral surface of the prism. To construct full development of a prism supplement obtained development by true figures of top and bottom bases. Construct development of ABCDEF — prism. A 1 C 1 D 1 F 1 A 2 B 21 2 2 2 A ‘ 2 C ‘ 2 B ‘ 2 A 0 B 0 C 0 A 01 1 D 0 F 0 E 0 D 0 2 01 0 3 0 1 0 D ‘ 2 E ‘ 2 2 23 21 2 E 2 F 2 D 2 D 0 A 0 B 1 E 1 C 2 3 2Ƴ 2 1 3 1 F ‘", null, "Development of a conical surface. Triangulation method. 1. Substitute (approximate) the conical surface by polyhedral pyramidal surface. 2. Define true lengths of lateral edges (method of revolution is used). 3. Construct true sizes of faces. 4. Obtained points of cone base join by a smooth curve. Size and number of straight line segments which approximate a curve directrix of a cone depend on curvature of a curve and size of a cone. Approximate developments. 1 1 S 2 S 1 2 1 3 1 5 16 17 1 8 11 ‘ 2 1 ‘ 1 6 ‘ 1 5 ‘ 1 7 ‘ 1 4 ‘ 1 8 ‘ 1 3 ‘ 1 2 ‘ 16 ‘ 2 5 ‘ 2 7 ‘ 2 4 ‘ 2 8 ‘ 2 3 ‘ 2 2 ‘ 23 ‘ 2 4 ‘ 2 5 ‘ 26 ‘ 27 ‘ 28 ‘ 21 ‘ 2 r =\| 1 1 2 1 \| R =\|S 2 2 2 \|", null, "Conventional developments. Development of Toroidal surface— Zone Method Each parallel of a surface is spaced an equal distance, m , apart along the surface. Cones are passed through the surface so that they pass through two parallels at the outer surface of the toroidal. The largest cone with element R 1 is found by extending it through where the equator and the next parallel intersect on the surface in the front view until R 1 intersects the extended centerline of the toroidal surface. Elements R 2 and R 3 are found by repeating this process. The development is begun by laying out the largest zone, using R 1 as the radius, on the arc that represents the base of an imaginary cone. The breadth of the zone is found by laying off distance m from the front view to the development and drawing the upper portion of the zone with a radius equal to R 1 -m using the same center. The next zone is drawn using radius R 2 with its center located on a line through the center of arc R 1. The last zone will appear as a circle with R 3 as its radius. The lengths of the arcs can be established by dividing the top view with vertical cutting planes that radiate through the poles — meridians. Preliminary nondevelopable surface is approximated by segments of developable surfaces. Then construct development of every segment, and the whole totality of developments gives conventional development of nondevelopable surface. b c", null, "" ]	[ null, "https://present5.com/wp-content/plugins/kama-clic-counter/icons/ppt.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_0.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_1.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_2.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_3.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_4.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_5.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_6.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_7.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_8.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_9.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_10.jpg", null, "https://present5.com/docs//ed_-_3_images/ed_-_3_11.jpg", 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https://www.arxiv-vanity.com/papers/hep-ph/0611158/	[ "# Connection between the Sivers function and the anomalous magnetic moment\n\nZhun Lu Ivan Schmidt Departamento de Física, Universidad Técnica Federico Santa María, Casilla 110-V, Valparaíso, Chile\nand Center of Subatomic Physics, Valparaíso, Chile\n###### Abstract\n\nThe same light-front wave functions of the proton are involved in both the anomalous magnetic moment of the nucleon and the Sivers function. Using the diquark model, we derive a simple relation between the anomalous magnetic moment and the Sivers function, which should hold in general with good approximation. This relation can be used to provide constraints on the Sivers single spin asymmetries from the data on anomalous magnetic moments. Moreover, the relation can be viewed as a direct connection between the quark orbital angular momentum and the Sivers function.\n\n###### pacs:\n13.40.Em, 13.60.-r, 13.88.+e\npreprint: USM-TH-198\n\nThe quark orbital angular momentum seh74 (or quark transverse motion) plays an important role for understanding the spin and quark structure of the nucleon, since as shown by many studies emc88 ; jaf90 ; ji97a ; hag98 ; ma98 ; har99 , it is the missing block of the total nucleon spin. Also many interesting phenomena or observables require the presence of quark orbital motion, among which the Sivers function sivers has attracted a lot of interest, since it is an essential piece in our understanding of the single spin asymmetries (SSA) observed in semi-inclusive deeply inelastic scattering (SIDIS). These SSAs have been measured recently by both the HERMES Airapetian:2004tw ; hermes05 and COMPASS compass ; compass06 Collaborations. Denoted as , the Sivers function describes the unpolarized distribution of the quark inside a transversely polarized nucleon, which comes from a correlation of the nucleon transverse spin and the quark transverse momentum. Although this is a -odd type correlation, it has been found that final state interaction bhs (FSI) between the struck quark and the spectator system can produce the necessary phase for a non-zero Sivers function, and its QCD definition collins02 ; jy ; belitsky ; bmp03 has just been worked out. Besides the single spin asymmetry, another important feature of the Sivers function is that it encodes the parton’s orbital angular momentum () inside the nucleon. This comes from the fact that the Sivers function requires the nucleon helicity to be flipped from the initial to the final state, while the quark helicity remains unchanged. A convenient tool to study this kind of single spin asymmetry (or the corresponding Sivers function) is the light-front formalism bro98 , which can express the Sivers function as the overlap integration of light-front wavefunctions differing by bhs ; bg06 . The same kind of overlap integration bl80 ; bhms ; bdh of light-front wavefunctions (with in the initial and final states) also appears in the anomalous magnetic moment of the nucleon, which apparently encodes the quark orbital angular momentum bhms . Therefore, it is interesting to find relations between the Sivers function and the anomalous magnetic moment of the nucleon, which is the main goal of this work. With such a relation one can constrain the Sivers function and the related asymmetries from data on nucleon anomalous magnetic moments, and viceversa. Also, the relation can be viewed as a direct connection between the quark orbital angular momentum distribution and the Sivers function.\n\nThe proton state can be expanded in a series of Fock states with coefficients , which are the light-front wavefunctions of the proton:\n\n Ψp(P+,P−,0⊥)=∑nψn(xi,k⊥,λi)\|n,xiP+,k⊥,λi⟩. (1)\n\nHere is the light-front momentum fraction of the quark, denotes the helicity, and its transverse momentum. The wavefunctions are Lorentz-invariant functions of the kinematics of the constituents of nucleon, and , with and .\n\nAs pointed out before, the Sivers function requires that the nucleon wavefunctions in the initial and final state differ by one unit of orbital angular momentum, and final state interactions play a crucial role. It describes the interference of two amplitudes which have different initial proton spin but couple to the same final state: Im. This can be realized by a gluon exchange between the struck quark and the spectator system. There have been already attempts jmy02 ; bg06 , using the proton light-front wavefunctions, to find a formula to calculate the Sivers functions. In those papers the final state interaction phase needed for Sivers functions has been incorporated into the wavefunctions. Another possibility is to express the Sivers function as the product of wavefunctions and the final state interactions term:\n\n f⊥1T ∝ ∑nψ↑∗n(xi,k⊥,λi)G(xi,x′i,k⊥,i,k′⊥,i)ψ↓n(x′i,k′⊥,λ′i), (2)\n\nwhere is the final state interaction term, and the light-front wavefunctions in this equation are the usual ones which do not contain the final state interactions phase.\n\nIn the scalar diquark model, the calculation gets simplified and some instructive result can be derived. In this model the kinematics is determined by the momentum fraction and the transverse momentum of the struck quark , since those of the spectator diquark can be related to by , . The advantage of the diquark model is that it is simple, and it works quite well in describing other properties of the proton, such as the helicity distribution ma96 , the transversity ma00 and the form factors ma02 of the proton. The light-front wavefunctions of this two-body Fock state have the simple forms bl80 ; bhs :\n\n ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩ψ↑+12(x,k⊥)=(M+mx)φ(x,k⊥) ,ψ↑−12(x,k⊥)=−(+k1+ik2)xφ(x,k⊥) . (3)\n\nfor and\n\n ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩ψ↓+12(x,k⊥)=(+k1−ik2)xφ(x,k⊥) ,ψ↓−12(x,k⊥)=(M+mx)φ(x,k⊥) . (4)\n\nfor . Here\n\n φ(x,k⊥)=g/√1−xM2−(k⊥+m2)/x−(k⊥+M2d)/(1−x), (5)\n\nand , and are the masses of the proton, the quark and the diquark, respectively.\n\nAccording to Fig. 1, a formula to calculate the Sivers function can be given by the overlap integration of the proton light-front wavefunction:\n\n kL2Mf⊥q1T(x,k⊥) = i∑n∫d2k′⊥dx′16π3ψ↑⋆n(x,k⊥,λi)G(x,x′,k⊥,k′⊥)ψ↓n(x′,k′⊥,λ′i), (6)\n\nwhere , and is the kernel which contain the final state interaction. In the scalar diquark model, the kernel has a simple form bhs ; bh04 :\n\n G(x,x′,k⊥,k′⊥)=iCFαsδ(x−x′)2π((k⊥−k′⊥)2+λ2g), (7)\n\nwhich can be calculated in the eikonal approximation. Final state interactions arise from gluonic exchanges between the struck quark and the spectator system, which are necessary in order to insure gauge invariance of the -dependent distributions. In our case we use a one-gluon exchange (with momentum ) approximation, which is commonly used in most model calculations bhs ; jy ; gg02 ; bbh03 ; yuan ; bsy04 ; lm04a . Here is the mass of the gluon, which is needed in general to regularize the infrared divergence in the integration, while in our case it can be set to . The sign in Eq. (7) is according to the Trento conventions tc . With Eqs. (3), (4) and (6), one can obtain the result for Sivers function that has given in jy ; bbh . An expression similar to (6) has been given in Ref. bh04 , where the Sivers function (or the SSA) is expressed as a product of a generalized parton distribution (GPD) and a FSI term in impact parameter space, while our result is expressed in momentum space, which can be connected with the result in Ref. bh04 by Fourier transformation.", null, "Figure 1: Diagram to calculate the Sivers function. The arrows ↑/↓ show the polarizations of the nucleon.\n\nThe same set of wavefunctions (spin-flipped in the initial and final states) appearing in (2) also appears in the expression of the Pauli form factor or anomalous magnetic momentum of the proton. In light-front formalism the Pauli form factor is identified from the helicity-flip vector current matrix elements of the current bhms\n\n ⟨P+q,↑∣∣∣J+(0)2P+∣∣∣P,↓⟩=−qLF2(q2)2M. (8)\n\nFurthermore, from Fig. 2, one can express the Pauli form factor in terms of the light-front wavefunctions as\n\n −qLF2(q2)2M = ∑n∫d2k⊥dx16π3∑jejψ↑⋆n(xi,k′⊥,λi)ψ↓n(xi,k⊥,λi), (9)\n\nwhere\n\n k′⊥=k⊥+(1−x)q⊥, (10)\n\nhere is momentum of the virtual photon, the momentum transfer between the initial and final nucleon.\n\nThe anomalous magnetic moment is defined from the Pauli form factor in the limit of : . After some algebra calculation, the anomalous magnetic moment can be expressed in terms of a local matrix element at zero momentum transfer:\n\n κM = −∑jej∑n∫d2k⊥dx16π3∑i≠jψ↑⋆n(xi,k⊥,λi)xi(∂∂k1+i∂∂k2)ψ↓n(xi,k⊥,λi). (11)\n\nFrom (2) and (11) we see that the same set of light-front amplitudes, with the orbital angular momenta differing by between the initial and final state, appears in the calculation of the Sivers function and the anomalous magnetic moment. Recently a study burkardt06 has shown how the non-vanishing anomalous magnetic moment constrains the quark orbital angular momentum.\n\nIt is clear from the previous expressions that the relation we are seeking between the Sivers function and the anomalous magnetic moment will necessarily have to be approximate. This can be also seen from the fact that the Sivers function, although leading twist, has () scale evolution dependence, which is not the case for the anomalous magnetic moment.\n\nIn the scalar diquark model, the calculation of the anomalous magnetic moment at the quark level has been given in bhms as\n\n κM = ∑n=±1/2∫d2k⊥dx16π3ψ↑⋆n(x,k⊥)(1−x)(∂∂k1+i∂∂k2)ψ↓n(x,k⊥). (12)\n\nUsing the wavefunctions given in (3) and (4), one gets the result\n\n κ = Mg28π2∫10dx(1−x)2(m+xM)(1−x)m2+xm2d−x(1−x)M2 (13) = ∫10dx(1−x)AB,\n\nwhere we have defined:\n\n A = g28π2(1−x)(xM+m); (14) B = (1−x)m2+xM2d−x(1−x)M2. (15)", null, "Figure 2: Diagram to calculate the Pauli form factor (or anomalous magnetic moment). The arrows ↑/↓ show the polarizations of the nucleon.\n\nOne can calculate the lowest -moment of defined as ()\n\n f⊥,q1T(x)=∫d2k⊥f⊥,q1T(x,k2⊥). (16)\n\nAccording to Eq. (6), which gives , and the wavefunctions given in (3) and (4), we directly get\n\n f⊥,q1T(x)=−π2CFαsA12B. (17)\n\nDefining\n\n κq(x)=(1−x)AB (18)\n\nwhich satisfies the normalization condition\n\n ∫10dxκq(x)=κq, (19)\n\nwe arrive at a simply relation between and in the quark-diquark model:\n\n f⊥,q1T(x)=−π2CFαs12(1−x)κq(x). (20)\n\nThe term in this equation is consistent with the result in Ref. bg06 . From this we can get:\n\n ∫10dx(1−x)f⊥,q1T(x) = −∫10dxπ2CFαs12κq(x) (21) = −π2CFαs12κq.\n\nAn approximate relation can also be obtained from the above equation, as\n\n ⟨1−x⟩F⊥,q1T = −π2CFαs12κq, (22)\n\nwhere is the first -moment of defined as , and is the average value of in the proton. Therefore we obtain a interesting relation that provides a constraint on the Sivers function from anomalous magnetic moment data. It suggests that the moment of the Sivers function is proportional to the anomalous magnetic moment contributed by the same quark. Although this relation as been obtained in the diquark model, it should hold in general with good approximation. This can be seen from the general relations for the anomalous magnetic moment and for the Sivers function in terms of light-front wavefunctions, as given by Eqs. (6) and (11), respectively. Since both quantities calculated from the same wavefunctions, and these are the ones that vary more strongly in the range of values covered by the integrals, it is appropriate to apply the mean value theorem, and therefore get the same type of relation as the one we have found here. In this sense our approximate result can be considered more general than the diquark model, and thus be applied to each quark individually.\n\nA more fundamental object than the anomalous magnetic moment is the spin-flipped generalized parton distribution (GPD) dm ; ji97a ; ji97b ; dvcs . Actually the function is at the forward limit ( and ):\n\n κq(x)=Eq(x,0,0). (23)\n\nThus in the scalar diquark model, is proportional to . According to Ji’s sum rule ji97a ( is the total angular momentum carried by quark flavor ):\n\n ∫dxx(Hq(x,ξ,t)+Eq(x,ξ,t))=12Jq(t), (24)\n\nwhich also holds in the forward limit. We see that the Sivers function is related to the angular momentum of the parton inside the nucleon, and therefore it is in fact sensitive to the orbital angular momentum of the quark. There is then the possibility to get information of the quark orbital angular momentum from the Sivers functions.\n\nAlthough the relation given in (20) and (21) is a simple result based on the approximation of the scalar diquark model, we can still apply the relation to given some prediction on the Sivers asymmetry of the meson production in SIDIS processes, such as the ratio of the asymmetries between different final mesons, since in this case the model dependence is reduced. The Sivers asymmetry is proportional to\n\n ASivUT∝⟨∑ae2af⊥a1TDa1⟩⟨∑ae2afa1Da1⟩, (25)\n\nwhich can be extracted by the weighting function , here and denote respectively the azimuthal angles of the produced hadron and of the nucleon spin polarization, with respect to the lepton scattering plane, is the unpolarized fragmentation function. We will focus on the large regime that the valence quark contribution dominates, and the disfavored fragmentation function can be ignored.\n\nSince the Sivers function and the anomalous magnetic moment “share” the same set of the proton wavefunctions, as shown in Figs. 1 and 2, one can start from the data of the anomalous magnetic moment to provide constraints on the proton wavefunctions, and then on the Sivers function. Similar methods have been used in Ref. bg06 , where a small Sivers asymmetry on a deuteron target has been predicted, and Ref. burkardt02 , where the sign of the Sivers asymmetries for different hadron targets combining different fragmenting hadrons has been classified.\n\nAs figured out in Ref. bg06 , the quark contribution dominates over the gluon contribution in the case of Sivers functions Jacques , which is also the result of an argument based on large considerations pobylista . There are also phenomenological supports of this conclusion from the SIDIS experiment from COMPASS of CERN compass , as pointed out in Ref. bg06 , and the analysis on hadron production of given in Ref. ans06 . Therefore in this work we only consider the quark contribution to the Sivers functions and the corresponding asymmetry.\n\nOne can constrain the proton wavefunctions by normalizing each and quarks contributions to the anomalous moments , . Isospin symmetry implies\n\n κd/n = κu/p, (26) κu/n = κd/p. (27)\n\nIn the valence quark approximation we have:\n\n κp = (2)(2/3)κu/p+(−1/3)κd/p, (28) κn = (2)(−1/3)κu/p+(2/3)κd/p. (29)\n\nThus one has , . In the following we use and to represent and , respectively.\n\nThen we can write the ratio of the asymmetries between and at large :\n\n ASivUT(π+)ASivUT(π−)≈2e2uf⊥u1TDπ+/u1e2df⊥d1TDπ−/d1≈2e2uκue2dκd=−3.3. (30)\n\nAlso we have\n\n ASivUT(π0)ASivUT(π−) ≈ 2e2uf⊥u1TDπ0/u1+e2df⊥d1TDπ0/d1e2df⊥d1TDπ−/d1 (31) ≈ 2e2uκu+e2dκd2e2dκd=−1.15, ASivUT(K+)ASivUT(K0) ≈ 2e2uf⊥u1TDK+/u1e2df⊥d1TDK0/d1≈4e2uκue2dκd=−6.6. (32)\n\nFor the above result we used isospin symmetry for the quark fragmentation functions:\n\n Dπ+/u1 = Dπ−/d1=2Dπ0/u1=2Dπ0/d1, (33) DK+/u1 = 2DK0/d1. (34)\n\nThe results show that in the large region, the Sivers asymmetry of is 3 times larger than that of and with opposite sign, which is consistent with the recent HERMES result hermes05 where at a four times larger asymmetry of is measured; the asymmetries of and are similar in size; the asymmetry of is much larger than that of , nearly one order of magnitude; and since in valence approximation.\n\nIn summary, both the formula for calculating Sivers function and that for calculating the anomalous magnetic moment of the proton, can be expressed in terms of the same set of the light-front wavefunctions, with helicity flipped from initial state to final states. Using the overlap representations of both Sivers functions as well as the anomalous magnetic moment, we give a simple relation between these two observables, in the approximation of the scalar diquark model. This relation is applied to provide constraints on the Sivers single spin asymmetries in the valence regime from data on anomalous magnetic moments. Also, the relation can be viewed as a connection between the quark orbital angular momentum and the Sivers function.\n\nAcknowledgements. This work is supported by Fondecyt (Chile) under Project No. 3050047, and by the ”Center of Subatomic Physics” (Chile) ." ]	[ null, "https://media.arxiv-vanity.com/render-output/5663010/x1.png", null, "https://media.arxiv-vanity.com/render-output/5663010/x2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84839106,"math_prob":0.9899034,"size":17639,"snap":"2022-40-2023-06","text_gpt3_token_len":4606,"char_repetition_ratio":0.17975616,"word_repetition_ratio":0.06836461,"special_character_ratio":0.27456206,"punctuation_ratio":0.18126217,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9904864,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T04:49:05Z\",\"WARC-Record-ID\":\"<urn:uuid:d330bc88-5617-4d7e-a6cd-4bd8d8d7fc26>\",\"Content-Length\":\"540050\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:616d271b-2586-4c2b-9b67-ced8f3722fc0>\",\"WARC-Concurrent-To\":\"<urn:uuid:15251f25-f615-447f-b710-1734a4696a86>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-ph/0611158/\",\"WARC-Payload-Digest\":\"sha1:FNVN5GJNNKFG33PIZS5GKZRTPKDTYT3L\",\"WARC-Block-Digest\":\"sha1:PQDKK564JM5HDNQSYOU5JHCSL5KREJUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499801.40_warc_CC-MAIN-20230130034805-20230130064805-00071.warc.gz\"}"}
http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Ackermann.html	[ "Wilhelm Ackermann\n\nBorn: 29 March 1896 in Schönebeck (Kr. Altena), Germany Died: 24 December 1962 in Lüdenscheid, Germany", null, "Click the picture above\nto see a larger version\n\nWilhelm Ackermann was a mathematical logician who worked with David Hilbert in Göttingen but spent his career as a high school teacher.\n\nAckermann entered the University of Göttingen in 1914 to study mathematics, physics and philosophy. However soon after he began his studies Word War I started. Ackermann was drafted into the army in 1915 and continued to serve until 1919 when he was able to return to his studies in Göttingen. He received his doctoral degree in 1925 with a thesis Begründung des \"tertium non datur\" mittels der Hilbertschen Theorie der Widerspruchsfreiheit written under David Hilbert's supervision. It provided a proof of the consistency of arithmetic without induction. It was intended to be a consistency proof for elementary analysis although this proof contained significant errors. Richard Zach explains the background to the thesis in :-\n\nAckermann's 1924 dissertation is of particular interest since it is the first non-trivial example of what Hilbert considered to be a finitistic consistency proof. Von Neumann's paper of 1927, the only other major contribution to proof theory in the 1920s, does not entirely fit into the tradition of the Hilbert school, and we have no evidence of the extent of Hilbert's involvement in its writing. Later consistency proofs, in particular those by Gentzen and Kalmár, were written after Gödel's incomplete ness results were already well-known and their implications understood by proof theorists. Ackermann's work, on the other hand, arose entirely out of Hilbert's research project, and there is ample evidence that Hilbert was aware of the range and details of the proof.\nAfter submitting his dissertation, Ackermann went to Cambridge, England, where he spent the first half of 1925. He was awarded a Rockefeller Scholarship to support this trip and Hilbert wrote in support of his application showing his high opinion of Ackerman's work :-\nIn his thesis \"Foundation of the 'tertium non datur' using Hilbert's theory of consistency,\" Ackermann has shown in the most general case that the use of the words \"all\" and \"there is,\" of the \"tertium non datur,\" is free from contradiction. The proof uses exclusively primitive and finite inference methods. Everything is demonstrated, as it were, directly on the mathematical formalism. Ackermann has here surmounted considerable mathematical difficulties and solved a problem which is of first importance to the modern efforts directed at providing a new foundation for mathematics.\nAckermann was also the main contributor to the development of the logical system known as the epsilon calculus, originally due to Hilbert. This formalism formed the basis of Bourbaki's logic and set theory.\n\nFrom 1929 until 1948 he taught as a teacher at the Arnoldinum Gymnasium in Burgsteinfurt and in Luedenscheid. He was corresponding member of the Akademie der Wissenschaften in Göttingen, and was honorary professor at the Universität Münster.\n\nIn 1928, Ackermann observed that A(x, y, z), the z-fold iterated exponentiation of x with y, is an example of a recursive function which is not primitive recursive. A(x, y, z) was simplified to a function P(x, y) of two variables by Rozsa Peter whose initial condition was simplified by Raphael Robinson. It is the latter which occurs as Ackermann's function in today's textbooks. Also in 1928 the often reprinted book Grundzüge der Theoretischen Logik by Hilbert and Ackermann appeared.\n\nAmong Ackermann's later work are consistency proofs for set theory (1937), full arithmetic (1940) and type free logic (1952). Further there was a new axiomatization of set theory (1956), and a book Solvable cases of the decision problem (North Holland, 1954. Second edition, 1962).\n\nThe new axiomatization of set theory was presented by Ackermann in Zur Axiomatik der Mengenlehre (1956). Dana Scott writes :-\n\nA remarkably simple axiomatization of a system of set theory is presented which the reviewer feels deserves serious consideration. The system is formalized in an applied first-order calculus with identity using a binary predicate e (membership) and a singulary predicate M (being a set). It is quite essential for the consistency of the system that M is not definable in terms of E. The axiom of extensionality is assumed so that all the individuals can be considered as collections of individuals, but it is easily proved from the axioms that there are collections that are not sets and even contain non-sets. The necessity for the existence of such improper collections in the theory makes comparison with the standard systems of set theory somewhat difficult.\nRudolf Grewe, who wrote his doctoral dissertation on Ackermann's set theory in 1966, gives models for the theory in . Several other authors have studied this system and references are given in .\n\nIn 1957 Ackermann published Philosophische Bemerkungen zur mathematischen Logik und zur mathematischen Grundlagenforschung and its English translation Philosophical observations on mathematical logic and on investigations into the foundations of mathematics. This paper, written for non-experts in the subject, gives an excellent overview of how Ackermann viewed mathematical logic. John Van Heijenoort writes :-\n\nAn objection to mathematical logic is that it is not the same as the philosophical logic which forms the foundation of our thought and which alone is necessary for thinking. Ackermann remarks that the traditional modes of inference are included in mathematical logic, besides many others, like the statement logic or the logic of relations. One has the illusion of getting by with \"Aristotelian logic\" in mathematics just as long as mathematical reasonings are insufficiently analyzed.\n\nA further objection to mathematical logic is that it is incomplete, in the sense that, by Gödel's result of 1931 (the text says 1932), intuitively correct theorems cannot be proved within a given system. But intuitive thinking is not consistent; paradoxes arise on the basis of modes of inference which from the naive intuitive point of view have to be considered correct. If we restrict intuitive inference so that paradoxes are eliminated, we come to a formal system, and incompleteness is the price we have to pay for consistency.\n\nMathematical logic is of use in clarifying the concepts of \"necessity\" and \"possibility,\" the distinction between \"analytic\" and \"synthetic.\" Discussing the \"triviality\" of logic, the author presents the decision problem.\n\nMathematics is today viewed as an investigation cf structures; but the obviousness connected with the concept of natural number is independent of any axiomatically introduced structure. Ackermann presents intuitionism, which constructs a mathematics with a minimum of logic, and the Frege-Russell analysis of the number concept. He defends this analysis against certain objections (circularity, necessity of an axiom of infinity). He remarks that Brouwer reduced the number intuition to two concepts: unity and possibility of repeatedly distinguishing a unity from another. In these concepts \"there is nothing foreign to logic.\" He concludes that \"in the theory of natural numbers we have a domain which is capable of an intuitive foundation with a minimum of logic in the sense of [Brouwer] and also of attainment purely by logical definitions if we presuppose an extensive logic.\"\n...\nThe paper ends with remarks on geometry, as a science conveying knowledge of the external world. Beyond any axiomatic treatment of geometry there are intuitive geometrical impressions which force themselves upon us with compelling power. In order to obtain a systematic and complete whole we may add principles in which free constructions occur. But, although limits may be difficult to trace, there remains a nucleus of obligatory intuitive elements. Thus the author sketches a position distinct, on the one hand, from a comprehensive apriorism in the sense of Kant and, on the other hand, from a thoroughgoing conventionalism.\n\nArticle by: J J O'Connor, E F Robertson, and Walter Felscher, Tübingen\n\nClick on this link to see a list of the Glossary entries for this page\n\nList of References (6 books/articles)\n\nMathematicians born in the same country\n\nOther Web sites\n1. Encyclopedia of Philosophy (The epsilon-calulus)\n2. World History\n3. Stanford Encyclopedia of Philosophy (The epsilon-calulus)\n4. Mathematical Genealogy Project" ]	[ null, "http://www-groups.dcs.st-and.ac.uk/~history/Thumbnails/Ackermann.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9494775,"math_prob":0.6776578,"size":7924,"snap":"2019-43-2019-47","text_gpt3_token_len":1669,"char_repetition_ratio":0.1260101,"word_repetition_ratio":0.008143323,"special_character_ratio":0.19535588,"punctuation_ratio":0.09084249,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9572479,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T17:15:27Z\",\"WARC-Record-ID\":\"<urn:uuid:0fda6757-3951-403d-9f50-791cd38e0154>\",\"Content-Length\":\"19146\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:35966516-9f3c-4263-b003-aefe8f8562a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b724f75e-6a77-48c5-9ebb-ed76f6973ed1>\",\"WARC-IP-Address\":\"138.251.192.92\",\"WARC-Target-URI\":\"http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Ackermann.html\",\"WARC-Payload-Digest\":\"sha1:VAX744XSG26ASSRWPI45YBRSZVBSXMDA\",\"WARC-Block-Digest\":\"sha1:CBQTX3ARUZTO3NNRD7VJSE3VHL3FJFQJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986669057.0_warc_CC-MAIN-20191016163146-20191016190646-00339.warc.gz\"}"}
https://akjournals.com/search?access=all&page=15&pageSize=10&q=%22thermodynamic+functions%22&sort=relevance	[ "# Search Results\n\n## You are looking at 141 - 150 of 215 items for :\n\n• \"thermodynamic functions\"\n• Refine by Access: All Content\nClear All\n\n## Solvent extraction of thorium from nitric acid solutions using di- N -butyl sulfoxide (DBSO) in xylene\n\nJournal of Radioanalytical and Nuclear Chemistry\nAuthors:\n,\nSyed Hasany\n, and\nAkbar Ali\n\n## Abstract\n\nThe extraction of thorium(IV) from nitric acid solutions by di-n-butyl sulfoxide (DBSO) in xylene has been investigated as a function of acid, extractant and the metal concentration. The effect of contact time and diverse ions on the extraction has been examined. Phosphate, fluoride, oxalate and perchlorate reduce the extraction to some extent. The extraction of other metal ions, especially impurities associated with thorium in ores, has been measured under optimised conditions selected for thorium extraction. Na(I), K(I), Ca(II), Sr(II), Mn(II), Fe(II), Ni(II), Zn(II), Pb(II), Al(III), Ti(IV) and Hf(IV) are not extracted. Among the stripping solutions employed for back-extraction, deionized water is found to be the best and more than 99% thorium can be back-extracted in three stages. The extracted species is supposed to be Th(NO3)4·2DBSO. The extraction is found to be almost independent of the thorium concentration in the range between 4.3·10–4–4.3·10–2M and inversely dependent upon the temperature. The values of thermodynamic functions", null, "H,", null, "G and", null, "S for extraction equilibrium have been evaluated to be –19.6±2.9 kJ·mole–1, –18.1±2.0 kJ·mole–1 and –5.0±2.9 J·mole–1·K–1, respectively.\n\nRestricted access\n\n## Molar heat capacities and standard molar enthalpy of formation of 2-amino-5-methylpyridine\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nJ. Zhang\n,\nZ. Tan\n,\nQ. Meng\n,\nQ. Shi\n,\nB. Tong\n, and\nS. Wang\n\n## Abstract\n\nThe heat capacities (C p,m) of 2-amino-5-methylpyridine (AMP) were measured by a precision automated adiabatic calorimeter over the temperature range from 80 to 398 K. A solid-liquid phase transition was found in the range from 336 to 351 K with the peak heat capacity at 350.426 K. The melting temperature (T m), the molar enthalpy (Δfus H m 0), and the molar entropy (Δfus S m 0) of fusion were determined to be 350.431±0.018 K, 18.108 kJ mol−1 and 51.676 J K−1 mol−1, respectively. The mole fraction purity of the sample used was determined to be 0.99734 through the Van’t Hoff equation. The thermodynamic functions (H T-H 298.15 and S T-S 298.15) were calculated. The molar energy of combustion and the standard molar enthalpy of combustion were determined, ΔU c(C6H8N2,cr)= −3500.15±1.51 kJ mol−1 and Δc H m 0 (C6H8N2,cr)= −3502.64±1.51 kJ mol−1, by means of a precision oxygen-bomb combustion calorimeter at T=298.15 K. The standard molar enthalpy of formation of the crystalline compound was derived, Δr H m 0 (C6H8N2,cr)= −1.74±0.57 kJ mol−1.\n\nRestricted access\n\n## Thermodynamic investigation of several natural polyols\n\n### Part III. Heat capacities and thermodynamic properties of erythritol\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nB. Tong\n,\nZ. Tan\n,\nJ. Zhang\n, and\nS. Wang\n\n## Abstract\n\nThe low-temperature heat capacity C p,m of erythritol (C4H10O4, CAS 149-32-6) was precisely measured in the temperature range from 80 to 410 K by means of a small sample automated adiabatic calorimeter. A solid-liquid phase transition was found at T=390.254 K from the experimental C p-T curve. The molar enthalpy and entropy of this transition were determined to be 37.92±0.19 kJ mol−1 and 97.17±0.49 J K−1 mol−1, respectively. The thermodynamic functions [H T-H 298.15] and [S T-S 298.15], were derived from the heat capacity data in the temperature range of 80 to 410 K with an interval of 5 K. The standard molar enthalpy of combustion and the standard molar enthalpy of formation of the compound have been determined: Δc H m 0(C4H10O4, cr)= −2102.90±1.56 kJ mol−1 and Δf H m 0(C4H10O4, cr)= − 900.29±0.84 kJ mol−1, by means of a precision oxygen-bomb combustion calorimeter at T=298.15 K. DSC and TG measurements were performed to study the thermostability of the compound. The results were in agreement with those obtained from heat capacity measurements.\n\nRestricted access\n\n## Rapid synthesis, kinetics and thermodynamics of binary Mn0.5Ca0.5(H2PO4)2 · H2O\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nBanjong Boonchom\nand\nChanaiporn Danvirutai\n\n## Abstract\n\nThe binary manganese and calcium dihydrogen phosphate monohydrate Mn0.5Ca0.5(H2PO4)2 · H2O was synthesized by a rapid and simple co-precipitation method using phosphoric acid, manganese metal, and calcium carbonate at ambient temperature. Thermal transformation shows complex processes and the final decomposed product was the binary manganese calcium cyclotetraphosphate MnCaP4O12. The activation energies of some decomposed steps were calculated by Kissinger method. Activated complex theory has been applied to each step of the reactions and the thermodynamic functions are calculated. These values for transformation stages showed that they are non-spontaneous processes without the introduction of heat. The differences of physical and chemical properties of the synthesized compound and its decomposed product are compared with M(H2PO4)2 · H2O and M2P4O12 (M = Mn and Ca), which indicate the effects of the presence of Ca ions in substitution of Mn ions and confirm the formation of solid solution.\n\nRestricted access\n\n## Kinetic and thermodynamic studies of MgHPO4 · 3H2O by non-isothermal decomposition data\n\nJournal of Thermal Analysis and Calorimetry\nAuthor:\nBanjong Boonchom\n\n## Abstract\n\nThe thermal decomposition of magnesium hydrogen phosphate trihydrate MgHPO4 · 3H2O was investigated in air atmosphere using TG-DTG-DTA. MgHPO4 · 3H2O decomposes in a single step and its final decomposition product (Mg2P2O7) was obtained. The activation energies of the decomposition step of MgHPO4 · 3H2O were calculated through the isoconversional methods of the Ozawa, Kissinger–Akahira–Sunose (KAS) and Iterative equation, and the possible conversion function has been estimated through the Coats and Redfern integral equation. The activation energies calculated for the decomposition reaction by different techniques and methods were found to be consistent. The better kinetic model of the decomposition reaction for MgHPO4 · 3H2O is the F 1/3 model as a simple n-order reaction of “chemical process or mechanism no-invoking equation”. The thermodynamic functions (ΔH, ΔG and ΔS*) of the decomposition reaction are calculated by the activated complex theory and indicate that the process is non-spontaneous without connecting with the introduction of heat.\n\nRestricted access\n\n## Low-temperature heat capacities and thermodynamic properties of 2,2-dimethyl-1,3-propanediol\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nB. Tong\n,\nZ. Tan\n,\nX. Lv\n,\nL. Sun\n,\nF. Xu\n,\nQ. Shi\n, and\nY. Li\n\n## Abstract\n\nThe molar heat capacities C p,m of 2,2-dimethyl-1,3-propanediol were measured in the temperature range from 78 to 410 K by means of a small sample automated adiabatic calorimeter. A solid-solid and a solid-liquid phase transitions were found at T-314.304 and 402.402 K, respectively, from the experimental C p-T curve. The molar enthalpies and entropies of these transitions were determined to be 14.78 kJ mol−1, 47.01 J K−1 mol for the solid-solid transition and 7.518 kJ mol−1, 18.68 J K−1 mol−1 for the solid-liquid transition, respectively. The dependence of heat capacity on the temperature was fitted to the following polynomial equations with least square method. In the temperature range of 80 to 310 K, C p,m/(J K−1 mol−1)=117.72+58.8022x+3.0964x 2+6.87363x 3−13.922x 4+9.8889x 5+16.195x 6; x=[(T/K)−195]/115. In the temperature range of 325 to 395 K, C p,m/(J K−1 mol−1)=290.74+22.767x−0.6247x 2−0.8716x 3−4.0159x 4−0.2878x 5+1.7244x 6; x=[(T/K)−360]/35. The thermodynamic functions H TH 298.15 and S TS 298.15, were derived from the heat capacity data in the temperature range of 80 to 410 K with an interval of 5 K. The thermostability of the compound was further tested by DSC and TG measurements. The results were in agreement with those obtained by adiabatic calorimetry.\n\nRestricted access\n\n## Low-temperature heat capacities and standard molar enthalpy of formation of the complex\n\n### Zn(Val)SO4·H2O(s) (Val=L-α-valine)\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nY. Y. Di\n,\nZ. C. Tan\n,\nL. W. Li\n,\nS. L. Gao\n, and\nL. X. Sun\n\n## Abstract\n\nLow-temperature heat capacities of a solid complex Zn(Val)SO4·H2O(s) were measured by a precision automated adiabatic calorimeter over the temperature range between 78 and 373 K. The initial dehydration temperature of the coordination compound was determined to be, T D=327.05 K, by analysis of the heat-capacity curve. The experimental values of molar heat capacities were fitted to a polynomial equation of heat capacities (C p,m) with the reduced temperatures (x), [x=f (T)], by least square method. The polynomial fitted values of the molar heat capacities and fundamental thermodynamic functions of the complex relative to the standard reference temperature 298.15 K were given with the interval of 5 K.\n\nEnthalpies of dissolution of the [ZnSO4·7H2O(s)+Val(s)] (Δsol H m,l 0) and the Zn(Val)SO4·H2O(s) (Δsol H m,2 0) in 100.00 mL of 2 mol dm−3 HCl(aq) at T=298.15 K were determined to be, Δsol H m,l 0=(94.588±0.025) kJ mol−1 and Δsol H m,2 0=–(46.118±0.055) kJ mol−1, by means of a homemade isoperibol solution–reaction calorimeter. The standard molar enthalpy of formation of the compound was determined as: Δf H m 0 (Zn(Val)SO4·H2O(s), 298.15 K)=–(1850.97±1.92) kJ mol−1, from the enthalpies of dissolution and other auxiliary thermodynamic data through a Hess thermochemical cycle. Furthermore, the reliability of the Hess thermochemical cycle was verified by comparing UV/Vis spectra and the refractive indexes of solution A (from dissolution of the [ZnSO4·7H2O(s)+Val(s)] mixture in 2 mol dm−3 hydrochloric acid) and solution A’ (from dissolution of the complex Zn(Val)SO4·H2O(s) in 2 mol dm−3 hydrochloric acid).\n\nRestricted access\n\n## Low-temperature thermodynamics of Ln(Me2dtc)3(C12H8N2) (Me2dtc = dimethyldithiocarbamate, Ln = La, Pr, Nd, Sm)\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nJian Wu\n,\nSan-Ping Chen\n,\nYou-Ying Di\n, and\nSheng-Li Gao\n\n## Abstract\n\nThe heat capacities of Ln(Me2dtc)3(C12H8N2) (Ln = La, Pr, Nd, Sm, Me2dtc = dimethyldithiocarbamate) have been measured by the adiabatic method within the temperature range 78–404 K. The temperature dependencies of the heat capacities, C p,m [La(Me2dtc)3(C12H8N2)] = 542.097 + 229.576 X − 27.169 X 2 + 14.596 X 3 − 7.135 X 4 (J K−1 mol−1), C p,m [Pr(Me2dtc)3(C12H8N2)] = 500.252 + 314.114 X − 17.596 X 2 − 0.131 X 3 + 16.627 X 4 (J K−1 mol−1), C p,m [Nd(Me2dtc)3(C12H8N2)] = 543.586 + 213.876 X − 68.040 X 2 + 1.173 X 3 + 2.563 X 4 (J K−1 mol−1) and C p,m [Sm(Me2dtc)3(C12H8N2)] = 528.650 + 216.408 X − 16.492 X 2 + 12.076 X 3 + 4.912 X 4 (J K−1 mol−1), were derived by the least-squares method from the experimental data. The heat capacities of Ce(Me2dtc)3(C12H8N2) and Pm(Me2dtc)3(C12H8N2) at 298.15 K were evaluated to be 617.99 and 610.09 J K−1 mol−1, respectively. Furthermore, the thermodynamic functions (entropy, enthalpy and Gibbs free energy) have been calculated using the obtained experimental heat capacity data.\n\nRestricted access\n\n## Thermodynamic investigation of room temperature ionic liquid\n\n### The heat capacity and thermodynamic functions of BMIPF6\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nZ. Zhang\n,\nT. Cui\n,\nJ. Zhang\n,\nH. Xiong\n,\nG. Li\n,\nL. Sun\n,\nF. Xu\n,\nZ. Cao\n,\nF. Li\n, and\nJ. Zhao\n\n## Abstract\n\nThe molar heat capacities of the room temperature ionic liquid 1-butyl-3-methylimidazolium hexafluoroborate (BMIPF6) were measured by an adiabatic calorimeter in temperature range from 80 to 390 K. The dependence of the molar heat capacity on temperature is given as a function of the reduced temperature (X) by polynomial equations, C P,m (J K−1 mol−1) = 204.75 + 81.421X − 23.828 X 2 + 12.044X 3 + 2.5442X 4 [X = (T − 132.5)/52.5] for the solid phase (80–185 K), C P,m (J K−1 mol−1) = 368.99 + 2.4199X + 1.0027X 2 + 0.43395X 3 [X = (T − 230)/35] for the glass state (195 − 265 K), and C P,m (J K−1 mol−1) = 415.01 + 21.992X − 0.24656X 2 + 0.57770X 3 [X = (T − 337.5)/52.5] for the liquid phase (285–390 K), respectively. According to the polynomial equations and thermodynamic relationship, the values of thermodynamic function of the BMIPF6 relative to 298.15 K were calculated in temperature range from 80 to 390 K with an interval of 5 K. The glass transition of BMIPF6 was measured to be 190.41 K, the enthalpy and entropy of the glass transition were determined to be ΔH g = 2.853 kJ mol−1 and ΔS g = 14.98 J K−1 mol−1, respectively. The results showed that the milting point of the BMIPF6 is 281.83 K, the enthalpy and entropy of phase transition were calculated to be ΔH m = 20.67 kJ mol−1 and ΔS m = 73.34 J K−1 mol−1.\n\nRestricted access\n\n## Thermodynamic properties of rubidium niobium tungsten oxide\n\nJournal of Thermal Analysis and Calorimetry\nAuthors:\nAleksandr Knyazev\n,\nMirosław Mączka\n,\nNataliya Kuznetsova\n,\nJerzy Hanuza\n, and\nAleksey Markin\n\n## Abstract\n\nIn the present work temperature dependence of heat capacity of rubidium niobium tungsten oxide has been measured first in the range from 7 to 395 K and then between 390 and 650 K, respectively, by precision adiabatic vacuum and dynamic calorimetry. The experimental data were used to calculate standard thermodynamic functions, namely the heat capacity\n\\documentclass{aastex} \\usepackage{amsbsy} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{bm} \\usepackage{mathrsfs} \\usepackage{pifont} \\usepackage{stmaryrd} \\usepackage{textcomp} \\usepackage{upgreek} \\usepackage{portland,xspace} \\usepackage{amsmath,amsxtra} \\pagestyle{empty} \\DeclareMathSizes{10}{9}{7}{6} \\begin{document} $$C_{\\text{p}}^{\\text{o}} (T),$$ \\end{document}\nenthalpy\n\\documentclass{aastex} \\usepackage{amsbsy} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{bm} \\usepackage{mathrsfs} \\usepackage{pifont} \\usepackage{stmaryrd} \\usepackage{textcomp} \\usepackage{upgreek} \\usepackage{portland,xspace} \\usepackage{amsmath,amsxtra} \\pagestyle{empty} \\DeclareMathSizes{10}{9}{7}{6} \\begin{document} $$H^{\\text{o}} ({\\rm T}) - H^{\\text{o}} (0)$$ \\end{document}\n, entropy\n\\documentclass{aastex} \\usepackage{amsbsy} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{bm} \\usepackage{mathrsfs} \\usepackage{pifont} \\usepackage{stmaryrd} \\usepackage{textcomp} \\usepackage{upgreek} \\usepackage{portland,xspace} \\usepackage{amsmath,amsxtra} \\pagestyle{empty} \\DeclareMathSizes{10}{9}{7}{6} \\begin{document} $$S^{\\text{o}} (T) - S^{\\text{o}} \\left( 0 \\right)$$ \\end{document}\n, and Gibbs function\n\\documentclass{aastex} \\usepackage{amsbsy} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{bm} \\usepackage{mathrsfs} \\usepackage{pifont} \\usepackage{stmaryrd} \\usepackage{textcomp} \\usepackage{upgreek} \\usepackage{portland,xspace} \\usepackage{amsmath,amsxtra} \\pagestyle{empty} \\DeclareMathSizes{10}{9}{7}{6} \\begin{document} $$G^{{^{\\text{o}} }} ({\\rm T}) - H^{{^{\\text{o}} }} (0)$$ \\end{document}\n, for the range from T→0 to 650 K. The high-temperature X-ray diffraction and the differential scanning calorimetry were used for the determination of temperature and decomposition products of RbNbWO6.\nRestricted access" ]	[ null, "https://akjournals.com/view/journals/10967/198/2/xxlarge916.gif", null, "https://akjournals.com/view/journals/10967/198/2/xxlarge916.gif", null, "https://akjournals.com/view/journals/10967/198/2/xxlarge916.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.894281,"math_prob":0.92622924,"size":11108,"snap":"2023-40-2023-50","text_gpt3_token_len":3342,"char_repetition_ratio":0.14787464,"word_repetition_ratio":0.11014994,"special_character_ratio":0.30266476,"punctuation_ratio":0.11875267,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99035436,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T02:14:11Z\",\"WARC-Record-ID\":\"<urn:uuid:c726db6e-abf6-43b8-8320-ae4a03a75371>\",\"Content-Length\":\"371352\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99a1da54-3569-497a-b231-5bd74a2a623a>\",\"WARC-Concurrent-To\":\"<urn:uuid:d92f3a53-80a5-4b07-be1c-4fe1714dc130>\",\"WARC-IP-Address\":\"34.251.65.102\",\"WARC-Target-URI\":\"https://akjournals.com/search?access=all&page=15&pageSize=10&q=%22thermodynamic+functions%22&sort=relevance\",\"WARC-Payload-Digest\":\"sha1:IZZK3DCX5UWDBOORW6T2CWJDFEB3V5DP\",\"WARC-Block-Digest\":\"sha1:HFH3CPCR2WR37FO5LU3N6FU4KEET4KMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506676.95_warc_CC-MAIN-20230925015430-20230925045430-00409.warc.gz\"}"}
https://ininet.org/language-specification-version-0-notice.html?page=34	[ "Language Specification Version 0\u0004\u0005 Notice\n\n Page 34/85 Date 29.01.2017 Size 3.2 Mb. #10878\n\n7.8Shift operators\n\nThe << and >> operators are used to perform bit shifting operations.\n\nshift-expression:\n\nFor an operation of the form x << count or x >> count, binary operator overload resolution (§7.2.4) is applied to select a specific operator implementation. The operands are converted to the parameter types of the selected operator, and the type of the result is the return type of the operator.\n\nWhen declaring an overloaded shift operator, the type of the first operand must always be the class or struct containing the operator declaration, and the type of the second operand must always be int.\n\nThe predefined shift operators are listed below.\n\n• Shift left:\n\nint operator <<(int x, int count);\nuint operator <<(uint x, int count);\nlong operator <<(long x, int count);\nulong operator <<(ulong x, int count);\n\nThe << operator shifts x left by a number of bits computed as described below.\n\nThe high-order bits outside the range of the result type of x are discarded, the remaining bits are shifted left, and the low-order empty bit positions are set to zero.\n\n• Shift right:\n\nint operator >>(int x, int count);\nuint operator >>(uint x, int count);\nlong operator >>(long x, int count);\nulong operator >>(ulong x, int count);\n\nThe >> operator shifts x right by a number of bits computed as described below.\n\nWhen x is of type int or long, the low-order bits of x are discarded, the remaining bits are shifted right, and the high-order empty bit positions are set to zero if x is non-negative and set to one if x is negative.\n\nWhen x is of type uint or ulong, the low-order bits of x are discarded, the remaining bits are shifted right, and the high-order empty bit positions are set to zero.\n\nFor the predefined operators, the number of bits to shift is computed as follows:\n\n• When the type of x is int or uint, the shift count is given by the low-order five bits of count. In other words, the shift count is computed from count & 0x1F.\n\n• When the type of x is long or ulong, the shift count is given by the low-order six bits of count. In other words, the shift count is computed from count & 0x3F.\n\nIf the resulting shift count is zero, the shift operators simply return the value of x.\n\nShift operations never cause overflows and produce the same results in checked and unchecked contexts.\n\nWhen the left operand of the >> operator is of a signed integral type, the operator performs an arithmetic shift right wherein the value of the most significant bit (the sign bit) of the operand is propagated to the high-order empty bit positions. When the left operand of the >> operator is of an unsigned integral type, the operator performs a logical shift right wherein high-order empty bit positions are always set to zero. To perform the opposite operation of that inferred from the operand type, explicit casts can be used. For example, if x is a variable of type int, the operation unchecked((int)((uint)x >> y)) performs a logical shift right of x.\n\n7.9Relational and type-testing operators\n\nThe ==, !=, <, >, <=, >=, is and as operators are called the relational and type-testing operators.\n\nrelational-expression:\nshift-expression\nrelational-expression < shift-expression\nrelational-expression > shift-expression\nrelational-expression <= shift-expression\nrelational-expression >= shift-expression\nrelational-expression is type\nrelational-expression as type\n\nequality-expression:\nrelational-expression\nequality-expression == relational-expression\nequality-expression != relational-expression\n\nThe is operator is described in §7.9.10 and the as operator is described in §7.9.11.\n\nThe ==, !=, <, >, <= and >= operators are comparison operators. For an operation of the form x op y, where op is a comparison operator, overload resolution (§7.2.4) is applied to select a specific operator implementation. The operands are converted to the parameter types of the selected operator, and the type of the result is the return type of the operator.\n\nThe predefined comparison operators are described in the following sections. All predefined comparison operators return a result of type bool, as described in the following table.\n\n Operation Result x == y true if x is equal to y, false otherwise x != y true if x is not equal to y, false otherwise x < y true if x is less than y, false otherwise x > y true if x is greater than y, false otherwise x <= y true if x is less than or equal to y, false otherwise x >= y true if x is greater than or equal to y, false otherwise\n\n7.9.1Integer comparison operators\n\nThe predefined integer comparison operators are:\n\nbool operator ==(int x, int y);\n\nbool operator ==(uint x, uint y);\nbool operator ==(long x, long y);\nbool operator ==(ulong x, ulong y);\n\nbool operator !=(int x, int y);\n\nbool operator !=(uint x, uint y);\nbool operator !=(long x, long y);\nbool operator !=(ulong x, ulong y);\n\nbool operator <(int x, int y);\n\nbool operator <(uint x, uint y);\nbool operator <(long x, long y);\nbool operator <(ulong x, ulong y);\n\nbool operator >(int x, int y);\n\nbool operator >(uint x, uint y);\nbool operator >(long x, long y);\nbool operator >(ulong x, ulong y);\n\nbool operator <=(int x, int y);\n\nbool operator <=(uint x, uint y);\nbool operator <=(long x, long y);\nbool operator <=(ulong x, ulong y);\n\nbool operator >=(int x, int y);\n\nbool operator >=(uint x, uint y);\nbool operator >=(long x, long y);\nbool operator >=(ulong x, ulong y);\n\nEach of these operators compares the numeric values of the two integer operands and returns a bool value that indicates whether the particular relation is true or false.\n\n7.9.2Floating-point comparison operators\n\nThe predefined floating-point comparison operators are:\n\nbool operator ==(float x, float y);\n\nbool operator ==(double x, double y);\n\nbool operator !=(float x, float y);\n\nbool operator !=(double x, double y);\n\nbool operator <(float x, float y);\n\nbool operator <(double x, double y);\n\nbool operator >(float x, float y);\n\nbool operator >(double x, double y);\n\nbool operator <=(float x, float y);\n\nbool operator <=(double x, double y);\n\nbool operator >=(float x, float y);\n\nbool operator >=(double x, double y);\n\nThe operators compare the operands according to the rules of the IEEE 754 standard:\n\n• If either operand is NaN, the result is false for all operators except !=, for which the result is true. For any two operands, x != y always produces the same result as !(x == y). However, when one or both operands are NaN, the <, >, <=, and >= operators do not produce the same results as the logical negation of the opposite operator. For example, if either of x and y is NaN, then x < y is false, but !(x >= y) is true.\n\n• When neither operand is NaN, the operators compare the values of the two floating-point operands with respect to the ordering\n\n–∞ < –max < ... < –min < –0.0 == +0.0 < +min < ... < +max < +∞\n\nwhere min and max are the smallest and largest positive finite values that can be represented in the given floating-point format. Notable effects of this ordering are:\n\n• Negative and positive zeros are considered equal.\n\n• A negative infinity is considered less than all other values, but equal to another negative infinity.\n\n• A positive infinity is considered greater than all other values, but equal to another positive infinity.\n\n7.9.3Decimal comparison operators\n\nThe predefined decimal comparison operators are:\n\nbool operator ==(decimal x, decimal y);\n\nbool operator !=(decimal x, decimal y);\n\nbool operator <(decimal x, decimal y);\n\nbool operator >(decimal x, decimal y);\n\nbool operator <=(decimal x, decimal y);\n\nbool operator >=(decimal x, decimal y);\n\nEach of these operators compares the numeric values of the two decimal operands and returns a bool value that indicates whether the particular relation is true or false. Each decimal comparison is equivalent to using the corresponding relational or equality operator of type System.Decimal.\n\n7.9.4Boolean equality operators\n\nThe predefined boolean equality operators are:\n\nbool operator ==(bool x, bool y);\n\nbool operator !=(bool x, bool y);\n\nThe result of == is true if both x and y are true or if both x and y are false. Otherwise, the result is false.\n\nThe result of != is false if both x and y are true or if both x and y are false. Otherwise, the result is true. When the operands are of type bool, the != operator produces the same result as the ^ operator.\n\n7.9.5Enumeration comparison operators\n\nEvery enumeration type implicitly provides the following predefined comparison operators:\n\nbool operator ==(E x, E y);\n\nbool operator !=(E x, E y);\n\nbool operator <(E x, E y);\n\nbool operator >(E x, E y);\n\nbool operator <=(E x, E y);\n\nbool operator >=(E x, E y);\n\nThe result of evaluating x op y, where x and y are expressions of an enumeration type E with an underlying type U, and op is one of the comparison operators, is exactly the same as evaluating ((U)x) op ((U)y). In other words, the enumeration type comparison operators simply compare the underlying integral values of the two operands.\n\n7.9.6Reference type equality operators\n\nThe predefined reference type equality operators are:\n\nbool operator ==(object x, object y);\n\nbool operator !=(object x, object y);\n\nThe operators return the result of comparing the two references for equality or non-equality.\n\nSince the predefined reference type equality operators accept operands of type object, they apply to all types that do not declare applicable operator == and operator != members. Conversely, any applicable user-defined equality operators effectively hide the predefined reference type equality operators.\n\nThe predefined reference type equality operators require one of the following:\n\n• Both operands are reference-type values or the value null. Furthermore, a standard implicit conversion (§6.3.1) exists from the type of either operand to the type of the other operand.\n\n• One operand is a value of type T where T is a type-parameter and the other operand is the value null. Furthermore T does not have the value type constraint.\n\nUnless one of these conditions are true, a compile-time error occurs. Notable implications of these rules are:\n\n• It is a compile-time error to use the predefined reference type equality operators to compare two references that are known to be different at compile-time. For example, if the compile-time types of the operands are two class types A and B, and if neither A nor B derives from the other, then it would be impossible for the two operands to reference the same object. Thus, the operation is considered a compile-time error.\n\n• The predefined reference type equality operators do not permit value type operands to be compared. Therefore, unless a struct type declares its own equality operators, it is not possible to compare values of that struct type.\n\n• The predefined reference type equality operators never cause boxing operations to occur for their operands. It would be meaningless to perform such boxing operations, since references to the newly allocated boxed instances would necessarily differ from all other references.\n\n• If an operand of a type parameter type T is compared to null, and the runtime type of T is a value type, the result of the comparison is false.\n\nThe following example checks whether an argument of an unconstrained type parameter type is null.\n\nclass C\n\n{\nvoid F(T x) {\nif (x == null) throw new ArgumentNullException();\n...\n}\n}\n\nThe x == null construct is permitted even though T could represent a value type, and the result is simply defined to be false when T is a value type.\n\nFor an operation of the form x == y or x != y, if any applicable operator == or operator != exists, the operator overload resolution (§7.2.4) rules will select that operator instead of the predefined reference type equality operator. However, it is always possible to select the predefined reference type equality operator by explicitly casting one or both of the operands to type object. The example\n\nusing System;\n\nclass Test\n{\nstatic void Main() {\nstring s = \"Test\";\nstring t = string.Copy(s);\nConsole.WriteLine(s == t);\nConsole.WriteLine((object)s == t);\nConsole.WriteLine(s == (object)t);\nConsole.WriteLine((object)s == (object)t);\n}\n}\n\nproduces the output\n\nTrue\nFalse\nFalse\nFalse\n\nThe s and t variables refer to two distinct string instances containing the same characters. The first comparison outputs True because the predefined string equality operator (§7.9.7) is selected when both operands are of type string. The remaining comparisons all output False because the predefined reference type equality operator is selected when one or both of the operands are of type object.\n\nNote that the above technique is not meaningful for value types. The example\n\nclass Test\n\n{\nstatic void Main() {\nint i = 123;\nint j = 123;\nSystem.Console.WriteLine((object)i == (object)j);\n}\n}\n\noutputs False because the casts create references to two separate instances of boxed int values.\n\n7.9.7String equality operators\n\nThe predefined string equality operators are:\n\nbool operator ==(string x, string y);\n\nbool operator !=(string x, string y);\n\nTwo string values are considered equal when one of the following is true:\n\n• Both values are null.\n\n• Both values are non-null references to string instances that have identical lengths and identical characters in each character position.\n\nThe string equality operators compare string values rather than string references. When two separate string instances contain the exact same sequence of characters, the values of the strings are equal, but the references are different. As described in §7.9.6, the reference type equality operators can be used to compare string references instead of string values.\n\n7.9.8Delegate equality operators\n\nEvery delegate type implicitly provides the following predefined comparison operators:\n\nbool operator ==(System.Delegate x, System.Delegate y);\n\nbool operator !=(System.Delegate x, System.Delegate y);\n\nTwo delegate instances are considered equal as follows:\n\n• If either of the delegate instances is null, they are equal if and only if both are null.\n\n• If the delegates have different runtime type they are never equal.\n\n• If both of the delegate instances have an invocation list (§15.1), those instances are equal if and only if their invocation lists are the same length, and each entry in one’s invocation list is equal (as defined below) to the corresponding entry, in order, in the other’s invocation list.\n\nThe following rules govern the equality of invocation list entries:\n\n• If two invocation list entries both refer to the same static method then the entries are equal.\n\n• If two invocation list entries both refer to the same non-static method on the same target object (as defined by the reference equality operators) then the entries are equal.\n\n• Invocation list entries produced from evaluation of semantically identical anonymous-function-expressions with the same (possibly empty) set of captured outer variable instances are permitted (but not required) to be equal.\n\n7.9.9Equality operators and null\n\nThe == and != operators permit one operand to be a value of a nullable type and the other to be the null literal, even if no predefined or user-defined operator (in unlifted or lifted form) exists for the operation.\n\nFor an operation of one of the forms\n\nx == null null == x x != null null != x\n\nwhere x is an expression of a nullable type, if operator overload resolution (§7.2.4) fails to find an applicable operator, the result is instead computed from the HasValue property of x. Specifically, the first two forms are translated into !x.HasValue, and last two forms are translated into x.HasValue.\n\n7.9.10The is operator\n\nThe is operator is used to dynamically check if the run-time type of an object is compatible with a given type. The result of the operation E is T, where E is an expression and T is a type, is a boolean value indicating whether E can successfully be converted to type T by a reference conversion, a boxing conversion, or an unboxing conversion. The operation is evaluated as follows, after type arguments have been substituted for all type parameters:\n\n• If E is an anonymous function, a compile time error occurs\n\n• If E is a method group or the null literal, of if the type of E is a reference type or a nullable type and the value of E is null, the result is false.\n\n• Otherwise, let D represent the dynamic type of E as follows:\n\n• If the type of E is a reference type, D is the run-time type of the instance reference by E.\n\n• If the type of E is a nullable type, D is the underlying type of that nullable type.\n\n• If the type of E is a non-nullable value type, D is the type of E.\n\n• The result of the operation depends on D and T as follows:\n\n• If T is a reference type, the result is true if D and T are the same type, if D is a reference type and an implicit reference conversion from D to T exists, or if D is a value type and a boxing conversion from D to T exists.\n\n• If T is a nullable type, the result is true if D is the underlying type of T.\n\n• If T is a non-nullable value type, the result is true if D and T are the same type.\n\n• Otherwise, the result is false.\n\nNote that user defined conversions, are not considered by the is operator.\n\n7.9.11The as operator\n\nThe as operator is used to explicitly convert a value to a given reference type or nullable type. Unlike a cast expression (§7.6.6), the as operator never throws an exception. Instead, if the indicated conversion is not possible, the resulting value is null.\n\nIn an operation of the form E as T, E must be an expression and T must be a reference type, a type parameter known to be a reference type, or a nullable type. Furthermore, at least one of the following must be true, or otherwise a compile-time error occurs:\n\n• An identity (§6.1.1), implicit reference (§6.1.6), boxing (§6.1.7), explicit reference (§6.2.4), or unboxing (§6.2.5) conversion exists from the type of E to T.\n\n• The type of E or T is an open type.\n\n• E is the null literal.\n\nThe operation E as T produces the same result as\n\nE is T ? (T)(E) : (T)null\n\n• except that E is only evaluated once. The compiler can be expected to optimize E as T to perform at most one dynamic type check as opposed to the two dynamic type checks implied by the expansion above.\n\nNote that some conversions, such as user defined conversions, are not possible with the as operator and should instead be performed using cast expressions.\n\nIn the example\n\nclass X\n{\n\npublic string F(object o) {\n\nreturn o as string; // OK, string is a reference type\n}\n\npublic T G(object o) where T: Attribute {\n\nreturn o as T; // Ok, T has a class constraint\n}\n\npublic U H(object o) {\n\nreturn o as U; // Error, U is unconstrained\n}\n}\n\nthe type parameter T of G is known to be a reference type, because it has the class constraint. The type parameter U of H is not however; hence the use of the as operator in H is disallowed." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83623147,"math_prob":0.99123263,"size":18808,"snap":"2022-05-2022-21","text_gpt3_token_len":4155,"char_repetition_ratio":0.2089449,"word_repetition_ratio":0.1518162,"special_character_ratio":0.23745215,"punctuation_ratio":0.13815261,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961444,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T08:20:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1828fddf-b582-44ef-8d2a-95a190079cb9>\",\"Content-Length\":\"42233\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f742fa6-1d98-46bc-8119-ed2fe800ad22>\",\"WARC-Concurrent-To\":\"<urn:uuid:1660040b-3e8e-4c86-b440-a67b4b9100db>\",\"WARC-IP-Address\":\"176.9.102.205\",\"WARC-Target-URI\":\"https://ininet.org/language-specification-version-0-notice.html?page=34\",\"WARC-Payload-Digest\":\"sha1:EUNAENKJAFC3CV3NQYJ6OGFBV24L5CYY\",\"WARC-Block-Digest\":\"sha1:VC4KDE7HOTYL5OXKNY4W6RQVEVXLEEQF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302740.94_warc_CC-MAIN-20220121071203-20220121101203-00272.warc.gz\"}"}
https://mathhelpboards.com/threads/automorphism-of-order-2-fixing-just-identity-prove-that-g-is-abelian.2833/	[ "# automorphism of order 2 fixing just identity. Prove that G is abelian.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nLet $G$ be a finite group, $T$ an automorphism of $G$ with the property that $T(x)=x$ if and only if $x=e$. Suppose further that $T^2=I$, that is, $T(T(x))=x$ for all $x\\in G$. Show that $G$ is abelian.\n\nI approached this problem using the permutation representation afforded by $T$ on $G$. Its easy to deduce that the cycle representation of the permutation of $G$ caused by $T$ has $(n-1)/2$ disjoint transpositions, where $n=\|G\|$. We know, from this, that $n$ is odd but so what? I am not able to exploit the homomorphism property of $T$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nhint: (i don't want to spoil your fun because this is a beautiful theorem)\n\ndefine $f:G \\to G$ by:\n\n$f(g) = g^{-1}T(g)$\n\nshow $f$ is injective (and thus bijective).\n\nnow...what is $T\\circ f(g)$?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8914682,"math_prob":0.9993056,"size":978,"snap":"2022-27-2022-33","text_gpt3_token_len":291,"char_repetition_ratio":0.12525667,"word_repetition_ratio":0.6347305,"special_character_ratio":0.297546,"punctuation_ratio":0.104477614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999887,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T01:04:05Z\",\"WARC-Record-ID\":\"<urn:uuid:ebace681-ebb4-4c38-ad2f-3df202ae19d6>\",\"Content-Length\":\"55760\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f3aaf1d-d2d3-4982-a72c-3cdf93a40ef6>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b9d026d-1dff-412b-b0ab-f683a74e42f4>\",\"WARC-IP-Address\":\"50.31.99.218\",\"WARC-Target-URI\":\"https://mathhelpboards.com/threads/automorphism-of-order-2-fixing-just-identity-prove-that-g-is-abelian.2833/\",\"WARC-Payload-Digest\":\"sha1:UONZ6CREEGMRQKTMLTGJEQPO26LOQV5N\",\"WARC-Block-Digest\":\"sha1:KUNLEUJQQ65T7WGGKWWP6KFFEAECQRRB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104683020.92_warc_CC-MAIN-20220707002618-20220707032618-00028.warc.gz\"}"}
https://www.farmrio.com/collections/midi-dresses-sale	[ "# Midi Dresses Sale", null, "Color\n• MULTI\nSize\n• XS\n• S\n• M\n3 left\n• L\n2 left\n• XL\n2 left\nColor\n• WHITE\nSize\n• XS\n• S\n• M\n• L\n1 left\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n2 left\n• L\n• XL\n2 left\nColor\n• WHITE\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n3 left\n• S\n• M\n2 left\n• L\n3 left\n• XL\nColor\n• Yellow MULTI\nSize\n• XS\n• S\n3 left\n• M\n1 left\n• L\n3 left\n• XL\n3 left\nColor\n• Red MULTI\nSize\n• XS\n• S\n• M\n2 left\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• OFF-WHITE\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• PINK MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• Red MULTI\nSize\n• XS\n• S\n• M\n• L\n1 left\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• BLUE MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n• L\n• XL\nColor\n• MULTI\nSize\n• XS\n• S\n• M\n1 left\n• L\n• XL\nColor\n• Red MULTI\nSize\n• XS\n• S\n• M\n1 left\n• L\n• XL" ]	[ null, "https://cdn.shopify.com/s/files/1/0077/6673/6963/collections/banner_subcategory_sale_final_1920x.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6074913,"math_prob":0.41638303,"size":238,"snap":"2020-45-2020-50","text_gpt3_token_len":76,"char_repetition_ratio":0.23504274,"word_repetition_ratio":0.074074075,"special_character_ratio":0.44117647,"punctuation_ratio":0.0862069,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99046123,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T22:45:20Z\",\"WARC-Record-ID\":\"<urn:uuid:66bcd2e2-adc3-4be2-bf9c-7e4b8bcab709>\",\"Content-Length\":\"523503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78958c1a-8531-4e62-9793-d291d6a08c06>\",\"WARC-Concurrent-To\":\"<urn:uuid:6faa9424-84fd-40d3-9b54-c9321870572b>\",\"WARC-IP-Address\":\"23.227.38.74\",\"WARC-Target-URI\":\"https://www.farmrio.com/collections/midi-dresses-sale\",\"WARC-Payload-Digest\":\"sha1:O2KOIDMCYUJEUJ6VKCTNRWHCJKQIJHEF\",\"WARC-Block-Digest\":\"sha1:ZO6J74ZG7N4UQITFOEKDZX7HHLJRNIEH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141732835.81_warc_CC-MAIN-20201203220448-20201204010448-00032.warc.gz\"}"}
https://nl.mathworks.com/help/images/ref/deconvblind.html	[ "# deconvblind\n\nDeblur image using blind deconvolution\n\n## Syntax\n\n``[J,psfr] = deconvblind(I,psfi)``\n``[J,psfr] = deconvblind(I,psfi,iter)``\n``[J,psfr] = deconvblind(I,psfi,iter,dampar)``\n``[J,psfr] = deconvblind(I,psfi,iter,dampar,weight)``\n``[J,psfr] = deconvblind(I,psfi,iter,dampar,weight,readout)``\n``[J,psfr] = deconvblind(___,fun)``\n\n## Description\n\n````[J,psfr] = deconvblind(I,psfi)` deconvolves image `I` using the maximum likelihood algorithm and an initial estimate of the point-spread function (PSF), `psfi`. The `deconvblind` function returns both the deblurred image `J` and a restored PSF, `psfr`.To improve the restoration, `deconvblind` supports several optional parameters, described below. Use `[]` as a placeholder if you do not specify an intermediate parameter.```\n````[J,psfr] = deconvblind(I,psfi,iter)` specifies the number of iterations, `iter`.```\n````[J,psfr] = deconvblind(I,psfi,iter,dampar)` controls noise amplification by suppressing iterations for pixels that deviate a small amount compared to the noise, specified by the damping threshold `dampar`. By default, no damping occurs.```\n\nexample\n\n````[J,psfr] = deconvblind(I,psfi,iter,dampar,weight)` specifies which pixels in the input image `I` are considered in the restoration. The value of an element in the `weight` array determines how much the pixel at the corresponding position in the input image is considered. For example, to exclude a pixel from consideration, assign it a value of `0` in the `weight` array. You can adjust the weight value assigned to each pixel according to the amount of flat-field correction.```\n````[J,psfr] = deconvblind(I,psfi,iter,dampar,weight,readout)` specifies the additive noise (such as background and foreground noise) and the variance of the read-out camera noise, `readout`.```\n````[J,psfr] = deconvblind(___,fun)`, where `fun` is a handle to a function that describes additional constraints on the PSF. `fun` is called at the end of each iteration. For more information about function handles, see Create Function Handle.```\n\n## Examples\n\ncollapse all\n\nCreate a sample image with noise.\n\n```% Set the random number generator back to its default settings for % consistency in results. rng default; I = checkerboard(8); PSF = fspecial('gaussian',7,10); V = .0001; BlurredNoisy = imnoise(imfilter(I,PSF),'gaussian',0,V);```\n\nCreate a weight array to specify which pixels are included in processing.\n\n```WT = zeros(size(I)); WT(5:end-4,5:end-4) = 1; INITPSF = ones(size(PSF));```\n\nPerform blind deconvolution.\n\n`[J P] = deconvblind(BlurredNoisy,INITPSF,20,10*sqrt(V),WT);`\n\nDisplay the results.\n\n```subplot(221);imshow(BlurredNoisy); title('A = Blurred and Noisy'); subplot(222);imshow(PSF,[]); title('True PSF'); subplot(223);imshow(J); title('Deblurred Image'); subplot(224);imshow(P,[]); title('Recovered PSF');```", null, "## Input Arguments\n\ncollapse all\n\nBlurry image, specified as a numeric array of any dimension. You can also specify the image as a cell array to enable interrupted iterations. For more information, see Tips.\n\nData Types: `single` \| `double` \| `int16` \| `uint8` \| `uint16`\n\nInitial estimate of PSF, specified as a numeric array. The PSF restoration is affected strongly by the size of the initial guess `psfi` and less by the values it contains. For this reason, specify an array of `1`s as your `psfi`.\n\nYou can also specify `psfi` as a cell array to enable interrupted iterations. For more information, see Tips.\n\nData Types: `single` \| `double` \| `int16` \| `uint8` \| `uint16`\n\nNumber of iterations, specified as a positive integer.\n\nData Types: `double`\n\nThreshold for damping, specified as a numeric scalar. Damping occurs for pixels whose deviation between iterations is less than the threshold. `dampar` has the same data type as `I`.\n\nWeight value of each pixel, specified as a numeric array with values in the range [0, 1]. `weight` has the same size as the input image, `I`. By default, all elements in `weight` have the value `1`, so all pixels are considered equally in the restoration.\n\nData Types: `double`\n\nNoise, specified as a numeric scalar or numeric array. The value of `readout` corresponds to the additive noise (such as noise from the foreground and background) and the variance of the read-out camera noise. `readout` has the same data type as `I`.\n\nFunction handle, specified as a handle. `fun` must accept the PSF as its first argument. The function must return one argument: a PSF that is the same size as the original PSF and that satisfies the positivity and normalization constraints.\n\n## Output Arguments\n\ncollapse all\n\nDeblurred image, returned as a numeric array or a 1-by-4 cell array. `J` (or `J{1}` when `J` is a cell array) has the same data type as `I`. For more information about returning `J` as a cell array for interrupted iterations, see Tips.\n\nRestored PSF, returned as an array of positive numbers or a 1-by-4 cell array. `psfr` has the same size as the initial estimate of the PSF, `psfi`, and it is normalized so the sum of elements is 1. For more information about returning `psfr` as a cell array for interrupted iterations, see Tips.\n\nData Types: `double`\n\n## Tips\n\n• You can use `deconvblind` to perform a deconvolution that starts where a previous deconvolution stopped. To use this feature, pass the input image `I` and the initial guess at the PSF, `psfi`, as cell arrays: `{I}` and `{psfi}`. When you do, the `deconvblind` function returns the output image `J` and the restored point-spread function, `psfr`, as cell arrays, which can then be passed as the input arrays into the next `deconvblind` call. The output cell array `J` contains four elements:\n\n`J{1}` contains `I`, the original image.\n\n`J{2}` contains the result of the last iteration.\n\n`J{3}` contains the result of the next-to-last iteration.\n\n`J{4}` is an array generated by the iterative algorithm.\n\n• The output image `J` could exhibit ringing introduced by the discrete Fourier transform used in the algorithm. To reduce the ringing, use `I = edgetaper(I,psfi)` before calling `deconvblind`.\n\n D.S.C. Biggs and M. Andrews, Acceleration of iterative image restoration algorithms, Applied Optics, Vol. 36, No. 8, 1997.\n\n R.J. Hanisch, R.L. White, and R.L. Gilliland, Deconvolutions of Hubble Space Telescope Images and Spectra, Deconvolution of Images and Spectra, Ed. P.A. Jansson, 2nd ed., Academic Press, CA, 1997.\n\n Timothy J. Holmes, et al, Light Microscopic Images Reconstructed by Maximum Likelihood Deconvolution, Handbook of Biological Confocal Microscopy, Ed. James B. Pawley, Plenum Press, New York, 1995." ]	[ null, "https://nl.mathworks.com/help/examples/images/win64/DeblurAnImageUsingBlindDeconvolutionExample_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69539076,"math_prob":0.96439165,"size":5775,"snap":"2023-40-2023-50","text_gpt3_token_len":1476,"char_repetition_ratio":0.12372206,"word_repetition_ratio":0.0654102,"special_character_ratio":0.23047619,"punctuation_ratio":0.16264522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98006237,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T17:30:18Z\",\"WARC-Record-ID\":\"<urn:uuid:f79a20ba-8f76-4e53-83e6-8ef37c023e38>\",\"Content-Length\":\"107372\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:141e75bf-840c-46bd-9abc-fba3131544b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:b791b871-5bac-4fa7-ab23-56ed1d118224>\",\"WARC-IP-Address\":\"23.34.160.82\",\"WARC-Target-URI\":\"https://nl.mathworks.com/help/images/ref/deconvblind.html\",\"WARC-Payload-Digest\":\"sha1:T2J2UW3EHFAZP5ZPIBWHHNQRRE3HGL4C\",\"WARC-Block-Digest\":\"sha1:ER3PQGAWOFIW474SQRGWWGQSVILRSV52\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510697.51_warc_CC-MAIN-20230930145921-20230930175921-00184.warc.gz\"}"}
http://st.tongbuedu.com/counts-33114.html	[ "", null, "同步学习网是一个提供了大量的中考真题下载,高考真题下载,中考试题解析,高考试题解析,各个学习段的模拟试题下载的网站,我们希望通过在线做题的方式能够切实的帮助广大学生查缺补漏,提高学习成绩！", null, "1. 已知全集", null, "，集合", null, "", null, "，则", null, "A.{0} B.{-3-4} C.{-4-2} D.", null, "2. 已知集合", null, "", null, "，则", null, "A.", null, "B.", null, "C.", null, "D.", null, "3. 设函数", null, "在区间（12）内有零点，则实数a的取值范围是（\n\nA.", null, "", null, "B.", null, "", null, "C.", null, "", null, "D.", null, "", null, "5. 下面几种推理中是演绎推理的序号为（\n\nA. 由金、银、铜、铁可导电，猜想：金属都可导电；\n\n6. 在下列命题中，真命题是（\n\nA.x=2,x23x+2=0的否命题； B.b=3,b2=9的逆命题;\n\nC. ac>bc,a>b; D.相似三角形的对应角相等的逆否命题\n\n7. 函数", null, "的图象大致是", null, "8. 已知命题p", null, "x1x2", null, "R(f(x2)", null, "f(x1)(x2", null, "x1)0，则", null, "p是（\n\nA.", null, "x1x2", null, "R(f(x2)", null, "f(x1)(x2", null, "x1)0\n\nB.", null, "x1x2", null, "R(f(x2)", null, "f(x1)(x2", null, "x1)0\n\nC.", null, "x1x2", null, "R(f(x2)", null, "f(x1)(x2", null, "x1)<0\n\nD.", null, "x1x2", null, "R(f(x2)", null, "f(x1)(x2", null, "x1)<0\n\nA.", null, "B.", null, "C.", null, "D.", null, "10. 定义在", null, "上的奇函数", null, "，当", null, "时，", null, "，则关于", null, "的函数", null, "的所有零点之和为（\n\nA.", null, "B.", null, "C.", null, "D.", null, "11. 若函数", null, "的单调递增区间是", null, "，则", null, "_________.\n\n12. 曲线", null, "在点", null, "处的切线斜率为 .\n\n13. 已知函数", null, "，则f (4) = .\n\n14. 把数列", null, "依次按第一个括号一个数，第二个括号两个数，第三个括号三个数，第四个括号四个数，第五个括号一个数，第六个括号两个数，……，循环下去，如：（3），（57），（91113），（15171921），（23），（2527），……，则第104个括号内各数字之和为________.\n\n15.", null, "是定义在", null, "上且周期为2的函数，在区间", null, "上，", null, "其中", null, ".", null, "，则", null, "的值为___________.\n\n16.（本题满分8分）已知命题", null, "，命题", null, "（其中m > 0），且", null, "的必要条件，求实数m的取值范围。\n\n17.（本题满分8分）\n\nx=1x=2是函数f(x)=alnx+bx2+x的两个极值点\n\n(1)a,b的值；\n\n(2)f(x)的单调区间。\n\n18.（本题满分10分）\n\n19.（本题满分12分）\n\n20.（本题满分13分）", null, "" ]	[ null, "http://st.tongbuedu.com/images/stsy_topzy.png", null, "http://st.tongbuedu.com/images/ad0000.gif", null, 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http://tool.114la.com/bus/shenzhen/s_R_65.html	[ "8路德兴花园 --- 火车站西广场\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n9路德兴花园总站 --- 皇岗口岸\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\nN16路德兴花园总站 --- 火车站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n57路鹤围村 --- 长岭(总站)\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n58路南头火车西站 --- 德兴花园总站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n38.\n\n39.\n\n40.\n\n41.\n\n42.\n\n43.\n\n44.\n\n45.\n\n46.\n\n47.\n\n61路德兴花园 --- 火车站西广场\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n79路东角头总站 --- 清水河总站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n38.\n\n39.\n\n40.\n\n41.\n\n42.\n\n43.\n\n44.\n\n45.\n\n46.\n\n47.\n\n48.\n\n85路盐田检查站 --- 德兴花园\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n38.\n\n39.\n\n40.\n\n41.\n\nK105路蛇口码头 --- 德兴花园总站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n38.\n\n39.\n\n40.\n\n123路深圳湾口岸 --- 德兴花园\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n38.\n\n39.\n\n40.\n\n41.\n\n42.\n\n43.\n\n219路石厦南总站 --- 清水河总站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n222路德兴花园 --- 世界之窗总站\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\n34.\n\n35.\n\n36.\n\n37.\n\n385路福田保税区 --- 元平学校\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n7.\n\n8.\n\n9.\n\n10.\n\n11.\n\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20.\n\n21.\n\n22.\n\n23.\n\n24.\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\nB616路清水河 --- 紫星花园\n\n1.\n\n2.\n\n3.\n\n4.\n\n5.\n\n6.\n\n### 热门车站\n\n 深圳晴转多云 32℃ ～25℃" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9724755,"math_prob":0.92402774,"size":1098,"snap":"2020-34-2020-40","text_gpt3_token_len":1034,"char_repetition_ratio":0.22029251,"word_repetition_ratio":0.0,"special_character_ratio":0.5364299,"punctuation_ratio":0.17786561,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9653503,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T23:52:53Z\",\"WARC-Record-ID\":\"<urn:uuid:81769b66-907a-4e9e-b592-eea4fc7a27e9>\",\"Content-Length\":\"56694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f1be624-4256-4641-a872-62db49240017>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0afcf8c-a1ef-477c-94d6-1bae097aaea4>\",\"WARC-IP-Address\":\"39.107.146.250\",\"WARC-Target-URI\":\"http://tool.114la.com/bus/shenzhen/s_R_65.html\",\"WARC-Payload-Digest\":\"sha1:OAMNAH4L7Z3LLPR7TJYX4OFLX3QKIXQO\",\"WARC-Block-Digest\":\"sha1:5GKOTPHNC72N5NY3XJY5NPRZRNQFET4X\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402132335.99_warc_CC-MAIN-20201001210429-20201002000429-00014.warc.gz\"}"}
http://lib.mexmat.ru/books/127579	[ "Электронная библиотека Попечительского советамеханико-математического факультета Московского государственного университета\n Главная Ex Libris Книги Журналы Статьи Серии Каталог Wanted Загрузка ХудЛит Справка Поиск по индексам Поиск Форум", null, "Авторизация", null, "Поиск по указателям", null, "", null, "", null, "", null, "", null, "Richard Haberman — Mathematical models", null, "Читать книгубесплатно\n\nСкачать книгу с нашего сайта нельзя\n\nОбсудите книгу на научном форуме", null, "Нашли опечатку?\nВыделите ее мышкой и нажмите Ctrl+Enter\n\nНазвание: Mathematical models\n\nАвтор: Richard Haberman\n\nАннотация:\n\nMathematics is a grand subject in the way it can be applied to various problems in science and engineering. To use mathematics, one needs to understand the physical context. The author uses mathematical techniques along with observations and experiments to give an in-depth look at models for mechanical vibrations, population dynamics, and traffic flow. Equal emphasis is placed on the mathematical formulation of the problem and the interpretation of the results. In the sections on mechanical vibrations and population dynamics, the author emphasizes the nonlinear aspects of ordinary differential equations and develops the concepts of equilibrium solutions and their stability. He introduces phase plane methods for the nonlinear pendulum and for predator-prey and competing species models.\n\nЯзык:", null, "Статус предметного указателя: Неизвестно\n\ned2k: ed2k stats\n\nГод издания: 1987\n\nКоличество страниц: 421\n\nДобавлена в каталог: 29.11.2013\n\nОперации: Положить на полку \| Скопировать ссылку для форума \| Скопировать ID", null, "Предметный указатель", null, "Реклама", null, "", null, "", null, "", null, "", null, "© Электронная библиотека попечительского совета мехмата МГУ, 2004-2019", null, "\|", null, "\|", null, "О проекте" ]	[ null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/img/main/2.jpg", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/covers/default.gif", null, "http://dxdy.ru/80x15.png", null, "http://lib.mexmat.ru/img/ico/en.png", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://lib.mexmat.ru/z.gif", null, "http://d5.cf.bb.a0.top.mail.ru/counter", null, "http://mc.yandex.ru/watch/1659949", null, "http://lib.mexmat.ru/img/ico/libmexmat_80x15.png", null, "http://lib.mexmat.ru/img/ico/valid_html401.png", null, "http://lib.mexmat.ru/img/ico/valid_css.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84231883,"math_prob":0.6233256,"size":916,"snap":"2019-13-2019-22","text_gpt3_token_len":177,"char_repetition_ratio":0.13486843,"word_repetition_ratio":0.0,"special_character_ratio":0.16048035,"punctuation_ratio":0.0945946,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9642235,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T14:27:24Z\",\"WARC-Record-ID\":\"<urn:uuid:5d3c6cc3-2686-4423-b81b-3d37eb946fda>\",\"Content-Length\":\"15814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d159a986-af70-4a7f-8ec6-f030df220e1c>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d06992f-0602-46b5-987b-121ac60a4457>\",\"WARC-IP-Address\":\"158.250.33.208\",\"WARC-Target-URI\":\"http://lib.mexmat.ru/books/127579\",\"WARC-Payload-Digest\":\"sha1:QD5PIBXFB7I2HFT3FMXROBFMPXWOFDUU\",\"WARC-Block-Digest\":\"sha1:ZGOXIIWW5SLUYVQGN6M7OPADDSDPMO6R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232262600.90_warc_CC-MAIN-20190527125825-20190527151825-00462.warc.gz\"}"}
http://fenixdocs.com/index.php/False	[ "This wiki is out of date, use the continuation of this wiki instead\n\n# FALSE\n\n(Redirected from False)\n\n## Definition\n\nINT FALSE\n\nFalse is a constant integer, equal to the value 0. It is used to state that something is false and not true.\n\nChecking whether a variable is false, is the same as checking if it's zero. In older versions of Fenix, it was the same as checking whether a variable is even. This has been changed because it's more commonly used.\n\n## Example\n\n```Program example;\nPrivate\nint b = false;\nBegin\n\n// comparison with the constant FALSE\nif(b == false)\nsay(\"b was FALSE! so b==0\");\nelse\nsay(\"b was not FALSE! so b!=0\");\nend\n\n// checking the integer itself\nif(!b)\nsay(\"b was false! so b==0\");\nelse\nsay(\"b was true! so b!=0\");\nend\n\nLoop\nframe;\nEnd\n\nEnd\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8848531,"math_prob":0.90625787,"size":775,"snap":"2019-13-2019-22","text_gpt3_token_len":199,"char_repetition_ratio":0.12970169,"word_repetition_ratio":0.014814815,"special_character_ratio":0.2683871,"punctuation_ratio":0.13772455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98561335,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-26T19:14:16Z\",\"WARC-Record-ID\":\"<urn:uuid:0ee7ae3d-b4eb-4f46-ab60-3eaf18d0eed9>\",\"Content-Length\":\"11185\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b5a87cd-62fc-4a37-ad54-607bdecade4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:5333f08f-5a12-438e-99af-b04a9b9d2c4c>\",\"WARC-IP-Address\":\"91.195.200.80\",\"WARC-Target-URI\":\"http://fenixdocs.com/index.php/False\",\"WARC-Payload-Digest\":\"sha1:Y74RLWGZX5WSBX52AI573BNVVBWELYO4\",\"WARC-Block-Digest\":\"sha1:RMFY4YKTHEPZNCZMEEA3P5NWXGW4IPSM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912205600.75_warc_CC-MAIN-20190326180238-20190326202238-00428.warc.gz\"}"}
https://www.scirp.org/journal/paperinformation.aspx?paperid=64342	[ "Thermal Decomposition and a Kinetic Study of Poly(Para-Substituted Styrene)s\n\nAbstract\n\nThe thermal decompositions of polystyrene (PS), poly(p-methyl styrene) (PMS), poly(p-bromo styrene) (PBrS), and poly(p-chloro styrene) (PClS) were investigated through thermogravimetric analysis (TGA). For this aim, Flynn-Wall-Ozawa method was applied to derivative thermogravimetric (DTG) curves. Continuous distribution kinetics was employed with a stoichiometric kernel to determine the rate coefficients for decomposition reactions. TGA data for the polymers were investigated by nonlinear fitting procedures that yielded activation energies and frequency factors for the combined chemical reactions. The reaction order values of PS derivatives are just about 1 in the nonisothermal decomposition process. Ea values for PS, PMS, and PClS increase with % conversion individually as they decrease in the order of PS/PMS/PClS which is consistent with the molecular weight increase. On the other hand, PBrS has the highest activation energy. Also its activation energy decreases with the % conversion. Thus it is suggested that PBrS degrades with somehow different mechanism.\n\nKeywords\n\nShare and Cite:\n\nŞenocak, A. , Alkan, C. and Karadağ, A. (2016) Thermal Decomposition and a Kinetic Study of Poly(Para-Substituted Styrene)s. American Journal of Analytical Chemistry, 7, 246-253. doi: 10.4236/ajac.2016.73021.\n\nReceived 8 January 2016; accepted 6 March 2016; published 9 March 2016", null, "1. Introduction\n\nOverview and Background\n\nPolystyrene (PS) is a thermoplastic material having so many using areas like home appliances, construction, packaging, and other industrial applications. It is cheap and easily processable. However, it has restricted mechanical properties and is very flammable. Due to its advantages, different kinds of derivatives have been produced. To enhance PS’s competition capacity in engineering resin applications, to protect its properties like thermal stability is crucial. In this work, poly(para-substituted styrene)s have been investigated using nonisothermal degradation kinetics.\n\nUsually, polystyrene, exposed to thermal degradation, softens and melts at 160˚C. Up to 275˚C, volatility of fused polymer with high molecular weight is lowered and this molten version of the polymer keeps its stability. However, polymer fragments with lower molecular weight have higher volatility form and the decomposition is completed after the temperature reaches to 470˚C . In addition, molecular weight rapidly decreases in the beginning of the degradation and the rate of this decline slows down at later stages . This case applies to not only polystyrene but also some of its substituted derivatives. Literature suggests that styrene is the major decomposition product in the complete depolymerization taking place below 500˚C. After this temperature, further fragmentation carries out and the oligomers turn into gaseous species having lower molecular weight .\n\nFor kinetic interpretations in this study, poly(p-substituted styrene)s’ thermal degradation data were collected in nitrogen atmosphere. With intent to understand the origin of the equations to determine the kinetic parameters, a brief derivation is included below. For a simple degradation step, the rate of conversion is described as follows:", null, "(1)\n\nIn this equation, a is the degree of chemical conversion, n is the reaction order, k is the reaction rate, and d/dt is the time derivative. A relation between the reaction rate and temperature is usually assumed:", null, "(2)\n\nHere, Z is the preexponential constant, R is the ideal gas constant, Ea is the activation energy, and T is the temperature in Kelvin. This is known as the Arrhenius relationship. Equation (3) is attained by combining Equations (1) and (2):", null, "(3)\n\nIn order to obtain a precise solution, one of three variables in Equation (3) either may be held as a constant or may be related to another variable when the reaction process is carried out. In this way, the reaction process can be defined. For years, a differential or an integral version of Equation (3) has been applied by researchers to calculate the kinetic parameters consisting of Ea, Z, and n. While some of the techniques rely on the functional version of conversion, f(a), in the literature - , activation energy values have also been indicated to be estimated without any preceding knowledge of the functional form.\n\nThere is a linear relationship between temperature and time as following:", null, "(4)\n\nand", null, "(5)\n\nwhere T is the current temperature, T(0) is the initial temperature, t is time, and b is the scanning rate. Combining Equation (3) with Equation (5) yields:", null, "(6)\n\nThis equation shows the theoretical shape of the TGA curve that is determined in the Kinetics Software. Logarithmic form of the equation is", null, "(7)\n\nA multilinear regression is performed by using ln(bda/dT), −l/RT, and ln(1−a) which have been appraised from TGA curve, and solving for Z, Ea, and n.\n\nKinetic parameters of degradation are determined upon the observations of Ozawa and Flynn and Wall . Their statements involve an approximation that has been discussed by Doyle and by Flynn and Wall , among others. This approximation comprises the so-called “exponential integral” and methods lowering error to a negligible level and by this way improving the results. Zsako and Zsako have suggested a closed form approximation for the exponential integral . Although the determination of the activation energy is straightforward, one cannot carry out further kinetic computations without further data or assumptions as to the form of the kinetic equation. It has often been observed that the initial portion of a TG curve can usually be well-fitted by a first-order reaction equation. Since the interest in lifetime or stability studies is most often concerned with the initial portion (low conversion level), it has become a procedural practice to assume first-order kinetics and compute the rate constants, pre-exponential factors, etc. on this assumption. When the value of n = 1 is used (first order kinetics), the method is essentially the same as that used by ASTM Standart Test Method E 1641.\n\nIn this work, the mathematical treatment followed in calculating the parameters are called as Standard method. A head start of this method is to obtain the activation energy exactly independently of any assumptions regarding the relation between the reaction rate and the percent reacted. The quality of the statistical fit can be judged from the 95% confidence limits and from observing the lnβ vs 1/T plot. In the ideal case, the data points fall on parallel straight lines, the slope of which is used to calculate the activation energy.\n\n2. Experimental\n\nApparatus\n\nLow heating rate degradation experiments were conducted in a Perkin Elmer PYRIS Diamond TG/DTA (DSC) apparatus in a dynamic nitrogen atmosphere which was calibrated with calcium oxalate. All TGA experiments were carried out with a linear heating rate of 5˚C - 25˚C/min up to 600˚C under nitrogen atmosphere. Kinetic parameters of degradation were determined by the Scanning Kinetics Software.\n\nPolystyrene with Mw = 230,000 g/mol (430,102), poly(p-methylstyrene) with Mw = 72000 (182,273), poly(p-chlorostyrene) with Mw = 75,000 (434,124), and poly(p-bromostyrene) with Mw = 65,000 (181,366) beads were all supplied from Aldrich company and used without any further process. TG analysis was carried out on approximately 10 mg samples.\n\n3. Result and Discussion\n\n3.1. Effect of Heating Rate on p-Substituted PSs\n\nNonisothermal TGA of the samples was carried out at 5, 10, 15, 20, and 25 K/min. Thermogravimetric curves for PS, PMS, PClS, and PBrS are represented in Figure 1 and Figure 2. The kinetic analysis of the degradation of samples was performed by utilizing the thermograms obtained from the TGA. Increase in the heating rate results in a shift toward higher temperatures in the thermograms (Figure 1 and Figure 2). The reason for this behavior is that the sample needs shorter time to catch a given temperature for a faster heating rate.\n\nSame behavior with thermograms was observed for the peak temperatures. An empirical relation between the peak volatilization temperature (Tpv in ˚C) and the heating rate (R in ˚C/min) was reported by Liu et al. .", null, "(8)\n\nThe temperature at which the decomposition rate of the sample is maximum is called as peak volatilization temperature. Tpv values for the PS and the para-substituted derivatives at different heating rates were calculated by using this relationship in the current study, and compared with the TGA experimental data. Experimental values are quite close with that of Liu et al.’s model (Table 1).\n\nln(k) versus 1/T plots for poly(para-substituted styrene)s are shown in Figure 3. The lines obtained at different conversion rates are observed to be roughly parallel to one another and the correlation coefficient of the li-\n\nFigure 1. TGA scans of poly(p-substituted styrene)s at different heating rates in nitrogen.\n\nFigure 2. DTG curves of poly(p-substituted styrene)s at different heating rates in nitrogen.\n\nFigure 3. Kinetic analysis using Ozawa’s method for poly(p-substituted styrene)s degradation in nitrogen environment.\n\nTable 1. Comparison of experimental poly(p-substituted styrene)s peak volatilization decomposition temperatures (Tpv) measured and expected values using Liu’s model in nitrogen at different heating rates.\n\nnearity for the calculation of each activation energy are larger than 0.98. Also there is no systematic divergence from linearity. Therefore it is suggested that the method fitted well for the nonisothermal degradation of poly(p- substituted styrene)s and rate order of degradation is 1 in all poly(p-substituted styrene)s.\n\n3.2. Effect of Substituent Group on Activation Energy and lnZ Values for the Decomposition of p-Substituted PSs\n\nThe activation energies and lnZ values of PS, PMS, PClS, and PBrS at the percent conversion levels of 3.0, 5.0, 8.0, 12.0, 18.0, and 26.0 are tabulated in Table 2. Activation energy is observed to increase with the degree of conversion for PS, PMS, and PClS, which should be attributed to the nature of random scission degradation and the differences in deviation from stationary reaction state at the different heating rates .\n\nThe electronegativity values of methyl and hydrogen which are both bonded to the same group are close to each other. The bonding character of methyl group in PMS is most likely to bonding character of hydrogen in PS. The only difference between the two polymers is the weight of the repeating unit in the backbone. The repeating segment in PMS is slightly heavier in comparison with PS. Therefore the activation energies of PMS are slightly lower than the activation energies of PS. In the case of PClS, the weight of the repeating elements with Cl atom is much higher than the repeating units of PS and PMS and so the activation energy of the PClS is much lower. However PBrS exhibits completely different attitude. Its activation energy is considerably higher than PS and it decreases with the weight loss value. Therefore it can be suggested that the degradation mechanism of PBrS is different.\n\nIn order to understand the dependency of activation energies to the substituent in poly(p-substituted styrene)s, their activation energy values were graphed with respect to molecular weight of the repeating unit in Figure 4.\n\nTable 2. Kinetic analysis results for poly(p-substituted styrene)s.\n\nFigure 4. Activation energies of poly(p-substituted styrene)s at different weight loss percentages.\n\nThe outcome of Figure 4 is as follows;\n\n1) The activation energies of degradation of PS and PMS are keeping up with its values upon different weight loss values as the activation energies of PClS and PBrS are considerably affected by the weight loss ratio.\n\n2) The activation energies of PS, PMS, and PClS increases with the weight loss ratio, as the activation energy value of PBrS decreases.\n\n3) The activation energy of the substituted styrenes decreases with the molecular weight of the repeating units, PBrS shows an extraordinary behavior. Its activation energy values are higher than the others up to high values of the weight losses.\n\nUpon handling individually, PS degrades with chain scission mostly. The bonding character of methyl group in PMS is very similar to bonding character of H in PS. The electronegativity values of methyl and H which are both bonded to the same group are close to each other. The only difference between the two polymers is the weight of the repeating unit in the backbone. The weight of repeating segment of PMS is slightly higher than that of PS. Therefore the activation energies of PMS are slightly lower than PS. In the case of PClS, the weight of the repeating elements with Cl atom is much higher than the repeating units of PS and PMS and the activation energy of the PClS is much lower.\n\nThe bonds between the substituent and the polystyrene backbone are C-H in PS, C-C in PMS, C-Cl in PClS, and C-Br in PBrS. The trend in the electronegativity values of the substituent of polystyrenes is as H(2.20) < C(2.55) < Br(2.96) < Cl(3.16) and, the bonding character of Br is much close to H and C than Cl. As a result, it is seen from the Figure 4 that the activation energy values of the degradation of PBrS is more close to the PS and PMS for any of the weight loss rates. On the other hand the character of Br substituent makes the chemical bonds between repeating groups stronger, leading to a high activation energy values.\n\n4. Conclusion\n\nThermal decomposition kinetics of poly(p-substituted styrene)s was examined in nitrogen atmosphere. Five different heating rates were applied to the samples and as heating rate increased, the degradation shifted to higher temperatures accompanied with an increase in the rate of weight loss. From the nonisothermal degradation data, kinetic parameters were calculated using Flynn-Wall-Ozawa method and the values of the reaction order of poly(p-substituted styrene)s were deduced as 1 with the correlation factor higher than 0.98. Activation energies of thermal degradations are increasing individually in PS, PMS, PClS with weight loss as it is decreasing in PbrS with weight loss. In general, the decrease in the activation energies of poly(p-substituted styrene)s with PS/PMS/ PClS trend is consistent with the increase in repeating unit molecular weight except for PBrS. Therefore PBrS is suggested to decompose with somehow different mechanism.\n\nNOTES\n\n*Corresponding author.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.", null, "", null, "", null, "", null, "", null, "[email protected]", null, "+86 18163351462(WhatsApp)", null, "1655362766", null, "", null, "Paper Publishing WeChat", null, "" ]	[ null, "https://html.scirp.org/file/2-2201342x6.png", null, "https://html.scirp.org/file/2-2201342x7.png", null, "https://html.scirp.org/file/2-2201342x8.png", null, "https://html.scirp.org/file/2-2201342x9.png", null, "https://html.scirp.org/file/2-2201342x10.png", null, "https://html.scirp.org/file/2-2201342x11.png", null, "https://html.scirp.org/file/2-2201342x12.png", null, "https://html.scirp.org/file/2-2201342x13.png", null, "https://html.scirp.org/file/2-2201342x14.png", null, "https://www.scirp.org/images/Twitter.svg", null, "https://www.scirp.org/images/fb.svg", null, "https://www.scirp.org/images/in.svg", null, "https://www.scirp.org/images/weibo.svg", null, "https://www.scirp.org/images/emailsrp.png", null, "https://www.scirp.org/images/whatsapplogo.jpg", null, "https://www.scirp.org/Images/qq25.jpg", null, "https://www.scirp.org/images/weixinlogo.jpg", null, "https://www.scirp.org/images/weixinsrp120.jpg", null, "https://www.scirp.org/Images/ccby.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9338143,"math_prob":0.938752,"size":13242,"snap":"2022-40-2023-06","text_gpt3_token_len":2927,"char_repetition_ratio":0.15364859,"word_repetition_ratio":0.07398568,"special_character_ratio":0.20714393,"punctuation_ratio":0.09506173,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9821797,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,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\":\"2023-01-28T03:13:51Z\",\"WARC-Record-ID\":\"<urn:uuid:cdfe9cf0-4f91-4bb8-8fdb-6310963cac6b>\",\"Content-Length\":\"110157\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9737a08e-6743-4ae6-b8df-787839f8a27a>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d0abb06-85b3-49a5-9791-fc3cfd668991>\",\"WARC-IP-Address\":\"107.191.112.46\",\"WARC-Target-URI\":\"https://www.scirp.org/journal/paperinformation.aspx?paperid=64342\",\"WARC-Payload-Digest\":\"sha1:IEXISXZWJVACX4Q4NGKNSZPYP64OQUPO\",\"WARC-Block-Digest\":\"sha1:WXBLUOIRQTYICYFOXMM7TIVJSNT5ALYT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499470.19_warc_CC-MAIN-20230128023233-20230128053233-00581.warc.gz\"}"}
https://www.hackmath.net/en/math-problem/10801	[ "# Coal mine\n\nThe towing wheel has a diameter of 1.7 meters. How many meters does the elevator cage lower when the wheel turns 32 times?\n\nCorrect result:\n\nl = 171 m\n\n#### Solution:", null, "We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!", null, "#### You need to know the following knowledge to solve this word math problem:\n\nWe encourage you to watch this tutorial video on this math problem:\n\n## Next similar math problems:\n\n• Angles 1", null, "It is true neighboring angles have not common arm?\n• Rhombus angles", null, "If one angle in the rhombus is 159°, what is it neighboring angle in rhombus?\n• Circle", null, "What is the radius of the circle whose perimeter is 6 cm?\n• The diameter", null, "The diameter of a circle is 4 feet. What is the circle's circumference?\n• Circle from string", null, "Martin has a long 628 mm string . He makes circle from it. Calculate the radius of the circle.\n• Semicircle", null, "Calculate the length of a semicircle with a radius of 6cm.\n• Round table", null, "A tablecloth should be sewn on a round table with a diameter of 78 cm, which should extend around the table by 10 cm. How many cms of ribbons need to be bought for edging?\n• Bicycle wheel", null, "After driving 157 m bicycle wheel rotates 100 times. What is the radius of the wheel in cm?\n• Circle - simple", null, "The circumference of a circle is 198 mm. How long in mm is its diameter?\n• Circle - easy 2", null, "The circle has a radius 6 cm. Calculate:\n• Clock hands", null, "The second hand has a length of 1.5 cm. How long does the endpoint of this hand travel in one day?\n• Velocipede", null, "The front wheel of velocipede from year 1880 had a diameter 1.8 m. If the front wheel turned again one then rear wheel 6 times. What was the diameter of the rear wheel?\n• Bicycle wheel", null, "Bicycle wheel diameter is 62 cm. How many times turns the bicycle on the road 1 km long?\n• Central angle", null, "What is the length of the arc of a circle with a diameter of 46 cm, which belongs to a central angle of 30°?\n• Well", null, "Rope with a bucket is fixed on the shaft with the wheel. The shaft has a diameter 50 cm. How many meters will drop bucket when the wheels turn 15 times?\n• Athlete", null, "How long length run athlete when the track is circular shape of radius 120 meters and an athlete runs five times in the circuit?\n• Mine", null, "Wheel in traction tower has a diameter 4 m. How many meters will perform an elevator cabin if wheel rotates in the same direction 89 times?" ]	[ null, "https://www.hackmath.net/img/1/koleso.jpg", null, "https://www.hackmath.net/hashover/images/avatar.png", null, "https://www.hackmath.net/thumb/4/t_304.jpg", null, "https://www.hackmath.net/thumb/71/t_71.jpg", null, "https://www.hackmath.net/thumb/85/t_785.jpg", null, "https://www.hackmath.net/thumb/13/t_21813.jpg", null, "https://www.hackmath.net/thumb/63/t_1963.jpg", null, "https://www.hackmath.net/thumb/41/t_27041.jpg", null, "https://www.hackmath.net/thumb/41/t_30141.jpg", null, "https://www.hackmath.net/thumb/93/t_4493.jpg", null, "https://www.hackmath.net/thumb/21/t_1921.jpg", null, "https://www.hackmath.net/thumb/67/t_1967.jpg", null, "https://www.hackmath.net/thumb/21/t_5621.jpg", null, "https://www.hackmath.net/thumb/19/t_1519.jpg", null, "https://www.hackmath.net/thumb/65/t_1965.jpg", null, "https://www.hackmath.net/thumb/51/t_24351.jpg", null, "https://www.hackmath.net/thumb/52/t_1152.jpg", null, "https://www.hackmath.net/thumb/24/t_1924.jpg", null, "https://www.hackmath.net/thumb/90/t_90.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90220445,"math_prob":0.9396939,"size":2275,"snap":"2020-45-2020-50","text_gpt3_token_len":568,"char_repetition_ratio":0.14707178,"word_repetition_ratio":0.009111618,"special_character_ratio":0.2487912,"punctuation_ratio":0.0955414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948521,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T01:52:59Z\",\"WARC-Record-ID\":\"<urn:uuid:9066261d-163b-4c4c-9e71-a918fe444721>\",\"Content-Length\":\"42027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1d82312d-46a0-4508-984b-5d4a0e6db8db>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d4bbef7-b5ee-4f00-b8b3-4b4332ce005d>\",\"WARC-IP-Address\":\"172.67.143.236\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/math-problem/10801\",\"WARC-Payload-Digest\":\"sha1:IGRFGES4YVMPOUCPNWJXVKONQOYSMLYD\",\"WARC-Block-Digest\":\"sha1:T7PZJIQSK7I4LNJW5DDDPF2L2TFPTBEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107874637.23_warc_CC-MAIN-20201021010156-20201021040156-00363.warc.gz\"}"}
http://the-dusty-deck.blogspot.com/2011/02/thinking-formally.html	[ "Labels\n\nAI (1) Haskell (1) Java 8 (2) JavaScript (1) Logic (10) Methodology (8) Refactoring (1) Scala (1) Security (2) Tools (9) Xtend (2) Xtext (1)\n\nTuesday, February 15, 2011\n\nThinking Formally\n\nIf there is no alternative to clear thinking about our programs, we should investigate what we can learn from the science of organized thought. Logic has a long history, with equal roots in philosophy and mathematics. In some sense, computer programs breath life into abstract logic and give logic the abilitiy to do real things. And, just like Pinocchio, they often misbehave!\n\nOne of the fundamental concerns of logic is the study of proofs: how can we convince ourselves in a precise and mechanical way of the truth of some claim. The desire to do so resulted from the need to distinguish between valid and invalid mathematical proofs, expressed using prose. Sometimes, hidden assumptions were not recognized, leading to faulty proofs. For example, consider the first axiomatic proof system, Euclidean geomerty, studied in middle school. It has five axioms, including the famous \"parallel postulate\". It is an amazing achievement, which is still an excellent example of how to reason formally, and was the source of many important advances in mathematics. But Euclidean geometry has many hidden assumptions, which are glossed over by the use of informal drawings. About 2200 years after Euclid, several mathematicians tried to completely formalize geometrical proofs. David Hilbert's axiomatization of geometry includes twenty axioms instead of Euclid's five. One of these states that between every two different points on a straight line there is a third point on the same line. This was so obvious to Euclid that he didn't even think to mention it in his list of axioms. But axioms are exactly those self-evident propositions on which the theory is based, and the proof isn't logically sound unless all these are stated explicitly.\n\nLogic is closely tied with computation, although the connection was made explicit only towards the 20th century. A formal proof is a series of deductions, each of which is based on axioms or previously-proven propositions, using a set of deduction rules. It is supposed to be so simple that it can easily be checked for correctness, without any understanding of its subject matter, whether it is geometry, number theory, or abstract algebra. In other words, it should be easy to write a program that checks a formal proof; all you need to do is verify that each step follows from the axioms and from previous steps according to the rules.\n\nBut wait! Is there a limit to the complexity of the axioms or of the rules used in the proof? Without such limits, the whole exercise falls apart. For example, imagine a system of axioms that contains all true statements in number theory. This is, of course, an infinite set. How difficult is it to write a program to check whether a given statement is an axiom? Well, it turns out to be impossible! (This is a consequence of Gödel's first incompleteness theorem.) So, if we are to be able to write a program to check formal proofs, we need to restrict the possible sets of axioms. For example, in Gödel's theorems, the set of axioms is required to be recursively enumerable; this means that there is a (finite) program that can generate all axioms and never generates anything that is not an axiom. Of course, this program will never terminate if the set is infinite. That's acceptable, as long as the program is guaranteed to produce every axiom if you wait long enough.\n\nSo now, a concept about computability has entered the study of logic itself. It is only fair, then, to apply logic to computation; and, indeed, the study of logics of programs is an important field in computer science. There are even computer programs called theorem provers, which have been used to prove difficult mathematical theorems as well as to verify the correctness of other programs. However, this has still not affected the day-to-day work of most software developers. In the next post, I will discuss why this is the case and what formal tools can still be used for software development today." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9614732,"math_prob":0.8515834,"size":3981,"snap":"2019-43-2019-47","text_gpt3_token_len":822,"char_repetition_ratio":0.103595674,"word_repetition_ratio":0.006006006,"special_character_ratio":0.19542828,"punctuation_ratio":0.1131579,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9841093,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T19:26:04Z\",\"WARC-Record-ID\":\"<urn:uuid:b7a89826-0031-45d6-8eaa-f12990198091>\",\"Content-Length\":\"61633\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0bbea447-ad28-4644-91dd-2fdc3fe8c563>\",\"WARC-Concurrent-To\":\"<urn:uuid:e975d315-536c-41b6-8178-ff878afe58d9>\",\"WARC-IP-Address\":\"172.217.7.225\",\"WARC-Target-URI\":\"http://the-dusty-deck.blogspot.com/2011/02/thinking-formally.html\",\"WARC-Payload-Digest\":\"sha1:BVTK6OBZ6ASOZE44JICMBIXZBKKAADYE\",\"WARC-Block-Digest\":\"sha1:R56XV2DUT7SO7R4KXMTUNL4VZLDYONBX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660231.30_warc_CC-MAIN-20191015182235-20191015205735-00162.warc.gz\"}"}
https://www.mecharithm.com/fundamentals-of-robot-motions-configurations-introduction/	[ "Blog Post\n\n# Fundamentals of Robot Motions: Configurations (Introduction)\n\nThis post is part one of the series of lessons on the fundamentals necessary to represent the robot’s configuration, and it gives an introduction to what we mean when we are talking about representing a robot’s configuration.\n\nIn previous lessons, we learned that the robot’s configuration answers the question of where the robot is, and we saw that there are two ways to represent the robot’s configuration: Implicit representation, where the configuration is represented by embedding the curved space in higher-dimensional Euclidean space subject to constraints and explicit representation where configuration is represented with a minimum number of coordinates. You can watch the lesson on configuration and configuration space before preceding this lesson.\n\nWe also advise reading the preliminaries lesson since the lessons are related to each other, and each one complements the other lesson.\n\nFor the full comprehension of the Fundamentals of Robot Motions and the necessary tools to represent the configurations, velocities, and forces causing the motion, please read the following lessons (note that more lessons will be added in the future):\n\nhttps://www.mecharithm.com/category/fundamentals-of-robotics/fundamentals-of-robot-motions/\n\n## Robot’s Configuration on Plane\n\nWe start this lesson with a planar example to see what we mean to express the configuration of a robot. Suppose a toy car with its motion confined to the plane and two coordinate frames {s}, and {b} with their corresponding unit axes (a hat notation shows a unit vector):", null, "A toy car on a plane with two coordinate frames {s} and {b} with the unit axes. The configuration of the toy car can be expressed by finding the position and orientation of the body fixed-frame with respect to the space frame {s}.\n\n{b} is called a body frame since it is a fixed frame attached instantaneously to the moving body. Therefore, to find the configuration of the toy car, we should express the position and orientation of the body’s fixed-frame coordinates in the base frame coordinates.\n\nOne way to represent the orientation of the body coordinates in terms of the base coordinates is using a rotation matrix:\n\n$R = \\begin{pmatrix} \\hat{x}_b.\\hat{x}_s & \\hat{y}_b.\\hat{x}_s \\\\ \\hat{x}_b.\\hat{y}_s & \\hat{y}_b.\\hat{y}_s \\end{pmatrix} = \\begin{pmatrix} cos\\alpha & -sin\\alpha \\\\ sin\\alpha & cos\\alpha \\end{pmatrix}_{2\\times2}$\n\nThe columns of this matrix are the coordinate axes of the {b} frame expressed in the coordinate axes of the {s} frame. The dot represents the dot product between the coordinate axes, and since they are unit vectors, the dot product represents the cosine of the angle between the two vectors.\n\nRecall that if a and b are two vectors with known length and angle between them:", null, "The dot product of two vectors is the multiplication of the length of the two vectors and cosine of the angle between the two vectors.\n\nThen, the dot product of the two vectors can be expressed as the projection of one vector onto the other multiplied by the length of the other vector (in other words, it is the multiplication of the length of the two vectors and cosine of the angle between the two vectors):\n\n$a.b = \|\|a\|\| \|\|b\|\| cos \\theta$\n\nThe 2 by 2 rotation matrix belongs to the four dimensional space subject to the three constraints as follows:\n\n• Each column of the rotation matrix is a unit vector. In other words, the lengths of the column vectors are 1.\n• Two columns are orthogonal to each other. In other words, the dot product of the columns is zero.\n\nThus, we have one degree of freedom which is parameterized by the angle α to represent the orientation. The rotation matrix is an implicit way to represent the orientation.\n\nNow that we found a way to represent the orientation of the toy car on the plane, it’s time to represent its position. The position of the origin of the body frame in base frame coordinates can be expressed as the vector p from the origin of the {s} frame to the origin of the {b} frame:\n\n$p = p_x \\hat{x}_s + p_y \\hat{y}_s = \\begin{pmatrix} p_x\\\\ p_y \\end{pmatrix} \\in R^2$\n\nThus the configuration of the toy car can be represented by the pair (R,p), which is the description of the orientation and position of {b} with respect to {s}. This means that base frame {s} can be coincident with the body frame {b} by rotating the base frame coordinates around the z-axis by α and then translating the origin by p.\n\n## Robot’s Configuration in Space\n\nUsing a similar approach, we can express the configuration of a robot in space. We can use frames to represent the configuration of a robot in space. One simple way to represent the position and orientation (configuration) of a robot in space is to:\n\n• Fix a frame to the body of the robot according to the Right-Hand Rule (RHR) (sometimes this frame is attached to an important point on the rigid body like the center of mass, but this is not required), and\n• Fix a frame in space", null, "The position and orientation of a robot in space can be represented by the position of the origin of the body frame expressed in the space frame coordinates and the directions of the coordinate axes of the body frame expressed in the space frame coordinates.\n\nThen the robot’s configuration can be represented by the position of the origin of the body frame expressed in the space frame coordinates and the directions of the coordinate axes of the body frame expressed in the space frame coordinates. Note that the body frame is the stationary frame that coincides with the frame attached to the body at a particular instant in time.\n\nThere are several implicit and explicit representations that are used to represent the rotations and configurations in general that we will discuss in the coming lessons. In the next lesson, we will start talking about different ways to represent the orientations in robotics. The next lesson will be dedicated to Rotation matrices which are the implicit representations of orientations.\n\nThe video version of the current lesson can be watched in the link below:\n\nThanks for reading this post. You can also find the other posts on the Fundamentals of Robotics Course in the link below:\n\nhttps://www.mecharithm.com/category/fundamentals-of-robotics/\n\n###### References:\n\n📘 Textbooks:\n\n• Modern Robotics: Mechanics, Planning, and Control by Frank Park and Kevin Lynch\n• A Mathematical Introduction to Robotic Manipulation by Murray, Lee, and Sastry\n\n✍️ Logo design by Minro Art Group\n\n#### If you enjoyed this post, please consider contributing to help us with the website’s running costs and keep making awesome content for you. We deeply thank you for your generous contribution!", null, "Be sure to let us know your thoughts and questions about this post as well as the other posts on the website. You can either contact us through the “Contact” tab on the website or email us at support[at]mecharithm.com." ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8849478,"math_prob":0.96589345,"size":6227,"snap":"2021-31-2021-39","text_gpt3_token_len":1340,"char_repetition_ratio":0.18190584,"word_repetition_ratio":0.049951028,"special_character_ratio":0.20957123,"punctuation_ratio":0.07455013,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994054,"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\":\"2021-09-28T03:41:40Z\",\"WARC-Record-ID\":\"<urn:uuid:ad200129-2d48-4099-a293-bfafbdfdbfad>\",\"Content-Length\":\"146323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f26029b6-7060-4d12-a7cf-26629ff055fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:861981f9-0684-4a14-a5e7-3c402b99c594>\",\"WARC-IP-Address\":\"162.214.210.109\",\"WARC-Target-URI\":\"https://www.mecharithm.com/fundamentals-of-robot-motions-configurations-introduction/\",\"WARC-Payload-Digest\":\"sha1:METFTB3IMIYBZ32W7VIBW65BE2DW3EGS\",\"WARC-Block-Digest\":\"sha1:I33A74S3AWSDHTXRBRU2EHDTBQ4WKCRN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060201.9_warc_CC-MAIN-20210928032425-20210928062425-00342.warc.gz\"}"}