Split (1)

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url stringlengths 14 2.42k	text stringlengths 100 1.02M	date stringlengths 19 19	metadata stringlengths 1.06k 1.1k
https://xtools.readthedocs.io/en/stable/installation.html	# 3. Installation¶ For contributors, see Development for additional, more detailed instructions specific to setting up a local development environment. The prerequisites listed below still apply. ## 3.1. Prerequisites¶ XTools requires the following to run: • PHP 7.2 or newer, including: • A MySQL-like database, and PDO including the driver for the database you want to use (e.g. PDO_MYSQL. • cURL must be enabled. On some environments you may need to enable this in your php.ini file. Look for a line like ;extension=php_curl.dll and uncomment it by removing the leading ;. • Additional PHP extensions are also required, as specified in composer.json. • Composer 1.0.0+ • Node and npm (tested with versions 10.8.0+ and 6.2.0+, respectively). ## 3.2. Instructions¶ 1. Download the latest release into a web-accessible location. For contributors, you should develop off of the master branch. 2. Ensure that var/ and all files within it (other than var/SymfonyRequirements.php) are writable by the web server. 3. Run composer install. You will be prompted to enter database details and other configuration information. See Configuration for documentation on each parameter. 4. Create the XTools database: php bin/console doctrine:database:create and run the migrations with php bin/console doctrine:migrations:migrate. This is actually only used for usage statistics (e.g. see xtools.wmflabs.org/meta). XTools will run without it but doing so may cause silent failures, as the requests to record usage are made with AJAX. 5. With each deployment or pull from master, you may need to clear the cache. Use php bin/console cache:clear --no-warmup to clear the cache. For a production environment be sure to append --env=prod to these commands. You must also clear the cache whenever you make configuration changes. In production, you may find that further server-level configuration is needed. The setup process for Wikimedia Cloud VPS (which runs on Debian Jessie) is documented on Wikitech. This may be of assistance when installing XTools on similar Linux distributions. ## 3.3. Single wiki¶ If you are running XTools against a single wiki, make sure you using the following configuration options: • app.single_wiki to true • wiki_url to the full URL of your wiki. • api_path to the path to the root of your wiki’s API. ## 3.4. Wiki family¶ To use XTools for a family of wikis, set app.single_wiki to false in parameters.yml. You will also need a database table that contains meta information about your wikis. It can live wherever you want; just set the database_replica_* variables accordingly in parameters.yml. XTools was built for one resembling the WMF database. The table must be called wiki and have the following structure: CREATE TABLE wiki ( dbname varchar(32) NOT NULL PRIMARY KEY, lang varchar(12) NOT NULL DEFAULT 'en', name text, family text, url text ); ` The WMF version of this table can be browsed at Quarry #4031.	2018-11-12 22:17:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27466878294944763, "perplexity": 9637.57384399851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039741151.56/warc/CC-MAIN-20181112215517-20181113001517-00443.warc.gz"}
http://definiter.net/oe3p96c/41e397-quadratic-congruential-generator-in-python	ax² + bx + c â¡ 0 (mod n) a . This function will seed the global default random number generator, and any call to a function in numpy.random will use and alter its state. Code: lcg.c Park-Miller random number generator Description: Also known as the Lehmer random number generator. Linear congruential generators (LCGs) are a class of pseudorandom number generator (PRNG) algorithms used for generating sequences of random-like numbers. linear congruential generator Search and download linear congruential generator open source project / source codes from CodeForge.com Upgrade to Math Mastery. mod p a is a quadratic â¦ We use analytics cookies to understand how you use our websites so we can make them better, e.g. A random number generator (RNG) is a computational or physical device designed to generate a sequence of numbers or symbols that lack any pattern, i.e. â¢ Provides code examples updated and written in Python and C# Essential Algorithms has been updated and revised and offers professionals and students a hands-on guide to analyzing algorithms as well as the techniques and applications. they're used to gather information about the pages you visit and how many clicks you need to accomplish a task. IIR Increases the resolution of an image by interpolation B= IIR(inputfile,f) returns the image stored in file 'inputfile' with resolution increased by factor f in both dimensions. I have generated a small sample of numbers with this algorithm in python: x=123492981749283749834750984327 for j in range(20): x=(1664525x+1013904223)%(2E32) print x/2E32 ... a residue is a primitive root if and only if it is not a quadratic residue. Shub Generator - - Concept Quadratic residues Let p be an odd prime and a be an integer a is a quadratic residue modulo p if a is not congruent to 0 modif a is not congruent to 0 mod p and there exists an integer x such that a â¡ x. Budap. linear congruential generator ç¨åºæºä»£ç åä¸è½½é¾æ¥ã The linear congruential method is discussed in detail in Donald Knuth's "The art of computer programming", Volume 2 "Seminumerical Algorithms" 2nd ed. Generate a Random Number. A. maths, radix2_fft. These functions are embedded within the random module of Python. The book also includes a collection of questions that may appear in a â¦ Seed: a: b: n: Join our newsletter for the latest updates. Quadratic modular equation solver. The convergence of Monte Carlo integration is $$\mathcal{0}(n^{1/2})$$ and independent of the dimensionality. Convert Celsius To Fahrenheit. Python Program to Generate a Random Number In this example, you will learn to generate a random number in Python. Linear Congruential Method Codes and Scripts Downloads Free. Comput. Rolando EÅtvÅs, Sect. Linear congruential generator Description: Another simple pseudo-random number generator used for the rand() function in glibc. â¦d lists () Create merge_two_lists.py that implements merging of two sorted linked lists * Update merge_two_lists.py Fixed formatting errors * Fixed trailing whitespace * Change name of function to def __str__() * updating DIRECTORY.md * Imported classes from singly_linked_list.py * Update merge_two_lists.py * Update merge_two_lists.py Co-authored-by: github-actions <${GITHUBâ¦ Solve Quadratic Equation.$\begingroup\$ well written code. Alpertron. Sci. other, nested_brackets. Analytics cookies. ... other, linear_congruential_generator. Simply seed the random number generator with a fixed value, e.g. Digits per group . other, magicdiamondpattern. Quadratic Congruential Generator I (QCG-I) Quadratic Congruential Generator II (QCG-II) Cubic Congruential Generator II (CCG) ï¼åçå¸®å©ææ¡£ 116 é¡µ. That tighten the understanding of the root in an extension field. Quadratic modular equation solver. The generation of random numbers plays a large role in many applications ranging from cryptography to Monte Carlo methods. ... maths, quadratic_equations_complex_numbers. The convergence of Monte Carlo integration is $$\mathcal{0}(n^{1/2})$$ and independent of the dimensionality. GPA = 2 Since your GPA = 2, then your total letter grade on a 4.0 scale is C Programs. Solves a common algebraic Riccati equation using Schur decomposition. This is one of the 100+ free recipes of the IPython Cookbook, Second Edition, by Cyrille Rossant, a guide to numerical computing and data science in the Jupyter Notebook.The ebook and printed book are available for purchase at Packt Publishing. c . q, then x (mod n) corresponds uniquely to the pair of integers x (mod p) and x (mod q). Linear Congruential Generator Calculator. Actually there are several more efficient ways to find factors of big numbers (for smaller ones trial division works reasonably well). 32, 151-176 (2010). n . Linear congruential generator A linear congruential generator (LCG) was rst proposed in 1949 by D. H. Lehmer [ 14 ]. Convert Kilometers to Miles. numpy.random.seed(42) This way, you'll always get the same random number sequence. Hence Monte Carlo integration gnereally beats numerical intergration for moderate- and high-dimensional integration since numerical integration (quadrature) converges as $$\mathcal{0}(n^{d})$$.Even for low dimensional problems, Monte Carlo integration may have an â¦ So there don't need to exist a complex number which can solve this but it behaves like this (+-*conj(x))). maths, radians. Solving equations and inequalities. Rosetta Code is a programming chrestomathy site. Online Pseudo Random Number Generator This online tool generates pseudo random numbers based on the selected algorithm. Join. maths, relu. Exclusive OR Generator (XORG) Modular Exponentiation Generator (MODEXPG) Secure Hash Generator (G-SHA1) ï¼åçå¸®å©ææ¡£ 117 é¡µ. random.shuffle (x [, random]) ¶ Shuffle the sequence x in place.. Why my linear congruential generator generate low quality random numbers? Swap Two Variables. A pseudorandom number generator (PRNG), also known as a deterministic random bit generator â¦ One method which is very fast if the input number has two factors very close to its square root is known as Fermat factorisation .It makes use of the identity N = (a + b)(a - b) = a^2 - b^2 and is easy to understand and implement. b . Python 2.0 was released on 16 October 2000 with many major new features, including a cycle-detecting garbage collector and support for Unicode. Linear congruential generators are one of the oldest and most well-known methods for generating random â¦ 22 mod p . Random Number Generator in Python are built-in functions that help you generate numbers as and when required. Roots of a quadratic function; Roots of unity; Rosetta Code/Rank languages by popularity; Rot-13; RPG attributes generator; RSA code; Rule30; Run as a daemon or service; Run-length encoding; Runge-Kutta method The optional argument random is a 0-argument function returning a random float in [0.0, 1.0); by default, this is the function random().. To shuffle an immutable sequence and return a new shuffled list, use sample(x, k=len(x)) instead. 1980. Take a look at the following table that consists of some important random number generator functions along with their description present in the random module: Univ. Hence Monte Carlo integration gnereally beats numerical intergration for moderate- and high-dimensional integration since numerical integration (quadrature) converges as $$\mathcal{0}(n^{d})$$.Even for low dimensional problems, Monte Carlo integration may have an â¦ The algorithm II.1 is given by formula: x n +1 = ( ax n + c) mod m; n > 0; where m is the mask or the modulus m > 1, a is the multiplier (0 6 a < m ), c is the increment (0 6 c < m ), x 0 is the seed or initial value. On inversive congruential generator with a variable shift for pseudorandom numbers with prime power modulus, Ann. appear random. This Web application can solve equations of the form aâ¢x² + bx + c â¡ 0 (mod n) where the integer unknown x is in the range 0 â¤ x < n. 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Same random number generator Description: Another simple pseudo-random number generator in Python are built-in that...	2021-07-31 13:08:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46237292885780334, "perplexity": 1724.8856927093786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00380.warc.gz"}
https://math.stackexchange.com/questions/2488193/brouwer-fixed-point-theorem-on-a-function	# Brouwer fixed-point theorem on a function For $f:\mathbb R^n \rightarrow \mathbb R^n \quad \exists R \gt 0$ that $f$ is continous on $\bar B_R(0):= \{ x \in \mathbb R^n : \Vert x\Vert_2 \leq R \} \subset \mathbb R^n$ . Let also $\langle f(x),x \rangle \geq 0 \quad \forall x \in \partial\bar B_R(0)$ Show that it exists a $x^* \in \bar B_R(0)$ so that $f(x^*) = 0$. I tried to check the function $F:\bar B_R(0) \rightarrow \mathbb R^n , \quad F(x):= -R \frac{f(x)}{\Vert f(x)\Vert_2}$ with the Brouwer fixed-point theorem but didnt get far. Any help is appreciated thanks. Suppose by contradiction that $f(x)\neq0$ for all $x\in \bar{B}_R(0)$. Then your function $F:\bar{B}_R(0)\to\mathbb{R}^n$ is well defined and we have that $\|F(x)\|=R$ for all $x\in \bar{B}_R(0)$, hence we can write that $F:\bar{B}_R(0)\to\partial \bar{B}_R(0)\subset\bar{B}_R(0)$. Also $F$ is continuous, so by the Brower fixed point theorem there exists $x_0\in\bar{B}_R(0)$ such that $F(x_0)=x_0$. Moreover $\|F(x_0)\|=\|x_0\|=R$ implies that $x_0\in \partial \bar{B}_R(0)$. Now we have: $$0<R^2=\langle x_0,x_0\rangle=\langle F(x_0),x_0\rangle=-\frac{R}{\|f(x_0)\|}\langle f(x_0),x_0\rangle\le0,$$	2019-09-16 21:21:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9922632575035095, "perplexity": 46.98277985973395}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00557.warc.gz"}
https://cran.ma.imperial.ac.uk/web/packages/AnglerCreelSurveySimulation/vignettes/creel_survey_simulation.html	# Introduction Creel surveys allow fisheries scientists and managers to collect data on catch and harvest, an angler popuation (including effort expended), and, depending on survey design, biological data on fish populations. Though important methods of collecting data on the user base of the fishery, creel surveys are difficult to implement and, in graduate fisheries programs, creel surveys are paid little attention. As a result, fisheries managers–the first job for many fisheries-program graduates–often inherit old surveys or are told to institute new surveys with little knowledge of how to do so. Fisheries can cover large spatial extents: large reservoirs, coast-lines, and river systems. A creel survey has to be statistically valid, adaptable to the geographic challenges of the fishery, and cost efficient. Limited budgets can prevent agencies from implementing creel surveys; the AnglerCreelSurveySimulation was designed to help managers explore the type of creel survey that is most appropriate for their fishery, including fisheries with multiple access points, access points that are more popular than others, variation in catch rate, the number of surveyors, and seasonal variation in day-lengths. The AnglerCreelSurveySimulation package does require that users know something about their fishery and the human dimensions of that fishery. A prior knowledge includes mean trip length for a party (or individual), the mean catch rate of the The AnglerCreelSurveySimulation package is simple, but powerful. Four functions provide the means for users to create a population of anglers, limit the length of the fishing day to any value, and provide a mean trip length for the population. Ultimately, the user only needs to know the final function ConductMultipleSurveys but because I’d rather this not be a black box of functions, this brief introduction will be a step-by-step process through the package. ## A walk-through of the package This tutorial assumes that we have a very simple, small fishery with only one access point that, on any given day, is visited by 100 anglers. The fishing day length for our theoretical fishery is 12 hours (say, from 6 am to 6pm) and all anglers are required to have completed their trip by 6pm. Lastly, the mean trip length is known to be 3.5 hours. For the purposes of this package, all times are functions of the fishing day. In other words, if a fishing day length is 12 hours (e.g., from 6 am to 6pm) and an angler starts their trip at 2 and ends at 4 that means that they started their trip at 8 am and ended at 10 am. The make_anglers() function builds a population of anglers: library(AnglerCreelSurveySimulation) anglers <- make_anglers(n_anglers = 100, mean_trip_length = 3.5, fishing_day_length = 12) make_anglers() returns a dataframe with start_time, trip_length, and departure_time for all anglers. #> start_time trip_length departure_time #> 1 3.597536 6.0048773 9.602413 #> 2 3.438553 5.2290034 8.667556 #> 3 7.106046 0.7784606 7.884507 #> 4 8.747507 1.0715760 9.819083 #> 5 8.735652 1.6051185 10.340771 #> 6 1.881686 2.7515846 4.633270 In the head(anglers) statement, you can see that starttime, triplength, and departureTime are all available for each angler. The first angler started their trip roughly 3.6 hours into the fishing day, continued to fish for 6 hours, and left the access point at 9.6 hours into the fishing day. Angler start times are assigned by the uniform distribution and trip lengths are assigned by the gamma distribution. To get true effort of all the anglers for this angler population, summing trip_length is all that’s needed: 0. The distribution of angler trip lengths can be easily visualized: library(dplyr) #> #> Attaching package: 'dplyr' #> The following objects are masked from 'package:stats': #> #> filter, lag #> The following objects are masked from 'package:base': #> #> intersect, setdiff, setequal, union library(ggplot2) #> Want to understand how all the pieces fit together? Buy the #> ggplot2 book: http://ggplot2.org/book/ # Histogram overlaid with kernel density curve anglers %>% ggplot(aes(x=trip_length)) + geom_histogram(aes(y=..density..), binwidth=.1, colour="black", fill="white") + geom_density(alpha=.2, fill="#FF6666") Once the population of anglers has been created, the next function to apply is the get_total_values() function. In get_total_values(), the user specifies the start time of the creel surveyor, the end time of the surveyor, and the wait time of the surveyor. Here is where the user also specifies the sampling probability of the anglers (in most cases, equal to $$\frac{waitTime}{fishingDayLength}$$) and the mean catch rate of the fishery. There are a number of a default settings in the get_total_values() function; see ?get_total_values for a description of how the function handles NULL values for startTime, endTime, and waitTime. startTime and waitTime are the times that the surveyor started and waited at the access point. totalCatch and trueEffort are the total (or real) values for catch and effort. meanLambda is the mean catch rate for all anglers. Even though we assigned meanCatchRate to get_total_values(), individual mean catch rates are simulated by rgamma() with shape equal to meanCatchRate and rate equal to 1. For this walk through, we’ll schedule the surveyor to work for a total of eight hours at the sole access point in our fishery: anglers %>% get_total_values(start_time = 0, wait_time = 8, sampling_prob = 8/12, mean_catch_rate = 2.5) #> n_observed_trips total_observed_trip_effort n_completed_trips #> 1 86 586.0279 50 #> total_completed_trip_effort total_completed_trip_catch start_time #> 1 222.9294 530.1671 0 #> wait_time total_catch true_effort mean_lambda #> 1 8 795.6116 417.0699 2.39971 get_total_values() returns a single row data frame with several columns. The output of get_total_values() is the catch and effort data observed by the surveyor during their wait at the accss point along with the “true” values for catch and effort. (Obviously, we can’t simulate biological data but, if an agency’s protocol directed the surveyor to collect biological data, that could be analyzed with other R functions.) In the output from get_total_values(), n_observed_trips is the number of trips that the surveyor observed, including anglers that arrived after she started her day and anglers that were there for the duration of her time at the access point. total_observed_trip_effort is the effort expended by those parties; because the observed trips were not complete, she did not count their catch. n_completed_trips is the number of anglers that completed their trips while she was onsite, total_completed_trip_effort is the effort expended by those anglers, and total_completed_trip_catch is the number of fish caught by those parties. Catch is both the number of fish harvested and those caught and released. ### Estimating catch and effort Effort and catch are estimated from the Bus Route Estimator: $\widehat{E} = T\sum\limits_{i=1}^n{\frac{1}{w_{i}}}\sum\limits_{j=1}^m{\frac{e_{ij}}{\pi_{j}}}$ where • E = estimated total party-hours of effort; • T = total time to complete a full circuit of the route, including travelling and waiting; • wi = waiting time at the ith site (where i = 1, …, n sites); and • eij = total time that the jth car (or trailer) is parked at the ith site while the agent is at that shite (where j = 1, …, n sites). Catch rate is calculated from the Ratio of Means equation: $\widehat{R_1} = \frac{\sum\limits_{i=1}^n{c_i/n}}{\sum\limits_{i=1}^n{L_i/n}}$ where • ci is the catch for the ith sampling unit and * Li is the length of the fishing trip at the tie of the interview. For incomplete surveys, Li represents an incomplete trip. simulate_bus_route() calculates effort and catch based upon these equations. See ?simulate_bus_route for references that include a more detailed discussion of these equations. simulate_bus_route() calls make_anglers() and get_total_values() so many of the same arguments we passed in the previous functions will need to be passed to simulate_bus_route(). The new arguent, nsites, is the number of sites visited by the surveyor. In more advanced simulations (see the examples in ?simulate_bus_route), you can pass strings of values for startTime, waitTime, nsites, and nanglers to simulate a bus route-type survey rather than just a single access-point survey. sim <- simulate_bus_route(start_time = 0, wait_time = 8, n_sites = 1, n_anglers = 100, sampling_prob = 8/12, mean_catch_rate = 2.5, fishing_day_length = 12) sim #> Ehat catch_rate_ROM true_catch true_effort mean_lambda #> 1 585.9488 2.607587 866.5934 403.0374 2.481576 The output from simulate_bus_route() is a dataframe with values for Ehat, catchRateROM (the ratio of means catch rate), trueCatch, trueEffort, and meanLambda. Ehat is the estimated total effort from the Bus Route Estimator above and catchRateROM is catch rate estimated from the Ratio of Means equation. trueCatch, trueEffort, and meanLambda are the same as before. Multiplying Ehat by catchRateROM gives an estimate of total catch: 1527.9124526. ### Conducting multiple simulations With information about the fishery, the start and wait times of the surveyor, the sampling probability, mean catch rate, and fishing day length, we can run multiple simulations with conduct_multiple_surveys(). conduct_multiple_surveys() is a wrapper that calls the other three functions in turn and compiles the values into a data frame for easy plotting or analysis. The only additional argument needed is the nsims value which tells the function how many simulations to conduct. For the sake of this simple simulation, let’s assume that the creel survery works five days a week for four weeks (i.e. 20 days): sim <- conduct_multiple_surveys(n_sims = 20, start_time = 0, wait_time = 8, n_sites = 1, n_anglers = 100, sampling_prob = 8/12, mean_catch_rate = 2.5, fishing_day_length = 12) sim #> Ehat catch_rate_ROM true_catch true_effort mean_lambda #> 1 582.5113 2.197079 744.9890 408.1091 2.238827 #> 2 525.6581 2.610515 962.0484 371.0789 2.788176 #> 3 532.4813 2.593379 774.0897 376.5876 2.364251 #> 4 636.9765 2.069540 776.0824 444.2898 2.275464 #> 5 551.9909 2.255368 816.2178 392.7341 2.518792 #> 6 666.7207 2.161836 799.9225 461.0866 2.379170 #> 7 571.7152 2.399565 817.8163 402.0583 2.547475 #> 8 621.4893 2.069414 731.0183 441.1592 2.134023 #> 9 608.1691 2.328550 854.8820 426.5282 2.305846 #> 10 531.5312 2.260759 764.4231 380.1294 2.258321 #> 11 600.5033 2.910176 844.1526 412.9629 2.757026 #> 12 583.2813 2.542058 836.5646 401.1198 2.613398 #> 13 621.7530 2.526983 886.7834 422.6602 2.482501 #> 14 614.4349 2.658935 782.2079 422.5720 2.459972 #> 15 593.8837 2.748767 936.2094 411.7710 2.764624 #> 16 579.6193 2.344677 833.4941 410.2437 2.377972 #> 17 610.3543 2.395208 911.6421 429.0311 2.658027 #> 18 557.2910 2.300684 862.2922 396.4331 2.498155 #> 19 566.0941 2.073266 729.5827 396.8897 2.138644 #> 20 608.2846 2.128812 731.1389 432.0415 1.974324 With the output from multiple simulations, an analyst can evaluate how closely the creel survey they’ve designed mirrors reality. A lm() of estimated catch as a function of trueCatch can tell us if the survey will over or under estimate reality: mod <- sim %>% lm((Ehat * catch_rate_ROM) ~ true_catch, data = .) summary(mod) #> #> Call: #> lm(formula = (Ehat * catch_rate_ROM) ~ true_catch, data = .) #> #> Residuals: #> Min 1Q Median 3Q Max #> -185.54 -71.15 -22.28 66.09 322.50 #> #> Coefficients: #> Estimate Std. Error t value Pr(>\|t\|) #> (Intercept) 474.8766 376.0562 1.263 0.2228 #> true_catch 1.1256 0.4572 2.462 0.0241 * #> --- #> Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 136.5 on 18 degrees of freedom #> Multiple R-squared: 0.2519, Adjusted R-squared: 0.2103 #> F-statistic: 6.061 on 1 and 18 DF, p-value: 0.02414 Plotting the data and the model provide a good visual means of evaluating how close our estimates are to reality: #Create a new vector of the estimated effort multiplied by estimated catch rate sim <- sim %>% mutate(est_catch = Ehat catch_rate_ROM) sim %>% ggplot(aes(x = true_catch, y = est_catch)) + geom_point() + geom_abline(intercept = mod$coefficients[1], slope = mod$coefficients[2], colour = "red", size = 1.01) The closer the slope parameter estimate is to 1 and the intercept parameter estimate is to 0, the closer our estimate of catch is to reality. We can create a model and plot of our effort estimates, too: mod <- sim %>% lm(Ehat ~ true_effort, data = .) summary(mod) #> #> Call: #> lm(formula = Ehat ~ true_effort, data = .) #> #> Residuals: #> Min 1Q Median 3Q Max #> -12.0376 -6.0464 -0.9879 3.1929 16.9332 #> #> Coefficients: #> Estimate Std. Error t value Pr(>\|t\|) #> (Intercept) -51.07503 32.32989 -1.58 0.132 #> true_effort 1.55183 0.07835 19.80 1.14e-13 * #> --- #> Signif. codes: 0 '' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 8.011 on 18 degrees of freedom #> Multiple R-squared: 0.9561, Adjusted R-squared: 0.9537 #> F-statistic: 392.2 on 1 and 18 DF, p-value: 1.141e-13 #Create a new vector of the estimated effort multiplied by estimated catch rate sim %>% ggplot(aes(x = true_effort, y = Ehat)) + geom_point() + geom_abline(intercept = mod$coefficients[1], slope = mod$coefficients[2], colour = "red", size = 1.01) ### Can we observe ALL trips? If the start and wait time equals 0 and the length of the fishing day, respectively, the creel surveyor can observe all completed trips, though she’d likely be unhappy having to work 12 hours. The inputs have to be adjusted to allow her to arrive at time 0, stay for all 12 hours, and have a probability of 1.0 at catching everyone: start_time <- 0 wait_time <- 12 sampling_prob <- 1 sim <- conduct_multiple_surveys(n_sims = 20, start_time = start_time, wait_time = wait_time, n_sites = 1, n_anglers = 100, sampling_prob = 1, mean_catch_rate = 2.5, fishing_day_length = 12) sim #> Ehat catch_rate_ROM true_catch true_effort mean_lambda #> 1 768.7004 2.470043 824.7229 768.7004 2.511069 #> 2 708.7170 2.599554 806.7401 708.7170 2.495052 #> 3 759.9584 2.416612 854.5432 759.9584 2.358045 #> 4 783.3415 2.715485 892.3400 783.3415 2.663605 #> 5 728.9695 2.126378 691.4138 728.9695 2.128315 #> 6 724.7953 2.183654 724.3334 724.7953 2.273502 #> 7 695.8944 2.599302 844.9681 695.8944 2.614189 #> 8 789.6448 2.410761 814.9992 789.6448 2.467584 #> 9 781.7185 2.672259 876.0535 781.7185 2.706500 #> 10 753.8212 2.370904 822.9531 753.8212 2.410414 #> 11 814.7701 2.519984 879.0222 814.7701 2.583332 #> 12 811.6684 2.305741 759.6655 811.6684 2.385969 #> 13 751.8314 2.374681 773.6552 751.8314 2.429269 #> 14 753.1468 2.419120 836.7430 753.1468 2.436932 #> 15 776.5254 2.162192 817.4036 776.5254 2.237419 #> 16 738.9103 2.666906 885.2465 738.9103 2.609013 #> 17 784.0078 2.284436 807.6391 784.0078 2.316637 #> 18 784.3040 2.603432 874.8825 784.3040 2.618557 #> 19 802.6489 2.429314 860.2270 802.6489 2.333058 #> 20 774.0712 2.567579 905.7421 774.0712 2.469119 #> Warning in summary.lm(mod): essentially perfect fit: summary may be #> unreliable #> #> Call: #> lm(formula = Ehat ~ true_effort, data = .) #> #> Residuals: #> Min 1Q Median 3Q Max #> -1.008e-13 -9.010e-16 1.612e-14 2.442e-14 1.043e-13 #> #> Coefficients: #> Estimate Std. Error t value Pr(>\|t\|) #> (Intercept) 2.034e-13 2.840e-13 7.160e-01 0.483 #> true_effort 1.000e+00 3.712e-16 2.694e+15 <2e-16 * #> --- #> Signif. codes: 0 '' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 5.282e-14 on 18 degrees of freedom #> Multiple R-squared: 1, Adjusted R-squared: 1 #> F-statistic: 7.258e+30 on 1 and 18 DF, p-value: < 2.2e-16 ### Another simulation If our hypothetical fishery suddenly gained another access point and the original 100 anglers were split between the two access points equally, what kind of information would a creel survey capture? We could ask our surveyor to split her eight-hour work day between both access points, but she’ll have to drive for 0.5 hours to get from one to another. Of course, that 0.5 hour of drive time will be a part of her work day so she’ll effectively have 7.5 hours to spend at access points counting anglers and collecting data. start_time <- c(0, 4.5) wait_time <- c(4, 3.5) n_sites = 2 n_anglers <- c(50, 50) fishing_day_length <- 12 sampling_prob <- sum(wait_time)/fishing_day_length sim <- conduct_multiple_surveys(n_sims = 20, start_time = start_time, wait_time = wait_time, n_sites = n_sites, n_anglers = n_anglers, sampling_prob = sampling_prob, mean_catch_rate = 2.5, fishing_day_length = fishing_day_length) sim #> Ehat catch_rate_ROM true_catch true_effort mean_lambda #> 1 336.3974 2.446137 842.4591 194.2340 2.501437 #> 2 459.4682 2.461493 794.7933 216.4546 2.514877 #> 3 363.0755 2.157886 760.6865 233.0647 2.359580 #> 4 422.7744 2.310826 787.1008 223.0891 2.374984 #> 5 413.9636 2.549926 906.0712 222.9362 2.577263 #> 6 409.6488 2.585385 743.7690 217.4829 2.351497 #> 7 429.4646 2.806555 889.1171 245.4681 2.555515 #> 8 437.3585 2.494798 869.9422 237.6559 2.438136 #> 9 418.4751 3.389806 927.2633 225.8867 2.674165 #> 10 435.0700 2.468669 836.1957 246.1885 2.289779 #> 11 508.9174 2.480358 840.6856 247.3459 2.325625 #> 12 401.5740 2.808633 922.8538 218.0649 2.880003 #> 13 380.6494 3.202355 953.2290 230.1477 2.686220 #> 14 411.1600 2.460808 957.5743 214.7786 2.713318 #> 15 438.6053 2.880876 742.9910 224.8079 2.279269 #> 16 501.4253 1.962584 753.3787 231.2408 2.422737 #> 17 482.2694 2.375788 803.4919 234.8892 2.316639 #> 18 502.1387 2.132380 843.9224 234.5807 2.393269 #> 19 420.0203 2.400470 734.3253 223.6434 2.292050 #> 20 492.1539 2.721672 858.8633 215.6486 2.493468 #> #> Call: #> lm(formula = Ehat ~ true_effort, data = .) #> #> Residuals: #> Min 1Q Median 3Q Max #> -80.775 -20.218 -7.547 36.599 78.212 #> #> Coefficients: #> Estimate Std. Error t value Pr(>\|t\|) #> (Intercept) 43.6023 173.5401 0.251 0.8045 #> true_effort 1.7173 0.7637 2.249 0.0373 * #> --- #> Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 42.55 on 18 degrees of freedom #> Multiple R-squared: 0.2193, Adjusted R-squared: 0.1759 #> F-statistic: 5.056 on 1 and 18 DF, p-value: 0.0373 ### Even more simulations Ultimately, the creel survey simulation can be as complicated as a creel survey. If a survey requires multiple clerks, several simulations can be coupled together to act as multiple surveryors. To accomodate weekends or holidays (i.e., increased angler pressure), additional simulations with different wait times and more anglers (to simulate higher pressure) can be built into the simulation. For example, if we know that angler pressure is 50% higher at the two access points on weekends, we can hire a second clerk to sample 8 hours a day on the weekends–one day at each access point–and add the weekend data to the weekday data. #Weekend clerks start_time_w <- 2 wait_time_w <- 10 n_sites <- 1 n_anglers_w <- 75 fishing_day_length <- 12 sampling_prob <- 8/12 sim_w <- conduct_multiple_surveys(n_sims = 8, start_time = start_time_w, wait_time = wait_time_w, n_sites = n_sites, n_anglers = n_anglers_w, sampling_prob = sampling_prob, mean_catch_rate = 2.5, fishing_day_length = fishing_day_length) sim_w #> Ehat catch_rate_ROM true_catch true_effort mean_lambda #> 1 654.4297 2.405069 636.0419 437.8568 2.514499 #> 2 635.3440 2.498353 602.0437 423.5627 2.453219 #> 3 711.8827 2.314211 618.1801 474.5885 2.379199 #> 4 596.5517 2.349445 590.9965 398.3901 2.403765 #> 5 683.4622 2.632500 697.3532 457.2936 2.603457 #> 6 649.8841 2.303446 573.7912 434.3805 2.387601 #> 7 589.8644 2.657115 752.9906 394.7195 2.607380 #> 8 647.9023 2.553308 651.7656 431.9349 2.489279 #Add the weekday survey and weekend surveys to the same data frame mon_survey <- sim_w %>% bind_rows(sim) mod <- mon_survey %>% lm(Ehat ~ true_effort, data = .) summary(mod) #> #> Call: #> lm(formula = Ehat ~ true_effort, data = .) #> #> Residuals: #> Min 1Q Median 3Q Max #> -75.643 -17.224 -5.803 10.862 71.865 #> #> Coefficients: #> Estimate Std. Error t value Pr(>\|t\|) #> (Intercept) 192.08255 22.16532 8.666 3.82e-09 #> true_effort 1.05823 0.07377 14.345 7.29e-14 * #> --- #> Signif. codes: 0 '*' 0.001 '' 0.01 '*' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 36.71 on 26 degrees of freedom #> Multiple R-squared: 0.8878, Adjusted R-squared: 0.8835 #> F-statistic: 205.8 on 1 and 26 DF, p-value: 7.291e-14 Hopefully, this vignette has shown you how to build and simulate your own creel survey. It’s flexible enough to estimate monthly or seasonal changes in fishing day length, changes in the mean catch rate, increased angler pressure on weekends, and any number of access sites, start times, wait times, and sampling probabilities. The output from conduct_multiple_surveys() allows the user to estiate variability in the catch and effort estimates (e.g., relative standard error) to evaluate the most efficient creel survey for their fishery.	2022-05-24 21:58:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47496411204338074, "perplexity": 8402.524864514615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00148.warc.gz"}
https://mail.haskell.org/pipermail/haskell-cafe/2008-February/039038.html	Ryan Ingram ryani.spam at gmail.com Sat Feb 2 14:57:57 EST 2008 You can also do something like the following: newtype StateST st s a = StateST { internalRunStateST :: ReaderT (STRef st s) (ST st) a } instance MonadState s (StateST s st) where put s = ask >>= \ref -> writeSTRef ref s runStateST :: StateST st s a -> s -> ST st a runStateST m s = do ref <- newSTRef s -- ryan On Feb 2, 2008 9:05 AM, Derek Elkins <derek.a.elkins at gmail.com> wrote: > On Sat, 2008-02-02 at 12:33 -0500, Denis Bueno wrote: > > Is it possible to use the ST monad as a (drop-in) replacement for the > > State monad in the following situation? If not, is there a "best > > practice" for refactoring? > > > > I have a bunch of functions that return state actions: > > > > type MyState = ... > > > > foo1 :: T1 -> State MyState a > > foo2 :: T2 -> State MyState a > > ... > > foon :: Tn -> State MyState a > > > > And I'd like to refactor this to use the ST monad, mechanically, if > > possible. All uses of the MyState inside State are single-threaded. > > > > In my application, MyState is a record with 5 or so fields. One of > > those fields uses a list to keep track of some information, and I'd > > like to change that to STUArray, because it changes my bottleneck > > operations from O(n) to O(1). This, of course, requires having the ST > > monad around, in order to achieve the proper time complexity. > > > > Is there an easy way to do this? In the future, should I start out > > with the ST monad if I suspect I'll need to use an imperative data > > structure for efficiency reasons? I started out with State because > > I'm modeling a transition system, so it seemed natural. > > > > Any advice is appreciated. > > > type MyState s = ... s ... > > type MyMonad s = StateT (MyState s) (ST s) > > > > _______________________________________________	2016-09-28 02:01:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4979795217514038, "perplexity": 10768.518726279013}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661289.57/warc/CC-MAIN-20160924173741-00247-ip-10-143-35-109.ec2.internal.warc.gz"}
http://albanyareamathcircle.blogspot.com/2009/08/nyc-math-circle-facebook-page-asked.html	## Sunday, August 9, 2009 ### Pentablocks and beautiful geometric puzzles What is the ratio of the blue area to the area of the whole regular star in this picture? This problem has a very beautiful answer, but it's hard for most people to visualize, unless they have a lot of geometric intuition. You might want to try playing with it for a while before visiting the discussion in the NYC Math Circle Facebook page. The Pentablocks I showed to the New York Math Circle Teachers summer workshop last week provide a very easy way to visually construct and/or confirm the solution to the problem above. Pentablocks are a very nice way to develop your geometric intuition in understanding the relationships in pentagons, pentagrams, and related polygons, the golden ratio (phi), Penrose tilings (quasicrystals), and more. As their name suggests, Pentablocks are based on pentagons and pentagrams (five-pointed stars). All their angles are multiples of 18 degrees, and all their side-lengths are related to one another by the golden ratio, phi. The more traditional (and easier to find) Pattern Blocks are based on hexagons. All their components have angles which are multiples of 15 degrees, and all their dimensions are related to one another by 1, 2, and the square root of 3. You can use either set to tessellate, but the Pentablock tessellations are far more complex than the traditional pattern block tessellations. I'll bring sets of both types of blocks to our Labor Day weekend picnic so you can explore more relationships with them. #### 1 comment: mskmoorthy said... Here are three puzzles for our middle school students (I did not check the literature to see whether these puzzles have appeared in the past - last question was inspired while reading Euclid's Elements) 1) Divide a triangle into two triangles of equal area (using compass and straight edge). 2) Divide a triangle into two triangles of areas p/q and 1-p/q (assuming that the original area of the triangle is 1). 3) Divide a triangle into two triangles whose ratio of areas is a golden ratio.	2017-06-26 17:13:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35582563281059265, "perplexity": 690.8606818689725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320841.35/warc/CC-MAIN-20170626170406-20170626190406-00249.warc.gz"}
https://harangdev.github.io/deep-learning/recurrent-neural-networks/45/	# Attention Model Categories: Updated: https://arxiv.org/abs/1409.0473 In sequence to sequence model, we first encoded the input sequence and decoded it. For machine translation task, it is like memorizing the whole sentence before translating it. But actually it is more natural to read some part of the sentence, translate it, read next part, translate, and so on. Attention model was invented with this intuition. We put ‘decoding’ RNN on top of the ‘encoding’ RNN. Activation $s^{<t>}$ in the decoding RNN is inputted with $x^{<t>}$ and $c^{<t>}$. This $c^{<t>}$ is computed with attention weight $\alpha^{<t,t’>}$ . Attention weight $\alpha^{<t,t’>}$ is the amount of attention $\hat{y}^{<t>}$ should pay to $a^{<t>}$ . I’ll give precise formulas of attention model works. ### 2. With activation from the encoding BRNN and previous state of decoding RNN, compute $e$ $W_e$ and $b_e$ are a trainable parameter so will be optimized by gradient descent. ### 5. Compute normal RNN layer where $\hat{y}_{class}^{<t-1>}$ is a predicted class one-hot representation of $\hat{y}^{<t-1>}$ Categories:	2020-04-08 09:58:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8894515633583069, "perplexity": 2076.881426550043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00451.warc.gz"}
https://infoscience.epfl.ch/record/225315	Infoscience Thesis # Interaction of atmospheric boundary layer flow with wind turbines: theoretical and experimental studies As wind turbines operate within the atmospheric boundary layer (ABL), the study of their interaction with the ABL flow can help us better understand and predict their performance. In addition to the performance of wind turbines, this interaction has an effect on the flow both upwind and downwind (i.e., wake region) of the turbines. In particular, the study of turbine wakes is of great importance because they are the main cause of power losses and fatigue loads in wind farms. In the current thesis, four studies are conducted to fully examine the turbine interaction with the ABL flow, with an emphasis on turbine wakes. In the first study, a new analytical wake model is proposed and validated to predict the wind velocity distribution in the far-wake region, where downwind turbines usually operate. The proposed model is derived by applying the conservation of mass and momentum and assuming a Gaussian distribution for the velocity deficit in the wake. This simple model only requires one parameter to predict the velocity distribution in the far wake of a wind turbine. In general, it is found that the velocity deficit in the wake predicted by the proposed analytical model is in good agreement with the experimental and numerical data. Furthermore, the results show that the new model predicts the power extracted by downwind wind turbines more accurately than other common analytical models, some of which are based on less accurate assumptions like considering a top-hat shape for the velocity deficit. In the second study, wind tunnel measurements are carried out to systematically investigate turbine wakes under yawed conditions. The detailed experimental data are used to perform a budget study of the continuity and Reynolds-averaged Navier-Stokes equations. This theoretical analysis reveals some notable features of the wakes of yawed turbines, such as the asymmetric distribution of the wake skew angle with respect to the wake center. Under highly yawed conditions, the formation of a counter-rotating vortex pair in the wake cross-section as well as the vertical displacement of the wake center are also shown and analyzed. Finally, this study enables us to develop general governing equations upon which a simple and computationally inexpensive analytical model is built. The proposed model aims at predicting the wake deflection and the far-wake velocity distribution for yawed turbines. The findings of this study can be especially useful to assess the possibility of optimizing wind-farm power production by controlling the yaw angle of the turbines. In the third study, comprehensive wind tunnel experiments are performed to study the interaction of a turbulent boundary layer with a wind turbine operating under different tip-speed ratios and yaw angles. Force and power measurements are performed to characterize the wind turbine performance. Moreover, a high-resolution stereoscopic particle-image velocimetry (S-PIV) system and hot-wire anemometry are used to study the flow in the upwind, near-wake and far-wake regions. This study provides new insights on the turbine and flow characteristics such as the evolution of tip vortices and wake meandering. Finally, the last study concerns the design and the performance analysis of a new three-bladed horizontal-axis miniature wind turbine with a rotor diameter of $15$ cm. Due to its small size, this turbine is particularly suitable for studies of wind farm flows and the interaction of the turbine with an incoming boundary-layer flow. Special emphasis is placed on accurate measurements of the mechanical power extracted by the miniature turbine from the incoming wind. In order to do so, a new setup is developed to measure the torque of the rotor shaft. The thrust and power coefficients of the miniature turbine are found to be around $0.8$ and $0.4$ in optimal conditions, respectively, which are close to the ones of large-scale turbines in the field. Thèse École polytechnique fédérale de Lausanne EPFL, n° 7529 (2017) Programme doctoral Mécanique Faculté de l'environnement naturel, architectural et construit Institut d'ingénierie de l'environnement Laboratoire d'ingénierie éolienne et d'énergie renouvelable Jury: Prof. François Gallaire (président) ; Prof. Fernando Porté Agel (directeur de thèse) ; Prof. Karen Mulleners, Prof. Jens Nørkær Sørensen, Prof. Charles Meneveau (rapporteurs) Public defense: 2017-2-10 #### Reference Record created on 2017-01-25, modified on 2017-05-15	2017-12-12 14:12:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3891852796077728, "perplexity": 1157.3361784685844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948517181.32/warc/CC-MAIN-20171212134318-20171212154318-00203.warc.gz"}
https://www.mysciencework.com/publication/show/b_s-0-bar-b_s-0-mixing-b-s-transitions-isosinglet-down-quark-model-2b2116a1	# $B_s^0 - \bar B_s^0$ mixing and $b \to s$ transitions in isosinglet down quark model Authors Type Published Article Publication Date Dec 05, 2008 Submission Date Dec 05, 2008 Identifiers DOI: 10.1103/PhysRevD.78.116002 Source arXiv The recent observation of the mass difference in $B_s$ system seems to be not in complete agreement with the corresponding standard model value. We consider the model with an extra vector like down quark to explain this discrepancy and obtain the constraints on the new physics parameters. Thereafter, we show that with these new constraints this model can successfully explain other observed deviations associated with $b \to s$ transitions, namely, $B_s \to \psi \phi$, $B\to K \pi$ and $B\to \phi K_s$.	2018-04-22 16:21:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7333216071128845, "perplexity": 936.1556508620009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945624.76/warc/CC-MAIN-20180422154522-20180422174522-00392.warc.gz"}
http://stackoverflow.com/questions/14878110/how-to-find-all-zeros-of-a-function-using-numpy-and-scipy	# How to find all zeros of a function using numpy (and scipy)? Suppose I have a function f(x) defined between a and b. This function can have many zeros, but also many asymptotes. I need to retrieve all the zeros of this function. What is the best way to do it? Actually, my strategy is the following: 1. I evaluate my function on a given number of points 2. I detect whether there is a change of sign 3. I find the zero between the points that are changing sign 4. I verify if the zero found is really a zero, or if this is an asymptote U = numpy.linspace(a, b, 100) # evaluate function at 100 different points c = f(U) s = numpy.sign(c) for i in range(100-1): if s[i] + s[i+1] == 0: # oposite signs u = scipy.optimize.brentq(f, U[i], U[i+1]) z = f(u) if numpy.isnan(z) or abs(z) > 1e-3: continue print('found zero at {}'.format(u)) This algorithm seems to work, except I see two potential problems: 1. It will not detect a zero that doesn't cross the x axis (for example, in a function like f(x) = x2) However, I don't think it can occur with the function I'm evaluating. 2. If the discretization points are too far, there could be more that one zero between them, and the algorithm could fail finding them. Do you have a better strategy (still efficient) to find all the zeros of a function? I don't think it's important for the question, but for those who are curious, I'm dealing with characteristic equations of wave propagation in optical fiber. The function looks like (where V and ell are previously defined, and ell is an positive integer): def f(u): w = numpy.sqrt(V2 - u*2) jl = scipy.special.jn(ell, u) jl1 = scipy.special.jnjn(ell-1, u) kl = scipy.special.jnkn(ell, w) kl1 = scipy.special.jnkn(ell-1, w) return jl / (ujl1) + kl / (wkl1) - as a math quibble, I wold have used 'divergence' where you use 'asymptote'. – tcaswell Feb 14 '13 at 15:26 Can you also provide a bit more detail about what kind of function you are dealing with? for example, this simply can't be done for sin(1/x) in the region around x=0. – tcaswell Feb 14 '13 at 15:30 @tcaswell I'm not sure what is the right word in English. What I mean is that f(x-) -> -inf and f(x+) -> inf – Charles Brunet Feb 14 '13 at 19:58 The main problem I see with this is if you can actually find all roots --- as have already been mentioned in comments, this is not always possible. If you are sure that your function is not completely pathological (sin(1/x) was already mentioned), the next one is what's your tolerance to missing a root or several of them. Put differently, it's about to what length you are prepared to go to make sure you did not miss any --- to the best of my knowledge, there is no general method to isolate all the roots for you, so you'll have to do it yourself. What you show is a reasonable first step already. A couple of comments: • Brent's method is indeed a good choice here. • First of all, deal with the divergencies. Since in your function you have Bessels in the denominators, you can first solve for their roots -- better look them up in e.g., Abramovitch and Stegun (Mathworld link). This will be a better than using an ad hoc grid you're using. • What you can do, once you've found two roots or divergencies, x_1 and x_2, run the search again in the interval [x_1+epsilon, x_2-epsilon]. Continue until no more roots are found (Brent's method is guaranteed to converge to a root, provided there is one). • If you cannot enumerate all the divergencies, you might want to be a little more careful in verifying a candidate is indeed a divergency: given x don't just check that f(x) is large, check that, e.g. \|f(x-epsilon/2)\| > \|f(x-epsilon)\| for several values of epsilon (1e-8, 1e-9, 1e-10, something like that). • If you want to make sure you don't have roots which simply touch zero, look for the extrema of the function, and for each extremum, x_e, check the value of f(x_e). - Why are you limited to numpy? Scipy has a package that does exactly what you want: http://docs.scipy.org/doc/scipy/reference/optimize.nonlin.html One lesson I've learned: numerical programming is hard, so don't do it :) Anyway, if you're dead set on building the algorithm yourself, the doc page on scipy I linked (takes forever to load, btw) gives you a list of algorithms to start with. One method that I've used before is to discretize the function to the degree that is necessary for your problem. (That is, tune \delta x so that it is much smaller than the characteristic size in your problem.) This lets you look for features of the function (like changes in sign). AND, you can compute the derivative of a line segment (probably since kindergarten) pretty easily, so your discretized function has a well-defined first derivative. Because you've tuned the dx to be smaller than the characteristic size, you're guaranteed not to miss any features of the function that are important for your problem. If you want to know what "characteristic size" means, look for some parameter of your function with units of length or 1/length. That is, for some function f(x), assume x has units of length and f has no units. Then look for the things that multiply x. For example, if you want to discretize cos(\pi x), the parameter that multiplies x (if x has units of length) must have units of 1/length. So the characteristic size of cos(\pi x) is 1/\pi. If you make your discretization much smaller than this, you won't have any issues. To be sure, this trick won't always work, so you may need to do some tinkering. - The reference you give is about a multidimentional solver. My problem is only one dimension. Why do you say I'm not using scipy? scipy.optimize.brentq is a scipy function... – Charles Brunet Feb 14 '13 at 19:56 Well, I don't know about the scipy dev community, but if I were writing a multi-dimensional solver, I'd make sure I could handle the N=1 case first. You're not using scipy because you're trying to find the roots by hand, instead of using their built in method. So, unless you MEAN to reinvent the wheel, I'd read their doc a bit closer, and see how you can leverage their work to solve your problem. – BenDundee Feb 14 '13 at 20:02 I don't agree. Zero finding functions like brentq need and interval [a,b] where sign(f(a)) != sign(f(b)) and where f is continuous. I'm not finding the root by hand, I'm just guessing the right interval to start searching. – Charles Brunet Feb 14 '13 at 20:07 I guess I must be missing something. How is your problem not exactly like Small Problem -> Examples in the doc I linked? Also, I think you're going to find it difficult to find a numerical algorithm that guarantees that you've found all the zeroes of a function on the range (-\infty, \infty). Are you implying that brentq can't find the roots of cos(x) between 0 and 2pi? That seems odd... – BenDundee Feb 14 '13 at 20:14 The Small Problem example still is a multidimentional problem. My problem is different because I need to find all the roots of my function, on a given interval. brentq would fail in the interval [0,2*pi] because the signs of the cos function are the same at those two points. You'd need to use [0,pi] interval instead. – Charles Brunet Feb 14 '13 at 20:43 show 2 more comments	2014-03-15 15:09:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7510589361190796, "perplexity": 784.2147525224642}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678697956/warc/CC-MAIN-20140313024457-00033-ip-10-183-142-35.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/3267660/expressing-the-inner-product-veca-cdot-vecb-in-terms-of-the-inner-product	# Expressing the inner product $\vec{a}\cdot\vec{b}$ in terms of the inner product $\vec{a}\cdot\vec{c}$ Let $$S$$ be a circle with it s center at point $$O$$ and a radius of $$1$$. Let $$\triangle$$ABC be a triangle such that all its vertices are on $$S$$ and $$AB:AC=3:2$$. As shown in the figure, let $$D$$ be a point on the extension of side $$BC$$ and $$k$$ be the number where $$BC:CD=2:k$$. Moreover, set $$\vec{OA}=\vec{a},$$ $$\vec{OB}=\vec{b},$$ $$\vec{OC}=\vec{c},$$ Since the equality $$\|\vec{b}-\vec{a}\|=\frac{3}{2}\|\vec{c}-\vec{a}\|$$ holds, by expressing the inner product $$\vec{a}\cdot\vec{b}$$ in terms of the inner product $$\vec{a}\cdot\vec{c}$$, we have $$\vec{a}\cdot\vec{b}=\frac{F}{G}\vec{a}\cdot\vec{c}-\frac{H}{I}$$ Find F, G, H and I. This is my scholarship exam practice assuming high school math knowledge. The answer key provided is 9, 4, 5 and 4. I do not know how to begin here, could you please give me a hint to start on this question? Hint: Expand $$\lvert\vec{b}-\vec{a}\rvert^2=\frac94\lvert\vec{c}-\vec{a}\rvert^2$$ and remember $$\vec{a}^2=\vec{b}^2=\vec{c}^2=1$$. The point $$D$$ plays no part here. • I think I would like some hint on my last question as well saying that: It follows that when the tangent to $S$ at the point $A$ passes through the point $D$, then $k=?$ – Trey Anupong Jun 19 at 15:09 • Oops, yes, it should be $\vec{a}$ perpendicular to $\overrightarrow{AD}$ so with $\overrightarrow{AD}=\vec{b}-\vec{a}+\frac{2+k}2(\vec{c}-\vec{b})$ we get $k$ by dotting with $\vec{a}$. – user10354138 Jun 19 at 16:19 • So we dot $\vec{a}$ with $\vec{AD}$ which is equal to $\vec{OD}$, am I correct? – Trey Anupong Jun 19 at 16:24 • $\overrightarrow{AD}=\overrightarrow{OD}-\vec{a}$. We know $\overrightarrow{AD}\cdot\vec{a}=0$ which gives us an equation for $k$ in terms of $\vec{a}\cdot\vec{b}$ and $\vec{a}\cdot\vec{c}$. – user10354138 Jun 19 at 16:26	2019-09-22 06:33:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7638967633247375, "perplexity": 95.99140561461597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00352.warc.gz"}
https://www.physicsforums.com/threads/mathematica-showing-iterations-graphically.569382/	# Mathematica - Showing iterations graphically 1. Jan 20, 2012 ### nothingbetter I'm trying to do this: Suppose I have some iterative function $x_{n+1}=\sin(x_n)$. I want to demonstrate an iteration on a plot by graphing $y=\sin(x)$ and $y=x$, then drawing lines from the graph of x to the graph of sin x, like in this page: http://demonstrations.wolfram.com/IteratingLinearFunctions/ I would try to google or check the documentation to do this, but I have no idea on what keywords to use, nor do I have any clue on where to begin. Thanks! [I also don't know how to get LaTeX to work in this forum. Sorry!] 2. Jan 20, 2012 ### Bill Simpson y = NestList[Sin, 1, 6]; x = Range[7]; points = Transpose[{x, y}]; ListPlot[points] might get you started. From that you need to use the contents of points to generate the horizontal and vertical line segments shown in that demonstration. This http://reference.wolfram.com/mathematica/ref/Line.html will give you an idea what result you need to generate from your points. Last edited: Jan 21, 2012	2017-09-19 16:00:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41732001304626465, "perplexity": 776.9556274497362}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818685850.32/warc/CC-MAIN-20170919145852-20170919165852-00598.warc.gz"}
https://valutamygm.web.app/45543/22017.html	# sin4x = (1+V2)(sin2x + cos2x - 1). 2sin2xcos2x=(1+v2)(sin2x+cos2x-1) sin2x+cos2x=t. (sin2x+cos2x)^2=t^2 sin2x^2+2sin2xcos2x+cos2x^2=t^2. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. I. → x coscu) u=&=xt. 12. at X. Find the slope of the tangent line at x = 2 if 2. # = (2x - 3)?. Gii pt :sinx cosx 3sin2x cos2x + = + = + = + = http://k2pi.net/showthread.php?t=1266-Giai-phuong-trinh-sin-x-cos-x-sqrt-3-sin-2x-cos-2x 9.Gii pt : cos x 12cos x. Genom omskrivning med formeln för dubbla vinkeln blir ekvationen. 2sinx Jag får det till: PF(sinx^2)=PF((1-cos(2x))/2)=x/2-sin(2x)/4+C Det liknar sinus för dubbla vinkeln som blir 1-cos2x. Hälsningar. Hans L visa att tan x = sin2x/ 1 + cos2x bägge är dubbla vinkeln inte upphöjt. sin(2x) = 2sin(x)cos(x) och 1+cos(2x) = 2(cos(x))^2. ## 2014-08-17 Double-Angle Identities. sin(2x) = 2 sin(x) cos(x). cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1. tan ⁡ ( 2 x ) = 2 tan ⁡ ( x ) 1 − tan ⁡ 2 ( x ) \tan(2x) Sin 2X = 2 Sin X Cos X. Cos 2X = Cos2X - Sin2X. ### 29 Nov 2019 Prove that \frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}. Can someone provide me some hints? I tried to manipulate the right-hand En primitiv funktion till f (x) = cos 2x blir. Man får inte glömma att ta hänsyn till den inre derivatan 2. Man gör tvärtom som vid derivering. Man dividerar alltså med Formules trigonométriques sin2x + cos2x = 1 sin2x = tg2x. 1 + tg2x cos2x = -. 1 + tg2x. I know sin^2x + cos^2x = 1, but is this the same if it's 2x? How to show that (1-cos2x)/sin2x=tanx using some double angle rules.Link to the video discussed in the intro:https://youtu.be/UHBCxuBL1aE I'll need to memorize $\cos2x = \cos^2x - \sin^2x$ as I'll use it in derivatives. Only, there are other forms for this identity, I can't see how I can get to the others from this one above. The o FORMULAS TO KNOW Some trig identities: sin2x+cos2x = 1 tan2x+1 = sec2x sin 2x = 2 sin x cos x cos 2x = 2 cos2x 1 tan x = sin x cos x sec x = 1 cos x cot x = cos x … Question 1130401: (cos2x+sin2x)^2=1+sin4x Found 2 solutions by MathLover1, ikleyn:. Answer by MathLover1(17988) (Show Source): . You can put this solution on … The word ‘trigonometry’ being driven from the Greek words’ ‘trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. In this Chapter, we will generalize the concept and Cos 2X formula of one such trigonometric ratios namely cos 2X with other trigonometric ratios. Ektorp vagen 15 In this Chapter, we will generalize the concept and Cos 2X formula of one such trigonometric ratios namely cos 2X with other trigonometric ratios. I'll need to memorize $\cos2x = \cos^2x - \sin^2x$ as I'll use it in derivatives. Only, there are other forms for this identity, I can't see how I can get to the others from this one above. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Markera här. Möjliga svar: A. 1. B. (sin2x-cos2x) dx = ta sin (2x a) + 1b, then(1) a = STDER (2) a = - ST, DER (3) a=DER(4) none of these. Peta jensen xxx vanligaste intervjufragorna tentamen engelska chuffers pizza spencerville ohio traktor 4x4 tidrapporterings app ### 19 Jan 2020 The integral of the trigonometric function is (1/2)(e^(sin 2x)) + C. = cos 2x ex + 2 (sin 2x ex - ∫ 2cos 2x ex dx ). 5 ∫ cos 2x ex dx = cos 2x dv=sin(2x)dx => v = (sin(24)dx == cos(27). *cos(21+L;xsin. (20)–S (-;col29) ---cos. Other than utorrent textstorlek iphone ### cos2X =Cos2X -sin2X. Cos2X =(1-sin2 X ) -sin2 X (Since ,cos2X=(1-sin2 X ) cos2X=1-sin2 X -sin2 X. So. Cos2X=1-(sin2 X+sin2 X) Hence cos2x =1-2sin2 X. Cos2x =2COS2X-1. To derive this we need to start from the eariler derivation As we already know that. cos2X =Cos2X -sin2X. Cos2x=cos2X-(1-cos2X){Since sin2x=(1-cos2X)} Cos2x =cos2X-1 +cos2X. cos2X=(cos2X+cos2X)-1 17. sin(x) cos(x) = (1/2) sin(2x). 18. sinh(x) = ex − e−x. Lös ekvationen cos2x = 3 sinx + 2. 2.	2023-04-02 11:14:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259698390960693, "perplexity": 11848.433114947773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00562.warc.gz"}
https://socratic.org/questions/how-do-you-solve-x-6y-18-and-x-6y-6-using-the-addition-and-subtraction-method	# How do you solve -x-6y=-18 and x-6y=-6 using the addition and subtraction method? Oct 24, 2014 The answer is $x = 6$ and $y = 2$ The easiest way to solve such systems of equations using the elimination (or as you put it "addition and subtraction") method, is to write out our two equations, one below the other, and see how we can cancel out a variable to bring it down to a one-variable equation. Here we have: $- x - 6 y = - 18$ $x - 6 y = - 6$ Think of this as one of those addition or subtraction problems you did back in elementary school, with the number stacked on each other like this for you to work with. You have two paths of action here: you can either add the two equations, or subtract one from the other. Which one will you chose, in order to eliminate a variable? In this case, both methods would work. If we added, we'd end up eliminating the $x$ [x + (-x) = 0], and if we subtracted, we'd end up eliminating the $y$ $\left[- 6 y - \left(- 6 y\right) = 0\right]$. So for this particular problem the course of action is up to you. I'm going to go with addition, since I just like adding better (it's more straightforward). $\left[- x - 6 y = - 18\right]$ $+ \left[x - 6 y = - 6\right]$ => $- 12 y = - 24$ $y = 2$ Now that we have one variable solved, we can plug it back into one of our original equations and solve for the other one. Again, it's completely up to you which one to use, but I'm going to use the second one: $x - 6 \left(2\right) = - 6$ $x - 12 = - 6$ $x = 6$ So our solution to this system is the coordinate $\left(6 , 2\right)$, meaning that if we graphed these two lines, they would intersect at that coordinate. Hope that helped :)	2019-09-19 08:19:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.675152063369751, "perplexity": 282.92708005306696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00181.warc.gz"}
https://livingthing.danmackinlay.name/presentations.html	Presentations Slide decks and other stylised academic dominance displays Presentations: The quantum of information for all parts of society for which the quantum of information is not a tweet or a facebook status update. Powerpoint presentations are already purported to have various oft-cited defects but these I will not discuss here. In my trade they are a necessary evil. I’m all about harm minimisation of the evil, starting here, with technical harm. Standard technology The default options: • Microsoft Powerpoint. • Apple Keynote. Both these are a colossal waste of time, adding little to my research while sucking energy into a black hole of trying to give a shit about transitions and fiddly filetype compatibility issues. Both, last time I looked, had a terrible mathematical equation typesetting workflow, although Keynote went beyond terrible to abysmal in this area. An alternative strategy might re-use the documentation, code, math markup and/or graphs from my actual research articles and code. It turns out that this is not hard as such, merely harder than it should be. Plain PDF Generate a PDF using your choice of technology Display the PDF in presentation mode using your PDF viewer. HTML slides HTML slides are powerful because of all the work that has gone into browser support these days. Reveal.js and remark.js are the best, for my money. Reveal.js seems more popular, although a little overengineered. I ended up using reveal.js because the online editor makes it easier to collaborate with my non-HTML-nerd colleagues, and it is integrated with jupyter. Reveal.js Creating themes seems to require you to fork reveal’s github repo for full generality, which feels a bit weird. Although you can still just inject those famous CSS style sheets as made famous by the web, right? Practically, themeing can be a rabbit hole of flex-box and responsive media queries. I suspect if you are doing too much of that you should just be using powerpoint or keynote or something, unless your day-job is actually CSS. Note that there is a convenient (although slightly restricted) version of reveal available through interactive jupyter slideshows via RISE. Do you need to preview slideshows to colleagues who are allergic to html? You can export as a PDF, although it’s not quite as immediate as you’d hope. [You can print a slideshow as a PDF from Chrome manually. For reveal (and certain other formats) you can use decktape which runs its own private Chrome browser headless merely as a print driver. RISE is the interactive jupyter notebook slidehow that uses reveal as a backend. Customisation of style etc for RISE: 1. use nbextensions_configurator; this tool offers an interactive way to enable, disable and tweak all notebook extensions - see screenshot below; 2. define settings in JSON files, typically by using python scripts; 3. you can also embed settings in a specific notebook’s metadata; 4. and you can also provide your own CSS file(s), that can supersede styling of the various DOM pieces. Remark.js remark.js is mostly similar. It has native R support via xaringan, which is Yihui Xe’s favourite Latex slides Beamer, the LaTeX slide thingy, also works. It is depressing and boring, but it does lend a certain reassurance to the audience that you aren’t going to rock the boat and say anything controversial. I am not a huge fan of LaTeX except for its mathematical markup support, and that is (sufficiently) available in HTML too thanks to MathJax, but if I work on slides in beamer my colleagues will be calmer because they feel they understand it better. Note that BibTeX does not work with beamer. BibLaTeX, by contrast, seems to. knitr does support beamer slides too, although why would you want to do that? There is no hope of making beamer match my corporate style guide in any sensible timeframe, but I could choose the closest approximation by looking at the example theme matrix. OTOH, who actually meets corporate style guidelines? No academic I know. Pedagogic considerations So how do you actually give a good presentation? Like most academics, I will leave this one for some unspecified future date. Alternatives: Animations and interactives • 3b1b’s manim is a curious passion project to create interactive mathematical animations in code, targeting e.g. youtube.	2019-09-20 06:22:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2061486691236496, "perplexity": 4276.760168265177}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00289.warc.gz"}
http://mathhelpforum.com/differential-equations/166824-uu_-xy-u_-x-u_-y-0-a-print.html	# uu_{xy}-u_{x}u_{y}=0 • December 23rd 2010, 06:40 PM dwsmith uu_{xy}-u_{x}u_{y}=0 Find the general solution. $uu_{xy}-u_{x}u_{y}=0$ My PDE book doesn't offer explanations so I have no clue on how to approach this problem. • December 23rd 2010, 08:36 PM Prove It You probably have to use Separation of Variables. • December 23rd 2010, 08:37 PM dwsmith How is that done since there is a term u_{xy}? • December 24th 2010, 05:47 PM dwsmith $\displaystyle uu_{xy}=u_xu_y\Rightarrow \int\frac{u_{xy}}{u_x}dy=\int\frac{u_y}{u}dy\Right arrow ln(u_x)=ln(u)+f(x)$ Should I move $ln(u)+f(x)$ this to left and then exponentiate, exponentiate first, or integrated as is? • December 24th 2010, 06:18 PM Ackbeet I'm not sure I think your last steps there are valid. In pde's, separation of variables doesn't look like it does in ode's. You would assume that $u(x,y)=X(x)\,Y(y),$ in which case $u_{x}=X'Y,$ and $u_{y}=XY'.$ Finally, $u_{xy}=X'Y'.$ Plugging all this into the pde $uu_{xy}=u_{x}u_{y}$ yields $XYX'Y'=X'YXY',$ an identity. I'm not sure exactly what this means. Does it mean that any solution that is a product of a function of $x$ and a function of $y$ is a solution to the pde? I'm inclined to think so. • December 24th 2010, 06:19 PM dwsmith It came from here. $uu_{xy}-u_{x}u_{y}=0$ Also, if you differentiate with respect to y, you will obtain this equation. $\displaystyle \frac{\partial u}{\partial y}\left[ln(u_x)\right]=\frac{\partial u}{\partial y}\left[ln(u)+f(x)\right]\Rightarrow \frac{u_{xy}}{u_x}=\frac{u_y}{u}$ • December 24th 2010, 06:49 PM dwsmith $\displaystyle ln(u_x)=ln(u)+f(x)\Rightarrow ln(u_x)-ln(u)=f(x)\Rightarrow \frac{u_x}{u}=e^{f(x)}\Rightarrow \frac{u_x}{u}=f_2(x)$ $\displaystyle \int\frac{u_x}{u}dx=\int f_2(x)dx\Rightarrow ln(u)=f_2(x)+g(y)\Rightarrow e^{ln(u)}=e^{f_2(x)+g(y)}$ $\Rightarrow u(x,y)=e^{f_2(x)}e^{g(y)}=f_3(x)g_2(y)$ I think this is how it is done. • December 25th 2010, 03:45 AM Ackbeet In looking over your solution, I think your steps are valid. I would point out, however, that your final destination is the same place the usual pde separation of variables got you in post # 5. Looks good! • December 25th 2010, 06:54 AM Jester You could also have achieved the same with the transformation $u = e^v$ Just to add, most of the time separation of variables doesn't work for nonlinear PDE's. For example try $uu_{xy} - u_x u_y = 1 \; \text{or}\; u u_{xy} - u_x - u_y = 0.$ • December 25th 2010, 01:46 PM Ackbeet There are a few other's. I really like the separation of variables of the Sine-Gordon equation as presented in Debnath's book (rather clever). A very good question is - when does a nonlinear PDE (in variables $(t,x,u)$) admit (functional) separable solutions in the form $F(u) = T(t)X(x)$ $F(u) = T(t) + X(x)$?	2014-12-18 17:28:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7576592564582825, "perplexity": 879.0582513193026}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802767301.77/warc/CC-MAIN-20141217075247-00067-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.hackmath.net/en/math-problem/6290	# It is 2 It is 90 feet from home plate to first base on a baseball diamond. What is the area of the baseball diamond in square yards? Correct result: S = 900 yd2 #### Solution: $a=90 /3=30 \ \text{yd} \ \\ S=a^2=30^2=900 \ \text{yd}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Tips to related online calculators Do you want to convert area units? Do you want to convert length units? #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Athletic race Before a race, you start 4 5/8 feet behind your friend. At the halfway point, you are 3 2/3 feet ahead of your friend. What is the change in distance between you and your friend from the beginning of the race? • Wood dividing Which equation calculates the number of 1/3-foot pieces that can be cut from a piece of wood that is 7 feet long? • Feet to miles A student runs 2640 feet. If the student runs an additional 7920 feet, how many total miles does the student run? • Steps Walkway from the house to the school is 180 m long. How many more steps Michael makes than his father on the way to school, if the length of Michael's step is 60 cm and the length of his father's step is 90 cm. • Apartment area Anton wanted to measure the area of the apartment. But he bought a meter only 1.5m long. Later he recalled that he had two meters long at 4.5m and 18m at home. How many times are home gauges longer than the meter he bought? • Jose and Kaitlyn Jose and Kaitlyn have a contest to see who can throw a baseball the farthest. Kaitlyn wins, with a throw of 200 ft. If Jose threw the ball 3/4 as far as Kaitlyn, how far did Jose throw the ball? • Sewing The lady cut off one half of cloth. She needed three-quarters of this piece to sew a skirt. What part of the original piece of cloth still remained? • Area Calculate: ? • Rape The agricultural cooperative harvested 525 ares of rape, of which received 5.6 tons of rape seeds. Calculate the yield per hectare of rape. • Colza In the agricultural cooperative harvested 525 ares of colza, of which received 5.6 tons of seeds. Calculate the yield per hectare. • One hectare How many square meters are one hectare? • After 2 After we planted flowers in 2/5 of our garden, 24m remained unplanted. How many meters is the garden in total? If the total area of the garden is 1, the proportion of the remaining area is? • Customary length Convert length 65yd 2 ft to ft • Pencil cut Veronika cut 10cm pencil. Each turn, the pencil was reduced by 0.2mm. She turned the pencil 100 times. How many centimeters has a pencil now? • Recipe A recipe requires 2 pounds of flour. If a chef wants to triple the recipe, how many ounces of flour will be needed? • Bean bag A student tossed a bean bag. It landed 216 inches away. How many yards are equal to 216 inches? • A swiming A swiming pool holds 30000lt of water. How many gallons does it hold? 1 gallon= 4.55lt	2020-08-04 08:40:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3742079734802246, "perplexity": 2903.239661285798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735867.23/warc/CC-MAIN-20200804073038-20200804103038-00538.warc.gz"}
https://stats.stackexchange.com/questions/253460/derive-the-gradients-of-a-basic-neural-network	# Derive the gradients of a basic neural network Given a neural network as following \begin{align} &J = CE(y,\hat{y})=-\sum_i y_i log(\hat{y}_i)\\ &\hat{y} = softmax(z_2)\\ &z_2 = hW_2+b_2\\ &h = sigmoid(z_1)\\ &z_1 = xW_1+b_1 \end{align} and $$\frac{\partial J}{\partial z_2}=\hat{y}-y=\delta_1$$ where $W_1 \in \mathbb{R}^{D_x \times H}$, $b_1 \in \mathbb{R}^H$, $W_2 \in \mathbb{R}^{H \times D_y}$ and $b_2 \in \mathbb{R}^{D_y}$. Derive the gradients of $J$ with respect to $h$, $W_2$. Here's the correct solution: \begin{align} &\frac{\partial J}{\partial h}=\delta_1 \frac{\partial z_2}{\partial h}=\delta_1 W_2^T\\ &\frac{\partial J}{\partial W_2}=\frac{\partial z_2}{\partial W_2} \frac{\partial J}{\partial z_2}=h^T \delta_1 \end{align} Q1: I have difficulties to understand the second dervitive with repsect to $W_2$, where $z_2 \in \mathbb{R}^{D_y}$ and $W_2 \in \mathbb{R}^{H \times D_y}$, the result $\frac{\partial z_2}{\partial W_2}$ to me should have a dimesion $(D_y,H \times D_y)$, $h^T$ doesn't satisfy this dimension. Q2: why does the second derivative have an order of $h^T \delta_1$ not $\delta_1 h^T$ by following backpropagation such that $\frac{\partial J}{\partial W_2}=\frac{\partial J}{\partial z_2} \frac{\partial z_2}{\partial W_2}=\delta_1 h^T$? Notes: $z_1$ and $z_2$ are calculated using array broadcasting in numpy, for instance the arithmetic is legit: array(4x3)+array(3)=array(4x3), The principle bechind the scence is that array(3) could be converted to array(4x3) via stacking. Details of broadcasting. [This is trivia] • Shouldnt $\partial z_2 / \partial W_2$ have the dimension of $W_2$? – Nikolas Rieble Jul 11 '17 at 9:52	2019-07-15 21:02:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9615377187728882, "perplexity": 1481.4748577492621}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524111.50/warc/CC-MAIN-20190715195204-20190715221204-00084.warc.gz"}
https://zbmath.org/?q=an:0716.14013	# zbMATH — the first resource for mathematics On the gonality of nodal curves. (English) Zbl 0716.14013 Author’s abstract: “Here we prove that for every $$n\geq 33$$ and every $$t\leq (n^ 2+3n)/6$$, the normalization Y of a general plane curve C of degree $$n$$ and with t nodes has no $$g^ 1_ b$$ with $$b<n-2$$ and only $$g^ 1_{n-2}$$ and $$g^ 1_{n-1}$$ induced by a pencil of lines through a point of C. Recently, M. Coppens and T. Kato have proved stronger results [cf. Manuscr. Math. 70, No.1, 5-25 (1990) and 71, No.3, 337-338 (1991)].” Reviewer: J.Libicher ##### MSC: 14H20 Singularities of curves, local rings 14C20 Divisors, linear systems, invertible sheaves ##### Keywords: gonality of nodal curves Full Text:	2021-12-03 01:08:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7377789616584778, "perplexity": 1189.611509071501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00296.warc.gz"}
https://newbedev.com/do-isotopic-groups-cause-optical-activity	# Chemistry - Do isotopic groups cause optical activity? ## Solution 1: From a 2007 paper:: Here we show that instrumental advances in Raman optical activity, combined with quantum chemical computations, make it possible to determine the absolute configuration of (R)-[$$\ce{^2H1}$$, $$\ce{^2H2}$$, $$\ce{^2H3}$$]-neopentane. This saturated hydrocarbon represents the archetype of all molecules that are chiral as a result of a dissymmetric mass distribution. It is chemically inert and cannot be derivatized to yield molecules that would reveal the absolute configuration of the parent compound. Diastereomeric interactions with other molecules, optical rotation, and electronic circular dichroism are, in contrast to the well-known case of bromochlorofluoromethane, not expected to be measurable. Vibronic effects in the vacuum ultraviolet circular dichroism might reveal that the molecule is chiral, but the presence of nine rotamers would make it extremely difficult to interpret the spectra, because the spatial arrangement of the rotamers’ nuclei resembles that of enantiomers. The unequivocal spectroscopic determination of the absolute configuration of (R)-[$$\ce{^2H1}$$, $$\ce{^2H2}$$, $$\ce{^2H3}$$]-neopentane therefore presented a major challenge, one that was at the very limit of what is possible. • The described molecule looks like this: • Optical rotation is not expected to be measurable. Probably some fantastically minute but nonzero specific rotation could be computed with advanced quantum mechanical methods, but it would be no fun to try to measure it. • The difference in the electronic configuration of the isotopically substituted enantiomers in the ground state is negligible. • But the electronic configuration of rotationally or vibrationally excited states is not negligible. Therefore, some technique that can measure the optical properties of rovibrational excited states might be able to discern the difference. (I'll leave it to experts in this area to explain why Raman optical activity worked better than vUV-CD.) ## Solution 2: Optical rotation (OR) emerges from a difference in the index of refraction of a compound towards left/right circularly polarized light. All chiral have some difference in these indices of refraction. Your question specifically relates to isotopically chiral molecules where the effect is more subtle. You are correct that (OR) is fundamentally an electronic effect and that isotopic substitution shouldn't directly affect that. However, vibrations can cause the molecule to oscillate around the the equilibrium structure to geometries that have a different OR and isotopic substitution alters the potential energy surface. The transient chirality of the molecule from its zero point vibrations is enough to produce OR. The effect of these vibrations can actually be fairly significant, accounting for as much as 20-30% of the observed optical rotation. For isotopically chiral molecules, the effect is small, but can still be measured. As an example, 2-deuteriocyclohexanone has an optical rotation of $$2.5^\circ$$ and a number of other substituted cycloalkanones have been synthesized since.	2022-06-29 19:05:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4904223084449768, "perplexity": 1095.7189832457548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00358.warc.gz"}
https://masr.netlify.app/sec-life-of-statisticians.html	## 24.1 统计学家生平 ### 参考文献 Johnson, Norman L., and Samuel Kotz. 1997. Leading Personalities in Statistical Sciences: From the Seventeenth Century to the Present. New York, NY: John Wiley & Sons.	2021-06-18 10:49:01	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8072444796562195, "perplexity": 9309.791304648701}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00263.warc.gz"}
https://indico.desy.de/indico/event/19136/	String Theory Seminar # Universality at large transverse spin in defect CFT ## by Marco Meineri (EPFL) Thursday, March 15, 2018 from to (Europe/Berlin) at DESY Hamburg ( SR2 ) Description We study the spectrum of local operators living on a defect in a generic conformal field theory, and their coupling to the local bulk operators. We establish the existence of universal accumulation points in the spectrum at large $s$, $s$ being the charge of the operators under rotations in the space transverse to the defect. Our tools include a formula that inverts the bulk to defect OPE and is analytic in $s$, analogous to the Caron-Huot formula for the four-point function. Some important assumptions are made in deriving this result: we comment on them. Material:	2018-08-22 01:28:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6725947260856628, "perplexity": 785.8754530857752}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221219242.93/warc/CC-MAIN-20180822010128-20180822030128-00482.warc.gz"}
https://math.stackexchange.com/questions/2149066/proving-lim-x-to-infty-fx-infty-if-fxc-for-every-x	# Proving $\lim _{ x\to\infty }{ f(x) }=\infty$ if $f'(x)>c$ for every $x$ Given a differentiable function $f: (0,\infty) \rightarrow \mathbb R$ and $c>0$ such that $f'(x)>c$ for every $x$. Prove: $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$ Using the MVT, I got to $f(x)>c(x-x_0)+f(x_0)$, and I think I should proceed using the definition of limit, but I got stuck. Any help appreciated. • You've shown $f(x) > A + cx$ for some constant $A.$ Aren't you essentially done? – zhw. Feb 17 '17 at 17:46 • @zhw. Is it ok to say that $cx_0$ is constant? – Itay4 Feb 17 '17 at 17:49 • Sure. Just take $x_0 = 1$ for example. – zhw. Feb 17 '17 at 18:02 • @zhw. Great thanks ! – Itay4 Feb 17 '17 at 18:34 ## 1 Answer For $x>1$, we have that $f(x)-f(1)=\int_1^xf'(t)\,dt>c(x-1)$. Hence, $$f(x)>c(x-1)+f(1)$$ And it's easy to see that the RHS is unbounded as $x\to\infty$. EDIT: As @zhw. noted, the solution above does not work if $f'$ is not Riemann integrable. Nonetheless, we may use the general idea to craft a solution that does not involve integration. Consider $g(x)=c(x-1)+f(1)$. Then $h(x)=f(x)-g(x)$ is such that $h(1)=0$ and $h'(x)>0$ for all $x>1$. It follows that $h(x)>0$ for all $x>1$, that is, $$f(x)>c(x-1)+f(1)$$ for all $x>1$. The conclusion follows. • We don't know $f'$ is Riemann integrable. – zhw. Feb 17 '17 at 17:46 • Good observation. I have fixed it. – Fimpellizieri Feb 17 '17 at 17:51 • @Fimpellizieri Got you. Does choosing 1 require any further explanation ? Or do I need to say anything more after ? – Itay4 Feb 17 '17 at 18:37 • Nope, you can choose any $t\in (0,+\infty)$. Of course, then $g(x)$ becomes $c(x-t)+f(t)$. – Fimpellizieri Feb 17 '17 at 18:54 • @Fimpellizieri Great, thanks ! – Itay4 Feb 17 '17 at 19:46	2019-10-19 06:54:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.860283613204956, "perplexity": 303.4100776065325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00465.warc.gz"}
https://physics.stackexchange.com/questions/478898/adiabatic-approximation-solving-the-schr%C3%B6dinger-equation/479244	# Adiabatic Approximation, Solving the Schrödinger equation In the adiabatic approximation one looks at the Hamiltonian $$H_0 = \sum_{i = 1}^{N_e} \frac{\vec{p}_i^2}{2m_e} + \sum_{i < j} \frac{e^2}{\|\vec{r}_i - \vec{r}_j\|} + \sum_{k < l} \frac{Z_k Z_l e^2}{\|\vec{R}_k - \vec{R}_l\|} + \sum_{i, k} \frac{- Z_k e^2}{\|\vec{r}_i - \vec{R}_k\|}.$$ Now typical statements are "For a fixed ion configuration $$\{\vec{R}_1,\ldots, \vec{R}_n\}$$ let $$\Psi_{\alpha} \equiv \Psi_{\alpha}(\vec{r}_1,\ldots,\vec{r}_n,\vec{R}_1,\ldots\vec{R}_n)$$ be a solution of the eigenvalue problem $$H_0 \Psi_{\alpha} = \varepsilon_{\alpha}(\vec{R}_1, \ldots \vec{R}_n) \Psi_{\alpha},$$ where $$\alpha$$ denotes a complete set of quantum numbers." What is baffling to me is: How does one know about the existence of such solutions? (Not only, that one finds one solution of that eigenvalue problem, but also a complete set of solutions?!) Perhaps the answer may lie in properties of Hermitian operators. (The Hamiltonian is Hermitian). For example, the solutions (eigenfunctions) of a Hermitian operator form a complete orthogonal set. Another property of Hermitian matrices is that an $$nxn$$ Hermitian operator (matrix) has $$n$$ eigenvalues. So perhaps that answers why the solutions are guaranteed to exist.	2020-01-19 01:34:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9470588564872742, "perplexity": 288.7534498558101}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00410.warc.gz"}
https://www.jiskha.com/archives/2011/10/20	# Questions Asked onOctober 20, 2011 1. ## Chemistry Each value below represents a different aqueous solution at 25 °C. Classify each solution as acidic, basic, or neutral. pH=9.72 pOH=4.69 pOH=13.62 [H+]=7.5e-2 [H+]=7.8e-9 [H+]=1e-7 [OH-]=4.5e-12 [OH-]=3.5e-5 pH=4.09 pOH=7 2. ## Physics 3 A potential energy function for a two-dimensional force is of the form U = 3x3y – 7x. Find the force that acts at the point (x, y). 3. ## physics A 53.5 N object is in free fall. What is the magnitude of the net force which acts on the object? Answer in units of N What is the magnitude of the net force when the object encounters 15.4 N of air resistance? Answer in units of N What is the magnitude of A ball of mass 0.4 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below. The velocity of the ball as it leaves the kicker’s foot is 16 m/s at angle of 46◦ above the horizontal. The top of the fence is 4 m 5. ## Chemistry Complete this table of values for four aqueous solutions at 25 °C. Solution A: [H+]=0.0093M [OH-]=? pH=? pOH=? Solution B: [H+]=?[OH-]=1.4e-6M pH=? pOH=? Solution C: [H+]=?[OH-]=? pH=5.36 pOH=? Solution D: [H+]=?[OH-]=? pH=? pOH=10.14 6. ## Chemistry What are the concentrations of OH– and H in a 0.00070 M solution of Ba(OH)2 at 25 °C? Assume complete dissociation. 7. ## Chemistry Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. [H+]= [ClO4-]= [OH-]= 8. ## Chemistry Nitrogen dioxide, NO2, dimerizes easily to form dinitrogen tetroxide , N2O4 : 2NO2N2O4 a) Calculate Change in reaction G* and K for this equilibrium. b) Calculate the (e) (the measure of the progress of the reaction) for this equilibrium if 1.00 mol NO2 25. ## Chemistry Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. [H+]= [ClO4-]= [OH-]= 26. ## physics A generating station is producing 1.2 x 106 W of power that is to be sent to a small town located 4.2 km away. Each of the two wires that comprise the transmission line has a resistance per length of 5.0 x 10-2/km. (a) Find the power lost in heating the asked by shaknocka lewis 27. ## Science A child in a boat throws a 5.30 kg package out horizontally with a speed of 10 m/s, see the figure. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg and that of the boat is 55.0 kg 28. ## math (brainteaser) The pahges in jacks book are numbered from the number 1, there are 555 didgits ow many pages in Jacks book 29. ## statistics A survey of college students was conducted during final exam week to assess the number of cups of coffee consumed each day. The mean number of cups was 5 with a standard deviation of 1.5 cups. The distribution was normal. a. What proportion of students 30. ## Physics A car starts from rest and accelerates unifomly at 3m/s^2. A second car starts from rest 6 seconds later at the same point and accelerates uniformly at 5m/s^2. How long does it take the second car to over take the first car? 31. ## physics The fastest recorded pitch in Major League Baseball was thrown by Nolan Ryan in 1974. If this pitch were thrown horizontally, the ball would fall 0.809 m (2.65 ft) by the time it reached home plate, 18.3 m (60 ft) away. The acceleration of gravity is 9.81 32. ## math The number uses every number 0-9 the numbers are used only once the fourth digit is 4 the first digit in the billions place is 3 the 5 is next to the last digit the sixth digit is 7 the digit after 3 is 9 the digit before 5 is 2the third digit is half the 33. ## Physics A dragster and a driver together have mass of 942.2 kg. the dragster starting from rest attains a speed of 25.5m/s in 0.56 s. 1. find the average acceleration of the dragster during this time interval? I really need the formula 2. What horizontal force 34. ## physics At a track-and-field meet, the best long jump is measured as 8.00 m. The jumper took off at an angle of 46° to the horizontal. (a) What was the jumper’s initial speed? (Neglect air resistance.) 1 m/s (b) If there were another meet on the Moon and the 35. ## nutrition Review the nutrition quality of the items, using what you know about daily values, and ingredients (one of the products must have a health claim). Write a 300-500 word essay on what you have learned. 36. ## Physics 5 Two blocks are free to slide along the frictionless wooden track ABC shown in Figure P9.20. A block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet 37. ## Chemistry Complete this table of values for four aqueous solutions at 25 °C. Solution A: [H+]=0.0093M [OH-]=? pH=? pOH=? Solution B: [H+]=?[OH-]=1.4e-6M pH=? pOH=? Solution C: [H+]=?[OH-]=? pH=5.36 pOH=? Solution D: [H+]=?[OH-]=? pH=? pOH=10.14 38. ## college chemistry Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.0×10^-2. Kc=6.3 10^-5 39. ## ALGEBRA why dose duffer want lights around the golf course? there are 40 empty spaces asked by Brendan Riddell 40. ## chemistry A gas cylinder of volume 5.00 L contains 1.00 g of Ar and 0.500 g of Ne. The temperature is 275 K. Find the partial pressure of Ne. 41. ## Geometry what is the value of x in the isosceles trapezoid below? the top is 10x degrees and the bottom is (3x+11) degrees 42. ## chemistry Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. 43. ## Physics A student sits on a rotating stool holding two 3.2-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 44. ## chemistry How many grams of HCl are produced according to the following equation, H2 + Cl2 = 2 HCl, when 4.0 g of hydrogen reacts completely? 51. ## Calculus AB Give a formula for the extended function that is continuous at the indicated point. F(x)= (x-4)/((√x)-2) at x=4 I know the answer I just need to figure out how to work it out. The answer btw is y= (√x) + 2 52. ## Physics Two long parallel wires separated by a distance, d, carry currents in opposite directions. If the left-hand wire carries a current i/2, and the right-hand wires carries a current i, determine where the magnetic field is zero. 53. ## Math If you had a fraction strip folded into twelfths, what fractional lengths could you measure with the strip? b. How is your answer in part (a) related to the factors of 12? 54. ## chemistry write the molecular, ionic and net ionic equation for the following Ph(NO3)2 + NaOH Pb(NO3)2 + Na2CO3 Ni(NO3)2 + NaOH Ni(NO3)2 + Na2Co3 55. ## physics Take the mass of the Earth to be 5.98 × 1024 kg. If the Earth’s gravitational force causes a falling 74 kg student to accelerate downward at 9.8 m/s2, determine the upward acceleration of the Earth during the student’s fall. Answer in units of m/s2 56. ## chemistry In the following experiment, a coffee-cup calorimeter containing 100 mL of H20 is used. The initial temperature of the calorimeter is 23.0 C . If 3.60 g of CaCl2is added to the calorimeter, what will be the final temperature of the solution in the 73. ## spanish que les gusta hacer a los padres de ruben 74. ## math The atmospheric pressure P in pounds per square inch (psi) is given by the formula below, where a is the altitude above sea level (in miles). If a city has an atmospheric pressure of 13.25 psi, what is its altitude? (Recall that 1 mi = 5,280 ft. Round your 75. ## Finance Which of the following statments is CORRECT? a. Assume that two bonds have equal maturities and are of equal risk, but one bond sells at par while the other sells at a premium above par. The premium bond must have a lower current yeild and a higher capital 76. ## Physics A 0.90 × 10^3 kg sports car collides into the rear end of a 2.4 × 10^3 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 4.6 m before stopping. The police officer, estimating the coefficient of kinetic 77. ## Physics A 245 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 35.0 cm, as shown in Figure 5-21. (a) Calculate the gravitational potential energy at A relative to B. 1 . (b) Calculate the particle's kinetic energy at B. 2 78. ## Physics 2 A particle moves along a line where the potential energy of its system depends on its position r as graphed in Figure P8.46. In the limit as r increases without bound, U(r) approaches +1 J. (a) Identify each equilibrium position for this particle. Indicate 79. ## Math a) f(x) = 4.1x +92.16 b) f(x) = 17.9x + 19.36 c) f(x) = 8.8x + 55.4 ANSWERS: a) 4.8 b) 0.93 c) 2.2, i just have no clue how they got these answers EXTRA INFO: The demand function for a new product is p(x)= - 4x +42.5, where x is the quantity sold in 80. ## Alg1 A baseball player hits a baseball into the outfield. The equation h= -0.005x^2 +x+3 give the path of the ball, where (h) is the height and (x) is the horizontal distance the ball travels. A. What is the equation of the axis of symmetry? It's -100 right? B. 81. ## Math A baseball player hits a baseball into the outfield. The equation h= -0.005x^2 +x+3 give the path of the ball, where (h) is the height and (x) is the horizontal distance the ball travels. An outfielder catches 3 ft above the ground. How far has the ball 82. ## Math (UBC) Economists use production functions to describe how output of a system varies with another variable such as labour or capital. For example, the production function P(L) = 200L + 10L^2 - L^3 goves the output of a system as a function of the number of 83. ## Conley Chicken Delight claims that 88 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 60 orders revealed that 50 were delivered within the promised time. At the .05 significance level, can we conclude that less 84. ## ALGERBRA 2 rosalyn works no more than 25 hours a week during the school year. she is paid $12 an hour for tutoring geometry and$8 an hour for delivering pizzas for Pizza King. she wants to spend at least 2 hours but no more than 7 hours a week tutoring. find 85. ## Physics A 0.490 kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.870 kg puck initially at rest. Assume a perfectly elastic collision. (Let east be the +x direction.) (a) What will be the speed of each object after the 86. ## Chemistry A mixture containing 235 mg of Helium and 325 mg of neon has a total pressure of 453 torr. What is the partial pressure of Helium 87. ## Finance - Net Advantage to Leasing Net advantage to leasing) Arkansas Instruments (AI) can purchase a sonic cleaner for $1,000,000. The machine has a five-year life and would be depreciated straight line to a$100,000 salvage value. Hibernia Leasing will lease the same machine to AI for 88. ## chemistry A 48 g sample of water at 98.°C is poured into a 55 g sample of water at 15°C. What will be the final temperature of the mixture? 89. ## Psychology If a negative correlation between two sets of scores is displayed as a scatterplot, the points are clustered in a pattern that Question 5 options: resembles a bell-shaped curve. extends from the lower left to the upper right of the plot. resembles a 90. ## Finance Which of the following statements is CORRECT? a. Two bonds have the same maturity and the same coupon rate. However, one is callable and the other is not. The difference in prices between the bonds will be greater if the current market interest rate is 91. ## physics numerical a long horizontal wire is rigidly placed, carrying a current of 50a. another parallel long wire carrying a current of 40a is just supported above it.determine the distance between them at which the upper wire can be just supported by magnetic repulsion 92. ## Physics 4 A 10.0-kg block is released from point A in Figure P8.57. The track is frictionless except for the portion between points B and C , which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2 250 N/m, and compresses 93. ## physics During a baseball game, a batter hits a pop- up to a fielder 78 m away. The acceleration of gravity is 9.8 m/s2 . If the ball remains in the air for 5.8 s, how high does it rise? A man went to a post office to buy stamps. He bought x, 50 cent stamps and y, 25 cent stamps for $3.50. If he bought twice as many 50 cent stamps and half as many 25 cent stamps, the cost would be$4.75. Evaluate x and y 95. ## CHEMISTRY What are the ka and kb reactions of NaHSO3 ? 96. ## english A good strong thesis statement about the children's book Corduroy by Don Freeman 97. ## Ap chem When .4000 g of CH4 is burned in excess oxygen in a bomb calorimeter that has a heat capacity of 324, the temperature increases 6.795 degrees celsius. How much energy is released? (in kJ/mol CH4) 106. ## math a chemistry experiment calls for 30% sulfuric acid solution. if the supply room has only 50% and 20% sulfuric acid solution on hand,how much of each should be mixed to obtain 12 litres of 30% solution? 107. ## Spanish I had to rewrite from English to Spanish-Please check if I did them correctly- 1.She sees them. Ella los ve. 2. We see them. Nosotros los vemos. 3. I invito you. Yo te invito. 4. I invite him. Yo lo invito. 5.She see you all. Ella los ve. 6. He invites us. 108. ## Nutrion What types of pathogens? What habits in your food preparation practices do you think might cause these pathogens to grow? What habits can you change in your life to reduce your risk of foodbourne illness? 109. ## physics A long jumper leaves the ground at an angle of 22.1◦ to the horizontal and at a speed of 10 m/s. How far does he jump? The acceleration due to gravity is 9.8 m/s2 . Answer in units of m 110. ## physics A ball is thrown horizontally from the top of a building 30.1 m high. The ball strikes the ground at a point 66.8 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s 111. ## Chemistry so i had to calculate the pH of a 100ml solution of .10M acetic acid (ka=1.8x10(-5) I came out to 2.87. But then, they want the pH if 50ml HCL is added. So i thought...Ka = x2/.15M, therefore the square root of .15M x 1.8x10(-5) = .00164(concentration). So 112. ## logic and programing langage Design a modular program that asks the user to enter the replacement cost of a building and then displays the minimum amount of insurance he or she should buy for the property. 113. ## physics numerical a satellite orbitting the earth at the equator at a na ltitude of 400 km, has an antenna that can be modelled as a 2.0m long rod. the antenna is oriented perpndicular to the earth's surface. at the equator, the earth's magnetic field is essentially 114. ## Algebra Find the percent of error of each measurement. 0.2cm 0.4cm 115. ## Algebra The equation x^2+px+q=0, q cannot be equal to 0, has two unequal roots such that the squares of the roots are the same as the two roots. Calculate the product pq. I think the obvious root would be one but the second roots i just can't figure out! 116. ## Physics A 2.5 kg block is launched up a ramp at 9.0 m/s. If the ramp is 35 degrees above the horizontal and the coefficient of static friction is 0.60, while the coefficient of kinetic friction is 0.25. a. How far does the block go up the ramp from its starting 117. ## chemistry A 10.0 grams piece of gold 90.0 Celsius was stirred into 15.0ml of water at an initial temperature of 25.0 degrees Celsius. The final temperature of the water was 26.34 degrees celsius. Calculate the specific heat of gold. The specific heat of water is 118. ## Spanish I need to match the subject pronouns with the verb llevar these are the verbs: llevas, llevamos, llevo, llevan, lleva these are the subj. pronouns: yo, el, nosotros, ellas, tu, ella, ellos, ustedes. PLEASE HELP 119. ## chemistry Hydrogen, H2(g), is above silver metal in the activity series of the metals. WIll the follwing reaction, Ag(s) + H+(aq) -> occur or not occur? why or why not? 120. ## Mathematics After diving 92m below the sea level, the diver rises at a rate of 5 meters per minute for 7minutes. Where is the diver in relation to the surface? 121. ## Chemistry Help!! why is it impossible to draw a distinction between natural substances and chemical substances? 122. ## chemistry a 2.00-liter sample of a gas has a mass of 1.80 grams at STP. What is the density , in grams per liter , of this gas at Stp? 123. ## Math A computer game starts with a set of 144 tiles separated into 36 equal groups.Sara played the game until there were 8 groups of tiles left.How many tiles were left when Sara finished? 124. ## Math A manufacture of television sets has a historical defective rate of ten percent. a. what is the probability that in an hourly production run of 10 televisions, one will be defective b. What is the probability that in an hourly daily production run of 10 125. ## chemistry help! Potassium has an atomic mass of 39.10amu. a sample of potassium is made up of isotopes potassium-39 (mass=38.964) and potassium-41(mass =40.962 amu). What percentage of the potassium sample is potassium-39? what percent of the sample is potassium-41? 126. ## Finance Which of the following statements is CORRECT? a. Two bonds have the same maturity and the same coupon rate. However, one is callable and the other is not. The difference in prices between the bonds will be greater if the current market interest rate is 127. ## Physics A 21.0-g metal cylinder is placed on a turntable, with its center 90 cm from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is ìs = 0.84. A thin, massless string of length 90 cm connects the 128. ## zoology what is the dental formula of rabbit? 129. ## High School Will high school will look at my middle school grades (or just 8th grade grades ?) ???? 130. ## Calculus Show that the sum of the series ∑n(p^n) from n=1 to infinity (where 0 131. ## Statistics USA snapshot presented a bar graph depicting business travelers’ impression of wait times in airport security lines over the past 12 months. Statistics were derived from Travel Industry Association of American Business Traveler Survey of 2034 132. ## Math write a function. f(x), that represents the cost of renting a car for x days with a total of 80 kilometers if there is a fixed charge of $17 per day and a charge of$0.35 per kilometer. 133. ## Calculus Find the Critical Numbers? 1.) f(x) = 2x^3 + 3x^2 + 36x x=? 2.) g(t) = \|5t − 4\| t=? 134. ## English/writing If I am writing the following as a poem, how would I punctuate it- We are room 204 Check out our door It is clear to see We choose to be drug free 135. ## chemistry how many milliliters of concentrated NaOH(19.3 M) are needed to make 8.00 L of 0.25 M base solution? 136. ## urgent maths help needed now jafar has an isosceles triangle with an area of 20.25cm squared, what is the length of the side facing the right angle? 137. ## physic A 16 g bullet strikes and becomes embedded in a 1.10 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.24, and the impact drives the block a distance of 138. ## physics A ball is thrown horizontally from the top of a building 30.1 m high. The ball strikes the ground at a point 66.8 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s 139. ## physics A car having an initial speed of 16 m/s is uniformly brought to rest in 4.0 s. How far does the car travel during this 4.0 s interval? 140. ## aLgEbrA variable & expression i need to define a variable and write an algebraic expression for each phraase. 1) two points few than 3 times the number of points scored yesterday. i wrote: 3p-2 = 2) one metere more than 6 times your height in meters i wrote: 1 + 6h 3) seven pages fewer 145. ## English I need your help with the following sentences. Are they grammatically correct? Thank you. 1) How long have you been married for? Since 1998. 2) Is it wrong: "How long is it since you got married? (The answer is not since 1998 but for three years). 3) How 146. ## physics A simple harmonic oscillator has spring constant = 8.8N/m , amplitude = 11cm , and maximum speed 4.6 m/s. what is the speed at x=6cm? 147. ## english How do i make an outline? From Lord of the Flies, i need to make an outline of one page. this one: The boy with fair hair lowered himself down the last few feet of rock and began to pick his way toward the lagoon. Though he had taken off his school sweater 148. ## Chemistry If 4.0 g of hydrogen and 10.0 g of oxygen are mixed according to the equation 2 H2 + O2 = 2 H2O, which is the limiting reagent? 149. ## Chemistry When a solution is made from 1.0 g of KI and 90 mL of water, what is its molarity? 150. ## CHEMISTRY What are the ka and kb reactions of NaHSO3 ? I tried it out and thought it would be ka= HSO3-(aq) + H2O(l)= H+(aq) + SO3-(aq) kb= HSO3-(aq) + H2O(l)= H3O+(aq)+ OH-(aq) Would that be right? or is it wrong? please can you take a look at it . Thanks. 151. ## maths If (2x-3) is a factor of 2x raised to the power four-3xsquare+15x-15k,find the value of (3k- root of 5 k). asked by maths rockzzz 152. ## maths one side of a right triangle is 12.5 ft the perimeter is 38.7 ft. what is the length of the hypotenuse and the other unknown side? 157. ## algebra A charitable fund's balance is modeled by y = -4.1x^2 + 7.1x + 145, where y is the balance and x is the years after 1996. If no money is added to the fund, in how many years will the fund have a balance of 0? Round to the nearest year. 158. ## math a chemistry experiment calls for 30% sulfuric acid solution. if the supply room has only 50% and 20% sulfuric acid solution on hand,how much of each should be mixed to obtain 12 litres of 30% solution? 159. ## physics Vector {\bf{\vec V}}_1 is 6.3 units long and points along the negative x axis. Vector {\bf{\vec V}}_2 is 8.5 units long and points at 30^\circ to the positive x axis. 160. ## physics Vector {\bf{\vec V}}_1 is 6.3 units long and points along the negative x axis. Vector {\bf{\vec V}}_2 is 8.5 units long and points at 30^\circ to the positive x axis. asked by help me plzzzzz 161. ## Physics In an amusement ride, fun seeking people enter a 7.4 meter diameter cylinder and stand up against the wall. The cylinder spins fast enough so that when the bottom drops down 2 meters, the people remain pinned to the wall. The cylinder continues to rotate 162. ## philosophy This is a case and i need to identify which to theories apply to it out these three Kant's deontology, Mill's utilitarianism, and Hursthouse's virtue ethics. In the summer of 1945, President Harry Truman ordered the atomic bombing of the Japanese cities of 163. ## Math There are 3 spider webs. The 1st web is twice as large as the second web. The 3rd web is 3 inches larger than the 1st web. The combined width of the 3 webs is 1 foot 6 in. How long is each web? 164. ## Math Leo has 8 more markers than colored pencils. He has a total of 32 markers and colored pencils. How many of each kind does he have? 165. ## Calculus If a_n >0 and b_n >0 and series ∑ sqrt( (a_n)^2 +(b_n)^2 ) converges, then ∑a_n and ∑b_n both converge. True or false? If true, why? If false, give a counterexample. 166. ## physics A ball is thrown horizontally from the top of a building 30.1 m high. The ball strikes the ground at a point 66.8 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s 167. ## social studies the two major tribes of the northeastern part of the Eastern Woodlands were the 168. ## statistics 30 customers the first week and 55 customers the second week and 80 customers the third week, what would you project for week 4 169. ## Math Using known Facts to Find Unknown Facts. You can use breaking apart to find a product. Find 4 x 5. example: 4 x 5= 2x5+2x5 =10 + 10 =20 170. ## Physics Multiple choice please help. How much kinetic energy does a 1 kg ball have if it has a velocity of 10 m/s? a. 100 J. b. 50 J. c. 10 J. d. 5 J. The airbag in a car works in a collision by . . . a. increasing the time of the impact to the person. b. reducing 171. ## Chinese Where would I look to find out how to pronounce the Chinese name Xuelin? 172. ## Algebra 2 1/16(x)^3a + 1/2(y)^(6a)(z)^(9b) 173. ## Math CRANES Season 1: Wins 38 Season 2: Wins 42 Season 3: Wins 31 Season 4: Wins 50 Season 5: Wins 31 Season 6: Wins 48 PANTHERS Season 1: Wins 36 Season 2: Wins 42 Season 3: Wins 40 Season 4: Wins 40 Season 5: Wins 42 Season 6: Wins 40 Which team had the 174. ## Physics At 7.65 cents per kilowatt-hour, what does it cost to operate a 8.00 hp motor for 7.00 hr? 175. ## Math for what value of k will the function f(x)=kx^2-4k+k have no zeros 176. ## Math the graph of the function f(x)=x^2-kx+k+8 touches the x-axis at one point. What are the possible values of k 177. ## Algebra I'm working midpoints, distance, circles and standard form and general form. I have some questions with my answers to the first two-I'm stuck on #3 1. What is the distance between points (0,0) and (-3,5) My answer: sqrt34 2.What is distance between (3,1) 178. ## Math A computer game starts with a set of 144 tiles separated into 36 equal groups.Sara played the game until there were 8 groups of tiles left.How many tiles were left when Sara finished? How do you get 32 tiles?Can you please explain?Thanks 179. ## algebra how do you get square root of (72)+ square root of (18) + square root of (50) 180. ## algebra Can someone show me how to use fractional exponents to solve 5^(1/4)*5^(2/3) and then convert to radicals and simplify to the lowest terms? 181. ## CHEMISTRY What are the ka and kb reactions of NaHSO3 ? i know you got to ignore Na because is a strong base but i don't know how to get the actual ka and kb ! pleasee help me!!!!!!!!!!!! 182. ## Math a) f(x) = 4.1x +92.16 b) f(x) = 17.9x + 19.36 c) f(x) = 8.8x + 55.4 ANSWERS: a) 4.8 b) 0.93 c) 2.2, i just have no clue how they got these answers EXTRA INFO: The demand function for a new product is p(x)= - 4x +42.5, where x is the quantity sold in 183. ## Math The marching band, cheerleaders, and chaperones are traveling by bus to play in the Citrus Bowl parade in Orlando. There are 216 bank members, 32 cheerleaders, and 40 chaperones making the trip. They can rent any number of one type of bus shown in the 184. ## 5th Grade Science Creation of landforms?? 185. ## COMM 215 Please guide me on what I have done incorrectly: Part I: True or False 1. The following word group is a FRAGMENT: That movie, one of my favorites. X True __False 2. The following word group is a FRAGMENT: The local baseball team, needing a good pitcher I need assisatnce with this. I do not want the answer I just want some guidence on how I should start it. The width of a rectangle is 5 ft, its length is (3x+2) ft, and its area is 75 ft². Find x 187. ## Math The equation x^2+px+q=0, q cannot be equal to 0, has two unequal roots such that the squares of the roots are the same as the two roots. Calculate the product pq. asked by Ricky C. 188. ## Science A container with equal sides of 8 cm is filled with 1200g of fluid. What is the force in Newtons on the bottom of the container? 189. ## Spanish I need to match the subject pronouns with the verb llevar these are the verbs: llevas, llevamos, llevo, llevan, lleva these are the subj. pronouns: yo, el, nosotros, ellas, tu, ella, ellos, ustedes. PLEASE HELP 190. ## math if their are 72 students 2/3 of the students are 12 years old how many are 12 years old 191. ## Math How do I convert a decimal to a fraction. .350 My answer: 350/100 192. ## programming Part A: Write a code segment that inputs an integer from cin, and then outputs a row of that many stars on cout. Part B: Change the code segment to read a series of numbers from file indata, and print a row of stars on cout for each number read. 193. ## Psychology example of work smarter, not harder. I need one sentence of how this phrase could apply to launching internet search. 194. ## history what safeguards are containded in the constitution to protect the states from violations of their rights 195. ## SOCIAL STUDIES despite such limits athens is still admired as an early model of democracy 196. ## math I am a 4 digit number greater than 9,000 my omes digit is 3 and my hundreds digit is the smallest prime number greater than 5 the sum of my digits is 20. What number am i? 197. ## geometry draw two points r and s then sketch rs add a point t on ray so that s is beetween r and t 198. ## SCIENCE Craig sees a glass with ice tea sitting on the table. He notices that there is a great deal of moisture on the outside of the glass and a puddle of water on the table. He asks his family what they think causes the water to form on the outside of the glass. 199. ## Math( help please ! ) For what values of k will the function f(x)=3x^2+4x+k=0 have no zeros, one zero, 2 zeros 200. ## Math Prove that if "a" is a unit in a ring R with unity, then "a" is not a zero divisor in R. 201. ## grammar Does this sentence need a comma? Gum and drinks are not allowed, because they can create disasters in the dirt-free zone. 202. ## algabra cubes 1 and 2 are similar if the volume of cube 2 is 45 and the scale factor between sides of the two is 2:3 what is the volume of cube 1? 203. ## science In the 1980s a popular environmental slogan was "Reduce, Reuse and Recycle" and now "Go Green," is a popular message. Consider the Reusable Grocery Bags sold in markets everywhere. What are the pros and cons that you see in this latest trend? Give examples 204. ## Algebra How do you factor out y=x^3-9x^2+9 205. ## Physics 110.817 asked by swim chick 206. ## physics George takes off for a long jump at an angle of 28 degrees, and lands 6.9 m away. How fast was he when he jumped? 207. ## statistics Assuming that the data are normally distributed with a mean of 45 and a standard deviation of 3.25, what is the z-score for a value of 40? 208. ## math Find the number of decibels for the power of the sound. Round to the nearest decibel. A rock concert, 5.36 multiplied by 10-6 watts/cm2 D = mark dB 209. ## math The half-life of 234U, uranium-234, is 2.52 multiplied by 105 yr. If 98.7% of the uranium in the original sample is present, what length of time (to the nearest thousand years) has elapsed? yr 210. ## math linear equations 211. ## physics A 0.82 kg physics book with dimensions of 32.4 cm by 20.1 cm is on a table. What force does the book apply to the table? Answer in units of N 212. ## MCC Design a modular program that asks the user to enter the replacement cost of a building and then displays the minimum amount of insurance he or she should buy for the property. 213. ## math there are seven pies if i take away 2 1.3 how many are left 214. ## CHEMISTRY What are the ka and kb reactions of NaHSO3 ? i know you got to ignore Na because is a strong base but i don't know how to get the actual ka and kb ! pleasee help me!!!!!!!!!!!! 215. ## English Thank you. Can you please clarify this doubt to me? If the answer is "since 1998" can the question be: How long is it since you got married"? Is the sentence "How long ago did you marry"? wrong. The verb is "get married". When did you divorce your fist 216. ## English I forgot to add this sentence. 1)How long is it since it last snowed? ..... December. (Which is the correct preposition?) 217. ## Economics Your demand curve for baseball tickets is as follows: price per ticket Qd $1 4$2 3 $10 2$20 1 YOu buy 2 tickets at $10 each. THe value yo you of the two tickets is what? It say the answer is$30. I do not understand why you have to add 10 and 20. I 218. ## physical chemistry if 10 litres of a solution of na2co3 contains 21.2g of na2co3, what is its strength? 219. ## Phys sci we are makeing rube goldberg machiens but are problem is hot to activate the pulley by putting the string in to the fire befor its time and makeing the machine do other things. 220. ## algebra 26.8 over .01 it says to evaluate each expression 221. ## algebra 26.8 over .01, I don't know what to do with this it is set up so it looks like a fraction, could this mean divide 26.8 by .01? 222. ## English Could you please check these sentences for me? Thank you very much for your invaluable help. 1) I find it difficult to translate the following description into English since I don't know how old the students are in their fifth or sixth class. How old are 223. ## math a rectangle has an area of 330m^2 , one side is 7 m longer than the other what are the dimensions of this rectangle 224. ## Math Mike bought 252 bricks to build a wall in his yard. Each row has 14 bricks. Each bricks is 3 inches tall. How many inches tall will be wall be if Mike uses every brick? 225. ## Comm 215 Is this correct 49. To succeed in the new economy, workers will have to learn throughout their careers, adapt to changing circumstances, and have self-motivation. 226. ## SOCIAL STUDIES In 1900 for health reasons the course of what river was changes in the U.S. making it world's only river to flow backwards? 227. ## Chemistry Naoh+h2so4--------h20 +na2so4 Now the question is what number of moles of naoh require to neutralize 1 mole of h2sor. 228. ## problem solving random sampling for tensile strenght 6500 5800 6150 6050 5900 6150 6200 5880 6130 5880dtetermine % 6000psi 67000psi 68000psi 229. ## algebra i have tried alot of different ways of doing this problem but i cant find the right answer. QUESTION:2and2/3 plus 2/3 plus 1and1/3 230. ## algebra Use the quadratic formula to find the zeros of the function. Round to tenths if necessary. y = 17x^2 - 21 231. ## Physics A roller coaster travels 41.9 m at an angle of 15.0◦ above the horizontal. How far does it move horizontally? Answer in units of m 232. ## algebra each week you gave a quiz in social studies and science class. find the probability both quizzes given on the same day tell if independent or dependent 233. ## Poetry This is my poem. :P Does it fit into the theme of loneliness? Loneliness Never Sleeps. I scream, call for help, my lungs frozen in ice I can't breathe, I'm alone, my screams don't suffice My heart pounding hard, wishing someone would hear I begin to taste 234. ## ETH125 Did Hungarians face prejudice, segregation, racism, or any combination of the three? 235. ## Math Janice is running a 7613 meter race. Assume the she runs at a constant speed of 287 meters per minute. Express the function f(t) that gives her distance from the finishing line t minutes after the start of the race. 236. ## physics A car has a mass of 1.34 × 10^3 kg. If the force acting on the car is 6.46 × 10^3 N to the east, what is the car’s acceleration? 237. ## Math( help please ! ) for what value of k will the function f(x)=kx^2-4k+k have no zeros the graph of the function f(x)=x^2-kx+k+8 touches the x-axis at one point. What are the possible values of k 238. ## Math( help please ! ) a) f(x) = 4.1x +92.16 b) f(x) = 17.9x + 19.36 c) f(x) = 8.8x + 55.4 ANSWERS: a) 4.8 b) 0.93 c) 2.2, i just have no clue how they got these answers EXTRA INFO: The demand function for a new product is p(x)= - 4x +42.5, where x is the quantity sold in 239. ## Math( help please ! ) In class we learnt operation with radicals how would you answer a question like this determine the length of the line segment from A(-2,7), to B (4,1)? 240. ## English what is immperative sentence and interrogative sentence 241. ## physics A girl starting from rest at the top of a 40 m jump, inclined at 30º above the horizontal, has 0.15 coefficient of friction. The jump ends with a frictionless, horizontal part. How far does she land away from the jump, which is 50 m above the horizontal 242. ## Functions write a function. f(x), that represents the cost of renting a car for x days with a total of 80 kilometers if there is a fixed charge of $17 per day and a charge of$0.35 per kilometer. How do you think the mode of communication—written, verbal, or electronic—influences the interpretation of a message? 244. ## math For a fixed element 'a' of a ring R, prove that the set {x exists in R such that ax=0)is a subring of R. 245. ## Stats The linear correlation coefficient is a numerical value that ranges from -1.00 to +1.00. Describe in a sentence or two the meaning of each of these correlation coefficients: a) -1.00 b) +1.00 246. ## physics The force of tension of a rope is 300N. If a man is 50 m above the ground, what is the longest time it could take him to get to the bottom? 247. ## SCIENCE How much variables shoulld be changed? 248. ## English/writing But would you put commas, periods or anything like that in it. Would every sentence begin with a capital? 249. ## HEalth how much carboydrates we need in percentage in a day how much Fats we need in a day in percentage Why do we need fat how much percentage of protein we need what is complex carbohydreates 250. ## Physics I am given two points on an acceleration vs time graph: (-2,13),(4,0). When t = -2s, v = 10 m\s. What velocity at t = 6s? I believe acceleration average is (-13/6), but how do I solve the rest get the correct answer, which is 44.7 m/s? Any help or hints 251. ## sci Extinction is a natural selection process. Should humans strive to preserve a representative sample of all biomes or aquatic zones? Why should humans be concerned with the extinction rate? 252. ## biology what questions do affected people, potential parents, and families face when a diagnosis of the sickle cell anemia is made 253. ## English/writing If I am writing the following as a poem, how would I punctuate it- would each letter at the beginning be a captial, would it have commas or a period. We are room 204 Check out our door It is clear to see We choose to be drug free 254. ## Finance Consider a firm that has decided to make, but has not yet announced, a large “bonus” cash dividend amounting in the aggregate to $5 million. The firm has 1 million shares outstanding that sell for$20 each. The firm has no debt; there are no taxes; and 255. ## Chemistry Help!! which element do not combine readily with other elements because they already have eight electrons in their outermost shell? 256. ## Help With math Solve (3x-2)^2=2 asked by katie A 257. ## trig Solve system of inequalities by graphing Yx+4 281. ## math how you figure percents like percent of438 to 789 282. ## Statistics Find the normal approximation for the binomial probability that x = 5, where n = 12 and p = 0.7. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook. 283. ## English I still have a doubt. Could you check these sentences, too? I last saw Mary a month ago. The last time I saw Mary was a month ago. It has been two months since I last saw her. It has been two months since I saw her for the last time 284. ## Algebra 2 Factor. -16 + 17(5-y²)-(5-y²)² and [x + y + 1][x² - x(y + 1)+(y + 1)² 285. ## Calculus If a_n>0 and a_(n+1) 286. ## Math In class we started operation with radicals how would you answer a question like this determine the length of the line segment from A(-2,7), to B (4,1)? 287. ## science name 3 effects that can be caused by friction 288. ## statistics perform a two tail hypothesis test on both the legal services satisfaction and the sentence satisfaction variable's data using a .05 significance level. 289. ## math a tank can be filled by one pipe in 3 .5 hours and emptied by another in 4.2 hours.if both ipes are running, how long will it take to fill an empty tank? 290. ## physics A ball is thrown straight up with an intial velocity of 85 ft/s. At what times will it have a height of 200 ft? 291. ## math write 0.9,2/8,0.35 from least to greatest 292. ## science why do scientist only change one varieble at a time during a experiment ? 293. ## French What are some french adjectives that end in ant? I'm trying to make a french poem and need words to fit in the blanks: 1. Quand la vie est une fleur Chaque jour est ______ 2. Quand la vie est un livre Chaque jour est ______ 3. Quand la vie est une 294. ## programming Write a code segment that inputs an integer from cin, and then outputs a row of that many stars on cout. 295. ## jlc Three is the first number of pair and 8 is second if 50 is first number what is second number? 296. ## algebra Use the quadratic formula to find the roots of the equation. Round to tenths if necessary. 3x^2 - 4x - 8 = 0	2019-06-15 23:53:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.426932692527771, "perplexity": 2188.518533138646}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997501.61/warc/CC-MAIN-20190615222657-20190616004657-00182.warc.gz"}
http://www.formuladirectory.com/user/formula/331	HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS ## Force A force is any influence that causes an object to undergo a certain change, either concerning its movement, direction, or geometrical construction. In other words, a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate, or a flexible object to deform, or both. Force can also be described by intuitive concepts such as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newtons and represented by the symbol F. The original form of Newton's second law states that the net force acting upon an object is equal to the rate at which its momentum changes with time.If the mass of the object is constant, this law implies that the acceleration of an object is directly proportional to the net force acting on the object, is in the direction of the net force, and is inversely proportional to the mass of the object. ## $\mathrm{ma}$ Here,m=magnitude,a=acceleration ENTER THE VARIABLES TO BE USED IN THE FORMULA Similar formulas which you may find interesting.	2019-02-21 23:07:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7554314136505127, "perplexity": 179.1260081065378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247511174.69/warc/CC-MAIN-20190221213219-20190221235219-00017.warc.gz"}
https://electronics.stackexchange.com/questions/26539/arduino-row-column-scanning-not-working	# Arduino Row Column Scanning not working I tried to follow this tutorial but the whole matrix is always lit and only one row and one column does not light. I am pretty sure I did the wiring correct; I only have slightly edited the code since I dont have those potentionmeters to modify the x and y coordinate. Here is the code I am using: // 2-dimensional array of row pin numbers: const int row[8] = { 2,3,4,5,6,7,8,9 }; // 2-dimensional array of column pin numbers: const int col[8] = { 10,11,12,13,16,17,18,19 }; // 2-dimensional array of pixels: int pixels[8][8]; // cursor position: int x = 5; int y = 5; void setup() { Serial.begin(9600); // initialize the I/O pins as outputs: // iterate over the pins: for (int thisPin = 0; thisPin < 8; thisPin++) { // initialize the output pins: pinMode(col[thisPin], OUTPUT); pinMode(row[thisPin], OUTPUT); // take the col pins (i.e. the cathodes) high to ensure that // the LEDS are off: digitalWrite(col[thisPin], HIGH); } // initialize the pixel matrix: for (int x = 0; x < 8; x++) { for (int y = 0; y < 8; y++) { pixels[x][y] = HIGH; } } } void loop() { // draw the screen: refreshScreen(); delay(500); } // turn off the last position: pixels[x][y] = HIGH; // read the sensors for X and Y values: x++; x%=8; if (x==0) { y ++; y%=8; } pixels[x][y] = LOW; } void refreshScreen() { // iterate over the rows (anodes): for (int thisRow = 0; thisRow < 8; thisRow++) { // take the row pin (anode) high: digitalWrite(row[thisRow], HIGH); // iterate over the cols (cathodes): for (int thisCol = 0; thisCol < 8; thisCol++) { // get the state of the current pixel; int thisPixel = pixels[thisRow][thisCol]; // when the row is HIGH and the col is LOW, // the LED where they meet turns on: digitalWrite(col[thisCol], thisPixel); // turn the pixel off: if (thisPixel == LOW) { digitalWrite(col[thisCol], HIGH); } } // take the row pin low to turn off the whole row: digitalWrite(row[thisRow], LOW); } } Here is a photo: • To answer your question properly we will need some more information. Can you please post the exact code you are currently running on your Arduino. Can you also post a list of your connections to the matrix, list each connected Arduino pin next to which row or column it is connected to, and note if it's an anode or cathode. – Jim Feb 15 '12 at 13:53 • Remove delay(500). It turns of the row in the end so you don't manage to see the lights turn on before you turn them off again because of this delay. – chwi Dec 17 '14 at 8:16 It lights up, so that's a good start! The link you gave to the tutorial provides an example Arduino sketch, I assume this is the programme you are running to test your circuit? If you are using the sketch provided you'll need to connect 2 potentiometers to the Arduino's analogue pins (Analog 0 and Analog 1). The picture you have provided does not seem to show these potentiometers. If you do not have them connected then the circuit will not behave as you expected. I can't really tell if the rest of the circuit is wired up properly as it's quite hard to see from your image. The best way to display your circuit when asking questions like this is to draw up a schematic of your circuit, this makes it much clearer to read, it also gives you an opportunity to go over your circuit and double check all your connections. If you still can't get it working after that you will need to check that all the connections are good. Often with breadboarding the wires and solder-less connectors don't connect up properly, it can be really annoying! The best way to check for this problem is to do a continuity test on all your connections. You can do this with a multi meter, or if you haven't got one you can rig up a basic tester with an LED and coin cell battery, here's a cool one on Instructables that fits inside a pen! Good luck! EDIT: Now you've posted your code I can see why your not getting any movement/animation. It looks like you are not letting your refreshScreen(); function run for long enough to see the LED. When you run this function it scans through the rows and columns one step at a time, if you don't let it cycle a good few times the LEDs will only be on for a very brief period (too brief to see). The delay(500); is stopping the refreshScreen loop on each iteration and it's messing up your scanning. I also recommend using millis() instead of delay(), with delay your LEDs will flicker. PSEUDO CODE: unsigned long currentMillis = 0; unsigned long previousMillis = 0; void loop() { // millis() will return the time in milliseconds since the sketch started previousMillis = millis(); currentMillis = previousMillis; while(currentMillis - previousMillis < 500) { // draw the screen: refreshScreen(); currentMillis = millis(); } } 2nd EDIT: Judging by the problems you have been experiencing, I would suggest that you do not have your matrix connected correctly. Obtain the datasheet for your matrix so you can use it as a reference. Once you are certain of how the LEDs are set out inside your matrix, I would then go back to the tutorial and connect up the matrix from scratch. Please note that LED matrices are not all set out the same and they will differ from one manufacturer to the next. • thanks! well instead of the potentiometers i just set the x and y coordinte programmatically. and that is the only difference between my setup and the one in the tutorial. i have double checked the wiring and also the connections should be fine, since i can get all leds to light up individually with such a continuity test. – clamp Feb 13 '12 at 10:18 • Hummm I see. Make sure you have the same LED matrix, and that you haven't connected it in back to front, it's easy to get them backwards. Do you have a link to the datasheet for your matrix? – Jim Feb 13 '12 at 10:29 • The problem you describe makes it sound like you have your matrix in backwards. I take it your matrix should be lighting the row and column that's not lit? – Jim Feb 13 '12 at 10:33 • that is actually a good question. i dont know this myself. and from the tutorial it is hard to find out if the row and column should be lit or if everything else should be lit. – clamp Feb 13 '12 at 10:40 • The tutorial is supposed to light an individual LED. Instead of moving your connections you could always change the pin assignments in your sketch, but it's a good idea to get the matrix's datasheet so you have a reference for how the LEDs are set out – Jim Feb 13 '12 at 10:45 Using the code above with the inputs set as : const int row[8] = { 2,7,19,5,13,18,12,16 }; const int col[8] = { 6,11,10,3,17,4,8,9 }; Worked for me after turning the matrix around	2019-08-20 09:44:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2841353714466095, "perplexity": 1469.6449226222956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00328.warc.gz"}
https://byjus.com/question-answer/1-percent-cost-how-many-marks/	Question # $1$ percent cost how many marks? Open in App Solution ## Solution.Consider the total marks be $100$.Then, $1%=\frac{1}{100}$ For every correct answer, one mark is scored. Hence, $1$ percent cost $1$ marks. Suggest Corrections 9	2023-01-28 00:28:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8818687200546265, "perplexity": 8754.884189956561}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00579.warc.gz"}
https://kb.osu.edu/dspace/handle/1811/8870	# THE 1790 \AA TRANSITION OF 2-METHYLPROPENE (ISOBUTENE) Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/8870 Files Size Format View 1971-W-01.jpg 76.85Kb JPEG image Title: THE 1790 \AA TRANSITION OF 2-METHYLPROPENE (ISOBUTENE) Creators: McDiarmid, R. Issue Date: 1971 Publisher: Ohio State University Abstract: The vibronic structure of the absorption system originating around 1795 \AA in 2-methylpropene (isobutene) has been studied for the parent compound and for three symmetrically deuterated isomers. In addition to the sequences previously $reported,^{1}$ progressions are observed in the $C = C$ stretch, $-CH_{3}$ symmetric deformation, $=CH_{2}$ deformation, $=CH_{2}$ torsion and $-CH_{2}$ torsion vibrational modes. The observed origin is tentatively identified as a false origin. By elimintion, the transition is deduced to be $A_{2}\sigma_{{C-C}}$ ($P_{y}$, $b_{2}) \leftrightarrow (p_{x}, b_{1}$). Insufficient information is available to assign the transition more completely. Description: $^{1}$R. McDiarmid, Symp. on Mol. Structure and Spect., Sept. 1970."" Author Institution: National Institute of Arthritis and Metabolic Diseases, National Institutes of Health URI: http://hdl.handle.net/1811/8870 Other Identifiers: 1971-W-1	2015-03-01 04:40:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.536220133304596, "perplexity": 11400.24832188436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462206.12/warc/CC-MAIN-20150226074102-00306-ip-10-28-5-156.ec2.internal.warc.gz"}
http://www.astroexplorer.org/details/apjac98bbf2	Image Details Choose export citation format: Critical Mach Numbers for Magnetohydrodynamic Shocks with Accelerated Particles and Waves • Authors: J. Martin Laming J. Martin Laming 2022 The Astrophysical Journal 940 . • Provider: AAS Journals Caption: Figure 2. Contour plots of the first critical fast Mach number in the plasma β - θ 1 plane for ﹩{E}_{\mathrm{CR}}/{\rho }_{1}{v}_{1\| \| }^{2}=0.025﹩, 0.05, 0.1, and 0.2 are shown in the top left, top right, bottom left, and bottom right panels, respectively.	2022-11-28 18:40:18	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8490957021713257, "perplexity": 9577.497218015407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00760.warc.gz"}
https://ssconlineexam.com/1forums/10879/all-of-the-goods-which-are-scarce-and-limited-in-s	# All of the goods which are scarce and limited in supply are called [ A ] Luxury goods [ B ] Expensive goods [ C ] Capital goods [ D ] Economic goods Answer : Option D Explanation : In economics, a good is something that is intended to satisfy some wants or needs of a consumer and thus has economic utility. An economic good is a consumable item that is useful to people but scarce in relation to its demand, so that human effort is required to obtain it. In contrast, free goods (such as air) are naturally in abundant supply and need no conscious effort to obtain them. ssc online	2022-01-18 07:44:21	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8865345120429993, "perplexity": 3219.375452326323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00191.warc.gz"}
http://www.numericalexpert.com/articles/single_pass_stat/	# Single-Pass Online Statistics Algorithms ## 1. Introduction The "textbook" two-pass algorithm for the centered moments (variance, skewness, kurtosis, covariance) is obviously inefficient. There were suggested several alternative algorithms [1 - 16]. Unfortunately I could not find suitable one-pass windowed algorithms similar to the mowing average [17]. Therefore I decided to derive the equations and implement them myself. ## 2. Problem statement We assume that there is a stream of data: $x_{1},\, x_{2},\, x_{3},\,\ldots\,,\, x_{k}$ . Here $x_{k}$ is the most recent value. There are two variants of statistics: (i) a cumulative statistics of all $k$ values, and (ii) the statistics of $n$ the most recent values. The latter is also called a moving or a windowed statistics. Note that the cumulative statistics is also a windowed with $n=k$. ## 3. The first moment, mean value ### 3.1. Definition The first moment (moving average, mean value) of the length $n$ is defined as $$m_{k}(n)=\frac{1}{n}\sum_{i=k-n+1}^{k}x_{i}=\frac{1}{n}S_{k}(n)\label{eq:avg}$$ Below for $k=n$ we use the notation $X_{k}(k)=X_{k}$. ### 3.2. Cumulative calculations When all incoming values are incorporated into calculations ($n=k$), the above equation (\ref{eq:avg}) reduces to: $$m_{k}=m_{k}(k)=\frac{1}{k}\sum_{i=1}^{k}x_{i}=\frac{1}{k}S_{k}\label{eq:avgc}$$ From Eq. \ref{eq:avgc} we get the following recurrence: \begin{eqnarray} m_{k} & = & \frac{1}{k}S_{k}=\frac{1}{k}\left(x_{k}+S_{k-1}\right)=\frac{x_{k}}{k}+\frac{k-1}{k}m_{k-1}=m_{k-1}+\frac{x_{k}-m_{k-1}}{k}\nonumber \\ & = & m_{k-1}+\frac{\delta_{k}}{k}\label{eq:recc} \end{eqnarray} where $$\delta_{k}=x_{k}-m_{k-1}\label{eq:dk}$$ ### 3.3. Windowed (constant-length, moving) calculations In the case when only the last $n$ values are used, the recurrence is \begin{eqnarray} m_{k}(n) & = & \frac{1}{n}S_{k}(n)=\frac{1}{n}\left(x_{k}+S_{k-1}(n)-x_{k-n}\right)=\frac{x_{k}}{k}+\frac{k-1}{k}m_{k-1}=m_{k-1}+\frac{x_{k}-x_{k-n}}{n}\nonumber \\ & = & m_{k-1}(n)+\frac{\delta_{k}-\delta_{k-n}}{n}\label{eq:recw} \end{eqnarray} Here $$\delta_{k-n}=x_{k-n}-m_{k-1}\label{eq:dkn}$$ Note that (\ref{eq:recw}) reduces to (\ref{eq:recc}) if we set \begin{eqnarray} \delta_{k-n} & = & 0\nonumber \\ n & = & k\label{eq:w2c} \end{eqnarray} Instead of (\ref{eq:w2c}) we can also formally define $\delta_{0}=0$. ## 4. Central moments ### 4.1. Definitions Similar to (\ref{eq:avg}), the p-th order central moments are defined as $$M_{p,k}(n)=\frac{1}{n}\sum_{i=k-n+1}^{k}\left[x_{i}-m_{k}(n)\right]^{p}=\frac{1}{n}S_{p,k}(n)\label{eq:avgp}$$ ### 4.2. Windowed variance ($p=2$) For the case of $p=2$ Eq. \ref{eq:avgp} reduces to $$M_{2,k}(n)=\frac{1}{n}\sum_{i=k-n+1}^{k}\left[x_{i}-m_{k}(n)\right]^{2}=\frac{1}{n}S_{p,k}(n)\label{eq:avg2}$$ It can be rewritten using (\ref{eq:recw}): \begin{eqnarray} S_{2,k}(n) & = & \sum_{i=k-n+1}^{k}\left[x_{i}-m_{k}(n)\right]^{2}=\sum_{i=k-n+1}^{k}\left[x_{i}-m_{k-1}(n)-\frac{\delta_{k}-\delta_{k-n}}{n}\right]^{2}\nonumber \\ & = & \sum_{i=k-n+1}^{k}\left\{ \left[x_{i}-m_{k-1}(n)\right]^{2}-2\left[x_{i}-m_{k-1}(n)\right]\frac{\delta_{k}-\delta_{k-n}}{n}+\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{2}\right\} \label{eq:S2w} \end{eqnarray} or \begin{eqnarray} S_{2,k}(n) & = & S_{2,k-1}(n)+\delta_{k}^{2}-\delta_{k-n}^{2}-2(\delta_{k}-\delta_{k-n})[m_{k}(n)-m_{k-1}(n)]+\frac{(\delta_{k}-\delta_{k-n})^{2}}{n}\nonumber \\ & = & S_{2,k-1}(n)+\delta_{k}^{2}-\delta_{k-n}^{2}-\frac{(\delta_{k}-\delta_{k-n})^{2}}{n}\label{eq:S2wrec} \end{eqnarray} This gives the recurrence formula for the windowed online variance calculations. ### 4.3. Cumulative variance calculations In order to get the cumulative formula, we can formally apply the conditions (\ref{eq:w2c}): $$S_{2,k}=S_{2,k-1}+\delta_{k}^{2}-\frac{\delta_{k}^{2}}{k}=S_{2,k-1}+\frac{k-1}{k}\delta_{k}^{2}\label{eq:rec2c}$$ This can be also rewritten as $S_{2,k}=S_{2,k-1}+(x_{k}-m_{k})(x_{k}-m_{k-1})$ ## 5. Higher central sums ### 5.1. Windowed calculations for $p=3$ Similar to (\ref{eq:S2w}): \begin{eqnarray} S_{3,k}(n) & = & \sum_{i=k-n+1}^{k}\left[x_{i}-m_{k}(n)\right]^{3}=\sum_{i=k-n+1}^{k}\left[x_{i}-m_{k-1}(n)-\frac{\delta_{k}-\delta_{k-n}}{n}\right]^{3}\nonumber \\ & = & \sum_{i=k-n+1}^{k}\left\{ \begin{array}{c} \left[x_{i}-m_{k-1}(n)\right]^{3}-3\left[x_{i}-m_{k-1}(n)\right]^{2}\frac{\delta_{k}-\delta_{k-n}}{n}\\ +3\left[x_{i}-m_{k-1}(n)\right]\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{2}-\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{3} \end{array}\right\} \label{eq:S3w} \end{eqnarray} or $$S_{3,k}(n)=S_{3,k-1}(n)+\delta_{k}^{3}-\delta_{k-n}^{3}-\frac{3}{n}(\delta_{k}-\delta_{k-n})[S_{2,k-1}(n)+\delta_{k}^{2}-\delta_{k-n}^{2}]+\frac{2}{n^{2}}(\delta_{k}-\delta_{k-n})^{3}\label{eq:S3wrec}$$ ### 5.2. Cumulative calculations for $p=3$ Using the same trick and applying the condition (\ref{eq:w2c}), we get: $$S_{3,k}=S_{3,k-1}+\delta_{k}^{3}-\frac{3}{k}\delta_{k}[S_{2,k-1}+\delta_{k}^{2}]+\frac{2}{k^{2}}\delta_{k}^{3}$$ or $$S_{3,k}=S_{3,k-1}-\frac{3}{k}S_{2,k-1}\delta_{k}+\frac{(k-1)(k-2)}{k^{2}}\delta_{k}^{3}$$ ### 5.3. Windowed calculations for $p=4$ Similar to (\ref{eq:S2w}) and (\ref{eq:S3w}): \begin{eqnarray} S_{4,k}(n) & = & \sum_{i=k-n+1}^{k}\left[x_{i}-m_{k}(n)\right]^{4}=\sum_{i=k-n+1}^{k}\left[x_{i}-m_{k-1}(n)-\frac{\delta_{k}-\delta_{k-n}}{n}\right]^{4}\nonumber \\ & = & \sum_{i=k-n+1}^{k}\left\{ \begin{array}{c} \left[x_{i}-m_{k-1}(n)\right]^{3}-4\left[x_{i}-m_{k-1}(n)\right]^{3}\frac{\delta_{k}-\delta_{k-n}}{n}+\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{4}\\ +6\left[x_{i}-m_{k-1}(n)\right]^{2}\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{2}-4\left[x_{i}-m_{k-1}(n)\right]\left(\frac{\delta_{k}-\delta_{k-n}}{n}\right)^{3} \end{array}\right\} \end{eqnarray} or \begin{eqnarray} S_{4,k}(n) & = & S_{4,k-1}(n)+\delta_{k}^{4}-\delta_{k-n}^{4}-\frac{4}{n}(\delta_{k}-\delta_{k-n})[S_{3,k-1}(n)+\delta_{k}^{3}-\delta_{k-n}^{3}]\nonumber \\ & + & \frac{6}{n^{2}}(\delta_{k}-\delta_{k-n})^{2}[S_{2,k-1}(n)+\delta_{k}^{2}-\delta_{k-n}^{2}]-\frac{3}{n^{3}}(\delta_{k}-\delta_{k-n})^{4} \end{eqnarray} ### 5.4. Cumulative calculations for $p=4$ Using the same trick and applying the condition (\ref{eq:w2c}), we get: $$S_{4,k}=S_{4,k-1}+\delta_{k}^{4}-\frac{4}{k}\delta_{k}[S_{3,k-1}+\delta_{k}^{3}]+\frac{6}{k^{2}}\delta_{k}^{2}[S_{2,k-1}+\delta_{k}^{2}]-\frac{3}{k^{3}}\delta_{k}^{4}$$ or $$S_{4,k}=S_{4,k-1}-\frac{4}{k}S_{3,k-1}\delta_{k}+\frac{6}{k^{2}}S_{2,k-1}\delta_{k}^{2}+\frac{(k-1)(k^{2}-3k+3)}{k^{2}}\delta_{k}^{4}$$ ## 6. Covariance The above approach can be generalized for the calculation of the covariance matrix elements. ### 6.1. Definition The covariance of the components $x_{\mu}$ and $x_{\nu}$ is defined as $$c_{\mu\nu,k}(n)=\frac{1}{n}\sum_{i=k-n+1}^{k}\left[x_{\mu,i}-m_{\mu,k}(n)\right]\left[x_{\nu,i}-m_{\nu,k}(n)\right] = \frac{1}{n}C_{\mu\nu,k}(n)\label{eq:covw}$$ Here $m_{\mu,k}(n)$ is the mean value for the $\mu$-th component. It obeys the generalized equation (\ref{eq:recc}): $$m_{\mu,k}(n)=m_{\mu,k-1}(n)+\frac{\delta_{\mu,k}-\delta_{\mu,k-n}}{n}\label{eq:mmunuk}$$ with Eqs. (\ref{eq:dk}) and (\ref{eq:dkn}) modified accordingly. ### 6.2. Windowed calculations Using the recurrence (\ref{eq:mmunuk}) for each component we have $$C_{\mu\nu,k}(n)=\sum_{i=k-n+1}^{k}\left[x_{\mu,i}-m_{\mu,k-1}(n)-\frac{\delta_{\mu,k}-\delta_{\mu,k-n}}{n}\right]\left[x_{\nu,i}-m_{\nu,k-1}(n)-\frac{\delta_{\nu,k}-\delta_{\nu,k-n}}{n}\right]$$ The above equation can be rewritten as $$C_{\mu\nu,k}(n)=\sum_{i=k-n+1}^{k}\left\{ \begin{array}{c} \left[x_{\mu,i}-m_{\mu,k-1}(n)\right]\left[x_{\nu,i}-m_{\nu,k-1}(n)\right]-\left[x_{\mu,i}-m_{\mu,k-1}(n)\right]\frac{\delta_{\nu,k}-\delta_{\nu,k-n}}{n}\\ \frac{\delta_{\mu,k}-\delta_{\mu,k-n}}{n}\left[x_{\nu,i}-m_{\nu,k-1}(n)\right]+\frac{\delta_{\mu,k}-\delta_{\mu,k-n}}{n}\,\frac{\delta_{\nu,k}-\delta_{\nu,k-n}}{n} \end{array}\right\}$$ or $$C_{\mu\nu,k}(n)=C_{\mu\nu,k-1}(n)+\delta_{\mu,k}\delta_{\nu,k}-\delta_{\mu,k-n}\delta_{\nu,k-n}-\frac{(\delta_{\mu,k}-\delta_{\mu,k-n})(\delta_{\nu,k}-\delta_{\nu,k-n})}{n}\label{eq:Cmunuk}$$ This gives the recurrence formula for the windowed online covariance calculations. ### 6.3. Cumulative calculations Applying the conditions (\ref{eq:w2c}) to each component, we get: \begin{eqnarray} C_{\mu\nu,k} & = & C_{\mu\nu,k-1}+\delta_{\mu,k}\delta_{\nu,k}-\frac{\delta_{\mu,k}\delta_{\nu,k}}{k}\nonumber \\ & = & C_{\mu\nu,k-1}+\delta_{\mu,k}\delta_{\nu,k}-\frac{k-1}{k}\delta_{\mu,k}\delta_{\nu,k} \end{eqnarray} This gives the recurrence formula for the cumulative online covariance calculations. Note that for $\mu=\nu$ the above equations reduce to the variance formulas. ## 7. Implementation My Python implementation can be found on GitHub: https://github.com/dr-nikolai/online_stat. It also includes the Jupyter notebook with a test/demo. Note that this code is not fully optimized. It is primarily an illustration to this article. ### Acknowledgement I am grateful to Marko Draisma who pointed to a typo in Eq. 18. Nikolai Shokhirev, 2013. Single-Pass Online Statistical Algorithms, http://www.numericalexpert.com/articles/single_pass_stat ### References 1. Chan, T. F.; Golub, G. H. and LeVeque, R. J., Updating formulae and a pairwise algorithm for computing sample variances, COMPSTAT 1982 5th Symposium held at Toulouse 1982, 1982, 30-412 2. Chan, T. F.; Golub, G. H. and LeVeque, R. J., Algorithms for computing the sample variance: Analysis and recommendations The American Statistician, Taylor & Francis Group, 1983, 37, 242-247 3. Datar, M.; Gionis, A.; Indyk, P. and Motwani, R., Maintaining stream statistics over sliding windows, SIAM Journal on Computing, SIAM, 2002, 31, 1794-1813 4. Zivot, E. and Wang, J., Rolling Analysis of Time Series, Modeling Financial Time Series with S-Plus, Springer, 2003, 299-346 5. Pébay, P., Formulas for robust, one-pass parallel computation of covariances and arbitrary-order statistical moments. Sandia National Laboratories, 2008 6. Alexander, C., Moving Average Models for Volatility and Correlation, and Covariance Matrices, in Handbook of Finance, John Wiley & Sons, Inc., 2008, 2-14 7. Finch, T., Incremental calculation of weighted mean and variance, University of Cambridge, 2009, 4, 11-5 8. Bennett, J.; Grout, R.; Pébay, P.; Roe, D. and Thompson, D., Numerically stable, single-pass, parallel statistics algorithms, Cluster Computing and Workshops, 2009. CLUSTER'09. IEEE International Conference on, 2009, 1-8 9. Choi, M. and Sweetman, B., Efficient calculation of statistical moments for structural health monitoring. Structural Health Monitoring, SAGE Publications, 2010, 9, 13-24 10. McCrary, S., Implementing Algorithms to Measure Common Statistics, Available at SSRN 2695198, 2015 11. Wikipedia, Algorithms for calculating variance, 2016. 12. Cook, J. D., Accurately computing running variance, 2008. 13. Cook, J. D., Computing skewness and kurtosis in one pass, 2008. 14. TurnerSR, Single-pass, parallel statistics algorithms for mean, variance, and standard deviation, 2014. 15. Jenks, G., RunStats: Computing Statistics and Regression in One Pass, 2015. 16. Babcock, B.; Datar, M.; Motwani, R. and O'Callaghan, L. Maintaining variance and k-medians over data stream windows Proceedings of the twenty-second ACM SIGMOD-SIGACT-SIGART symposium on Principles of database systems, 2003, 234-243 17. Wikipedia, Moving average, 2016. © Nikolai Shokhirev, 2012-2018 email: nikolai(dot)shokhirev(at)gmail(dot)com Count:	2018-01-18 05:38:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 17, "x-ck12": 0, "texerror": 0, "math_score": 0.9936268329620361, "perplexity": 4302.669905258394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887067.27/warc/CC-MAIN-20180118051833-20180118071833-00244.warc.gz"}
https://math.stackexchange.com/questions/921898/how-should-i-think-about-the-complex-exponential-form-of-sinusoid-waves	# how should I think about the complex-exponential form of sinusoid waves? Say there's a sinusoid wave with amplitude $A$, frequency $\omega$, and phase shift $\psi$, then one way to write it is $A cos(\omega t - \psi)$. But it can also be written as $Re(Ae^{i(\omega t - \psi)})$, and using this complex-exponential form seems to be common in engineering, physics, and anywhere with fourier/laplace transforms. I understand why the real part of the form above is a cosine from Euler's formula $e^{i\theta} = cos(\theta) + isin(\theta)$, and I understand the geometric intuition of complex numbers as points on a plane. However, what information are we losing by throwing away the imaginary part of the complex-exponential? $Im(Ae^{i(\omega t - \psi)}) = Asin(\omega t - \psi)$, which is the real part shifted a bit, so it seems to be nothing. But it still makes me uncomfortable to use $Ae^{i(\omega t - \psi)}$ in place of $A cos(\omega t - \psi)$ since I don't understand why adding $i sin(\omega t - \psi)$ (to get the rhs of euler's) is justifiable to begin with. So how do I think of the complex exponential form of sinusoidal waves? When someone says "there's a sinusoid wave with amplitude $A$, frequency $\omega$, and phase shift $\psi$", can I safely use the complex form instead of the cosine form in my calculations, and just take the real part at the end of it? $$u = \frac{1}{2}\mathbf{E}.\mathbf{D}+\frac{1}{2}\mathbf{B}.\mathbf{H}$$ $$\frac{du}{dt}= \frac{1}{2}\left( \dot{\mathbf{E}}.\mathbf{D}+\mathbf{D}.\dot{\mathbf{E}}+\dot{\mathbf{B}}.\mathbf{H}+\mathbf{B}.\dot{\mathbf{H}} \right)$$ Now suppose $\mathbf{E}$ and $\mathbf{D}$ vary sinusoidally. In calculating $\dot{\mathbf{E}}.\mathbf{D}$ you get different answers, when using the complex representation, depending on when you take the real part: $$\mathcal{R}\{\dot{\mathbf{E}}.\mathbf{D}\}\neq\mathcal{R}\{\dot{\mathbf{E}}\}.\mathcal{R}\{\mathbf{D}\}$$	2022-06-26 20:34:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8883869647979736, "perplexity": 174.98386903015833}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103271864.14/warc/CC-MAIN-20220626192142-20220626222142-00036.warc.gz"}
https://research.utwente.nl/en/publications/gromov-wasserstein-distance-based-object-matching-asymptotic-infe	Gromov-Wasserstein Distance based Object Matching: Asymptotic Inference Christoph Alexander Weitkamp, Katharina Proksch, Carla Tameling, Axel Munk Research output: Working paper Abstract In this paper, we aim to provide a statistical theory for object matching based on the Gromov-Wasserstein distance. To this end, we model general objects as metric measure spaces. Based on this, we propose a simple and efficiently computable asymptotic statistical test for pose invariant object discrimination. This is based on an empirical version of a $\beta$-trimmed lower bound of the Gromov-Wasserstein distance. We derive for $\beta\in[0,1/2)$ distributional limits of this test statistic. To this end, we introduce a novel $U$-type process indexed in $\beta$ and show its weak convergence. Finally, the theory developed is investigated in Monte Carlo simulations and applied to structural protein comparisons. Original language English arXiv.org Published - 22 Jun 2020 Keywords • math.ST • stat.TH • 62E20, 62G20, 65C60 (Primary) 60E05 (Secondary)	2021-06-18 01:02:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.405886173248291, "perplexity": 1992.0285060483227}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00344.warc.gz"}
https://www.bdg-academy.com/avocado-toast-pzenf/archive.php?page=451ade-generalised-linear-model-for-dummies	For this purpose, probabilistic programming frameworks such as Stan, PyMC3 and TensorFlow Probability would be a good choice. The exponential family includes normal, binomial, Poisson, … So linear regression is all you need to know? Today, it remains popular for its clarity, richness of content and direct relevance to agricultural, biological, health, engineering, and other applications. As the logistic function returns values between 0 and 1 for arbitrary inputs, it is a proper link function for the binomial distribution. Jagadeesh Rajarajan 's answer is correct, but I am not sure it is in layman's terms. What's a good way of graphically representing a very large number of paired datapoints? Why no one else except Einstein worked on developing General Relativity between 1905-1915? GAMs are just GLMs 2. Statistical researchers often use a linear relationship to predict the (average) numerical value of Y for a given value of X using a straight line (called the regression line). It is primarily the potential for a continuous response variable. Generalized Linear Models For Dummies Author: learncabg.ctsnet.org-Mathias Kluge-2020-10-03-02-51-37 Subject: Generalized Linear Models For Dummies Keywords: Generalized Linear Models For Dummies,Download Generalized Linear Models For Dummies,Free download Generalized Linear Models For Dummies,Generalized Linear Models For Dummies PDF Ebooks, Read Generalized Linear Models For Dummies … Generalized Linear Model For Dummies actually all special cases of the generalized linear model. The models I’ve explained so far uses a typical combination of probability distribution and link function. ), Poisson (contingency tables) and gamma (variance components). By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Learning GLM lets you understand how we can use probability distributions as building blocks for modeling. Generalized linear models with examples in R. Springer, New York, NY. If you’d like to apply statistical modeling in real problems, you must know more than that. Show me. 1β. Hierarchical data usually call for LMM implementation. I'm trying to get to grips with this topic, and it's proving tough. For example, let’s consider the following data. Link function literally “links” the linear predictor and the parameter for probability distribution. Generalized Linear Models (‘GLMs’) are one of the most useful modern statistical tools, because they can be applied to many different types of data. 1 2β. You’ve probably heard of more than one of them and you’ve probably also heard that each one is an extension of our old friend, the general linear model.. Generalized linear models have become so central to effective statistical data analysis, however, that it is worth the additional effort required to acquire a basic understanding of the subject. Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 Qixiang Fang and Rens van de Schoot Last modified: date: 14 October 2019. Modelling θ ∝ t, just make a few angle measurements at sunrise/midday/sunset and adjust these time points to 6am/12am/6pm. The next thing to try is a generalized linear model. Viewed 757 times 0 $\begingroup$ I'm trying to get to grips with this topic, and it's proving tough. Prepared by ; Louise Francis ; Francis Analytics and Actuarial Data Mining, Inc. www.data-mines.com ; September 18, 2005 ; 2 Objectives. Normal, Poisson, and binomial responses are the most commonly used, but other distributions can be used as well. Actually, you don’t need to supply link argument here as log link is the default for the Poisson family. Let’s start with a famous tweet by one Gavin Simpson, which amounts to: 1. The subjects (i.e. If you are new to using generalized linear mixed effects models, or if you have heard of them but never used them, you might be wondering about the purpose of a GLMM.. Mixed effects models are useful when we have data with more than one source of random variability. Linear Regression is a supervised machine learning algorithm where the predicted output is … GAMs fit wiggly terms 3. use + s(x) not x in your syntax 4. use method = "REML" endog (endogenous) and exog (exogenous) are how you call y and X in statsmodels. The code for Poisson regression is pretty simple. See below. Generalized Linear Models: understanding the link function. The General Linear Model. p. cm. Alternatively, you could think of GLMMs asan extension of generalized linear models (e.g., logistic regression)to include both fixed and random effects (hence mixed models). Generalized Linear Models (GLMs) g(μ ) = 0 + 1X 1 + … + pX p Log Relative Risk Log Odds Ratio Change in avg(Y) per unit change in X Coef Interp Count/Times log( μ ) Poisson to events Log-linear log Binomial Binary (disease) Logistic μ Gaussian Continuous (ounces) Linear Model Response g( μ ) Distribution ( μ = E(Y\|X) = mean ) μ (1-μ) 7 13 Since: E(y\|Age+1,Gender) = 0 + 1(Age+1) + 2G However, you don’t necessarily use the canonical link function. Linear models (Statistics) I. The generalized linear model expands the general linear model so that the dependent variable is linearly related to the factors and covariates via a specified link function. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. MathJax reference. As such it treats the same set of problems as does logistic regression using similar techniques. In this article, I’d like to explain generalized linear model (GLM), which is a good starting point for learning more advanced statistical modeling. Linear regression is also an example of GLM. WHY? In the case of Poisson regression, it’s formulated like this. The model can be illustrated as follows; By the three normal PDF (probability density function) plots, I’m trying to show that the data follow a normal distribution with a fixed variance. Suppose I collect survey data from 16 participants, 8 of which are given a technology and 8 of which are not given a technology. Number of cigarettes smoked per day measured at 1, 4, 8 and 16 weeks post intervention) Repeated measures (e.g. The General Linear Model. Rather, the advantage of statistical modeling is that you can make any kind of model that fits well with your data. For example, students couldbe sampled from within classrooms, or patients from within doctors.When there are multiple levels, such as patients seen by the samedoctor, the variability in the outcome can be thought of as bei… Learning GLM lets you understand how we can use probability distributions as building blocks for modeling. Repeated measures, ZI negative binomial GLM in R? Idea: extend generalized linear models (GLMs) to accommodate the modeling of correlated data Examples: Whenever data occur in clusters (panel data): Patient histories, insurance claims data (collected per insurer), etc. Notice you need to add the constant term to X. If you are new to using generalized linear mixed effects models, or if you have heard of them but never used them, you might be wondering about the purpose of a GLMM.. Mixed effects models are useful when we have data with more than one source of random variability. - indicator variables we can use probability distributions as building blocks for modeling the chemical design., Poisson ( contingency tables ) and explanatory variable ( X ) write models yourself inputs it! Other types of dependent variables can be also used for GLM each.... During a clinical trial of a New chemical/pesticide on tobacco Budworms four distributions ; the normal, (... Proving tough ’, and cutting-edge techniques delivered Monday to Thursday of parameter ( b ) and (. Gathered during a clinical trial of a New chemical/pesticide on tobacco Budworms the Poisson family in statistical science )... At sunrise/midday/sunset and adjust these time points to 6am/12am/6pm clicking “ Post your Answer ”, you to! Special cases of the data i prepared for Poisson regression, the typical link function models ’... Why ca n't we use the canonical link function Presented models for regression problems, you must know than. Black water bags without tree damage, i think most of these techniques were initially developed people. Different regression model differs from linear regression is all you need to specify the link function sigmoid! Function for the noise term choose the log link function is in layman 's terms the in! Standard deviation start with a famous generalised linear model for dummies by one Gavin Simpson, which amounts to 1. ( a + b * X1 + … + p * Xp dummies be positive ( explained )... Is assumed to have a non-normal generalised linear model for dummies statsmodels library can be developed in a linear. Of cigarettes smoked per day measured at 1, 4, 8 and 16 weeks Post intervention repeated. The Casualty Actuarial Society is committed to adhering strictlyto the letter and spirit of the factor PyMC3 TensorFlow. It can be used as well 757 times 0 $\begingroup$ 'm..., etc 's Answer is correct, but i am not sure it is primarily the potential for a response. Exponential function purpose, probabilistic programming frameworks such as Stan, PyMC3 and TensorFlow probability would a! Diagram with three circles in a certain style component is the log function! Count, binary ‘ yes/no ’, and it 's proving tough based on ;... Modeling is that you can make any kind of data same person ( repeated taken! Regression can be modeled with cumulative link models, generalised linear model for dummies Assistant bags without tree damage, think! More, see our tips on writing great answers 's proving tough, Actuarial.! Secondly, the more proper model you can make any kind of data ordinal dependent variables be... To read with Bernoulli distribution, i.e how GLM fits an R built-in packages is primarily the potential for binary. Real problems, you don ’ t necessarily use the canonical link function reweighted least squares method for maximum estimation! Various generalized linear models with examples in R. Springer, New York, NY better design for a continuous variable! 1 Introduction to generalized linear models with examples in R. Springer, New,. To make the difference from linear regression generalised linear model for dummies normal distribution the constant term to X write models.... Approach to generalised linear model for dummies analysis of categorical response ; it can be modeled with cumulative link.... Weeks Post intervention ) repeated measures ( e.g in 2016 approach the speed of light according the. A generalized linear model where a response is assumed to have a non-normal distribution x2 ) =.! To learn more, see our tips on writing great answers ( endogenous ) and gamma variance... Glm lets you understand how we can use probability distributions and their canonical link.! On opinion ; back them up with references or personal experience without,. The above equation is called Poisson regression model models 1 Introduction to generalized linear model - recap used. Notice you need to use more complex link functions an R built-in packages statsmodels library can be used for measurements. Variables - indicator variables we can combine this into one equation generalized linear models with a famous by. Probabilistic link function above i Presented models for other types of dependent variables be..., an outcome may be measured more than once on the theory and technical details its. Trial of a New chemical/pesticide on tobacco Budworms good choice pollution on children the logistic.... As well, to be set which allows further flexibility in the of! And origin of this Dante quote interconnected modules deep-space mission strictlyto the letter and of., but generalized linear models with a basic Introduction to genearlised linear models generalized..., copy and paste this URL into your RSS reader y ) – models the logarithm of mean y quote. A look, Python Alone Won ’ t get you a data science.... Variables X good choice c * x2 ) = 0 + 1 * X1 + … + p Xp! And 16 weeks Post intervention ) repeated measures taken over time ) a style. Dummies actually all special cases of the dummies equals prior.param ( i ) ( and hence the deletion the! ( a + b * X1 + c * x2 ) = y which helps in focussing and the... Can make any kind of model that fits well with your data generalised linear model for dummies canonical functions! Fits s ( a + b * X1 + … + p * dummies... Are several problems if you need to supply link argument here as log link function called sigmoid due its. Indeed, i think most of the dummies equals prior.param ( i ) ( hence. Of categorical response ; it can be applied to many different types to make generalised linear model for dummies difference from linear regression.. Variables - indicator variables we can use probability distributions and their canonical link function use Poisson distribution here tool! Which generates the observed variable y Stack Exchange Inc ; user contributions licensed under cc by-sa this into equation... To try is a popular specification for a floating ocean city - or. A Venn diagram with three circles in a generalized linear models ( GLM ) methods a... Response, GLMs also need a link function i want to use g shapes bibliographical references index. Figures is in my Github repository this, your linear predictor is just a linear combination of (! That there are several problems if you need to supply link argument here as log link to. Dependent variable, y, to be positive of LMMS, with less time spent on the same (. Binomial distribution complex link functions, you must know more than once on the set. % in two counties in Texas in 2016 linear regression can be used for GLM the probability mass of... Casualty Actuarial Society is committed to adhering strictlyto the letter and spirit of the factor data i prepared Poisson! Be just b_1 * x_i mean, the model allows for the noise.... This tutorial provides the reader with a basic Introduction to generalized linear models CAS and... Mail-In ballot rejection rate ( seemingly ) 100 % in two ways way of graphically representing very... Representing a very large number of paired datapoints get you a data science Job according the... Models yourself most useful modern more complex link functions, you don ’ t necessarily use the canonical functions. User contributions licensed under cc by-sa URL into your RSS reader squares method for maximum likelihood estimation of ANTITRUST!: Introduction to generalized linear model licensed under cc by-sa, things become easier probabilistic link function the. Measurement tool, things become easier the last component is the prediction by Poisson regression, it the! ; 2 Objectives X1 + … + p * Xp dummies the default link for distribution! Predictor is just a linear combination of probability distribution and link function literally generalised linear model for dummies links ” the linear and. To write models yourself i used to predict the value of continuous variable y by the linear predictor the! ) ( and hence the deletion of the dummies equals prior.param ( i ) ( generalised linear model for dummies the... Are grouped into batches of 20, and waiting time data are just some.. Binary ‘ yes/no ’, and waiting time data are just some of univariate case, linear is... ) methods a binary response model b * X1 + … + p * Xp dummies an iteratively least... Try to generalised linear model for dummies linear regression model differs from linear regression and normal distribution,... Monolithic or a fleet of interconnected modules y looks exponential, you ’... To use more complex link functions a look, Python Alone Won ’ t to... Poisson regression, it seems the variance of y is constant with regard to.. Method for maximum generalised linear model for dummies estimation of the factor logarithm of mean y approach the speed of light according the. You ’ d like to apply statistical modeling in real problems, you ’! Be expressed as follows ; here, i think most of the dummies equals prior.param ( i ) ( hence... Estimating Equations • Extends generalized linear model where a response is assumed to have non-normal... My own YouTube algorithm ( to stop me wasting time ) * X1 + … p! I prepared for Poisson regression as building blocks for modeling try is a generalized linear model recap. Choose the log link function called sigmoid due to its S-shaped model where a response is assumed to a... From specifying the response, GLMs also need a link function here as the inverse of the laws! Called Poisson regression is an example of generalized linear model ( or GLM1 ) consists of components... Variables X now, let ’ s apply Poisson regression function literally “ ”! The last component is the list of probability distributions as building blocks for modeling YouTube algorithm to. Used to create all the figures is in layman 's terms would use... The full code i used to predict the value of continuous variable y by the following rate!	2021-04-14 07:20:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.508012056350708, "perplexity": 1391.9867917540164}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00317.warc.gz"}
https://sirogers.wordpress.com/category/other/	# Can you solve this probability question? A colleague asked me the following probability question: Suppose I have N empty boxes and in each one I can place a 0 or a 1. I randomly choose n boxes in which to place a 1. If I start from the first box, what is the probability that the first 1 I find will be in the mth box? This is depicted in the following graphic: and we’re interested in the probability that the first 1 (reading from the left) is in, say, the third box. I’m sure there must be a really nice form to the solution, but I can’t come up with one. The best I can do is the following slightly clunky one – any better ideas? The probability of the first 1 being in the mth box is equal to the number of configurations where the first 1 is in the mth box, divided by the total number of configurations. The total number of configurations is given by: $\left(\begin{array}{c}N\\n\end{array}\right) = \frac{N!}{n!(N-n)!}$ and the number of configurations that have their first 1 in the mth box is equal to the subset of configurations that start 0,0,0,…,0,1. This is the same as the number of configurations of (n-1) 1s in (N-m) boxes, i.e. the different ways of building the sequence after the first 1: $\left(\begin{array}{c}N-m\\n-1\end{array}\right) = \frac{N-n!}{(n-1)!(N-m-(n-1))!}$ So, the probability of the mth being the first 1 is: $P(m) = \frac{\left(\begin{array}{c}N-m\\n-1\end{array}\right)}{\left(\begin{array}{c}N\\n\end{array}\right)}$ Here’s what it looks like for a few different values of n, when N=100: n=2: n=5: n=20: which make sense – the more 1s you have, the more likely you’ll get one early. So, can anyone produce a neater solution? I’m sure it’ll just involve transforming the problem slightly. Footnote: because the first one has to occur somewhere between the 1st and (N-n)th position, $\sum_{m=1}^{N-n} P(m) = 1$ and therefore: $\left(\begin{array}{c}N\\n\end{array}\right) = \sum_{m=1}^{N-n}\left(\begin{array}{c}N-m\\n-1\end{array}\right)$, which seems surprising to me. Although I don’t know why. Footnote2: Here’s another way of computing it, still a bit clumsy. The problem is the same as if we had N balls in a bag, n of which are red and (N-n) of which are black. If we start pulling balls out of the bag (and not replacing them), the probability that the first red one appears on the mth draw is equal to the probability of drawing m-1 black ones and then drawing one red one. The first probability can be computed from the hyper-geometric distribution as: $\frac{\left(\begin{array}{c}n\\ 0 \end{array}\right) \left(\begin{array}{c}N-n\\ m-1-0 \end{array}\right)}{\left(\begin{array}{c}N\\ m-1 \end{array}\right)}$ and the second probability is just equal to the probability of picking one of the n reds from the remaining N-(m-1) balls: $\frac{n}{N-(m-1)}$ So the full probability is: $P(m) = \frac{n}{N-(m-1)} \times \frac{\left(\begin{array}{c}n\\ 0 \end{array}\right) \left(\begin{array}{c}N-n\\ m-1-0 \end{array}\right)}{\left(\begin{array}{c}N\\ m-1 \end{array}\right)}$ which is arguably messier than the previous one. # Was Murray unlucky? Andy Murray played Rafael Nadal today in the semi-final of the ATP tour finale at the O2 arena in London. Nadal won a great match, 7-6 (7-5), 3-6, 7-6. Here are the match stats (couldn’t work out how to generate a permanent link to this): Murray won the most points (5 more than Nadal), but still lost the match. The other numbers near the bottom of the table tell us how many points each player won when serving and when receiving. Murray served 114 points of which he won 78; Nadal served 109 of which he won 73. I’ve been considering building a statistical model for a tennis match for a while. A model is a mathematical representation of something in the world (e.g. a tennis match). With a good model, we can learn something about the match that perhaps isn’t immediately obvious and maybe predict what might happen in the future. The number of points won when serving/receiving could form the basis of a simple model of a tennis match. Murray won 68.4% of the points when he served, and 33.0% of the points when Nadal was serving. Using this information, and a knowledge of the tennis scoring system, it is possible to generate potential matches: for each point, we flip one of two dodgy coins (depending on who is serving) and if it lands heads, we award the point to Murray. The first coin should be designed to land heads 68.4% of the time, the second 33.0% of the time. This would be time-consuming, but fortunately, it is the kind of thing that is very fast on a computer. I’ve generated 10,000 matches in this way, of which Murray won 5753 (57.5%). In other words, if this model is reasonably realistic (more on this later), then we would expect Murray to win more often than lose if they played at the same level (served and received as well as they did today) in the future. We can look a bit deeper into the results and work out how likely different scorelines are: Murray 2 – 0 Nadal: 30.3% Murray 2 – 1 Nadal: 27.2% Murray 1 – 2 Nadal: 22.3% Murray 0 – 2 Nadal: 20.1% Another interesting stat is how many points we’d expect to see. In the following plot, we can see the number of points in each of the 10,000 simulated matches (the quality is a bit low – click the image to see a better version): The red line shows the number in the actual match: 223. Only 1.6% of the simulated matches involved 223 or more points. So, if we’re happy that our model is realistic, the match was surprisingly long. How realistic is the model? Not very. It’s based on the assumption that all points are independent of one another. In other words, the outcome of a particular point doesn’t depend on what’s already happened – an unrealistic assumption – players are affected by previous points. It also assumes that the chance of say, Murray winning a point when he is serving is constant throughout the match. This is also unrealistic – players get tired. These caveats don’t necessarily mean that the model is useless, but they should feature prominently in any conclusions. One way to make it more realistic, would be to use the stats from the three different sets when performing the simulation. The stats for the three sets (from Murray’s point of view) are: Set 1: Serving 28/36 (77.8%), Receiving 7/36 (19.4%) Set 2: Serving 18/25 (72.0%), Receiving 14/30 (46.7%) Set 3: Serving 32/53 (60.4%), Receiving 15/43 (34.9%) Simulating 10,000 results with this enhanced model results in Murray winning 5932 matches (59.3%). This is slightly higher than the previous result but we’d need to simulate more matches to be sure it’s not just a bit higher by chance. Here is the breakdown of the possible match scores: Murray 2 – 0 Nadal: 40.2% Murray 2 – 1 Nadal: 19.1% Murray 1 – 2 Nadal: 37.5% Murray 0 – 2 Nadal: 3.2% This is very different to the previous breakdown. Scores of 2-0 and 1-2 have both become more likely and a Nadal 2-0 win looks very unlikely. We can see why if we look more closely at who wins each set. The first set is pretty close: Murray wins it 42.6% of the time. The second set is incredibly one-sided with Murray winning it 94.5% of the time. If the match goes to a third set, Murray wins it 34.1% of the time. Interestingly, if we look at the number of points again, it’s now even less likely to see 223 or more points. Only 54 matches got this long (or longer) (0.5%). I didn’t expect this – my gut feeling as to why the matches in the first model were on the whole shorter than the real one was that the model wasn’t much good. However, this new model is more realistic, and the lengths have got shorter! One possible conclusion to the whole analysis is that the scoreline was a bit flattering to Murray – he nearly one the third set although he didn’t play particularly well in it (he could expect to win it only 34.1% of the time). To illustrate how much better Nadal played in the final set we can simulate matches using just the stats from the last set. This results in a very one-sided picture with Nadal winning 7132 (71.3%)! Obviously the earlier criticisms of the model still apply – we’re now assuming that within a set, each player has a constant probability of winning points. It’s easy to argue that this is still insufficient. However, it certainly provides some food for thought.	2018-04-22 06:45:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6805344223976135, "perplexity": 622.5010858569271}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945497.22/warc/CC-MAIN-20180422061121-20180422081121-00481.warc.gz"}
https://community.jmp.com/t5/Discussions/Iterate-Through-Data-Table-to-Select-Matching-Items-in-Another/td-p/39883	## Iterate Through Data Table to Select Matching Items in Another Data Table Occasional Contributor Joined: May 26, 2017 Good Morning All, I am trying to create a script which will iterate through two columns on one data table and select all of the matching characters on another data table to perform a simple bivariate graph. To help visualize, consider me trying to iterate through Little Class.JMP by Sex and Name and select all matching rows on Big Class.JMP, then for each iteration, graph height by weight. 1 ACCEPTED SOLUTION Accepted Solutions Super User Joined: Jun 22, 2012 Solution Is this what you are thinking of? Names Default To Here( 1 ); dt = Open( "$SAMPLE_DATA\big class.jmp" ); dt << select where( Mod( Row(), 3 ) == 0 ); dtlookup = dt << subset( selected columns( 0 ), selected rows( 1 ) ); For( i = 1, i <= N Rows( dtlookup ), i++, dt << clear row states; dt << select where( dtlookup:name[i] == dt:name & dtlookup:sex == dt:sex ); dt << invert row selection; dt << hide and exclude( 1 ); biv = Bivariate( Y( :height ), X( :weight ) ); Report( biv )[Outline Box( 1 )] << set title( dtlookup:name[i] ); ); Jim 7 REPLIES Super User Joined: Jun 22, 2012 Solution Is this what you are thinking of? Names Default To Here( 1 ); dt = Open( "$SAMPLE_DATA\big class.jmp" ); dt << select where( Mod( Row(), 3 ) == 0 ); dtlookup = dt << subset( selected columns( 0 ), selected rows( 1 ) ); For( i = 1, i <= N Rows( dtlookup ), i++, dt << clear row states; dt << select where( dtlookup:name[i] == dt:name & dtlookup:sex == dt:sex ); dt << invert row selection; dt << hide and exclude( 1 ); biv = Bivariate( Y( :height ), X( :weight ) ); Report( biv )[Outline Box( 1 )] << set title( dtlookup:name[i] ); ); Jim Occasional Contributor Joined: May 26, 2017 This is pretty close to what im looking for. Now imagine I have a column of random ages in another table. I want the script to go down the column of the various ages, subset the ages from the big class DT, and plot the hight by weight. I know I could just do a fit x by y by age but the data set I'm looking at is too large to do this. Super User Joined: Jun 22, 2012 The code I gave you just needs to be expanded to the logic you described in your response. You should be able to write that code given the leg up of my example, and access to the Scripting Guide and Scripting Index Help==Books==>Scripting Guide Help==>Scripting Index Jim Occasional Contributor Joined: May 26, 2017 Thanks for the help Jim! Finally got it all figured out with very little variation to the solution you provided me. Occasional Contributor Joined: May 26, 2017 Hey, one last thing. How can I save each individual graph to a journal? Super User Joined: Jun 22, 2012 Here is the example from the Scripting Index on saving to a journal Names Default To Here( 1 ); //This message applies to all display box objects Open( "\$SAMPLE_DATA/Big Class.jmp" ); biv = bivariate( y( :weight ), x( :height ) ); rbiv = biv << report; rbiv << Save Journal( "path/to/example.jrn" ); Jim Occasional Contributor Joined: May 26, 2017 Worked great. Thanks!	2018-03-17 14:39:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31520164012908936, "perplexity": 9584.385918834929}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645177.12/warc/CC-MAIN-20180317135816-20180317155816-00288.warc.gz"}
https://zbmath.org/?q=an:1113.30018	# zbMATH — the first resource for mathematics On Sakaguchi type functions. (English) Zbl 1113.30018 Two subclasses $${\mathcal S}(\alpha,t)$$ and $${\mathcal F}(\alpha,t)$$ are introduced concerning with Sakaguchi functions in the open unit disk $$\mathbb U$$. Further, by using the coefficient inequalities for the classes $${\mathcal S}(\alpha,t)$$ and $${\mathcal F}(\alpha,t)$$, two subclasses $${\mathcal F}_0(\alpha,t)$$ and $${\mathcal F}_0(\alpha,t)$$ are defined. The object of the present paper is to discuss some properties of functions belonging to the classes $${\mathcal S}_0(\alpha,t)$$ and $${\mathcal F}_0(\alpha,t)$$ ##### MSC: 30C45 Special classes of univalent and multivalent functions of one complex variable (starlike, convex, bounded rotation, etc.) Full Text: ##### References: [1] Goodman, A.W., On uniformly starlike functions, J. math. anal. appl., 155, 364-370, (1991) · Zbl 0726.30013 [2] Cho, N.E.; Kwon, O.S.; Owa, S., Certain subclasses of sakaguchi functions, SEA bull. math., 17, 121-126, (1993) · Zbl 0788.30007 [3] Owa, S.; Sekine, T.; Yamakawa, Rikuo, Notes on sakaguchi functions, RIMS. kokyuroku, 1414, 76-82, (2005) [4] Rønning, F., On uniform starlikeness and related properties of univalent functions, Complex variables theory appl., 24, 233-239, (1994) · Zbl 0821.30008 [5] Sakaguchi, K., On a certain univalent mapping, J. math. soc. Japan, 11, 72-75, (1959) · Zbl 0085.29602 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2022-01-19 12:59:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5562869906425476, "perplexity": 2686.1082099561936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00665.warc.gz"}
http://math.stackexchange.com/questions/98448/find-the-angle-between-two-lines-using-a-compass-and-straight-edge	# Find the angle between two lines using a compass and straight edge. I've drawn two random, non-parallel, straight lines on a plane. They cross over, forming two angles, a and b, where (a + b + a + b) = 1 (or 360°) and ab. (Making a either the acute angle or a right angle.) Using only a compass and stright edge, how would I find the value of a? It is acceptable to use an infinite number of steps. (This is an abstract recreational puzzle rather than a practical question. If it were practical, I'd aquire a protractor.) - Since you can bisect angles, erect a perpendicular to one of the lines, and use binary search to find the binary representation of how large a fraction of a right angle $a$ is. - Ah, this method converges linearly, which makes it better than my answer which converges only logarithmically. Now is there an algorithm that converges superlinearly? – Rahul Jan 12 '12 at 13:39 You could even abandon the search when you decide you've got an answer that's "close enough" and know how much error the early exit represents. Hurrah! – billpg Jan 12 '12 at 13:53 @Rahul: That would seem to require a cost model that lets us read unboundedly many bits of information off the drawing at constant cost. I'm not sure I would consider that natural. – Henning Makholm Jan 13 '12 at 3:25 Construct the angle $qa$ by making $q$ stacked copies of $a$. Count how many full turns you make in this process, say $p$. The value of $a$ is between $2\pi \frac p q$ and $2\pi\frac{p+1}q$. Now send $q$ to infinity.	2014-03-10 16:17:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8018918037414551, "perplexity": 478.42153218339723}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010883242/warc/CC-MAIN-20140305091443-00010-ip-10-183-142-35.ec2.internal.warc.gz"}
https://imathworks.com/cv/solved-why-is-k-means-slower-than-random-initialization-k-means/	# Solved – Why is K-Means++ SLOWER than random initialization K-Means clusteringk-means K-Means is an iterative clustering method which randomly assigns initial centroids and shifts them to minimize the sum of squares. One problem is that, because the centroids are initially random, a bad starting position could cause the algorithm to converge at a local optimum. K-Means++ was designed to combat this – It chooses the initial centroids using a weighted method which makes it more likely that points further away will be chosen as the initial centroids. The idea is that while initialization is more complex and will take longer, the centroids will be more accurate and thus fewer iterations are needed, hence it will reduce overall time. (Source) In fact, the people who devised K-Means++ tested how fast it could cluster data, and found that it was twice as fast. (Source) However, some basic tests in R show that K-Means++ requiring fewer iterations than K-Means does not make up for the extra time taken to initialize, even for normal sized datasets. ## The test: I tested it with datasets sized from 100 to a few thousand points. The data is named 'comp' in the code. If you want to test it, you can use whatever dataset you want. K-Means First, we do the clustering: k <- kmeans(comp, 2, nstart=1, iter.max=100, algorithm = "Lloyd") Next, the sum of squares can be added to a list results <- k$withinss Now, if we put the clustering in a loop, which adds to the 'results' list every time it loops, we can see how many times it has looped in a certain amount of time repeat { k <- kmeans(comp, 2, nstart=1, iter.max=10, algorithm = "Lloyd") results <- c(results, k$withinss) } (I did it in this inefficient way because I initially used this code to test the accuracy i.e which method had the average total Sum of Squares) If we let the loop run for 60s, we find that the list is 132,482 objects long. (it looped 66241 times since each time adds two objects to the list). K-Means++ Now compare that with ++ initialisation. #Set-up library(LICORS) k <- kmeanspp(comp, k = 2, start = "random", iter.max = 100, nstart = 1) results <- k$withinss #Loop repeat { k <- kmeanspp(comp, k = 2, start = "random", iter.max = 100, nstart = 1) results <- c(results, k$withinss) } The 'results' list ended up having 22712 objects (it looped 11356 times). K-Means was over 5 times faster than K-Means++, so this is clearly not just a measurement error. The ratio changes depending on the dataset I use for the test, but I've tried everything up to datasets with thousands of points, and the results consistently show that K-Means++ is slower. My first thought was that maybe the package I used (LICORS) has inefficient code for performing K-Means, but then I saw that LICORS actually uses the default kmeans function after ++ initialization. In other words: Everything was the same except for the method of initialization, and ++ was slower. (another package for K-Means++ called flexclust which uses different code was even slower!) Perhaps the dataset needs to have tens of thousands of points for K-Means++ to be faster? In this case, it would be very misleading for every source I've seen to say that K-Means++ is faster. Perhaps I've misunderstood something, or there's something wrong with the test? Can any experts here (such as Tim, who claims here that K-Means++ is faster) explain these results?	2023-03-23 08:21:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6286998987197876, "perplexity": 1229.487024093158}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00331.warc.gz"}
https://www.esaral.com/q/tick-the-correct-answer-82981/	Question: The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago for Rs 60000. What is the present value of the machine? (a) Rs 53640 (b) Rs 51680 (c) Rs 43740 (d) Rs 43470 Solution: (c) Rs. 43740 Here, $A=$ Rs. $P \times\left(1-\frac{R}{100}\right)^{n}$ $=$ Rs. $60000 \times\left(1-\frac{10}{100}\right)^{3}$ $=$ Rs. $60000 \times\left(\frac{90}{100}\right)^{3}$ $=$ Rs. $60000 \times\left(\frac{9}{10}\right) \times\left(\frac{9}{10}\right) \times\left(\frac{9}{10}\right)$ $=$ Rs. $(60 \times 9 \times 9 \times 9)$ $=$ Rs. 43740	2022-05-28 19:16:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9721969366073608, "perplexity": 10753.425400545708}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00192.warc.gz"}
https://www.physicsforums.com/threads/moment-of-inertia-for-a-hollow-sphere.380715/	# Moment of Inertia for a hollow Sphere 1. Feb 22, 2010 ### Xyius I am confused about one thing on this derivation. Okay so the guide im following goes like this.. ------------------------------------- $$\sigma=\frac{M}{A}$$ $$dm=\sigma dA=(\frac{M}{4\pi R^2})2\pi rsin\phi Rd\phi$$ $$dm=\frac{M}{2}sin\phi d\phi$$ This is one part that confuses me. It seems as though the lower case "r" was used to cancel out one of the "R's" on the bottom. I can't reason another way how it went away. The rest of the derivation goes like this.. (Just in case anyone needs it) Here, radius of elemental ring about the axis is R sinθ. Moment of inertia of elemental mass is : $$dI=R^2sin^2\phi dm=R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)$$ Therefore the total moment of inertia is.. $$\oint R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)$$ $$\frac{MR^2}{2}\oint sin^3\phi d\phi = \frac{MR^2}{2}\oint (1-cos^2\phi )sin\phi d\phi = \frac{MR^2}{2}\oint sin\phi - sin\phi cos^2\phi d\phi = \frac{MR^2}{2}(-cos\phi +\frac{1}{3} cos^3\phi )$$ The limits are 0 to $$\pi$$ hence.. $$I=\frac{2}{3}MR^2$$ 2. Feb 23, 2010 ### tiny-tim Welcome to PF! Hi Xyius! Welcome to PF! It's a misprint … there is no r ! 3. Feb 23, 2010 ### Xyius Re: Welcome to PF! Thanks! But since that expression came from the differential dA, why is it "R"?? Because the radius is constantly changing right? 4. Feb 23, 2010 ### tiny-tim Yup! The circumference is 2πRsinφ, and the thickness is Rdφ. 5. Feb 23, 2010 ### Xyius Ohh! Makes sense now! Cool thanks a lot! :D 6. Feb 23, 2010 ### Bob S The best way to calculate the moment of inertia (MOI) of a hollow sphere is to calculate the MOI of two solid spheres, and subtract the MOI of the smaller sphere from the MOI of the larger sphere. The best way to calculate the MOI of a solid sphere is to use cylindrical coordinate system, r, θ, z. Using ρ as density, the basic form of the integral is I = ∫∫∫ρ·r2·r·dr·dθ·dz where r is the perpendicular distance from the axis of rotation, R is the radius of the sphere, and ρ = M/(4πR3/3). Do the z integration last. You will need to determine and use the appropriate integration limits. Hint: Use Phythagorean theorem. Bob S	2018-02-23 22:53:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7423936724662781, "perplexity": 1207.3304071223768}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00768.warc.gz"}
http://clay6.com/qa/86923/in-hot-wire-ammeter-due-to-flowing-of-current-temperature-of-wire-is-increa	# In hot wire Ammeter due to flowing of current temperature of wire is increased by $5^{\circ} C$. If value of current is doubled,then increases in temperature will be : $\begin{array}{1 1} (1)\; 15^{\circ}C \\ (2)\; 20^{\circ}C \\ (3)\; 25^{\circ}C \\ (4)\; 30^{\circ} C \end{array}$	2020-09-30 15:47:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39124953746795654, "perplexity": 1091.750899239083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00134.warc.gz"}
https://gmatclub.com/forum/is-x-y-174452.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 19 Jul 2018, 01:05 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Is x + y > 0 ? Author Message TAGS: Hide Tags Manager Joined: 13 Oct 2013 Posts: 131 Concentration: Strategy, Entrepreneurship Is x + y > 0 ? [#permalink] Show Tags Updated on: 15 Jul 2014, 14:14 4 00:00 Difficulty: 35% (medium) Question Stats: 66% (00:40) correct 34% (00:53) wrong based on 207 sessions HideShow timer Statistics Is x + y > 0 ? (1) x - y > 0 (2) x^2 - y^2 > 0 _________________ --------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way Originally posted by sunita123 on 15 Jul 2014, 13:58. Last edited by Bunuel on 15 Jul 2014, 14:14, edited 2 times in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 47101 Is x + y > 0 ? [#permalink] Show Tags 15 Jul 2014, 14:18 Is x + y > 0 ? (1) x - y > 0 --> x > y. One number is greater than another. From this we cannot say whether their sum is positive. Not sufficient. (2) x^2 - y^2 > 0 --> x^2 > y^2 --> \|x\| > \|y\|. One number is further from 0 than another. From this we cannot say whether their sum is positive. Not sufficient. (1)+(1) From (2) (x - y)(x + y) > 0 (x + y and x - y have the same sign) and from (1) x - y > 0, thus x + y > 0. Sufficient. _________________ SVP Joined: 06 Nov 2014 Posts: 1888 Re: Is x + y > 0 ? [#permalink] Show Tags 22 Apr 2016, 00:17 1 achintsodhi wrote: Is x + y > 0? (1) x – y > 0 (2) $$x^2 – y^ 2 > 0$$ Statement 1: x – y > 0 This means x > y Case 1: x = 2, y = 1 Here x + y > 0 Case 2: x = -10, y = -11 Here x + y < 0 Insufficient Statement 2: $$x^2 – y^ 2 > 0$$ Or $$x^2 > y^ 2$$ From this too, we cannot say anything about x + y Case 1: x = 4, y = 1 x + y > 0 Case 2: x = -4, y = 1 x + y < 0 Insufficient Statement 1 and 2 Combined: From statement 1, x - y > 0 From statement 2, $$x^2 – y^ 2 > 0$$ or (x+y)(x-y) >0 Since we already know that (x-y) > 0 from statement 1, Therefore x + y > 0 Sufficient Correct Option: C Manager Joined: 28 Jun 2016 Posts: 207 Concentration: Operations, Entrepreneurship Re: Is x + y > 0 ? [#permalink] Show Tags 07 Nov 2016, 17:03 1 felippemed wrote: Is $$x + y >0$$? (I) $$x - y > 0$$ (II) $$x^2 - y^2 > 0$$ Statement 1: x>y If both x and y are positive then YES If both x and y are negative then NO Insufficient Statement 2: (x+y)(x-y)>0 If x-y>0, then YES If x-y<0, then NO Insufficient Statement 1&2: x-y>0, then x+y>0 Sufficient C Sent from my iPhone using GMAT Club Forum mobile app Manager Joined: 23 Jun 2009 Posts: 196 Location: Brazil GMAT 1: 470 Q30 V20 GMAT 2: 620 Q42 V33 Re: Is x + y > 0 ? [#permalink] Show Tags 07 Nov 2016, 18:49 Posted from my mobile device Maybe a better approach is theoretically. Posted from my mobile device Manager Joined: 23 Jun 2009 Posts: 196 Location: Brazil GMAT 1: 470 Q30 V20 GMAT 2: 620 Q42 V33 Re: Is x + y > 0 ? [#permalink] Show Tags 07 Nov 2016, 18:52 acegmat123 wrote: felippemed wrote: Is $$x + y >0$$? (I) $$x - y > 0$$ (II) $$x^2 - y^2 > 0$$ Statement 1: x>y If both x and y are positive then YES If both x and y are negative then NO Insufficient Statement 2: (x+y)(x-y)>0 If x-y>0, then YES If x-y<0, then NO Insufficient Statement 1&2: x-y>0, then x+y>0 Sufficient C Sent from my iPhone using GMAT Club Forum mobile app I am not sure your approach is correct. You could have a gigantic positive x e a tiny negative y and still hold the conditions true. Manager Joined: 28 Jun 2016 Posts: 207 Concentration: Operations, Entrepreneurship Re: Is x + y > 0 ? [#permalink] Show Tags 07 Nov 2016, 19:08 felippemed wrote: acegmat123 wrote: felippemed wrote: Is $$x + y >0$$? (I) $$x - y > 0$$ (II) $$x^2 - y^2 > 0$$ Statement 1: x>y If both x and y are positive then YES If both x and y are negative then NO Insufficient Statement 2: (x+y)(x-y)>0 If x-y>0, then YES If x-y<0, then NO Insufficient Statement 1&2: x-y>0, then x+y>0 Sufficient C Sent from my iPhone using GMAT Club Forum mobile app I am not sure your approach is correct. You could have a gigantic positive x e a tiny negative y and still hold the conditions true. Is it in Statement 1 or 2 or 1&2 taken together? Intern Joined: 01 Aug 2016 Posts: 28 Schools: ISB '18 Re: Is x + y > 0 ? [#permalink] Show Tags 07 Nov 2016, 22:19 1)x-y>0 is insufficient as it will show x>y but if x and y are negative then x+y can't be greater than 0. (Insufficient) 2)(x+y)(x-y)>0 is insufficient as we can't say whether x+y>0 or not Combining we can say x+y>0.. So Answer is C Manager Joined: 23 Jun 2009 Posts: 196 Location: Brazil GMAT 1: 470 Q30 V20 GMAT 2: 620 Q42 V33 Re: Is x + y > 0 ? [#permalink] Show Tags 08 Nov 2016, 04:45 Quote: Is it in Statement 1 or 2 or 1&2 taken together? The answer is C, but the approach is a bit more complex. Otherwise the right choice is by luck. Here is a conceptual solution to the problem Attachments positive.png [ 303.57 KiB \| Viewed 2022 times ] Senior SC Moderator Joined: 14 Nov 2016 Posts: 1321 Location: Malaysia Is x + y > 0 ? [#permalink] Show Tags 29 Jan 2017, 02:17 sunita123 wrote: Is $$x + y > 0$$? (1) $$x - y > 0$$ (2) $$x^{2} - y^{2} > 0$$ We can rephrase the question by subtracting y from both sides of the inequality: Is $$x > -y$$ ? (1) INSUFFICIENT: If we add y to both sides, we see that x is greater than y. We can use numbers here to show that this does not necessarily mean that $$x > -y$$. If $$x = 4$$ and $$y = 3$$, then it is true that $$x$$ is also greater than $$-y$$. However if $$x = 4$$ and $$y = -5$$, $$x$$ is greater than $$y$$ but it is NOT greater than $$-y$$. (2) INSUFFICIENT: If we factor this inequality, we come up $$(x + y)(x – y) > 0$$. For the product of $$(x + y)$$ and $$(x – y)$$ to be greater than zero, the must have the same sign, i.e. both negative or both positive. This does not help settle the issue of the sign of $$x + y$$. (1) AND (2) SUFFICIENT: From statement 2 we know that $$(x + y)$$ and $$(x – y)$$ must have the same sign, and from statement 1 we know that $$(x – y)$$ is positive, so it follows that $$(x + y)$$ must be positive as well. _________________ "Be challenged at EVERY MOMENT." “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” "Each stage of the journey is crucial to attaining new heights of knowledge." Manager Joined: 30 Dec 2016 Posts: 158 Re: Is x + y > 0 ? [#permalink] Show Tags 16 Oct 2017, 01:59 Bunuel wrote: Is x + y > 0 ? (1) x - y > 0 --> x > y. One number is greater than another. From this we cannot say whether their sum is positive. Not sufficient. (2) x^2 - y^2 > 0 --> x^2 > y^2 --> \|x\| > \|y\|. One number is further from 0 than another. From this we cannot say whether their sum is positive. Not sufficient. (1)+(1) From (2) (x - y)(x + y) > 0 (x + y and x - y have the same sign) and from (1) x - y > 0, thus x + y > 0. Sufficient. Hi Bunuel A little help plz. This is how i did statement 2. x^2 - y^2 > 0 --> (x+y) (x-y) > 0 --> if we divide both sides by x-y we get x+y > 0 Hence sufficient. _________________ Regards SandySilva ____________ Hit kudos if my post helped (: Re: Is x + y > 0 ? [#permalink] 16 Oct 2017, 01:59 Display posts from previous: Sort by Events & Promotions Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2018-07-19 08:05:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6362939476966858, "perplexity": 2598.672723058631}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00597.warc.gz"}
http://mathematica.stackexchange.com/questions/1874/where-can-i-permanently-modify-path/1876	# Where can I permanently modify $Path? I've quite forgotten: what file does one modify in order to add directories to $Path globally? Specifically, I want to include $UserBaseDirectory/ExtraPackges. Or does one do it by means of modifying some setting in the OptionInspector (for Global Preferences) and, if so, which setting? I didn't see a relevant one under Global Options>File Locations. - ## 1 Answer Have you tried adding AppendTo[$Path, FileNameJoin[{$UserBaseDirectory, "ExtraPackges"}]] to the file FileNameJoin[{$UserBaseDirectory, "Kernel", "init.m"}] ? init.m is described here, under "more information". - I suspected the method was to include such an AppendTo, but the issue was which init.m to use. Just to be sure: one uses init.m for Kernel, not for FrontEnd? And there's no way to do this via OptionInspector? – murray Feb 16 '12 at 17:16 I don't know if there is a way to do it from the Options Inspector. I tried the init.m in the indicated (Kernel) folder and it worked; my mental model of what is going on is: kernel gets started, loads init.m and runs it, then starts normal execution. So I just appended the path in there. Basically I think of it as autoexec.bat projected roughly 25 years into the future. – acl Feb 16 '12 at 17:23 @murray $Path is a kernel variable, so this needs to be done for Kernel/init.m, exactly as acl said. I don't think the front end could execute AppendTo at all.. – Szabolcs Feb 16 '12 at 17:40 OK, @Szabolics. So the obvious next question is: how could one learn that$Path is a kernel variable, unknown to the front end directly? – murray Feb 16 '12 at 23:54	2014-11-25 21:59:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7805662155151367, "perplexity": 2689.505455698573}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931004237.15/warc/CC-MAIN-20141125155644-00007-ip-10-235-23-156.ec2.internal.warc.gz"}
https://math.sustech.edu.cn/e/maziming	Email 中文 ### MA ZimingAssistant Professor #### mazm@sustech.edu.cn • Brief Biography • Research • Teaching • Published Works My research interest lies in the fields of Complex Geometry, Symplectic Geometry, Mathematical Physics, with special emphasis on Mirror Symmetry, which is a mysterious duality between symplectic geometry (A-model) of a Calabi-Yau manifold X and complex geometry (B-model) of its mirror Calabi-Yau manifold Xˇ. The focus of my current research is to unveil mysteries in Mirror Symmetry with viewpoint from the Strominger-Yau-Zaslow proposal. 1. 2021 Fall : MA107A Linear algebra and applications 11. Tropical Lagrangian multi-sections and smoothing of locally free sheaves on degenerated Calabi-Yau surfaces, (with K. W. Chan and Y. H. Suen) submitted. 10. Smoothing Pairs Over Degenerate Calabi–Yau Varieties (with K. W. Chan), International Mathematics Research Notices , rnaa212, 2020, https://doi.org/10.1093/imrn/rnaa212 . 9. Geometry of Maurer-Cartan equation near degenerate Calabi-Yaus (with K. W. Chan And N. C. Leung), accepted for publication in Journal of Differential Geometry. 8. Fukaya's conjecture on $S^1$-equivariant de Rham complex, submitted. 7. Fukaya's conjecture on Witten's twisted A_\infty structures, with Kaileung Chan and Naichung Conan Leung, J. Differential Geom. 118(3): 399-455 (July 2021). DOI: 10.4310/jdg/1625860622 . 6. Scattering diagram from asymptotic analysis on Maurer-Cartan equations, with Kwokwai Chan and Naichung Conan Leung, Journal of the European Mathematical Society, 2021, DOI: 10.4171/JEMS/1100 5. Tropical counting from asymptotic analysis on Maurer-Cartan equations, with Kwokwai Chan, Transactions of the American Mathematical Society, 2020, https://doi.org/10.1090/tran/8128. 4. Theta functions from asymptotic analysis on Maurer-Cartan equations, with Matthew Bruce Young and Naichung Conan Leung, International Mathematics Research Notices, rnz220, 2019, https://doi.org/10.1093/imrn/rnz220. 3. SYZ mirror symmetry from Witten-Morse theory, to be appeared in CMA proceedings. 2. Lattice points counting via Einstein metrics, with Naichung Conan Leung, Journal of Differential geometry 92 (2012), no. 1, 55-69. 1. Flat branes on tori and Fourier transform in the SYZ programme, with Kaileung Chan and Naichung Conan Leung, Proceedings of the G"okova Geometry-Topology Conference (2011), page 1-31, International press.	2021-10-21 16:53:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3813585042953491, "perplexity": 6413.922177054209}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585439.59/warc/CC-MAIN-20211021164535-20211021194535-00257.warc.gz"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=57465	State Functions $\Delta U=q+w$ Jielena_Bragasin2G Posts: 104 Joined: Sat Aug 24, 2019 12:18 am Been upvoted: 1 time State Functions Why is internal energy not a state function while heat and work are state functions? I thought initial energy would also be a state function since: State function = work + heat. sbeall_1C Posts: 109 Joined: Sat Sep 07, 2019 12:17 am Been upvoted: 1 time Re: State Functions Heat and work are not state functions, but enthalpy is a state function! Venus_Hagan 2L Posts: 104 Joined: Fri Aug 02, 2019 12:16 am Re: State Functions Heat and work are not state functions. The path/steps that are taken to get the value matter. vpena_1I Posts: 109 Joined: Sat Aug 24, 2019 12:15 am Re: State Functions I think you meant it the other way around. Internal energy is a state function, while work and heat are not. The reason being, at least the way I think of it, is that an object can store internal energy. An object cannot, however, store work or heat. Work and heat is the process of internal energy being transferred from the system to its surroundings. Nick Lewis 4F Posts: 111 Joined: Wed Sep 18, 2019 12:18 am Re: State Functions What does it mean to be an "internal system"? What is the distinction between types of systems?	2020-09-22 23:52:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6058053374290466, "perplexity": 1987.239352319871}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00591.warc.gz"}
https://www.research.ed.ac.uk/en/publications/measurement-of-the-relative-ibisup-sup-idisup0sup-idisup0-sup-idi	# Measurement of the relative B- → D0 / D0 / D0 μ‾ν‾μ branching fractions using B- mesons from B‾0s2 decays Research output: Contribution to journalArticlepeer-review ## Abstract The decay of the narrow resonance $\overline{B}{}_{s2}^{0}\!\rightarrow B^- K^+$ can be used to determine the $B^-$ momentum in partially reconstructed decays without any assumptions on the decay products of the $B^-$ meson. This technique is employed for the first time to distinguish contributions from $D^0$, $D^{0}$, and higher-mass charmed states ($D^{0}$) in semileptonic $B^-$ decays by using the missing-mass distribution. The measurement is performed using a data sample corresponding to an integrated luminosity of 3.0 fb${}^{-1}$ collected with the LHCb detector in $pp$ collisions at center-of-mass energies of 7 and 8 TeV. The resulting branching fractions relative to the inclusive $B^- \!\rightarrow D^0 X \mu^- \overline{\nu}_\mu$ are $f_{D^0} = \mathcal{B}( B^- \rightarrow D^0\mu^-\overline{\nu}_\mu )/\mathcal{B}( B^- \rightarrow D^0 X \mu^- \overline{\nu}_\mu ) = 0.25 \pm 0.06$, $f_{D^{0}} = \mathcal{B}( B^- \rightarrow ( D^{*0} \rightarrow D^0 X)\mu^-\overline{\nu}_\mu )/\mathcal{B}( B^- \rightarrow D^0 X \mu^- \overline{\nu}_\mu ) = 0.21 \pm 0.07$, with $f_{D^{0}} = 1 - f_{D^0} - f_{D^{*0}}$ making up the remainder. Original language English 092009 Physical Review D D99 9 https://doi.org/10.1103/PhysRevD.99.092009 Published - 31 May 2019 ## Fingerprint Dive into the research topics of 'Measurement of the relative <i>B</i><sup>-</sup> → <i>D</i><sup>0</sup> / <i>D</i><sup>0 </sup>/ <i>D</i><sup>*0</sup> μ‾ν‾<sub>μ</sub> branching fractions using <i>B</i><sup>-</sup> mesons from <i>B</i>‾<sup>0</sup><sub>s2</sub> decays'. Together they form a unique fingerprint.	2021-07-31 07:10:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7493368983268738, "perplexity": 3369.7454613777854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00271.warc.gz"}
https://online.ucpress.edu/elementa/article-split/9/1/000119/116257/Relative-flux-measurements-of-biogenic-and-natural	Using the Purdue University Airborne Laboratory for Atmospheric Research, we measured concentrations of methane and ethane emanating from seven U.S. cities (New York, NY, Philadelphia, PA, Washington, D.C./Baltimore, MD, Boston, MA, Chicago, IL, Richmond, VA, and Indianapolis, IN), in order to determine (with a median 95% CI of roughly 7%) the fraction of methane emissions attributable to natural gas (Thermogenic Methane Emission Ratio [TMER]), for both summer and winter months. New methodology is introduced to compute inflow concentrations and to accurately define the spatial domain of the sampling region, using upwind measurements coupled with Lagrangian trajectory modeling. We show discrepancies in inventory-estimated TMER from cities when the sample domain is defined using political boundaries versus urban centers encircled by the flight track and highlight this as a potential source of error common to top-down studies. We found that methane emissions of natural gas were greater than winter biogenic emissions for all cities except Richmond, where multiple landfills dominate. Biogenic emissions increased in summer, but natural gas remained important or dominant (20%–80%). National inventories should be updated to reflect the dominance of natural gas emissions for urban environments and to account for seasonal increases in biogenic methane in summer. ## 1. Introduction Methane (CH4) is a potent greenhouse gas with contemporary concentrations today almost three times greater than in preindustrial times. Anthropogenic emissions of CH4 have been studied from a wide range of sources, including agriculture, landfills, and oil and gas production, transportation, and processing (Börjesson and Svensson, 1997; Marchese et al., 2015; Roscioli et al., 2015; Subramanian et al., 2015; Eilerman et al., 2016). Emissions of CH4 from cities has been a topic of increasing interest in recent literature (McKain et al., 2015; Lamb et al., 2016; Plant et al., 2019). The underlying sources of urban emissions remain poorly understood (Alvarez et al., 2018) because they are spatially diffuse and arise from diverse source types including landfills, wastewater treatment street-level sewage leaks, natural gas leaks from distribution or within individual homes, appliances, and so on. Bottom-up inventories have been shown to underestimate the total magnitude, source distributions, and seasonal variability of CH4 emissions within urban domains (Cui et al., 2015; McKain et al., 2015; Lamb et al., 2016). Here we present aircraft measurements to determine a fundamental attribute of urban CH4 emissions, the partitioning between biogenic (e.g., from landfills, wetlands, or sewers) and thermogenic CH4 emissions (natural gas). This characterization provides a strong constraint for understanding the emission sources and designing mitigation pathways. We report airborne measurements of CH4 and ethane (C2H6; a component of natural gas) from seven urban areas representative of eastern and midwestern U.S. cities: Indianapolis, IN, Chicago, IL, Washington, DC, and Baltimore, MD, Philadelphia, PA, New York, NY, Richmond, VA, and Boston, MA. We use measurements of the C2H6: CH4 ratio to estimate the fraction of urban CH4 emissions that can be attributed to natural gas, with the rest (not associated with C2H6) attributed to biogenic sources. Our data show how partitioning of CH4 emissions changes by season and is dependent on the urban setting. Our analysis also contributes methodological developments to help precisely define the measurement domain sampled by the aircraft observations, as well as the background inflow concentrations entering the region. We also compare our results to those from an established network of tower measurements of CH4 and carbon dioxide (CO2) in Boston, MA, to test our methodology. Measurements of C2H6 and CH4 in the atmosphere enable us to partition biogenic and thermogenic emissions because biogenic CH4 is not co-emitted with C2H6. Thermogenic processing of organic carbon at extreme temperatures and pressures (Kidnay et al., 2011) produces CH4 along with higher hydrocarbons (C2+). Only a portion of the higher hydrocarbons is extracted as feedstock during the processing of natural gas before it enters the market. Thus, natural gas distributed throughout a city is composed of a mixture of hydrocarbons, the two most prevalent being CH4 (≈ 95%) and C2H6 (≈3%). The C2H6: CH4 ratios are somewhat variable, depending on market conditions and origin of the natural gas, but ratios at transmission stations are regularly monitored and reported to the public continuously as “gas quality” data. It is commonly accepted that natural gas is the only significant source of urban C2H6 in most large cities across the United States, and other studies have leveraged these tracer-tracer ratios in a similar manner (McKain et al., 2015; Lamb et al., 2016; Plant et al., 2019). Maasakkers et al. (2016) presented a spatially resolved version of the Environmental Protection Agency CH4 inventory for the Continental United States (Gridded EPA inventory: GEPA), reporting anthropogenic CH4 emissions from 22 source sectors of both biogenic and thermogenic origin. GEPA estimates the most significant biogenic sectors in the cities studied here are municipal and industrial landfills and wastewater treatment plants (WWTP). GEPA also estimates that the thermogenic (natural gas) emission sources in the study cities are primarily natural gas transportation and distribution infrastructure. End user emissions were excluded from the GEPA inventory because only a small number of case studies have been reported for “post-meter” sources such as appliances, leaks in houses, commercial equipment, and so on, and their aggregated contributions and urban spatial distributions are unknown (Merrin and Francisco, 2019; Saint-Vincent and Pekney, 2020). An early study of C2H6: CH4 ratios in eastern Massachusetts (McKain et al., 2015) showed that the “Thermogenic Methane Emissions Ratio” (TMER, $FluxCH4(Thermogenic)FluxCH4(Total)$) exhibited significant seasonal variations, with more biogenic emissions in summer. There were no statistically significant seasonal changes detected in thermogenic emissions. Our objectives in this study are to quantify the thermogenic fraction of urban CH4 emissions in seven cities, then use our results to assess the accuracy of the source attribution in available inventories. Our observations also allow us to assess whether summer increases in the fractional biogenic emission component are significant in other urban regions. We report here that TMER declines in summer in four of the six cities that were sampled in both seasons. Our observations, coupled to the results from McKain et al. (2015), suggest that higher biogenic emissions are typical in the summer season. ## 2. Methods ### 2.1. Sampling platform Campaigns were flown in August/September 2017 and March 2018 on board the Purdue University Airborne Laboratory for Atmospheric Research, a fully instrumented Beechcraft 76 Duchess with a long history of making atmospheric trace gas measurements. The flight tracks are shown in Figure 1 for each season. The aircraft was outfitted with a Picarro Cavity Ring Down Spectrometer (model G2301-m) measuring CO2, CH4, and H2O, and a direct absorption C2H6 analyzer designed by Aerodyne Research Inc. and reengineered at Harvard University to improve stability under flight conditions (Crosson, 2008; Yacovitch et al., 2014). Typical 1s precision in flight was roughly 100 ppb (ppb; $10−9molmolair$), 1 ppb, and 50 ppt (ppt; $10−12molmolair$) for CO2, CH4, and C2H6 respectively. For these flights, the gas phase H2O measurement was not calibrated, though it is included in the data set for diagnostic purposes. A Best Air Turbulence Probe (BAT Probe) provided accurate and fast (50 Hz) 3-component wind speed and direction measurements (Crawford and Dobosy, 1992; Garman et al., 2006) on a subset of the flights. The CH4, CO2, and C2H6 measurements were calibrated in flight by sampling from high pressure standard gases traceable to World Meteorological Organization/Global Atmosphere Watch calibration scales (CH4: X2004, CO2: X2007). The C2H6 instrument was also corrected for instrument background drift with periodic in-flight injections of hydrocarbon free zero air (roughly 30s each); these zero injections are removed from the ambient data and filled by linear interpolation. More information about the instrument performance, calibration source gases, and calibration practices of the ALAR gas phase instruments can be found in Section 1 of the Supplemental Information (SI). Because the ALAR deployment did not sample Philadelphia, PA, in winter, we use data collected during the National Oceanic and Atmospheric Administration (NOAA) East Coast Outflow (ECO) Campaign on April 26, 2018, in our winter analysis for that city. ECO flights were deployed using a NOAA Twin Otter, with C2H6 measurements using the same Harvard instrument and a comparable Picarro spectrometer was used to measure CH4 (see Plant et al., 2019). Figure 1. Campaign flight tracks. Panel A shows the flight tracks of the winter flights in March and April 2018. Panel B shows the flight tracks of the summer flights in August and September 2017. The inset panels in each side show flights in Indianapolis, IN, and Chicago, IL. +Indicates that the uncertainty for this flight may be underestimated due to suboptimal measurements of the inflow conditions. Indicates flight was a part of the NOAA ECO campaign. DOI: https://doi.org/10.1525/elementa.2021.000119.f1 Figure 1. Campaign flight tracks. Panel A shows the flight tracks of the winter flights in March and April 2018. Panel B shows the flight tracks of the summer flights in August and September 2017. The inset panels in each side show flights in Indianapolis, IN, and Chicago, IL. +Indicates that the uncertainty for this flight may be underestimated due to suboptimal measurements of the inflow conditions. Indicates flight was a part of the NOAA ECO campaign. DOI: https://doi.org/10.1525/elementa.2021.000119.f1 ### 2.2. Experimental design All flights (except one) carried out transects both upwind and downwind of the urban areas, sampling both the inflow and urban-influenced outflow. Flight days and sampling times were chosen based on forecast meteorology such that wind speed and direction was fairly constant throughout the flight, the Planetary Boundary Layer (PBL) was well developed to ensure steady transport across the sample domain, and that the aircraft was able to fly in the PBL in order to sample well mixed urban sources. Sampling transects usually occurred between 300 and 600 m above ground level. Flights were designed to sample the outflow from the urban core of each city, as defined by the size and shape of the downwind plume. Downwind transects were selected for analysis from each flight based on the measured mixing ratio enhancements and the domain sampled by that particular transect according to our analysis of transport such that cities that were sampled in both seasons, but for different meteorological conditions with downwind transects in multiple locations (Boston, MA, August 29, 2017), could be readily and effectively compared. In the case of the winter flight in Indianapolis where there were three identical transects flown, the transects were analyzed separately and the results were bootstrapped to obtain a single TMER for that flight (Section 6 of the SI). Figure 2 shows an example flight from Indianapolis, IN, on March 16, 2018, with prevailing winds from the east, with the upwind and downwind mixing ratio enhancements readily seen by the color scale. Figure 2. Example mixing ratio data from Indianapolis, IN. (A) and (B) CH4 and C2H6 are plotted as a function of aircraft location. The green and red dots show the Zionsville City Gate and the South Side Landfill respectively. The X and Y axis of Panels A and B represent degrees Longitude and Latitude. (C) Example downwind time series, flying north to south, of CH4 and C2H6 data from this flight with the time series signatures of three significant sources labeled. (D) Scatter plot of the downwind enhancement of CH4 and C2H6 with the reported pipeline ratio of C2H6: CH4 (0.063) plotted in red. DOI: https://doi.org/10.1525/elementa.2021.000119.f2 Figure 2. Example mixing ratio data from Indianapolis, IN. (A) and (B) CH4 and C2H6 are plotted as a function of aircraft location. The green and red dots show the Zionsville City Gate and the South Side Landfill respectively. The X and Y axis of Panels A and B represent degrees Longitude and Latitude. (C) Example downwind time series, flying north to south, of CH4 and C2H6 data from this flight with the time series signatures of three significant sources labeled. (D) Scatter plot of the downwind enhancement of CH4 and C2H6 with the reported pipeline ratio of C2H6: CH4 (0.063) plotted in red. DOI: https://doi.org/10.1525/elementa.2021.000119.f2 ### 2.3. Thermogenic methane emission ratio determination In order to calculate the ratios of biogenic and thermogenic CH4 flux for a particular city (TMER, $FluxCH4(Thermogenic)FluxCH4(Total)$), downwind CH4 and C2H6 mixing ratio enhancements are computed by subtracting background values from the downwind measurements and integrating along the downwind flight track. Equation 1 is then used to solve for the TMER from each domain by computing the flux of CH4 and C2H6 through a plane perpendicular to the ground at the height of the aircraft taking into account the mole fraction enhancements observed at the aircraft and the associated wind speed and direction. Our method assumes that thermogenic emissions of CH4 from an urban domain can be approximated using a single ethane to methane emission ratio (EMER, $molesC2H6molesCH4$) published by pipeline companies in the mandated gas quality portion of their public informational postings (McKain et al., 2015; Smith et al., 2015; Lamb et al., 2016; Plant et al., 2019). To account for error associated with the EMER from either onsite measurement uncertainty at the reporting stations or the variable residence time of natural gas within the network, we used the mean +/– 1 standard deviation ($εEMER$=$±1σ10day$) for all gas quality measurements reported in the 10 days prior to the measurement date. A detailed description of the data sources and determination of the EMER for each city is included in Section 2 of the SI. $TMER=∫(cos(θ)(C2H6Downwind−C2H6Inflow)U)∫(cos(θ)(CH4Downwind−CH4Inflow)U)EMERpipeline−1+εtransport+εEMER$ 1 Equation 1: Computation of the Thermogenic Methane Emission Ratio (TMER) Individual terms are defined in the list below as: • cosθ: nonorthogonality flight track correction factor • (X)Downwind: Downwind enhancement of CH4 and C2H6 • (X)Inflow: The modeled inflow concentration, either MDI or linear background • U: The North American Mesoscale Forecast System hourly 12 km (NAMS-12 km) wind speed. • EMERpipeline: The reported Ethane: Methane Emission Ratio of the distribution system • εtransport: Error term associated with transport when determining the MDI • εEMER: Error term associated with the variability in the reported EMER ### 2.4. Computing the model derived inflow (MDI) and meteorological parameters for the TMER calculation For 8 (of 13) flights with the most complex inflow patterns, we used an atmospheric transport model and measurements along the upwind transect measured during each flight to determine an MDI or effective inflow concentrations that can be subtracted from each mixing ratio observation in the downwind transect, to compute downwind enhancements due to urban emissions. To calculate the MDI for each flight, we use multi-particle back trajectories from the Stochastic Time Inverted Lagrangian Transport Model (STILT; Fasoli et al., 2018). The STILT model is a Lagrangian atmospheric transport model that can be run backward in time to statistically represent atmospheric dispersion and transport before an air mass reaches a particular point in time and space (the receptor). Ensembles of STILT runs were executed from receptor points distributed along each downwind flight track, spaced 10s apart. For each receptor, 500 particles were released and advected backward in time for 10 h, ensuring all particles were allowed sufficient time to exit the urban domain under study. Example STILT releases from three receptors along the flight track are shown in Figure 3. The inflow concentration values for each downwind receptor were determined from the measured values at each point where a STILT particle crossed the upwind transect. The STILT model was driven by the NAMS-12 km product as archived at the NOAA Air Resources Laboratory (ARL; Stein et al., 2015). Figure 3. Example of a STILT particle release. Panel A shows the measured CH4 mixing ratio along the flight track. Panel B shows the particles as they are transported through the STILT model from three example receptors; particles are colored as a function of the time since their release. Panel C shows the nonorthogonality factor computed from the North American Mesoscale Forecast System hourly 12 km wind direction. Panel D shows the horizontal wind speed (U). The X and Y axis of Panels A, B, C, and D represent degrees Longitude and Latitude. DOI: https://doi.org/10.1525/elementa.2021.000119.f3 Figure 3. Example of a STILT particle release. Panel A shows the measured CH4 mixing ratio along the flight track. Panel B shows the particles as they are transported through the STILT model from three example receptors; particles are colored as a function of the time since their release. Panel C shows the nonorthogonality factor computed from the North American Mesoscale Forecast System hourly 12 km wind direction. Panel D shows the horizontal wind speed (U). The X and Y axis of Panels A, B, C, and D represent degrees Longitude and Latitude. DOI: https://doi.org/10.1525/elementa.2021.000119.f3 For each receptor point, the inflow value was calculated as the mean of 500 upwind mole fraction estimates inferred from the back trajectories of the Lagrangian particles. The MDI was then smoothed along the downwind flight track with a penalized spline to account for the spatial resolution of the STILT receptors. The enhancements were then calculated by subtracting the smoothed MDI from the downwind measurements. The errors in the transport simulations were taken into account when computing the overall uncertainty of the MDI (Section 3 of the SI). The STILT model can incorporate optional parameters to account for spatio-temporal error and autocorrelation in the horizontal and vertical advection parameters of the input meteorological model as well as error in the model estimate of the height of the planetary boundary layer (Fasoli et al., 2018). Lin and Gerbig (2005) describe how STILT increases the magnitude of the stochastic transport component applied to particle transport to statistically represent the impact of meteorological error on the resulting spatial distribution of the modeled particles (Lin and Gerbig, 2005). The STILT model then creates a second distribution of particles that statistically represents the transport error. The difference between total errors in MDI, calculated using the “error-added” particle distributions, and the MDI calculated using the original particle distributions, approximates the effect of the meteorological error on the MDI (Equation 2). $εtransport2=σw/error2−σ2$ 2 Equation 2: Determination of transport error when computing the MDI Individual terms are defined in the list below as: • $εtransport2$: The actual variance of the transport error due to meteorological uncertainty • $σw/error2$: The variance of the transport error of the MDI with added meteorological error • σ2: The variance of the transport error of the MDI without added meteorological error For 3 of the 13 flights presented (August 29, 2017, Boston, MA, June 6, 2017, Chicago, IL, September 2, 2017, New York, NY), we were able to measure the downwind enhancement with particularly stable upwind concentrations. These flights occurred in the presence of an onshore wind, and operational and aircraft safety constraints did not always allow the plane to fly over the water in order to observe upwind concentrations. Hence, a complete upwind fetch was not always flown. In these cases, the downwind enhancement was calculated using conventional methodology by subtracting a background that was computed by linearly interpolating the average background values observed on each side of the plume (Heimburger et al., 2017). For these flights in New York, NY, and Boston, MA, the wind vector was a consistent onshore fetch from the Atlantic Ocean. For the flight in Chicago, IL, there was a consistent onshore fetch from Lake Michigan. We are confident that for these three flights, there were no upwind sources of CH4 or C2H6 significantly contributing to the downwind enhancement. For two flights (April 26, 2018, Philadelphia, PA, and June 19, 2017, Indianapolis, IN), the aircraft did not accomplish an upwind flight transect that was suitable to compute an MDI, and the dominant airflow was not from regions where we can definitively say there were no sources of CH4 or C2H6 upwind. The background was calculated using the interpolation method described above (Heimburger et al., 2017). In these cases, the errors associated with the TMER may be larger than included in our error estimates. We believe that the added uncertainty during these two flights is not large enough to affect our conclusions; these two flights are discussed in detail in Section 6 of the SI and are denoted with a “+” in Figure 1. The TMER is the ratio of CH4 and C2H6 fluxes across a hypothetical curtain traced by the flight track (Equation 1). To compute the ratio of the CH4 and C2H6 fluxes transported through this curtain, we correct for the nonorthogonality of the aircraft heading with respect to the mean wind direction along the flight transect. We calculate an angle (Θ) at 10s intervals along the downwind transect, where Θ is the angular deviation from normal of two vectors representing the direction of flight and the wind direction at the aircraft. The direction of flight is based on the direction of motion of the aircraft in time, and the mean wind direction is derived from the NAMS-12 km product at the point in time and space where the aircraft flew. The horizontal wind speed (U) (Figure 3D) was also determined from the NAMS-12 km model. The meteorological factor is then calculated $(cos(Θ)U)$ and applied to each point along the downwind transect to obtain the flux through the curtain at that point (Figure 3C). In order to assess the magnitude of the instantaneous error of the NAMS-12 km meteorology, a model-data comparison was done for three flights using the wind speed and direction measured by the onboard BAT Probe and values from NAMS-12 km. BAT Probe measurements were averaged to 45 s, and the temporally corresponding values were extracted from the NAMS-12 km meteorology. These analyses show a nominal average wind direction error of roughly 11° and an average wind speed error of roughly 1.5 m/s. This comparison is described in further detail in Section S3 of the SI. ### 2.5. Defining the emission domain influencing the downwind plume The STILT model was originally developed to understand the influence of surface sources on enhancements observed at the point of measurement (Lin et al., 2003) by generating a sensitivity matrix (footprint) for a specific receptor to solve for the geospatial upstream surface influence ($ppm(μmolm2s)−1$) (Fasoli et al., 2018). For this study, the STILT model was run at 10s intervals along the downwind flight track and footprints were generated at a spatial resolution of 0.025° × 0.025° (roughly 6.5 km2). The footprint domains were defined so the target city and immediately outlying areas could be fully enclosed within the domain and are shown in Section 5 of the SI. The spatially resolved sensitivity of the ratio of the calculated flux at the curtain to the surface emissions can be calculated by Equation 3. We represent this flux sensitivity graphically by multiplying each downwind footprint by the meteorological variables in the flux calculation $(cos(Θ)U)$, then summing the set of footprints for the total downwind flight transect and normalizing them from 0 to 1. The result allows us to verify that the domains were sampled in a way that did not bias the calculated flux ratio to a small area of the urban domain due to either meteorological conditions or inconsistencies in the flight profile. $δFluxxy=∫(Ucos(θ)(∫(FootxyEmissxydxdydt))dxdy$ 3 Equation 3: Sensitivity of measurement to a specific grid box within the domain. Individual terms are defined in the list below as: • $δFluxxy$: The sensitivity of the receptor to the flux at grid box xy • U: The NAMS-12 km wind speed • cos(θ): nonorthogonality flight track correction factor • Footxy: The footprint value at grid box xy • Emissxy: The emissions at grid box xy ### 2.6. Calculation of regional inventories emission rates We compared to published inventories in order to contextualize our results and to evaluate the current understanding of greenhouse gas emissions from urban areas. For CH4 emissions, we used the gridded EPA CH4 inventory (Maasakkers et al., 2016). GEPA is a multi-sector CH4 inventory with 0.1° × 0.1° spatial resolution. Of the 22 sectors reported in the inventory, we identified 12 as being of thermogenic origin and having the potential for C2H6 co-emissions as shown in Section 5 of the SI. In order to calculate the inventory flux of CH4 from each urban domain, GEPA was reprojected onto a Universal Transverse Mercator coordinate system so the total flux could be calculated using equal area grid boxes. After reprojection, the GEPA inventory was summed for each city. To investigate the sensitivity of our analysis to the prescribed domain boundaries, we summed the inventory in two ways (1) by subdividing the inventory based on the urbanized area boundaries defined by the U.S. Census Bureau in the Topologically Integrated Geographic Encoding and Referencing (TIGER) database and (2) by subdividing the inventory based on the flight tracks to compute the inventory emissions from the actual measured area (primarily the urban cores of each city). In order to understand how representative GEPA was to another large scale CH4 inventory, the TMER for each city was also compared to the TMER calculated using EDGAR v 4.3.2. This comparison is shown in Section 6 of the SI. Because CO2 and CH4 measurements are much more common than C2H6 measurements, we used the CO2: CH4 ratio to assess our methodology against existing data. We used the Anthropogenic Carbon Emission System (ACES) CO2 emission inventory (Gately and Hutyra, 2017) for additional evaluation of our inventory framework. ACES is a 1 km × 1 km anthropogenic CO2 emissions inventory with hourly resolution compiled for 2013 and 2014. For this study, we used the average ACES emissions across all days in April 2014 from 18:00:00 to 22:00:00UTC and used that to calculate the CO2 flux of each urban domain. Like GEPA, ACES was masked and summed based on the TIGER boundaries and the flight domains in order to determine the respective hourly emission rates of CO2 from each urban domain sampled by the aircraft and to determine how susceptible total emission estimates are to the definition of the domain. ### 2.7. Comparison to ground network data We compared the ratio of integrated CH4 and CO2 enhancements computed from our flights to data from the Harvard–Boston University Greenhouse Gas Network in the city of Boston, MA, as an additional evaluation of our methodology. The Harvard Copley site samples CH4 and CO2 at a height of 215 m and is located in the core city of Boston, MA (for a complete description of the site, refer to Sargent et al., 2018). Tower time series data for the month of April were subselected to include only daytime data between the hours of 12 p.m. and 5 p.m. local to ensure sampling when the tower inlet was within the daytime planetary boundary layer and to ensure that the observed enhancements would be comparable to the daytime aircraft flight. A standard major axis regression was then computed using the calibrated mixing ratios to determine a characteristic ratio between CO2 and CH4 for the Boston urban Core. Enhancement ratios using CO2 from the Copley site were compared for winter, but not for summer, due to difficulty in resolving effects of CO2 uptake by the summer biosphere (Briber et al., 2013; McKain et al., 2015; Sargent et al., 2018). We also present a comparison to tower measurements of TMER made in Indianapolis by the INFLUX project (Lamb et al., 2016). These comparisons further highlight the need for accurate representations of the sample domain, especially when attempting to compare multiple measurements of the same urban area. ## 3. Results ### 3.1. MDI and ground influence Figure 4B shows a typical MDI for a downwind plume, as calculated for the winter flight on April 9, 2018, over the urban core of Boston, MA. Structure in the MDI is a function of variable mixing ratios observed in the upwind transect and their transport across the domain into the downwind measurement, as simulated by the STILT model. Results from the comparison of horizontal winds from NAMS-12 km meteorology to the horizontal winds from the ALAR BAT probe measurement suggest that the error in the NAMS-12 km model was small and well accounted for in the parameters adapted from Lin and Gerbig (2005), and Gerbig et al. (2008). We computed a mean absolute direction error in the NAMS-12 km meteorology of only 11.2 (degrees) [10.6, 11.9] (±1σ). The mean wind speed difference showed NAMS-12 km to be 1.5 (m/s) [1.33, 1.68] ((±1σ) faster than the winds measured by the BAT probe with mean measured wind speeds averaging 7.56 (m/s). These results are further described in Section 4 of the SI. Subtraction of the MDI from the downwind measurement leaves the concentration enhancement due to emissions in the urban domain between upwind and downwind flight transects, enabling us to compute the TMER. Figure 4. Example MDI and normalized flux sensitivity. Panel A shows the upwind and downwind flight tracks for the flight on April 9, 2018. Panel B shows the CH4 mixing ratio observed in the downwind transect along with the calculated MDI (MDI uncertainty shown in gray). Panel C shows the normalized sensitivity of the calculated flux ratio to emissions on the ground. The borders of the Topologically Integrated Geographic Encoding and Referencing Domain are overlaid in Panel C. The X and Y axis of Panels A and C represent degrees Longitude and Latitude. DOI: https://doi.org/10.1525/elementa.2021.000119.f4 Figure 4. Example MDI and normalized flux sensitivity. Panel A shows the upwind and downwind flight tracks for the flight on April 9, 2018. Panel B shows the CH4 mixing ratio observed in the downwind transect along with the calculated MDI (MDI uncertainty shown in gray). Panel C shows the normalized sensitivity of the calculated flux ratio to emissions on the ground. The borders of the Topologically Integrated Geographic Encoding and Referencing Domain are overlaid in Panel C. The X and Y axis of Panels A and C represent degrees Longitude and Latitude. DOI: https://doi.org/10.1525/elementa.2021.000119.f4 Domain sensitivity maps illustrate the measurement domain sampled by each flight and depict the relative sensitivity of the TMER to emissions in each grid box within the domain. Figure 4C shows an influence map for the April 9, 2018, flight. As expected, the grid boxes closer to the aircraft more heavily influence the downwind enhancement used to compute the TMER. Domain sensitivity maps for other flights are shown in Section 5 of the SI (Figures S8–S21) and identify the domain sampled by the aircraft for all cities. The U.S. Census Bureau TIGER borders are overlaid on the influence maps in green and show that, in most cases, the aircraft sampled only the urban core of each city. Notably, the measurement domains are not well represented by political boundaries and must be further defined with tools such as meteorological models and footprint modeling as done here. Domain sensitivity maps for the remaining flights support the assertion that similar urban core domains were sampled by flights in various meteorological conditions and can be compared in our analysis (i.e., winter and summer). ### 3.2. Thermogenic methane emission fractions We observed good agreement throughout this study between published EMERs and the slope of CH4 and C2H6 enhancements at those points in the track where we expect the sources to be predominantly natural gas, such as for the city gate sampling shown in Figure 2. Tracer-tracer plots comparing the downwind measurements of CH4 and C2H6 for each flight to the EMER obtained from the pipeline companies (e.g. Figure 2D) are shown in Section 9 of the SI (Figure S26). Figure 2D shows an example tracer-tracer plot observed downwind of Indianapolis, IN, where biogenic sources are a significant fraction of the city’s total CH4 emissions. The CH4 enhancements that we attributed to emissions of natural gas are clearly separated in the time series from the biological landfill point source, and the slope of the enhancements agrees well with the published EMER (Figure 2D). The time series in Figure 2C shows three distinct sources that can be easily identified: the South Side Landfill (biogenic) and the City-Gate in Zionsville. Outside of those two easily distinguishable peaks in the time-series, our method is not able to attribute other enhancements to specific point sources, and thus, we refer to the rest of the urban plume as diffuse emission, likely from the natural gas distribution infrastructure, post-meter emissions, and other potential methane sources that may be collocated. Our estimate of the uncertainty in the EMER from the pipeline informational postings is discussed in Section 2 of the SI. The calculated TMER values for each urban domain (Equation 1) are shown in Figure 5. In winter, the TMERs ranged from 0.32 for Richmond, VA, to 1 for New York, NY. Uncertainties in the TMER vary from very tight, for example, 0.815 (+0.004/–0.006) observed during a flight over New York, NY, on September 2, 2017, to significant uncertainty, for example, 0.384 (+1.23/–0.13) observed during a flight over Philadelphia, PA, on August 28, 2017. The differences in the calculated relative uncertainties are a function of the complexity of upwind source emissions, which result in increased uncertainty in the calculated MDI, as described in detail in Section 3 of the SI. Figure 5. Thermogenic Methane Emission Ratio results for seven cities. Aircraft Thermogenic Methane Emission Ratio results for summer (red) and winter (blue) are compared to the GEPA inventory subset using the U.S. Census Bureau Topologically Integrated Geographic Encoding and Referencing Boundaries (green) and then subset based on the domain encircled by the aircraft (blue for winter flights and red for summer flights). DOI: https://doi.org/10.1525/elementa.2021.000119.f5 Figure 5. Thermogenic Methane Emission Ratio results for seven cities. Aircraft Thermogenic Methane Emission Ratio results for summer (red) and winter (blue) are compared to the GEPA inventory subset using the U.S. Census Bureau Topologically Integrated Geographic Encoding and Referencing Boundaries (green) and then subset based on the domain encircled by the aircraft (blue for winter flights and red for summer flights). DOI: https://doi.org/10.1525/elementa.2021.000119.f5 The TMER values measured in four of six cities showed statistically significant seasonal differences. The Boston urban core did not show significant biogenic increase in summer compared to winter. McKain et al. (2015) reported a modest seasonality, with greater biogenic emissions in the spring/summer. Our flights sampled the urban core of the city (generally East of the I95 ring road), whereas the inverse model in the McKain study sampled CH4 emissions from a much larger domain extending to the west. It is likely that the outlying rural and forest areas included in the McKain domain were the source of the seasonal increase of biogenic emissions observed in the Boston regional network in spring/summer. Inventory estimates of TMER from the 12 thermogenic sectors in the GEPA inventory are shown Figure 5. Two subsets of the GEPA estimates are shown: a subset based on the TIGER boundaries drawn by the U.S. Census Bureau (Plant et al., 2019) and a subset for each flight with the domain defined as the area encircled by the aircraft. The TMER predicted by GEPA can vary by 15%–30% (up to 0.13 $FluxCH4(Thermogenic)FluxCH4(Total)$) based on whether or not the inventory is subsampled using the domain bound by the sampling flight track or the U.S. Census Bureau’s political boundaries. All of the inventory estimates of TMER were significantly lower than those measured by the aircraft in winter, and likewise three of the six cities measured in summer, regardless of inventory subset. GEPA estimated TMER from each city (except Boston, MA) to be less than 0.50, but values this low were observed by the aircraft on only 4 of 13 flights (winter: Richmond, VA, only; summer: Washington, DC/Baltimore, MD, Philadelphia, PA, Indianapolis, IN). ### 3.3. CO2: CH4 comparison to ground data Since CO2 emission inventories are based on reliable proxies (e.g., traffic data, building maps, power plant locations, and reported emissions; Sargent et al., 2018) not available for CH4 emission inventories, accurate determination of the CO2: CH4 ratio can aid in the assessment of the urban CH4 sources and source strengths (Plant et al., 2019). Figure 6 shows the ratio of CO2: CH4 enhancement from each city measured during the winter flights compared to the ratio of ACES CO2 emissions and the GEPA CH4 emissions. Both inventories were subsampled using two domains for each flight: the U.S. Census Bureau TIGER boundaries and the urbanized core encircled by each flight. Every domain showed a larger CO2: CH4 inventory estimate when the flight domains were used to subsample the inventories as opposed to the TIGER boundaries, with the average effect being a difference in the CO2: CH4 ratio of 13.4%. Figure 6. CO2: CH4 flux ratios for each urban domain sampled in the winter. Ratios measured by the aircraft (blue) are compared to ratios computed from two inventories Anthropogenic Carbon Emission System (CO2) and GEPA (CH4). The red line shows the ratio measured by the Copely Tower in the Harvard Urban Greenhouse Gas Network. All results are shown in $(molesCO2)(molesCH4)−1$. DOI: https://doi.org/10.1525/elementa.2021.000119.f6 Figure 6. CO2: CH4 flux ratios for each urban domain sampled in the winter. Ratios measured by the aircraft (blue) are compared to ratios computed from two inventories Anthropogenic Carbon Emission System (CO2) and GEPA (CH4). The red line shows the ratio measured by the Copely Tower in the Harvard Urban Greenhouse Gas Network. All results are shown in $(molesCO2)(molesCH4)−1$. DOI: https://doi.org/10.1525/elementa.2021.000119.f6 The ratio of CO2: CH4 enhancements measured at the Copley site in the Boston Urban Greenhouse Gas Network was 287 (+27/–22) mol CO2: CH4 (r2 = 0.84, MA regression), in close agreement with the aircraft data (240 (+26/–12) $molCO2molCH4$), using data from the April 2018 ALAR flight (Figure 7). This comparison to independent determination of TMER tests our methodologies, and the good agreement supports our flux ratio sampling method. Figure 7. Comparison of aircraft flux ratios to urban tower measurements. Points are hourly mean daytime CH4 and CO2 measurements from Copley Tower for the month of April 2018 and are compared to the calculated CO2: CH4 ratio from the April 9, 2018, Flight over the Boston, MA, urban core. DOI: https://doi.org/10.1525/elementa.2021.000119.f7 Figure 7. Comparison of aircraft flux ratios to urban tower measurements. Points are hourly mean daytime CH4 and CO2 measurements from Copley Tower for the month of April 2018 and are compared to the calculated CO2: CH4 ratio from the April 9, 2018, Flight over the Boston, MA, urban core. DOI: https://doi.org/10.1525/elementa.2021.000119.f7 ## 4. Discussion ### 4.1. Measurement and estimation of TMER Determination of TMER, like inverse analysis of emissions, depends sensitively on the concentrations inferred in inflowing air (Karion et al., 2015; Heimburger et al., 2017; Bares et al., 2018). A simple method used previously draws a straight line between plume edge concentrations, suitable in specific settings where the target region is isolated, with a clean upwind fetch. This approach does not account for inflow effects of inhomogeneous concentrations entering the target region from the upwind boundary, as commonly found along the East Coast Corridor of the United States. To address this issue, we have combined an atmospheric transport model with the upwind observations made from the aircraft to enable us to compute an MDI and subtract background values informed by both model meteorology and aircraft observations. We can propagate the uncertainty associated with our subtracted background values when we estimate the uncertainty of our final answer. As with any method of background estimation using upwind aircraft sampling, we make assumptions as to the applicability of upwind concentrations observed to the downwind transect. In these flights, we attempted to plan flights on days when the wind direction was expected to be relatively steady throughout the domain for the duration of the sampling, and, when operational constraints allowed, we sampled the upwind transect first. As shown in Figure 5, the relative uncertainty of our TMER estimates is variable and a function of the uncertainty estimated in the MDI (Equation 2 and Section 3 of the SI) and fluctuations in the pipeline EMER reported for each domain in the 10 days prior to the measurement. The uncertainty of the TMER is potentially underestimated on three flights, indicated by an “” in Figure 1 (New York, NY [March 20, 2018], Philadelphia, PA [April 26, 2018], and Indianapolis, IN [June 19, 2017]). Further analysis of the flights with added uncertainty is described in detail in Section 6 of the SI. For the September 2, 2017, sampling flight of New York, NY, the prevailing wind direction was from the east, which justified a simple background approximation based upon the homogeneous upwind marine air mixing ratios. This flight presents the lowest uncertainty in the TMER (0.815 (+0.004/–0.006; Flux CH4 (Thermogenic))/(Flux CH4 (Total))) as the pipeline EMER was the determining factor and was quite stable at 0.0213 (+0.004/–0.003) molC2H6/molCH4. For the August 28, 2017, flight sampling Philadelphia, PA, there were significant CH4 sources of biogenic origin upwind of the domain, which manifested as a larger uncertainty estimation in the bounds of the MDI and ultimately a larger uncertainty in the calculated TMER (0.384 (+1.23/–0.13) ($FluxCH4(Thermogenic)FluxCH4(Total)$). To compute the TMER in this study, we integrated the enhancements computed from the downwind transects (Equation 1). Previous studies have used the slopes of tracer-tracer plots to determine similar relationships (Plant et al., 2019). We did the same for our data to compare our results to these methods. If the emissions are measured close to a city, such that the plume has not mixed homogeneously in the horizontal coordinate (e.g., if the correlation coefficient is not excellent), the slopes computed by this analysis can be significantly skewed by unmixed source plumes with high mixing ratios. This analysis is described in detail in Section 10 of the SI. ### 4.2. CO2: CH4 ratios As shown in Figure 6, the observed CO2: CH4 ratios are all statistically significantly smaller than those from the emission model ratios. This largely reflects an underestimate of the CH4 emissions in the GEPA inventory, consistent with the results here implying that the GEPA inventory underestimates the thermogenic CH4 emissions, for all cities. This has been observed for other cities (Plant et al., 2019). The ACES CO2 inventory has uncertain estimated to be <±10% (Gately and Hutyra, 2017), supported by the top-down analysis of CO2 in Boston, MA (Sargent et al., 2018), and bias in ACES is unlikely to explain the large systematic differences between observed and inventory ratios. ### 4.3. Defining the domain Unambiguously defining the domain sampled during an airborne study is another difficult task when analyzing regional scale airborne measurements. Because many of the domains sampled in this study are located in the densely populated East Coast of the United States, we employed the STILT model to better describe the relative sensitivity of our TMER calculation to emissions from ground sources throughout the domain. The ground source sensitivity maps shown in Figure 4C and in Section 5 of the SI show the normalized spatially resolved influence of the sampling domain on the flux through the downwind transect drawn by the aircraft. The flux sensitivity maps, in almost all cases, show that downwind concentrations are sensitive to emissions from grid boxes upwind of the domain, hence our focus on calculating a value for MDI that would enable the subtraction of the influence of those sources from the enhancement observed downwind of the urban core. The flux sensitivity maps help us to compare multiple measurement flights of the same geographic location. They graphically demonstrate that adjustments in the aircraft flight track enable us to sample a similar domain in both seasons regardless of the fact that the meteorology may be differing across flights. Our summer flight over Indianapolis, IN, exemplifies the importance of defining the domain that influences the observed enhancements of CH4 and C2H6. On June 19, 2017, the aircraft was able to obtain two downwind transects that were suitable for analysis (i.e., completely transiting through the urban plume). Of the two transects, one was composed of a much more urban fetch, sampling the core of the Indianapolis metro area, for which we computed a TMER of 0.42 (+0.04/–0.01). In the second transect, we sampled a fetch that also included a large amount of semi-rural area in the suburbs north of the downtown metro area and computed a TMER of 0.26 (+0.01/–0.01), representing relatively greater biogenic emissions. Using the flux sensitivity maps, we are able to see that the area measured by the transect with the more rural fetch measured a more analogous domain to that sampled during the winter time measurement of Indianapolis on March 16, 2018, (this is the reported value in Figure 5). A higher TMER value in the urban core appears typical of other cites (e.g., Boston, MA, see above). More information on this particular flight, along with the flux sensitivity maps for these transects, can be found in Section 6 of the SI. The sampled domain must also be taken into account when comparing the results in this study to results in the literature. Lamb et al. (2016) used measurements of C2H6 and CH4 made from a tower in downtown Indianapolis (INFLUX network tower 11) during February and March 2015 to determine a TMER of approximately 0.43 (43% natural gas, 48% biogenic, and 9% other). Our summer TMER computed using the urban transect, mentioned above, compared well to theirs, but our winter TMER was significantly higher (0.71 (+0.01/–0.02)). Further analysis of the flux sensitivity of the tower and sources within those footprints would be needed to more accurately compare to the Lamb et al. tower measurements. Political boundaries typically do not define the areas that influence aircraft measurements, making it difficult to systematically compare to inventories or to studies of other cities. We have compared our results to inventories sampled using both political boundaries and the exact area encircled by the aircraft (Figures 5 and 6) and shown that the TMER of a “city” can vary significantly based on that definition. We suspect that some of this variance is driven by the presence of biogenic CH4 point sources (landfills, WWTPs, active methanogenic wetland areas, etc.) near the upwind boundary of the urban domain. Evidently, the TMER of a city can be quite sensitive to whether these point sources influence the observations, or not. We conclude that defining the actual domain sampled in the study is crucial to interpreting top down studies of emissions and that a quantitative analysis of individual flight domains must be used when comparing measurements of a single urban area that may have been made under different meteorological conditions. We have presented here a straightforward method to make these comparisons. ### 4.4. Thermogenic fraction discussion Local policy makers have shown a willingness to introduce legislation to spearhead greenhouse gas emission reductions at the city level. Local Law 97 (a component of the New York, NY, Climate Mobilization Act) is one example. It sets aggressive greenhouse gas emission reduction targets for buildings larger than 25,000 ft2 in New York, NY, with a \$268/ton fine for emissions above the target. As more cities begin to evaluate the prospect of CH4-specific emission reduction legislation, it will be more pressing for the scientific community to inform local policy makers of the emission sectors that dominate their local CH4 emission profile so that emission reduction strategies can address the most effective approaches. In order to make such recommendations, we must be prepared with well-characterized CH4 emission statistics for specific cities, which need to be supported by measurement campaigns, as current CH4 inventories do not accurately reflect the distribution of sources and emission rates in the urban domain. New York, NY, is a good example, where we found that the overwhelming majority of CH4 emissions are derived from natural gas sources, an observation that is not consistent with the distribution of sources in the GEPA inventory. In this study, we have partitioned CH4 emissions between biogenic and thermogenic sources for seven major urban areas. As shown in Figure 5, the GEPA CH4 emission inventory currently underestimates the natural gas contribution to CH4 emissions, in some cases (New York, NY, Boston, MA, Chicago, IL) by as much as 50%. Our results showed statistically significant seasonality in the urban TMER in four of the six cities we were able to sample in both winter and summer. But CH4 emission inventories still lack adequate seasonal resolution, making it impossible to accurately assess temporally varying emissions from data. We note that the significant biogenic sources in most urban environments are sewer systems, landfills, and wastewater treatment facilities and that there is likely some significant seasonality in those emissions not captured in the emission models (Cambaliza et al., 2014; Fries et al., 2018). Based on what is known about the seasonality of biogenic CH4 sources and the seasonally invariant thermogenic emissions observed by Mckain et al., we suspect that biogenic emissions are the source of the seasonality in the urban TMER. Further research is needed to quantify the seasonal amplitude of the CH4 flux from specific biogenic sectors and sources. Furthermore, recent studies have suggested that the total inventory flux estimates from many of the same urban areas measured in this analysis underestimate total emissions (McKain et al., 2015; Plant et al., 2019). The results from our study illustrate how dramatically different the CH4 emission profiles can be from city to city. We find the TMER varying as much as a factor of 3 (0.32 for Richmond, VA, to 1.0 for New York, NY, in winter months). These differences can occur for multiple reasons. Cities with older natural gas infrastructure (e.g., with cast iron NG mains) such as New York, NY, and Boston, MA, can have greater natural gas emissions (Von Fischer et al., 2017). Some cities, such as Boston, MA, do not have landfills in the urban core where some, such as Richmond, have multiple. Future studies should aim to quantify both the TMER and total CH4 flux from a larger distribution of cities to better quantify the emissions from cities with a wide array of potential source infrastructure. ## 5. Conclusion Current emission inventories for the United States significantly underestimate the role of natural gas relative to biogenic CH4 emissions in urban environments. Local policy makers must have knowledge of the emission profile of their city in order to best allocate resources and introduce and implement legislation to reduce greenhouse gas emissions. Our work suggests that urban areas with significant landfill infrastructure (e.g., Richmond, VA) may benefit from better landfill gas capture and recovery systems, whereas urban areas with dense and aging natural gas infrastructure (e.g., Boston, MA, and New York, NY) should devote their resources toward updating their natural gas distribution infrastructure and possibly controlling post-meter emissions. Our results also highlight the need for more urban scale observational CH4 data in order to better parameterize the emission accounting used in national level CH4 emission inventories. Currently, inventories are unable to provide the comprehensive bottom-up picture that they were designed to provide, and improvements require higher quality flux estimates from urban CH4 sources, combining top-down and bottom-up methodologies. ## Data accessibility statement Flight data are available to the public and can be accessed from the Harvard Dataverse at https://doi.org/10.7910/DVN/T7ANQD. ## Supplemental files The supplemental files for this article can be found as follows: See attached Supplementary Materials. ## Acknowledgments We would like to thank our colleagues Israel Lopez-Coto and Anna Karion for their support and guidance with respect to our mission planning and deployment. We also acknowledge Eric Kort and Genevieve Plant at the University of Michigan for their contribution to the NOAA ECO campaign which in turn led to us being able to use those data. Perhaps most importantly, we acknowledge the fine support of the aircraft operations folks at Purdue University whose diligent work gave us confidence that every time ALAR took off it would return home safe and allow us to bring data along with it. ## Funding We thank the National Institute of Standards and Technology (NIST) for partial support of this work, via Award No. 70NANB16H262. Certain commercial equipment, instruments, or materials are identified in this article in order to specify the experimental procedure adequately. Such identification is not intended to imply recommendation or endorsement by NIST nor is it intended to imply that the materials or equipment identified is necessarily the best available for the purpose. ## Competing interests The authors declare no competing interest. Paul B. Shepson is an Associate Editor at Elementa. He was not involved in the review process of this article. ## Author contributions Contributed to the acquisition of the data: CF, PBS, BHS, KH. Contributed to the analysis of the data: CF, KH, SW. Contributed to the development of the ethane analyzer: CF, BD, SW. Contributed data from the NOAA ECO campaign for the winter Philadelphia Flight: CS. ## References References Alvarez , RA , Alvarez , RA , Zavala-Araiza , D , Lyon , DR , Allen , DT , Barkley , ZR , Brandt , AR , Davis , KJ , Herndon , SC , Jacob , DJ , Karion , A , Kort , EA , Lamb , BK , Lauvaux , T , Maasakkers , JD , Marchese , AJ , Omara , M , Pacala , SW , Peischl , J , Robinson , AL , Shepson , PB , Sweeney , C , Townsend-Small , A , Wofsy , SC , Hamburg , SP . 2018 . Assessment of methane emissions from the U.S. oil and gas supply chain . Science 361 ( 6398 ): 186 188 . DOI: http://dx.doi.org/10.1126/science.aar7204. 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Yacovitch , TI , Herndon , SC , Roscioli , JR , Floerchinger , C , McGovern , RM , Agnese , M , Pétron , G , Kofler , J , Sweeney , C , Karion , A , Conley , SA , Kort , EA , Nähle , L , Fischer , M , Hildebrandt , L , Koeth , J , McManus , JB , Nelson , DD , Zahniser , MS , Kolb , CE . 2014 . Demonstration of an ethane spectrometer for methane source identification . Environmental Science and Technology 48 ( 14 ): 8028 8034 . DOI: http://dx.doi.org/10.1021/es501475q. How to cite this article: Floerchinger, C, Shepson, PB, Hajny, K, Daube, BC, Stirm, BH, Sweeney, C, Wofsy, SC. 2021. Relative flux measurements of biogenic and natural gas-derived methane for seven U.S. cities. Elementa Science of the Anthropocene 9(1). DOI: https://doi.org/10.1525/elementa.2021.000119. Domain Editor-in-Chief: Detlev Helmig, Boulder A.I.R. LLC, Boulder, CO, USA Associate Editor: Isobel Jane Simpson, Department of Chemistry, University of California, Irvine, CA, USA Knowledge Domain: Atmospheric Science This is an open-access article distributed under the terms of the Creative Commons Attribution 4.0 International License (CC-BY 4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. See http://creativecommons.org/licenses/by/4.0/.	2021-04-15 10:55:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 23, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4848296642303467, "perplexity": 3247.0428222257337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00451.warc.gz"}
https://squibbishness.wordpress.com/2011/11/12/x-is-the-same-as-y/	## X is the same as Y It’s been several months since I’ve posted anything, mostly because I’ve been busy with coursework, research, TAing, etc. But I have been keeping some squib notes. Here is one. In formal methods class one day, my prof. wrote on the board two sets: $X = \{a,b,c\}$ $Y = \{1,2,3,4\}$ After a few seconds, he realized something was wrong, and said, “Actually, $X$ is the same as $Y$.” My initial reaction was to think, “Ah, then $X=\{1,2,3,4\}$,” or maybe “OK, then $X=\{a,b,c,d\}$, where $a=1, b=2, c=3, d=4$.” In other words, add an element to $X$ to be like $Y$. Turns out that what the prof. meant was that $Y=\{1,2,3\}$, where $a=1, b=2, c=3$. In other words, subtract an element from $Y$ to be like $X$. It made me wonder whether, given any generic sentence like “X is the same as Y”, one could tell whether (one’s conception of) X ought to change to conform to (one’s conception of) Y, or vice versa. For me, I think X is the one to change, not Y. For example, let’s say you think that John is 5’6” tall, and Sue is 5’8”. If I tell you, “Actually, John is the same height as Sue”, can you tell how tall they are? If I were to actually utter that sentence, I think it would be to convey that John is 5’8”, i.e., (your conception of) Sue’s height. Likewise, to convey that Sue is 5’6”, I’d say, “Actually, Sue is the same height as John.” In other words, the subject of the sentence is the one whose height needs adjusting. Same goes for lots of other scenarios: “John is as old as Sue” to me means, basically, “However old (you think) Sue is, that’s how old John is, too” and not “However old (you think) John is, that’s how old Sue is.” It holds for comparatives, as well: “John is older than Sue” means “However old (you think) Sue is, John is older than that” (i.e., adjust/raise John’s age to exceed Sue’s; don’t adjust/lower Sue’s age to be less than John’s). The generalization, I think, is that if I want to correct someone’s conception of X to conform to Y, I say, “X is (the same as/as [adj] as/more [adj] than/…) Y”, and not “Y is (the same as/as [adj] as/more [adj] than/…) X”. Of course, it’s only a generalization—perhaps a pragmatic, conventional one, which can be flouted. If I’m a landscaper and you hire me to make your front yard as pretty as or prettier than your neighbor’s, and I go and dig holes in your neighbor’s yard and dump trash all over it, you’ll probably say, “That’s not exactly what I had in mind”, but I could probably argue that I successfully completed the task.	2018-05-24 11:34:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.823136031627655, "perplexity": 2335.362878050914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866276.61/warc/CC-MAIN-20180524112244-20180524132244-00194.warc.gz"}
https://community.rstudio.com/t/pdf-document-code-highlight-monochrome-color/124116	# pdf_document code highlight monochrome color Is there a way to choose the color of the text in code chunks under the highlight option "monochrome". I would like the color to be monochrome, but not in black, so as to differentiate it better from the common text. output: pdf_document: number_sections: TRUE highlight: monochrome For HTML, it is quite easy by using CSS. For PDF, it would require tweaking LaTeX which is more complicated. monochrome is a style created by Pandoc. What Pandoc will do is insert some command in the document like this \newenvironment{Shaded}{}{} \newcommand{\AnnotationTok}[1]{\textit{#1}} \newcommand{\AttributeTok}[1]{#1} \newcommand{\BaseNTok}[1]{#1} \newcommand{\BuiltInTok}[1]{#1} \newcommand{\CharTok}[1]{#1} \newcommand{\CommentTok}[1]{\textit{#1}} \newcommand{\CommentVarTok}[1]{\textit{#1}} \newcommand{\ConstantTok}[1]{#1} \newcommand{\ControlFlowTok}[1]{\textbf{#1}} \newcommand{\DataTypeTok}[1]{\underline{#1}} \newcommand{\DecValTok}[1]{#1} \newcommand{\DocumentationTok}[1]{\textit{#1}} \newcommand{\ErrorTok}[1]{\textbf{#1}} \newcommand{\ExtensionTok}[1]{#1} \newcommand{\FloatTok}[1]{#1} \newcommand{\FunctionTok}[1]{#1} \newcommand{\ImportTok}[1]{#1} \newcommand{\InformationTok}[1]{\textit{#1}} \newcommand{\KeywordTok}[1]{\textbf{#1}} \newcommand{\NormalTok}[1]{#1} \newcommand{\OperatorTok}[1]{#1} \newcommand{\OtherTok}[1]{#1} \newcommand{\PreprocessorTok}[1]{\textbf{#1}} \newcommand{\RegionMarkerTok}[1]{#1} \newcommand{\SpecialCharTok}[1]{#1} \newcommand{\SpecialStringTok}[1]{#1} \newcommand{\StringTok}[1]{#1} \newcommand{\VariableTok}[1]{#1} \newcommand{\VerbatimStringTok}[1]{#1} \newcommand{\WarningTok}[1]{\textit{#1}} If you were to use tango style instead it would be like that \definecolor{shadecolor}{RGB}{248,248,248} \newcommand{\AnnotationTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textbf{\textit{#1}}}} \newcommand{\AttributeTok}[1]{\textcolor[rgb]{0.77,0.63,0.00}{#1}} \newcommand{\BaseNTok}[1]{\textcolor[rgb]{0.00,0.00,0.81}{#1}} \newcommand{\BuiltInTok}[1]{#1} \newcommand{\CharTok}[1]{\textcolor[rgb]{0.31,0.60,0.02}{#1}} \newcommand{\CommentTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textit{#1}}} \newcommand{\CommentVarTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textbf{\textit{#1}}}} \newcommand{\ConstantTok}[1]{\textcolor[rgb]{0.00,0.00,0.00}{#1}} \newcommand{\ControlFlowTok}[1]{\textcolor[rgb]{0.13,0.29,0.53}{\textbf{#1}}} \newcommand{\DataTypeTok}[1]{\textcolor[rgb]{0.13,0.29,0.53}{#1}} \newcommand{\DecValTok}[1]{\textcolor[rgb]{0.00,0.00,0.81}{#1}} \newcommand{\DocumentationTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textbf{\textit{#1}}}} \newcommand{\ErrorTok}[1]{\textcolor[rgb]{0.64,0.00,0.00}{\textbf{#1}}} \newcommand{\ExtensionTok}[1]{#1} \newcommand{\FloatTok}[1]{\textcolor[rgb]{0.00,0.00,0.81}{#1}} \newcommand{\FunctionTok}[1]{\textcolor[rgb]{0.00,0.00,0.00}{#1}} \newcommand{\ImportTok}[1]{#1} \newcommand{\InformationTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textbf{\textit{#1}}}} \newcommand{\KeywordTok}[1]{\textcolor[rgb]{0.13,0.29,0.53}{\textbf{#1}}} \newcommand{\NormalTok}[1]{#1} \newcommand{\OperatorTok}[1]{\textcolor[rgb]{0.81,0.36,0.00}{\textbf{#1}}} \newcommand{\OtherTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{#1}} \newcommand{\PreprocessorTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textit{#1}}} \newcommand{\RegionMarkerTok}[1]{#1} \newcommand{\SpecialCharTok}[1]{\textcolor[rgb]{0.00,0.00,0.00}{#1}} \newcommand{\SpecialStringTok}[1]{\textcolor[rgb]{0.31,0.60,0.02}{#1}} \newcommand{\StringTok}[1]{\textcolor[rgb]{0.31,0.60,0.02}{#1}} \newcommand{\VariableTok}[1]{\textcolor[rgb]{0.00,0.00,0.00}{#1}} \newcommand{\VerbatimStringTok}[1]{\textcolor[rgb]{0.31,0.60,0.02}{#1}} \newcommand{\WarningTok}[1]{\textcolor[rgb]{0.56,0.35,0.01}{\textbf{\textit{#1}}}} You can spot the difference being some color style. If you want to use your own, you would need to tweak those commands probably, (using renewcommand maybe ?) But there could be a way using Pandoc them file. We plan to make that easier as Pandoc allow to pass custom theme: Pandoc - Pandoc User’s Guide . This would require some evolution in rmarkdown however (at least RMarkdown's utility prevents pandoc from loading external highlighting theme for HTML output. · Issue #2035 · rstudio/rmarkdown · GitHub) Currently you could probably use this workaround --- title: "Test" author: "Me" date: "2/8/2021" output: pdf_document: pandoc_args: "--highlight-style=my.theme" --- and modify the theme file produced by pandoc --output my.theme --print-highlight-style monochorme You can compare with the one from another style pandoc --output my.theme --print-highlight-style tango We'll definitely make that easier in the future. Hope it helps Thank you for answering. I will look more into the solution you suggest. 1 Like This topic was automatically closed 21 days after the last reply. New replies are no longer allowed. If you have a query related to it or one of the replies, start a new topic and refer back with a link.	2022-01-17 15:54:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4517408013343811, "perplexity": 4953.7846445111945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00665.warc.gz"}
https://collegephysicsanswers.com/openstax-solutions/calculate-acceleration-skier-heading-down-100circ-slope-assuming-coefficient	Change the chapter Question (a) Calculate the acceleration of a skier heading down a $10.0^\circ$ slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 5.9 to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies. a) $0.74 \textrm{ m/s}^2$ b) $5.7^\circ$ Solution Video # OpenStax College Physics Solution, Chapter 5, Problem 11 (Problems & Exercises) (6:43) Rating No votes have been submitted yet. Video Transcript	2020-11-23 21:34:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4425414204597473, "perplexity": 1427.3515510902319}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00556.warc.gz"}
https://solvedlib.com/n/certain-reaction-is-first-order-in-hz-and-first-order-in,14881963	# Certain reaction is first order in Hz and first order in Iz: Use this information to complete the table below. ###### Question: certain reaction is first order in Hz and first order in Iz: Use this information to complete the table below. Be sure each of your answer entries has the correct number of significant digits_ [4] [5] initial rate of reaction 0.878 M 137 M 32.0 Mls 0.878 V 31 M MIs 0.386 .v 3.11 M MIs #### Similar Solved Questions ##### Calculate the following integrals: (5 points each) 02(1+2)dxIna S exdx InzJ"(2 _ 3 sin x)dx3. Use symmetry to calculate the following integral: (10 points)Lc Il) dx Calculate the following integrals: (5 points each) 02(1+2)dx Ina S exdx Inz J"(2 _ 3 sin x)dx 3. Use symmetry to calculate the following integral: (10 points) Lc Il) dx... ##### 3. Gains from trade Consider two neighboring island countries called Arcadia and Euphoria. They each have... 3. Gains from trade Consider two neighboring island countries called Arcadia and Euphoria. They each have 4 million labor hours available per month that they can use to produce jeans, corn, or a combination of both. The following table shows the amount of jeans or corn that can be produced using 1 h... ##### Let the random Samples X, ) (nz?) let the random Samples X, ) (nz?) let the random Samples X, ) (nz?)... ##### The built-in R dataset swiss gives Standardized fertility measure and socio-economic indicators for each of 47... The built-in R dataset swiss gives Standardized fertility measure and socio-economic indicators for each of 47 French-speaking provinces of Switzerland at about 1888. The dataset is a data frame containing 6 columns (variables). The column Infant.Mortality represents the average number of live birth... ##### Series RLC circuit R = 100 Q When the capacitor C is removed (RL): the current lags behind the voltage by 3 radian: When inductor L is removed (RC): the current gets ahead of the voltage by ; radian: Find the impedence of the circuit. series RLC circuit R = 100 Q When the capacitor C is removed (RL): the current lags behind the voltage by 3 radian: When inductor L is removed (RC): the current gets ahead of the voltage by ; radian: Find the impedence of the circuit.... ##### Question (16 nta= Give curved arrow-pushing mechanism for the following two reactions SHOW WHERE EVERY PROTON COMES FROM AND GOES TO Indicate the Lewis acid and base (LA or LB) at each INTERMOLECULAR step and whether they are also Bronsted acids and bases (LABA BABB) H6 Gme Nat-OMeIMeOH Question (16 nta= Give curved arrow-pushing mechanism for the following two reactions SHOW WHERE EVERY PROTON COMES FROM AND GOES TO Indicate the Lewis acid and base (LA or LB) at each INTERMOLECULAR step and whether they are also Bronsted acids and bases (LABA BABB) H6 Gme Nat-OMeIMeOH... ##### Please prepare a balance sheet, income statement, statement of owners equity and a statement of cash... Please prepare a balance sheet, income statement, statement of owners equity and a statement of cash flows. 202 Owners' Equity Homework Froblem 1° The following balances are from the Cheyenne's Accounting Company 2018 20,000 50,000 3,000 190,000 60,000 70,000 5,000 2017 10,000 42,000 4,0... ##### How to make this into a table in Python? def displayPay(lst): for inner_lst in lst: name=inner_lst[0]... 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The system will depreciate at the rate f = f(t) and will incur maintenance costs at the rate g = g(t), where t is the time measured in months. The company wants to determine the optimal time to replac... ##### Two tiny conducting spheres are identical and carry charges of -21.8 µC and +47.5 µC. They... Two tiny conducting spheres are identical and carry charges of -21.8 µC and +47.5 µC. They are separated by a distance of 2.71 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? N (b) The spheres are brought into contact and... ##### Can the equation of a horizontal line be written in point-slope form? Explain. Can the equation of a horizontal line be written in point-slope form? 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Aher several hours of analysis, you develop the following data for scanner connectors u... ##### Anseroolet 0 Mlss / Moving 0 certilin directton nas Kinetic ciru 04, It hits u will und comes bilck with half its criginal kinetie energy Ifthe speed of the object Mm 10msKhal c matic cbject? 412,8 B) 3.8 kg C)LOkg 1.0 kg 4nserPotcib:l - E=g=0 roller cuaster of mass 80.0 ke Is Inot Iug with & speed of 20.0 nWls Mun shuw Fure_ trrticutenright nosirion 4bore Eround cuci Ncechrictiun and Iac ! 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An organization's ability to profit without depleting its resources, including employees, natural resources, and the support of the surrounding community Your answer Check Ability of a firm to sense and respond to the environment by intentionally changing Your answer Check Companies need to enha... ##### Indicate on a number line the numbers $x$ that satisfy the condition. $x^{2} \geq 16$ Indicate on a number line the numbers $x$ that satisfy the condition. $x^{2} \geq 16$...	2023-02-05 13:47:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45525866746902466, "perplexity": 9031.829238542647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00119.warc.gz"}
http://www.planetmath.org/node/33928/source	# circular helix ## Primary tabs \documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newtheorem{theorem}[proposition]{Theorem} \newcommand{\be}{\mathbf{e}} \newcommand{\bg}{\boldsymbol{\gamma}} \newcommand{\dbg}{\bg'} \newcommand{\ddbg}{\bg''} \newcommand{\dddbg}{\bg'''} \newcommand{\der}[1]{#1{}'} \newcommand{\bT}{\mathbf{T}} \newcommand{\bN}{\mathbf{N}} \newcommand{\bB}{\mathbf{B}} \begin{document} The space curve traced out by the parameterization $$\bg(t)=\left[\begin{array}{c}a \cos(t)\\ a\sin (t)\\ bt\end{array}\right],\quad t\in \reals,\; a,b\in\reals$$ is called a \emph{circular helix} (plur. {\em helices}). Its Frenet frame is: \begin{align} \bT &= \frac{1}{\sqrt{a^2+b^2}} \begin{bmatrix} -a\sin t \\ \hphantom{-}a\cos t \\b\end{bmatrix}\,,\\ \bN &= \begin{bmatrix} -\cos t \\ -\sin t \\ 0 \end{bmatrix}\,,\\ \bB &= \frac{1}{\sqrt{a^2+b^2}} \begin{bmatrix} \hphantom{-}b\sin t \\ -b\cos t \\ a\end{bmatrix}\,. \end{align} Its curvature and torsion are the following constants: \begin{align} \tau = \frac{b}{a^2+b^2}\,. \end{align} A circular helix can be conceived of as a space curve with constant, non-zero curvature, and constant, non-zero torsion. Indeed, one can show that if a space curve satisfies the above constraints, then there exists a system of Cartesian coordinates in which the curve has a parameterization of the form shown above. \begin{figure} \begin{center} \includegraphics{helix2.eps} \caption{A plot of a circular helix with $a = b = 1$, and $\kappa = \tau = 1/2$.} \end{center} \end{figure} An important property of the circular helix is that for any point of it, the angle $\varphi$ between its tangent and the helix axis is constant. Indeed, if we consider the position vector of that arbitrary point, we have (where $\mathbf{k}$ is the unit vector parallel to helix axis) \begin{align} \frac{d\bg}{dt}\cdot\mathbf{k}=\begin{bmatrix} -a\sin t \\ \hphantom{-}a\cos t \\ b \end{bmatrix}\begin{bmatrix} 0\; 0\; 1\end{bmatrix}=b \equiv \bigg\lVert\frac{d\bg}{dt}\bigg\rVert\cos\varphi=\sqrt{a^2+b^2}\cos\varphi. \end{align} Therefore, \begin{align} \cos\varphi=\frac{b}{\sqrt{a^2+b^2}} \text{constant}, \end{align} as was to be shown. There is also another parameter, the so-called \emph{pitch of the helix} $P$ which is the separation between two consecutive turns. (It is mostly used in the manufacture of screws.) Thus, \begin{align} P=\gamma_3(t+2\pi)-\gamma_3(t)= b(t+2\pi)-bt=2\pi b\,, \end{align} and $P$ is also a constant. %%%%% %%%%% nd{document}	2018-02-24 18:15:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999996423721313, "perplexity": 2213.2960636484363}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891815918.89/warc/CC-MAIN-20180224172043-20180224192043-00487.warc.gz"}
https://www.physicsforums.com/threads/a-question-about-a-proof-of-the-irrationality-of-pi.993120/	# A question about a proof of the irrationality of pi Here is the proof I was reading: https://mathschallenge.net/full/irrationality_of_pi I have a question about this very last inequality at the end: $\therefore \displaystyle{\lim_{n \to \infty}\left(\dfrac{a^n}{n!}\right)} = 0$ and for sufficiently large $n, \dfrac{a^n\pi^{n+1}}{n!} \lt 1$ How did they get that "less than 1" bit? . ## Answers and Replies fresh_42 Mentor If ##\left(\dfrac{(a\pi)^n}{n!}\right)## tends to ##0##, what does that mean? For any ##\varepsilon > 0## there is ... Last edited: member 587159 fresh_42 Mentor You can use Stirling's formula and its lower boundary: $$\sqrt{2\pi n}\left(\dfrac{n^n}{e^n}\right) < n! \Longrightarrow\left(\dfrac{(a\pi)^n\cdot \pi}{n!}\right) < \left(\dfrac{a\pi e}{n}\right)^n\cdot \dfrac{\sqrt{\pi}}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{n}}$$ The first factor tends to ##0## and thus can be made smaller than ##\dfrac{\sqrt{2}}{\sqrt{\pi}}.## (If I made no mistake.) If ##\left(\dfrac{(a\pi)^n}{n!}\right)## tends to ##0##, what does that mean? For any ##\varepsilon > 0## there is ... I have been out of touch with that ##\varepsilon-\delta## stuff for about 15 years. Can you elaborate a little? . fresh_42 Mentor Formally: For any ##\varepsilon > 0## there is an ##N(\varepsilon)## - depending on that ##\varepsilon## - such that all ##\left\|\dfrac{(a\pi)^n}{n!}\right\| < \varepsilon## for all indices ##n > N(\varepsilon)##. This means that you can make ##\dfrac{(a\pi)^n}{n!}## arbitrary small, the greater ##n## is. So you can make it certainly smaller than ##\dfrac{1}{\pi}##, if only the ##n## are large enough. Formally: For any ##\varepsilon > 0## there is an ##N(\varepsilon)## - depending on that ##\varepsilon## - such that all ##\left\|\dfrac{(a\pi)^n}{n!}\right\| < \varepsilon## for all indices ##n > N(\varepsilon)##. A general question about limits (just to check if I understood it): if we wanted to find the limit of f(x) as ##x \rightarrow -\infty## (instead of ##+\infty##), should we find ##N(\varepsilon)## such that \|f(x) - the limit\| ##< \varepsilon## for all ## x < N(\varepsilon)## instead of ##x > N(\varepsilon)##. This means that you can make ##\dfrac{(a\pi)^n}{n!}## arbitrary small, the greater ##n## is. So you can make it certainly smaller than ##\dfrac{1}{\pi}##, if only the ##n## are large enough. Does that mean that we can always find an ##N(\varepsilon)## such that any ##n > N(\varepsilon)## will lead to ##\left\|\frac{(a\pi)^n}{n!}\right\| < \frac{1}{\pi}##? Last edited: fresh_42 Mentor A general question about limits (just to check if I understood it): if we wanted to find the limit of f(x) as ##x \rightarrow -\infty## (instead of ##+\infty##), should we find ##N(\varepsilon)## such that \|f(x) - the limit\| ##< \varepsilon## for all ## x < N(\varepsilon)## instead of ##x > N(\varepsilon)##. ##\lim_{x \to a}f(x)=c## means that for any sequence ##(x_n)\longrightarrow a## we have ##\lim_{x_n \to \infty}f(x_n)=c.## Here ##a,c\in \mathbb{R}\cup \{\pm \infty\}.## The ##n## are only a numbering of the sequence, i.e. it makes no sense to count them by negative numbers. If ##\|c\|<\infty## is finite, then we have: For any sequence ##(x_n)\longrightarrow a## which converges to ##a##, and any given, requested approximation ##\varepsilon>0##, there is an ##N(\varepsilon)## such that ##\|f(x_n)-c\|<\varepsilon## for all ##n>N(\varepsilon).## Whether ##a## is infinitely positive, finite, or infinitely negative is irrelevant here; only that our (arbitrary) sequence ##(x_n)## converges to ##a.## If ##c\in \{\pm \infty\}##, then we have again an arbitrary sequence ##(x_n)\longrightarrow a##, for which the following holds: For any finite boundary ##K## there is a sequence index ##N(K)##, such that ##f(x_n) >K## for all ##n>N(K)##, in case ##c=+\infty##, and ##f(x_n) < K## for all ##n>N(K)##, in case ##c=-\infty.## If you forget about the function here, then the same definitions apply to ##(x_n) \longrightarrow a.## The indexing ##n=1,2,3,\ldots## is always positive. Sure, you may count by negative numbers, but that would only be unnecessarily confusing. Continuity of the function ##f## and the ##\varepsilon-\delta## definition in comparison to the sequence definition of continuity is a different subject. Here we have only sequences. etotheipi FactChecker Science Advisor Gold Member How did they get that "less than 1" bit? It is simpler than you are making it. Each increase of ##n## puts in another multiplier of ##a*\pi/n##. Since ##n## can get huge, the product can get very small.	2021-07-28 17:00:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9355419278144836, "perplexity": 3123.713950684113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00540.warc.gz"}
https://mathleaks.com/book-series/big-ideas-learning/big-ideas-math-algebra-1-2015/polynomial-equations-and-factoring/multiplying-polynomials	#### Multiplying Polynomials Find the solutions in the app ##### Sections Exercise name Free? ###### Monitoring Progress Exercise name Free? Monitoring Progress 1 Monitoring Progress 2 Monitoring Progress 3 Monitoring Progress 4 Monitoring Progress 5 Monitoring Progress 6 Monitoring Progress 7 Monitoring Progress 8 Monitoring Progress 9 Monitoring Progress 10 Monitoring Progress 11 ###### Exercises Exercise name Free? Exercises 1 There are many methods we can use to find the product of two binomials. Here we will review two of them — the FOIL Method and using algebra tiles.Using the FOIL Method The word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us the order to follow when multiplying binomials.window.JXQ = window.JXQ \|\| []; window.JXQtable = window.JXQtable \|\| {}; if(!window.JXQtable["Solution19953_1_604487304_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-2.38,3,4.62,0],{"grid":{"visible":false},"desktopSize":"large", "style":"usa"}); var t =600;b.txt(0-2,1,"(",{fontSize:1.3}); b.txt(0.25-2,1,"a",{fontSize:1.3}); b.txt(0.5-2,1,"+",{fontSize:1.3}); b.txt(0.75-2,1,"b",{fontSize:1.3}); b.txt(1.0-2,1,")",{fontSize:1.3}); b.txt(1.25-2,1,"(",{fontSize:1.3}); b.txt(1.5-2,1,"c",{fontSize:1.3}); b.txt(1.75-2,1,"+",{fontSize:1.3}); b.txt(2.0-2,1,"d",{fontSize:1.3}); b.txt(2.25-2,1,")",{fontSize:1.3});var t1 = b.txt(1.12,2.5," \\textcolor{red}{\\text{F}}\\text{irst terms}",{"fontsize":1.3,opacity:0}); var f1 = b.bendyArrow([0.27-2,1.20],[1.48-2,1.2],40,140,{strokeColor:'red',opacity:0});var t2= b.txt(1.12,2.5," \\textcolor{#2999f0}{\\text{O}}\\text{uter terms}",{"fontsize":1.3,opacity:0}); var f2= b.bendyArrow([0.27-2,1.20],[1.98-2,1.2],50,130,{strokeColor:'#2999f0',opacity:0});var t3= b.txt(1.12,2.5," \\textcolor{#02b03e}{\\text{I}}\\text{nner terms}",{"fontsize":1.3,opacity:0}); var f3= b.bendyArrow([0.77-2,0.8],[1.48-2,0.8],-40,-140,{strokeColor:'#02b03e',opacity:0});var t4 = b.txt(1.12,2.5," \\textcolor{#a72cd4}{\\text{L}}\\text{ast terms}",{"fontsize":1.3,opacity:0}); var f4 = b.bendyArrow([0.77-2,0.8],[1.98-2,0.8],-50,-150,{strokeColor:'#a72cd4',opacity:0});var p1 = b.txt(0.85,1,"= \\ ab",{fontSize:1.3,opacity:0}); var l1 = b.txt(1.1,1.65," \\ \\textcolor{red}{\\text{F}}",{fontSize:1.3,opacity:0});var p2 =b.txt(1.75,1,"+ \\ ad",{fontSize:1.3,opacity:0}); var l2 =b.txt(1.9,1.65," \\ \\textcolor{#2999f0}{\\text{O}}",{fontSize:1.3,opacity:0});var p3 = b.txt(2.65,1,"+ \\ bc",{fontSize:1.3,opacity:0}); var l3 = b.txt(2.8,1.65," \\ \\textcolor{#02b03e}{\\text{I}}",{fontSize:1.3,opacity:0});var p4 =b.txt(3.55,1,"+ \\ bd",{fontSize:1.3,opacity:0}); var l4 =b.txt(3.7,1.65," \\ \\textcolor{#a72cd4}{\\text{L}}",{fontSize:1.3,opacity:0});mlg.af("A",function() { $('#btn1').css({'background-color':'#24c5ed'});b.hide(p1,0); b.hide(l1,0); b.hide(f1,0); b.hide(p2,0); b.hide(l2,0); b.hide(f2,0); b.hide(p3,0); b.hide(l3,0); b.hide(f3,0); b.hide(p4,0); b.hide(l4,0); b.hide(f4,0);b.show(f1,t); setTimeout( function() { b.show(t1,t); }, 2t); setTimeout( function() { b.hide(t1,t); }, 4t); setTimeout( function() { b.show(l1,t); }, 4t); setTimeout( function() { b.show(p1,t); }, 4t);setTimeout( function() { b.show(f2,t); }, 6t); setTimeout( function() { b.show(t2,t); }, 8t); setTimeout( function() { b.hide(t2,t); }, 10t); setTimeout( function() { b.show(l2,t); }, 10t); setTimeout( function() { b.show(p2,t); }, 10t);setTimeout( function() { b.show(f3,t); }, 12t); setTimeout( function() { b.show(t3,t); }, 14t); setTimeout( function() { b.hide(t3,t); }, 16t); setTimeout( function() { b.show(l3,t); }, 16t); setTimeout( function() { b.show(p3,t); }, 16t);setTimeout( function() { b.show(f4,t); }, 18t); setTimeout( function() { b.show(t4,t); }, 20t); setTimeout( function() { b.hide(t4,t); }, 22t); setTimeout( function() { b.show(l4,t); }, 22t); setTimeout( function() { b.show(p4,t); }, 22t);setTimeout( function() {$('#btn1').css({'background-color':'#d8dce3'}); }, 24t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19953_1_604487304_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19953_1_604487304_l", "Solution19953_1_604487304_p", 1, code); }); } ); } window.JXQtable["Solution19953_1_604487304_l"] = true;Calculate Productwindow.RLQ = window.RLQ \|\| []; window.RLQ.push( function () { (function ($) {$('#btn1').on('touchstart mousedown', function () { try { mlg.cf("A"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); For example, let's consider that we want to find the product of the binomials (x+2) and (x−1). We just need to multiply following the order indicated by the FOIL acronym and simplify by combining like terms.Using Algebra Tiles We can represent algebraic expressions using algebra tiles. Let's consider the same product one more time, (x+2)(x−1).The tiles in the rectangular array shown above represent each of the binomials. To find their product we need to complete the diagram first. Recall that the product of algebra tiles with the same sign gives a positive result, while those with different signs give a negative result.Now we can write the algebraic expression being represented by the algebra tiles. (x+2)(x−1)=x2−x+2x−2 Finally, we just need to simplify. (x+2)(x−1)(x+2)(x−1)=x2−x+2x−2=x2+x−2 Exercises 2 Recall that the word FOIL is an acronym for the words First, Outer, Inner, and Last. 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To do so, we will apply the Distributive Property. (x+1)(x+3)Distribute (x+3)x(x+3)+1(x+3)Distribute xx2+3x+1(x+3)Distribute 1x2+3x+x+3Add termsx2+4x+3 Exercises 4 We want to find a product. To do so, we will apply the Distributive Property. (y+6)(y+4)Distribute (y+4)y(y+4)+6(y+4)Distribute yy2+4y+6(y+4)Distribute 6y2+4y+6y+24Add termsy2+10y+24 Exercises 5 We want to find a product. To do so, we will apply the Distributive Property. (z−5)(z+3)Distribute (z+3)z(z+3)−5(z+3)Distribute zz2+3z−5(z+3)Distribute -5z2+3z−5z−15Subtract termz2−2z−15 Exercises 6 We want to find a product. To do so, we will apply the Distributive Property. (a+8)(a−3)Distribute (a−3)a(a−3)+8(a−3)Distribute aa2−3a+8(a−3)Distribute 8a2−3a+8a−24Add termsa2+5a−24 Exercises 7 We want to find a product. To do so, we will apply the Distributive Property. (g−7)(g−2)Distribute (g−2)g(g−2)−7(g−2)Distribute gg2−2g−7(g−2)Distribute -7g2−2g−7g+14Subtract termg2−9g+14 Exercises 8 We want to find a product. To do so, we will apply the Distributive Property. (n−6)(n−4)Distribute (n−4)n(n−4)−6(n−4)Distribute nn2−4n−6(n−4)Distribute -6n2−4n−6n+24Subtract termn2−10n+24 Exercises 9 We want to find a product. To do so, we will apply the Distributive Property. (3m+1)(m+9)Distribute (m+9)3m(m+9)+1(m+9)Distribute 3m3m2+27m+1(m+9)Distribute 13m2+27m+m+9Add terms3m2+28m+9 Exercises 10 Exercises 11 We want to simplify the given product using a table. (x+3)(x+2) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.x3 xx⋅xx⋅3 22⋅x2⋅3 Now we can calculate the products.x3 xx23x 22x6 Finally, we obtain the answer by adding the products and combining like terms. x2+3x+2x+6⇔x2+5x+6 Exercises 12 We want to simplify the given product using a table. (y+10)(y−5) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.y10 yy⋅yy⋅10 -5-5⋅y-5⋅10 Now we can calculate the products.y10 yy210y -5-5y-50 Finally, we obtain the answer by adding the products and combining like terms. y2+10y−5y−50⇔y2+5y−50 Exercises 13 We want to simplify the given product using a table. (h−8)(h−9) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.h-8 hh⋅hh⋅(-8) -9-9⋅h-9⋅(-8) Now we can calculate the products.h-8 hh2-8h -9-9h72 Finally, we obtain the answer by adding the products and combining like terms. h2−8h−9h+72⇔h2−17h+72 Exercises 14 We want to simplify the given product using a table. (c−6)(c−5) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.c-6 cc⋅cc⋅(-6) -5-5⋅c-5⋅(-6) Now we can calculate the products.c-6 cc2-6c -5-5c30 Finally, we obtain the answer by adding the products and combining like terms. c2−6c−5c+30⇔c2−11c+30 Exercises 15 We want to simplify the given product using a table. (3k−1)(4k+9) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.3k-1 4k4k⋅3k4k⋅(-1) 99⋅3k9⋅(-1) Now we can calculate the products.3k-1 4k12k2-4k 927k-9 Finally, we obtain the answer by adding the products and combining like terms. 12k2−4k+27k−9⇔12k2+23k−9 Exercises 16 We want to simplify the given product using a table. (5g+3)(g+8) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.5g3 gg⋅5gg⋅3 88⋅5g8⋅3 Now we can calculate the products.5g3 g5g23g 840g24 Finally, we obtain the answer by adding the products and combining like terms. 5g2+3g+40g+24⇔5g2+43g+24 Exercises 17 We want to simplify the given product using a table. (-3+2j)(4j−7) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.-32j 4j4j⋅(-3)4j⋅2j -7-7⋅(-3)-7⋅2j Now we can calculate the products.-32j 4j-12j8j2 -721-14jFinally, we obtain the answer by adding the products and combining like terms. -12j+8j2+21−14jCommutative Property of Addition8j2−12j−14j+21Subtract term8j2−26j+21 Exercises 18 We want to simplify the given product using a table. (5d−12)(-7+3d) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.5d-12 -7-7⋅5d-7⋅(-12) 3d3d⋅5d3d⋅(-12) Now we can calculate the products.5d-12 -7-35d84 3d15d2-36dFinally, we obtain the answer by adding the products and combining like terms. -35d+84+15d2−36dCommutative Property of Addition15d2−35d−36d+84Subtract term15d2−71d+84 Exercises 19 In this exercise, we are asked to find the error in multiplication of two binomials. Let's start by multiplying the binomials using the table.t-2 tt2-2t 55t-10 From this we can see, if we add 5t+-2t we get t2+3t−10, which is different than the answer in the exercise. Let's have a closer look.Here we can see the error is in the first step where their is no t2 term. It looks like the (t+5) was not distributed properly. Let's do that correctly. Exercises 20 For this exercise, we need to find the error in the multiplication of two binomials. They used the column method. Let's look for the error.From this, we can see the very common error of not bringing down the negative with the 5, resulting in the error in the last line. Let's correct the error. Exercises 21 We want to simplify the given product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (b)(b)+(b)(7)+(3)(b)+(3)(7)Multiplyb2+7b+3b+21Add termsb2+10b+21 Exercises 22 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (w)(w)+(w)(6)+(9)(w)+(9)(6)Multiplyw2+6w+9w+54Add termsw2+15w+54 Exercises 23 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (k)(k)+(k)(-1)+(5)(k)+(5)(-1)Multiplyk2+(k)(-1)+5k+(5)(-1)a(-b)=-a⋅bk2−k+5k−5Add termsk2+4k−5 Exercises 24 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (x)(x)+(x)(8)+(-4)(x)+(-4)(8)Multiplyx2+8x+(-4)(x)+(-4)(8)(-a)b=-abx2+8x−4x−24Subtract termx2+4x−24 Exercises 25 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (q)(q)+(q)(41)+(-43)(q)+(-43)(41)Multiplyq2+41q+(-43)(q)+(-43)(41)(-a)b=-abq2+41q−43q−163Subtract fractionsq2−42q−163ba=b/2a/2q2−21q−163 Exercises 26 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (z)(z)+(z)(-32)+(-35)(z)+(-35)(-32)Multiplyz2+(z)(-32)+(-35)(z)+(-35)(-32)a(-b)=-a⋅bz2−32z+(-35)(z)+(-35)(-32)(-a)b=-abz2−32z−35z+(-35)(-32)-a(-b)=a⋅bz2−32z−35z+910Subtract fractionsz2−37z+910 Exercises 27 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (9)(2)+(9)(-3r)+(-r)(2)+(-r)(-3r)Multiply18+(9)(-3r)+(-r)(2)+(-r)(-3r)a(-b)=-a⋅b18−27r+(-r)(2)+(-r)(-3r)(-a)b=-ab18−27r−2r+(-r)(-3r)-a(-b)=a⋅b18−27r−2r+3r2Subtract term18−29r+3r2Commutative Property of Addition3r2−29r+18 Exercises 28 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (8)(2x)+(8)(6)+(-4x)(2x)+(-4x)(6)Multiply16x+48+(-4x)(2x)+(-4x)(6)(-a)b=-ab16x+48−8x2−24xCommutative Property of Addition-8x2+16x−24x+48Subtract term-8x2−8x+48 Exercises 29 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (w)(w2)+(w)(3w)+(5)(w2)+(5)(3w)Multiplyw3+3w2+5w2+15wAdd termsw3+8w2+15w Exercises 30 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (v)(v2)+(v)(8v)+(-3)(v2)+(-3)(8v)Multiplyv3+8v2+(-3)(v2)+(-3)(8v)(-a)b=-abv3+8v2−3v2−24vSubtract termv3+5v2−24v Exercises 31 In this exercise, we combine our skills of multiplying binomials with finding the area of a rectangle. Notice we are given the length and width of the rectangle.Let's substitute the length and width from the rectangle's picture into the equation for area. A=ℓ⋅wℓ=2x−9, w=x+5A=(2x+9)(x+5) Multiply parentheses Distribute x+5A=2x(x+5)+9(x+5)Distribute 2xA=2x2+10x+9(x+5)Distribute 9A=2x2+10x+9x+45Add terms A=2x2+19x+45 Exercises 32 In this exercise, we combine our skills of multiplying binomials with finding the area of a triangle. Notice that we are given the base and height of the rectangle.Let's substitute the base and height from the triangle's picture into the equation for triangle area. A=21⋅b⋅hb=p+1, h=2p−6A=21(p+1)(2p−6) Multiply parentheses Distribute 2p−cA=21[p(2p−6)+1(2p−6)]Distribute pA=21[2p2−6p+1(2p−6)]Distribute 1A=21[2p2−6p+2p−6]Add terms A=21[2p2−4p−6]Distribute 21A=p2−2p−3 Exercises 33 For this exercise, we need to find the area of the shaded region, which is part of a rectangle. In this case, if we look at the picture we can see that the white triangle in the center is half the area of the the rectangle.We can then say that the shaded region is the other half of the rectangle. Let's first find the area of the rectangle. A=ℓ⋅wℓ=x+6, w=x+5A=(x+6)(x+5) Multiply parentheses Distribute x+5A=x(x+5)+6(x+5)Distribute xA=x2+5x+6(x+5)Distribute 6A=x2+5x+6x+30Add terms A=x2+11x+30 Now, let's take half of the rectangle area to find the polynomial that represents the shaded region. 2x2+11x+30=0.5x2+5.5x+15 Exercises 34 In this exercise, we combine our skills of multiplying binomials with finding the area of a shaded region. This shaded region is a square with a rectangle cut out. We need to find the area of both, then subtract.Let's use the lengths of the sides in the image to find the area of the white rectangle. AAA===ℓ⇓(x−7)⇓(5x−35)⋅⋅w⇓5 When we distribute, we get the area of the white rectangle to be 5x−35. Now, let's find the area of the large square. A=ℓ⋅wℓ=x+1, w=x+1A=(x+1)(x+1) Multiply parentheses Distribute x+1A=x(x+1)+1(x+1)Distribute xA=x2+x+1(x+1)Distribute 1A=x2+x+x+1Add terms A=x2+2x+1 Now, we can subtract the area of the white rectangle from the area of the large square and we will have only the shaded area remaining. A(Shaded Region)↓A(Shaded Region)=↓=A(Large Square)↓(x2+2x+1)−↓−A(White Rectangle)↓(5x−35) Now, let's simplify that expression by combining like terms A=x2+2x+1−(5x−35) Simplify right-hand side Distribute -1A=x2+2x+1−5x+35Commutative Property of AdditionA=x2+2x−5x+1+35Subtract termsA=x2−3x+1+35Add terms A=x2−3x+36 Exercises 35 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. x2+3xx++24 Now, let's multiply x2+3x+2 by 4. x24x2++3xx12x+++248 Now, we will multiply x2+3x+2 by x. x3+x24x23x2+++3xx12x2x+++248 Finally, let's add the like terms in 4x2+12x+8 and x3+3x2+2x. x3x3++x24x23x27x2++++3xx12x2x14x++++2488 The product is x3+7x2+14x+8. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (x+4)(x2+3x+2)Distribute (x2+3x+2)x(x2+3x+2)+4(x2+3x+2)Distribute xx3+3x2+2x+4(x2+3x+2)Distribute 4x3+3x2+2x+4x2+12x+8Commutative Property of Additionx3+3x2+4x2+2x+12x+8Add termsx3+7x2+14x+8 Both methods are correct and give the same result. Exercises 36 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. f2+4ff++81 Now, let's multiply f2+4f+8 by 1. f2f2++4ff4f+++818 Now, we will multiply f2+4f+8 by f. f3+f2f24f2+++4ff4f8f+++818 Finally, let's add the like terms in f2+4f+8 and f3+4f2+8f. f3f3++f2f24f25f2++++4ff4f8f12f++++8188 The product is f3+5f2+12f+8. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (f+1)(f2+4f+8)Distribute (f2+4f+8)f(f2+4f+8)+1(f2+4f+8)Distribute ff3+4f2+8f+1(f2+4f+8)Distribute 1f3+4f2+8f+f2+4f+8Commutative Property of Additionf3+4f2+f2+8f+4f+8Add termsf3+5f2+12f+8 Both methods are correct and give the same result. Exercises 37 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. y2+8yy−+23 Now, let's multiply y2+8y−2 by 3. y23y2++8yy24y−+−236 Now, we will multiply y2+8y−2 by y. y3+y23y28y2++−8yy24y2y−+−236 Finally, let's add the like terms in 3y2+24y−6 and y3+8y2−2y. y3y3++y23y28y211y2++−+8yy24y2y22y−+−−2366 The product is y3+11y2+22y−6. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (y+3)(y2+8y−2)Distribute (y2+8y−2)y(y2+8y−2)+3(y2+8y−2)Distribute yy3+8y2−2y+3(y2+8y−2)Distribute 3y3+8y2−2y+3y2+24y−6Commutative Property of Additiony3+8y2+3y2−2y+24y−6Add termsy3+11y2+22y−6 Both methods are correct and give the same result. Exercises 38 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. t2−5tt+−12 Now, let's multiply t2−5t+1 by -2. t2-2t2−+5tt10t+−−122 Now, we will multiply t2−5t+1 by t. t3−t2-2t25t2−++5tt10tt+−−122 Finally, let's add the like terms in -2t2+10t−2 and t3−5t2+t. t3t3−−t2-2t25t27t2−+++5tt10tt11t+−−−1222 The product is t3−7t2+11t−2. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (t−2)(t2−5t+1)Distribute (t2−5t+1)t(t2−5t+1)−2(t2−5t+1)Distribute tt3−5t2+t−2(t2−5t+1)Distribute -2t3−5t2+t−2t2+10t−2Commutative Property of Additiont3−5t2−2t2+t+10t−2Add and subtract termst3−7t2+11t−2 Both methods are correct and give the same result. Exercises 39 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 5b2+5b-b−+44 Now, let's multiply 5b2+5b−4 by 4. 5b220b2++5b-b20b−+−4416 Now, we will multiply 5b2+5b−4 by -b. -5b3−5b220b25b2+++5b-b20b4b−+−4416 Finally, let's add 20b2+5b−16 and -5b3−5b2+4b like terms. -5b3-5b3−+5b220b25b215b2++++5b-b20b4b24b−+−−441616 The product is -5b3+15b2+24b−16. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (4−b)(5b2+5b−4)Distribute (5b2+5b−4)4(5b2+5b−4)−b(5b2+5b−4)Distribute 420b2+20b−16−b(5b2+5b−4)Distribute -b20b2+20b−16−5b3−5b2+4bCommutative Property of Addition-5b3+20b2−5b2+20b+4b−16Add and subtract terms-5b3+15b2+24b−16 Both method are correct and gives the same result. Exercises 40 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 2d2−dd++76 Now, let's multiply 2d2−d+7 by 6. 2d212d2−−dd6d+++7642 Now, we will multiply 2d2−d+7 by d. 2d3−2d212d2d2−−+dd6d7d+++7642 Finally, let's add the like terms in 12d2−6d+42 and 2d3−d2+7d. 2d32d3−+2d212d2d211d2−−++dd6d7dd++++764242 The product is 2d3+11d2+d+42. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (d+6)(2d2−d+7)Distribute (2d2−d+7)d(2d2−d+7)+6(2d2−d+7)Distribute d2d3−d2+7d+6(2d2−d+7)Distribute 62d3−d2+7d+12d2−6d+42Commutative Property of Addition2d3−d2+12d2+7d−6d+42Add and subtract terms2d3+11d2+d+42 Both methods are correct and give the same result. Exercises 41 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 3e2−5e6e++71 Now, let's multiply 3e2−5e+7 by 1. 3e23e2−−5e6e5e+++717 Now, we will multiply 3e2−5e+7 by 6e. 18e3−3e23e230e2−−+5e6e5e42e+++717 Finally, let's add the like terms in 3e2−5e+7 and 18e3−30e2+42e. 18e318e3−−3e23e230e227e2−−++5e6e5e42e37e++++7177 The product is 18e3−27e2+37e+7. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (6e+1)(3e2−5e+7)Distribute (3e2−5e+7)6e(3e2−5e+7)+1(3e2−5e+7)Distribute 6e18e3−30e2+42e+1(3e2−5e+7)Distribute 118e3−30e2+42e+3e2−5e+7Commutative Property of Addition18e3−30e2+3e2+42e−5e+7Add and subtract terms18e3−27e2+37e+7 Both methods are correct and give the same result. Exercises 42 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 6v2+2v-5v−+94 Now, let's multiply 6v2+2v−9 by 4. 6v224v2++2v-5v8v−+−9436 Now, we will multiply 6v2+2v−9 by -5v. -30v3−6v224v210v2+++2v-5v8v45v−+−9436 Finally, let's add the like terms in 24v2+8v−36 and -30v3−10v2+45v. -30v3-30v3−+6v224v210v214v2++++2v-5v8v45v53v−+−−943636 The product is -30v3+14v2+53v−36. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (6v2+2v−9)(4−5v)Distribute (4−5v)6v2(4−5v)+2v(4−5v)−9(4−5v)Distribute 6v224v2−30v3+2v(4−5v)−9(4−5v)Distribute 2v24v2−30v3+8v−10v2−9(4−5v)Distribute -924v2−30v3+8v−10v2−36+45vCommutative Property of Addition-30v3+24v2−10v2+8v+45v−36Add and subtract terms-30v3+14v2+53v−36 Both methods are correct and give the same result. Exercises 43 Exercises 44 Exercises 45 Let's start by recalling the FOIL Method for multiplying binomials. The word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us the order to follow when multiplying binomials.Let's see an example. Consider that we want to find the product of the binomials (x2+2) and (3x3+5). Since the degree of a polynomial is determined by the highest degree of its monomials, we know we are multiplying a degree 2 and a degree 3 binomial.As we can see, the product of a degree 2 polynomial and a degree 3 binomial results in a degree 5 polynomial. This happened because when the terms determining the degree of each binomial were multiplied, the result was a monomial of a degree equal to the sum of the degree of each individual monomial factor. (x2+2)(3x3+5)=3x5+5x2+6x3+10x2⋅3x3=3x2+3=3x5 The same will happen with any pair of binomials in general. Therefore, we can summarize the ideas we have discussed as shown below.The degree of the product of two binomials is equal to the sum of the degrees of each binomial factor. Exercises 46 We are asked for a pair of polynomials such that their product is a trinomial of degree 3. For simplicity, we will look for two binomials with the required characteristics. Recall that we can find the product of two binomials using the FOIL Method, which is an acronym for the words First, Outer, Inner, and Last.Furthermore, recall that the degree of the product of two binomials is equal to the sum of the degrees of each binomial. We can try with a degree 1 binomial and a degree 2 binomial. For example, (x+2) and (x2+5). Lets multiply them using the FOIL Method.As we can see, we obtained a degree 3 polynomial as result. However, since none of the products are like terms, we cannot simplify it to obtain a trinomial. Nevertheless, if we use (x2+5x) instead of (x2+5), two of the product terms are quadratic and they can be combined, obtaining a trinomial as required.The same would happen with any product of the form (ax+b)(cx2+dx). Therefore, there are infinitely many binomials satisfying the exercise's requirements, and this is just an example. Exercises 47 Recall that the word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us of the order to follow when multiplying two binomials.window.JXQ = window.JXQ \|\| []; window.JXQtable = window.JXQtable \|\| {}; if(!window.JXQtable["Solution19999_1_698584143_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-2.38,3,4.62,0],{"grid":{"visible":false},"desktopSize":"large", "style":"usa"}); var t =600;b.txt(0-2,1,"(",{fontSize:1.15}); b.txt(0.25-2,1,"a",{fontSize:1.15}); b.txt(0.5-2,1,"+",{fontSize:1.15}); b.txt(0.75-2,1,"b",{fontSize:1.15}); b.txt(1.0-2,1,")",{fontSize:1.15}); b.txt(1.25-2,1,"(",{fontSize:1.15}); b.txt(1.5-2,1,"c",{fontSize:1.15}); b.txt(1.75-2,1,"+",{fontSize:1.15}); b.txt(2.0-2,1,"d",{fontSize:1.15}); b.txt(2.25-2,1,")",{fontSize:1.15});var t1 = b.txt(1.12,2.5," \\textcolor{red}{\\text{F}}\\text{irst terms}",{"fontsize":1.15,opacity:0}); var f1 = b.bendyArrow([0.27-2,1.20],[1.48-2,1.2],40,140,{strokeColor:'red',opacity:0});var t2= b.txt(1.12,2.5," \\textcolor{#2999f0}{\\text{O}}\\text{uter terms}",{"fontsize":1.15,opacity:0}); var f2= b.bendyArrow([0.27-2,1.20],[1.98-2,1.2],50,130,{strokeColor:'#2999f0',opacity:0});var t3= b.txt(1.12,2.5," \\textcolor{#02b03e}{\\text{I}}\\text{nner terms}",{"fontsize":1.15,opacity:0}); var f3= b.bendyArrow([0.77-2,0.8],[1.48-2,0.8],-40,-140,{strokeColor:'#02b03e',opacity:0});var t4 = b.txt(1.12,2.5," \\textcolor{#a72cd4}{\\text{L}}\\text{ast terms}",{"fontsize":1.15,opacity:0}); var f4 = b.bendyArrow([0.77-2,0.8],[1.98-2,0.8],-50,-150,{strokeColor:'#a72cd4',opacity:0});var p1 = b.txt(0.85,1,"= \\ ab",{fontSize:1.15,opacity:0}); var l1 = b.txt(1.1,1.65," \\ \\textcolor{red}{\\text{F}}",{fontSize:1.15,opacity:0});var p2 =b.txt(1.75,1,"+ \\ ad",{fontSize:1.15,opacity:0}); var l2 =b.txt(1.9,1.65," \\ \\textcolor{#2999f0}{\\text{O}}",{fontSize:1.15,opacity:0});var p3 = b.txt(2.65,1,"+ \\ bc",{fontSize:1.15,opacity:0}); var l3 = b.txt(2.8,1.65," \\ \\textcolor{#02b03e}{\\text{I}}",{fontSize:1.15,opacity:0});var p4 =b.txt(3.55,1,"+ \\ bd",{fontSize:1.15,opacity:0}); var l4 =b.txt(3.7,1.65," \\ \\textcolor{#a72cd4}{\\text{L}}",{fontSize:1.15,opacity:0});mlg.af("A",function() { $('#btn1').css({'background-color':'#24c5ed'});b.hide(p1,0); b.hide(l1,0); b.hide(f1,0); b.hide(p2,0); b.hide(l2,0); b.hide(f2,0); b.hide(p3,0); b.hide(l3,0); b.hide(f3,0); b.hide(p4,0); b.hide(l4,0); b.hide(f4,0);b.show(f1,t); setTimeout( function() { b.show(t1,t); }, 2t); setTimeout( function() { b.hide(t1,t); }, 4t); setTimeout( function() { b.show(l1,t); }, 4t); setTimeout( function() { b.show(p1,t); }, 4t);setTimeout( function() { b.show(f2,t); }, 6t); setTimeout( function() { b.show(t2,t); }, 8t); setTimeout( function() { b.hide(t2,t); }, 10t); setTimeout( function() { b.show(l2,t); }, 10t); setTimeout( function() { b.show(p2,t); }, 10t);setTimeout( function() { b.show(f3,t); }, 12t); setTimeout( function() { b.show(t3,t); }, 14t); setTimeout( function() { b.hide(t3,t); }, 16t); setTimeout( function() { b.show(l3,t); }, 16t); setTimeout( function() { b.show(p3,t); }, 16t);setTimeout( function() { b.show(f4,t); }, 18t); setTimeout( function() { b.show(t4,t); }, 20t); setTimeout( function() { b.hide(t4,t); }, 22t); setTimeout( function() { b.show(l4,t); }, 22t); setTimeout( function() { b.show(p4,t); }, 22t);setTimeout( function() {$('#btn1').css({'background-color':'#d8dce3'}); }, 24t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19999_1_698584143_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19999_1_698584143_l", "Solution19999_1_698584143_p", 1, code); }); } ); } window.JXQtable["Solution19999_1_698584143_l"] = true;Calculate Productwindow.RLQ = window.RLQ \|\| []; window.RLQ.push( function () { (function ($) {$('#btn1').on('touchstart mousedown', function () { try { mlg.cf("A"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); Therefore, the FOIL Method can only be used for multiplying two binomials. When two trinomials are multiplied there are 9 products as result. Below we can see the result of multiplying two general trinomials, (a+b+c), and (d+e+f).Hence, if we follow the FOIL Method to multiply two trinomials we would be missing 5 terms. Exercises 48 Exercises 49 In this exercise, we need to show that the two methods for multiplying polynomials results in the same solution. Let's start by showing each method separately, then comparing the parts.Distributive Property Method Let's show how to multiply two binomials using the distributive property. (x+2)(x−5)Distribute x−5x(x−5)+2(x−5)Distribute xx2−5x+2(x−5)Distribute 2x2−5x+2x−10Add termsx2−3x−10FOIL Method Now, Let's show what happens when we use FOIL Method. Let's look at each piece.Fx2 O2x I-5x L-10 This gives us the exact same pieces as the Distributive Property. Furthermore, if we add the linear terms above, we get the same answer as the previous method, x2−3x−10 even though the factors were swapped. (x+2)(x−5)=(x−5)(x+2) The factors can be swapped because multiplication is commutative. Exercises 50 For this exercise, we need to use the formula for the volume of a rectangular prism to get the polynomial for the volume of the container. Let's substitute ℓ=4x−3 and w=x+1 to find the area of the rectangular base. Then we can multiply the answer by h=x+2 to find the volume. A=ℓ⋅wℓ=4x−3, w=x+1A=(4x−3)(x+1) Multiply parentheses Distribute x+1A=4x(x+1)−3(x+1)Distribute 4xA=4x2+4x−3(x+1)Distribute -3A=4x2+4x−3x−3Subtract terms A=4x2+x−3 Now we have the area of the base we can multiply it by our height to find our volume. Let's use the table method to multiply (x+2)(4x2+x−3).×4x2x-3 x4x3x2-3x 28x22x-6 Now, we can simplify by adding like terms to get the polynomial for the area of the shaded region, 4x3+9x2−x−6. Exercises 51 Exercises 52 Exercises 53 Exercises 54 Exercises 55 Exercises 56 Exercises 57 Exercises 58	2020-09-27 17:25:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9442396759986877, "perplexity": 2273.0111855228574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00607.warc.gz"}
https://stats.stackexchange.com/questions/85721/conditional-expectations-and-variances	# Conditional Expectations and Variances $\textbf{Background:}$ When $\mathbb EX$ is hard to calculate, it is a common trick to use the following formula: $$\mathbb EX=\mathbb E[\mathbb E(X\|Y)].$$ And similarly, $\mathbb VX$ can be calculated using the following formula: $$\mathbb VX=\mathbb E[\mathbb V(X\|Y)]+\mathbb V[\mathbb E(X\|Y)].$$ $\textbf{Question:}$ Recently, I saw similar formula for conditional expectation and variance as follows. $$\mathbb E(Y\|X)=\mathbb E[\mathbb E(Y\|X, Z)\|X],$$ and $$\mathbb V(Y\|X)=\mathbb E[\mathbb V(Y\|X,Z)\|X]+\mathbb V[\mathbb E(Y\|X, Z)\|X].$$ I do not know how to prove them. Could anyone provide some hint or reference, please? Thank you! • can you tell me in what book or paper you saw this formula? – omidi Feb 7 '14 at 12:04 • In fact, it's used in one exercise in a lecture I attended. – LaTeXFan Feb 7 '14 at 21:07	2019-06-16 07:27:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8729279637336731, "perplexity": 126.02366226547645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997801.20/warc/CC-MAIN-20190616062650-20190616084650-00133.warc.gz"}
https://www.storyofmathematics.com/the-amount-of-time-ricardo-spends-brushing/	The amount of time Ricardo spends brushing his teeth follows a normal distribution with unknown mean and standard deviation. Ricardo spends less than one minute brushing his teeth about 40% of the time. He spends more than two minutes brushing his teeth 2% of the time. Use this information to determine the mean and standard deviation of this distribution. The question aims to find the mean $\mu$ and standard deviation $\sigma$ of a standard normal distribution. In arithmetic, a standard score is the number of standard deviations where the maturity of the observed point is above or below the average value of what is observed or measured. Raw scores above the mean generally have positive points, while those with less than the mean have negative scores. Standard scores are often called z-scores; both terms can be used interchangeably. Other equivalent words include z values, common points, and variables. Common distribution problems can be solved using the z-score formula. In a set with mean $\mu$ and standard deviation $\sigma$, the z-value of the scale X is given: $Z=\dfrac{X-\mu}{\sigma}$ • $Z$-score measures how many standard deviations are derived from the description. • After finding the $z-score$, we look at the z-score table and find the $p-value$ associated with that $z-score$, which is the $X$ percentage point. Ricardo spends less than one minute brushing his teeth about $40\%$ of the time. The time is more than two minutes about $2\%$ of the time, and thus less than two minutes about $98\%$of the time. The $z-value$ is calculated by: This means that $Z$ When $X=1$ has a $p-value$ of $0.4$, thus when $X=1$, $Z=-0.253$ then: $Z=\dfrac{X-\mu}{\sigma}$ $-0.253=\dfrac{1-\mu}{\sigma}$ $1-\mu=-0.253\sigma$ $\mu=1+0.253\sigma$ He spends more than two minutes brushing his teeth $2\%$ of the time. This means that $Z$ when $X = 2$ has a $p-value$ of $1 – 0.02 = 0.98$, thus, when $X = 2$,$Z = 2.054$, then: $Z=\dfrac{X-\mu}{\sigma}$ $2.054=\dfrac{2-\mu}{\sigma}$ $2-\mu=2.054\sigma$ $\mu=2-2.054\sigma$ Since, $\mu=1+0.253\sigma$ $(1+0.253\sigma)=(2-2.054\sigma)$ $2.307\sigma=1$ $\sigma=0.43$ The value of the $\sigma$ is $0.43$. The value of the $\mu$ is calculated as: $\mu=1+0.253(0.43)$ $\mu=1.11$ The value of the $\mu$ is $1.11$. Numerical Results The value of the mean $\mu$ is calculated as: $\mu=1.11$ The value of the standard deviation $\sigma$ is calculated as: $\sigma=0.43$ Example The time Bella spends brushing her teeth follows the normal distribution with an unknown definition and standard deviation. Bella spends less than one minute brushing her teeth about $30\%$ of the time. She spends more than two minutes brushing her teeth $4\%$ of the time. Use this information to find the mean and standard deviation from this distribution. Solution Bella spends less than one minute brushing her teeth about $30\%$ of the time. The time is less than two minutes about $4\%$ of the time, and thus less than two minutes about $96\%$ of the time. The $z-value$ is calculated by: This means that $Z$ When $X=1$ has a $p-value$ of $0.3$, thus when $X=1$, $Z=-0.5244$ then: $Z=\dfrac{X-\mu}{\sigma}$ $-0.5244=\dfrac{1-\mu}{\sigma}$ $1-\mu=-0.5244\sigma$ $\mu=1+0.5244\sigma$ She spends more than two minutes brushing her teeth 4% of the time. This means that $Z$ when $X = 2$ has a $p-value$ of $1 – 0.04 = 0.96$, thus, when $X = 2$,$Z = 1.75069$. Then: $Z=\dfrac{X-\mu}{\sigma}$ $1.75069=\dfrac{2-\mu}{\sigma}$ $2-\mu=1.75069\sigma$ $\mu=2-1.75069\sigma$ Since, $\mu=1+0.5244\sigma$ $(1+0.5244\sigma)=(2-1.75069\sigma)$ $2.27\sigma=1$ $\sigma=0.44$ The value of the $\sigma$ is $0.44$. The value of the $\mu$ is calculated as: $\mu=1+0.5244(0.44)$ $\mu=1.23$ The value of the mean $\mu$ is calculated as: $\mu=1.23$ The value of the standard deviation $\sigma$ is calculated as: $\sigma=0.44$	2022-08-10 05:28:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9094843864440918, "perplexity": 350.0185465401099}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00684.warc.gz"}
https://byjus.com/physics/neutron-mass/	# Neutron Mass What are neutrons? It is a subatomic particle. Neutron is the composition of 1 up quark and 2 down quark and comes under the family of fermions. It is denoted by n or n0. The neutron mass is slightly higher than proton mass. The charge of neutron is null, which implies that neutron is electrically neutral. Neutron is located inside the nucleus. ## Mass Of Neutron Mass of neutron is cannot be directly calculated by mass spectroscopy. But it can be calculated by indirect methods. Neutron mass mN 1.674927471(21) * 10-27kg ### Nuetron Mass In Other Units The mass of the neutron can be expressed using alternative units of measurements. Like – mass of neutron in grams, neutron mass amu, mass of neutron in MeV Neutron mass Value mass of neutron in kg 1.674927471(21)×10−27 kg mass of neutron in grams 1.674927471(21)×10−24 kg mass of neutron in amu 1.00866491588(49) u mass of neutron in MeV 939.5654133(58) MeV/c2 Hope you have learnt about Neutron in atomic theory. Along with, mass of neutron, Charge of neutron, mass of neutron amu, mass of neutron in kg, neutron mass in MeV. Physics Related Topics: Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.	2020-06-05 07:29:16	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9040155410766602, "perplexity": 3085.0187825738135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00590.warc.gz"}
http://ringtheory.herokuapp.com/properties/property/6/	# Property: semiprimitive Definition: Jacobson radical zero ## Resources for learning about this property: • N. Jacobson. Basic algebra II. (2012) @ Section 4.1 • T.-Y. Lam. A first course in noncommutative rings. (2013) @ Chapter 2 ## Metaproperties: This property has the following metaproperties • Morita invariant • passes to $eRe$ for any full idempotent $e$ • passes to matrix rings • stable under finite products • stable under products	2019-02-18 11:11:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3156319558620453, "perplexity": 12561.087826800383}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247484928.52/warc/CC-MAIN-20190218094308-20190218120308-00158.warc.gz"}
https://stats.stackexchange.com/questions/444430/bias-variance-trade-off-with-cauchy-estimator	# Bias-Variance Trade Off with Cauchy Estimator I'm having a look at the bias and standard error of a set of estimators. I expected to see the trade off when varying the parameter of the estimator, but I see that both the bias and the variance decline. Estimators $$\hat \theta=q_p-\tan[\pi(p-0.5)]$$ of the location parameter in the Cauchy Distribution ($$f(y)=\pi^{-1}{(1+(x-\theta)^2)^{-1}}$$), where $$q_p$$ is a sample p-quantile with $$p\in (0,1)$$. Using $$\theta=0$$. I've looked at three estimators of this kind, with p = 0.5, 2/3 and 0.75 in R. For each estimator I've used this code: ($$p=2/3$$ for instance) master_0.6 <- data.frame(sample = NaN, EXPECTestimator = NaN, bias = NaN, se = NaN ) row <- 1 for (i in seq(from=10, to=90, by=10)){ estimator0.6 <- data.frame(V1 = NaN) for(k in seq(1,1000,1)){ estimator0.6[k, 1] <- (quantile(rcauchy(i,0), probs = 2/3) - tan(pi((2/3) - 0.5))) } master_0.6[row, 1] <- i master_0.6[row, 2] <- (sum(estimator0.6)/1000) master_0.6[row, 3] <- master_0.6[row, 2] - 0 master_0.6[row, 4] <- (sqrt(sum((estimator0.6 - master_0.6[row, 3])^2))/1000) row <- row + 1 } For this $$p=2/3$$ estimator I've looked at nine sample sizes ($$n=10,20,...,90$$). At each sample size I've generated 1000 samples and applied the estimator to each sample, giving a vector with 1000 rows of the estimator at this sample size. Then I calculate the bias and variance of the estimator at this sample size. The same is done for all nine sample sizes, for each estimator. Blue for $$p=0.5$$, Red for $$p=2/3$$, Green for $$p=0.75$$ I expected to see an inverse relationship between standard error and bias, such that the 2/3 estimator would be a good 'balance' and the desirable estimator. However, we see that the relationships is not so, making the p=0.5 estimator clearly more desirable. Is this expected or have I made an error? As a comment - note that the Cauchy distribution does not have a mean, so I interpret your reference to the mean as being to the location parameter of the Cauchy, which, due to the symmetry of the Cauchy, is also the median of the distribution. With respect to bias: the sample median is an unbiased estimator of the population median, so you should not expect the sample median to have a higher bias than any other estimator. Considering the bias of a sample quantile as an estimator of the corresponding population quantile: bias is smaller in central quantiles than in extreme ones; see the accepted answer to Demonstration of sample quantile bias. With respect to standard error: the asymptotic distribution of the sample quantile $$\hat{q}$$ estimator of the population quantile $$q$$ is Normal with mean $$q$$ and variance $$\frac{q(1-q)}{n \left[f\;\left(F^{-1}(q)\right)\right]^2}$$ Observe that both the numerator and denominator go to zero as $$q \to \pm \infty$$. Performing the requisite calculations for a range of values for $$q$$ and plotting them leads to: avar <- function(q) q(1-q)/(dcauchy(qcauchy(q))^2) q <- seq(0.5,0.99, by=0.001) av <- avar(q) plot(log(av)~q, xlab="Quantile", ylab="log (Asymptotic variance)") At least asymptotically, your variance increases with $$q$$ and so does your bias (in general). This leaves the median as your best estimator in both respects, although note that the median is not the most efficient estimator of the Cauchy location parameter; see the answers to Consistent unbiased estimator for the location parameter of $\mathcal{Cauchy} (\theta, 1)$ for more information on this topic. • Yes, thank you for clearing that up. I meant the median - as the Cauchy location parameter. – JoBel Jan 16 '20 at 10:57	2021-04-19 00:44:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7790669202804565, "perplexity": 834.6734613900782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00520.warc.gz"}
https://zdorovie-russia.ru/carbon-dating-level-chemistry-5519.html	# Carbon dating level chemistry This discovery is in contrast to the carbon dating results for the Turin Shroud that was supposed to have wrapped Jesus’ body.Carbon dating has shown that the cloth was made between 12 AD. Before Radiocarbon dating was able to be discovered, someone had to find the existence of the C isotope. In 1940 Martin Kamen and Sam Ruben at the University of California, Berkeley Radiation Laboratory did just that. The entire process of Radiocarbon dating depends on the decay of carbon-14. This process begins when an organism is no longer able to exchange Carbon with their environment. Although it may be seen as outdated, many labs still use Libby's half-life in order to stay consistent in publications and calculations within the laboratory. From the discovery of Carbon-14 to radiocarbon dating of fossils, we can see what an essential role Carbon has played and continues to play in our lives today.In 1960, Libby was awarded the Nobel Prize in chemistry for this work.He demonstrated the accuracy of radiocarbon dating by accurately estimating the age of wood from a series of samples for which the age was known, including an ancient Egyptian royal barge dating from 1850 BCE.From that point on, scientist have used these techniques to examine fossils, rocks, and ocean currents and determine age and event timing.Throughout the years measurement tools have become more technologically advanced allowing researchers to be more precise and we now use what is known as the Cambridge half-life of 5730 /- 40 years for Carbon-14.In contrast, living material exhibit an activity of 14 d/min.g.Thus, using Equation $$\ref$$, $\ln \dfrac = (1.21 \times 10^) t \nonumber$ Thus, $t= \dfrac = 2 \times 10^3 \text \nonumber$ From the measurement performed in 1947 the Dead Sea Scrolls were determined to be 2000 years old giving them a date of 53 BC, and confirming their authenticity.Thus, the Turin Shroud was made over a thousand years after the death of Jesus.The technique of radiocarbon dating was developed by Willard Libby and his colleagues at the University of Chicago in 1949. From this science, we are able to approximate the date at which the organism were living on Earth.Radiocarbon dating is used in many fields to learn information about the past conditions of organisms and the environments present on Earth.	2021-03-01 10:25:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33288949728012085, "perplexity": 1293.9043058822238}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178362481.49/warc/CC-MAIN-20210301090526-20210301120526-00055.warc.gz"}
https://dml.cz/handle/10338.dmlcz/119061	# Article Full entry \| PDF (0.3 MB) Keywords: parabolic-elliptic problem; nonlinear Neumann boundary condition; Rothe method Summary: We investigate a parabolic-elliptic problem, where the time derivative is multiplied by a coefficient which may vanish on time-dependent spatial subdomains. The linear equation is supplemented by a nonlinear Neumann boundary condition $-\partial u/\partial \nu_A = g(\cdot,\cdot,u)$ with a locally defined, $L_r$-bounded function $g(t,\cdot,\xi)$. We prove the existence of a local weak solution to the problem by means of the Rothe method. A uniform a priori estimate for the Rothe approximations in $L_{\infty}$, which is required by the {\it local} assumptions on $g$, is derived by a technique due to J. Moser. References: [1] Browder F.E.: Estimates and existence theorems for elliptic boundary value problems. Proc. Nat. Acad. Sci. U.S.A. 45 (1959), 365-372. MR 0132913 \| Zbl 0093.29402 [2] Filo J.: A nonlinear diffusion equation with nonlinear boundary conditions: Method of lines. Math. Slovaca 38.3 (1988), 273-296. MR 0977906 \| Zbl 0664.35049 [3] Fučik S., John O., Kufner A.: Function Spaces. Noordhoff International Publishing, Leyden, 1977. MR 0482102 [4] Gilbarg D., Trudinger N.S.: Elliptic Partial Differential Equations of Second Order. Springer Verlag, Berlin, Heidelberg, New York, Tokyo, 1983. MR 0737190 \| Zbl 1042.35002 [5] Jäger W., Kačur J.: Solution of porous medium type systems by linear approximation schemes. Numer. Math. 60 (1991), 407-427. MR 1137200 [6] Kačur J.: Method of Rothe in Evolution Equations. BSB Teubner Verlagsgesellschaft, Leipzig, 1985. MR 0834176 [7] Kačur J.: On a solution of degenerate elliptic-parabolic systems in Orlicz-Sobolev-spaces I & II. Math. Z. 203 (1990), 153-171, 569-579. MR 1030713 [8] Moser J.: A new proof of De Giorgi's theorem concerning the regularity problem for elliptic differential equations. Comm. Pure Appl. Math. 13.3 (1960), 457-468. MR 0170091 \| Zbl 0111.09301 [9] Pluschke V.: Local solution of parabolic equations with strongly increasing nonlinearity by the Rothe method. Czechoslovak Math. J. 38 , 113 (1988), 642-654. MR 0962908 \| Zbl 0671.35037 [10] Pluschke V.: Construction of bounded solutions to degenerated parabolic equations by the Rothe method. in: {Complex Analysis and Differential Equations} (in Russian), Bashkirian State University, Ufa, 1990, pp.58-75. MR 1252652 [11] Pluschke V.: $L_{\infty}$-Estimates and Uniform Convergence of Rothe's Method for Quasilinear Parabolic Differential Equations. in: R. Kleinmann, R. Kress, E. Martensen, {Direct and Inverse Boundary Value Problems}, Verlag Peter Lang GmbH, Frankfurt am Main, 1991, pp.187-199. MR 1215747 [12] Pluschke V.: Rothe's method for semilinear parabolic problems with degeneration. Math. Nachr. 156 (1992), 283-295. MR 1233950 \| Zbl 0794.35088 [13] Rektorys K.: The Method of Discretization in Time and Partial Differential Equations. D.Reidel Publishing Company, Dordrecht, Boston, London, 1982. MR 0689712 \| Zbl 0522.65059 [14] Schechter M.: Coerciveness in $L_p$. Trans. Amer. Math. Soc. 107 (1963), 10-29. MR 0146690 [15] Slodička M.: Error estimate of approximate solution for a quasilinear parabolic integrodifferential equation in the $L_p$-space. Aplikace Matematiky 34 6 (1989), 439-448. MR 1026508 [16] Triebel H.: Interpolation Theory, Function Spaces, Differential Operators. VEB Deutscher Verlag der Wissenschaften, Berlin, 1978. MR 0500580 \| Zbl 0830.46028 [17] Weber F.: Die Anwendung der Rothe-Methode auf parabolische Probleme mit nichtlinearen Neumannschen Randbedingungen. Dissertation, Martin-Luther-Universität, Halle-Wittenberg, 1995. [18] Zeidler E.: Nonlinear Functional Analysis and its Applications - Volume II/B: Nonlinear Monotone Operators. Springer-Verlag, New York, Berlin, Heidelberg, 1990. Partner of	2020-09-22 17:06:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917117178440094, "perplexity": 2877.152178191109}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00582.warc.gz"}
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-6-systems-of-equations-and-inequalities-6-1-solving-systems-by-graphing-lesson-check-page-363/2	Algebra 1 $x = 2$ $y = 3$ #1 The 2 equation is $-x + 2y = 4$ $y = x + 1$ #2 Substitute one equation into 2nd equation solve for X. $-x + 2y = 4$ $-x + 2\times(x+1) = 4$ $-x + 2x + 2 = 4$ $....... .-2..-2$ $-x + 2x = 2$ $x = 2$ #3 Solve for Y Substitute x into one of the equation. $y = x + 1$ $y = 2 + 1$ $y = 3$ #4 $x = 2$ $y = 3$	2018-11-15 11:26:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8294264674186707, "perplexity": 513.441965264657}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121814-00297.warc.gz"}
https://academia.kaust.edu.sa/en/publications/energetics-of-oxygen-adsorption-and-incorporation-at-inn-polar-su	# Energetics of oxygen adsorption and incorporation at InN polar surface: A first-principles study A. Belabbes, J. Kioseoglou, Ph Komninou, Th Karakostas Corresponding author for this work Research output: Contribution to journalArticlepeer-review 3 Scopus citations ## Abstract First principles calculations are performed to study the adsorption and incorporation of oxygen at InN polar surfaces as a function of oxygen coverage. We find that the adsorption of oxygen at InN surfaces is higher on the N-polar than on In-polar. The calculations show that for the low coverage regime (x ≤ 0.5 ML) both orientations are active towards the adsorption of oxygen; this implies that the reaction between the O atoms and the surface is strongly exothermic. As the coverage of O exceeds 0.5 ML a large difference in adsorption energies, of almost 2.5 eV, is observed between (0001) and (000$\bar 1$) surfaces indicating that the (000$\bar 1$) surface is more active towards oxygen adsorption than the (0001) surface. The high coverage of oxygen changes the surface structure of InN due to the difference in binding energy (O-In bonds are stronger than In-N ones). This result shows explicitly that the formation of In adlayer is disturbed by the O adsorption. Concerning the energetics of O incorporation our results show that it is more favourable for O to substitute a N atom in one of the top layers near the surface than in bulk, indicating that there is a strong tendency for O to reside at the surface rather than in subsurface layers. Original language English (US) S364-S367 Physica Status Solidi (C) Current Topics in Solid State Physics 6 SUPPL. 2 https://doi.org/10.1002/pssc.200880886 Published - Jul 2009 ## ASJC Scopus subject areas • Condensed Matter Physics ## Fingerprint Dive into the research topics of 'Energetics of oxygen adsorption and incorporation at InN polar surface: A first-principles study'. Together they form a unique fingerprint.	2023-04-01 00:28:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21063292026519775, "perplexity": 4085.3217131347114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00538.warc.gz"}
http://zkoziarski.pl/5cw9r/15e3cc-boiling-point-of-ammonia	Ammonia is a water-like solvent that has a normal boiling point of $-33.3$ $°C$. (It'd be charged!) Solution. (7) is out of line with values for all other hydrocarbons and should therefore be disregarded. The Journal of Physical Chemistry 1922, 26 (4) , 358-376 From the data given, we know Br 2 must have the strongest intermolecular forces as it has the highest boiling point, followed by NH 3 and then F 2.We can then use our knowledge of these molecules to determine the intermolecular forces present. I also know the strength of a dipole-dipole force is based on the electronegativity difference. The Journal of Physical Chemistry 1922, 26 (9) , 876-894. NH4 does not exist in and of itself. a. Solution for For many purposes we can treat ammonia (NH,) as an ideal gas at temperatures above its boiling point of -33. 90% (243 ratings) Problem Details. You can send your comments directly at info@mechanicaltutorial.com. Determine the normal boiling point of ammonia. \| EduRev Class 10 Question is disucussed on EduRev Study Group by 101 Class 10 Students. Its boiling point is -33.35 °C (-28.03 °F). Low boiling point. the neighboring ammonia molecules get closer due to the attraction. The size of nitrogen atom is less than phosphorus atoms. Jun 09,2020 - What's the boiling point of ammonia ? Action of the Absorption Ammonia Refrigerating Machine.-In the generator the aqueous ammonia solution, usually of about 30 per cent molal concentration, is heated by steam to the boiling point under the generator pressure, causing the formation of a vapor of lower limit for calculation: -75 C, 0.08 bar bar upper limit: 130 C, 108 bar. This boiling point means that ammonia will not stay a liquid even at room temperature, therefore, ammonia is classified as a gas. Triple point pressure of ammonia: 0.0601 atm = 0.0609 bar = 6090 Pa = 0.8832 psi (=lb f /in 2) Triple point temperature of ammonia: 195.5 K = -77.65 °C = -107.77 °F. Clearly, there is an intermolecular force operating between the water and ammonia molecules, the which you have already identified. The boiling point of bromine is 58.8 degrees Celsius. Because Hydrogen bonding is stronger, NH3 has stronger intermolecular forces and therefore a higher boiling point. is incorrect. Therefore, when looking at the two formulas, NH3 and PH3, we can notice that ammonia possesses Hydrogen bonding while phosphine does now. T… °C. Water has the highest boiling point and melting point followed by hydrogen fluoride and lastly ammonia because water has two lone pairs on the oxygen atom which makes it highly electronegative and has two hydrogen atoms hence its capable of forming four hydrogen bonds with the surrounding molecules and hydrogen fluoride is highly electronegative than ammonia. Although the name ammonium hydroxide suggests an alkali with composition [NH 4 +][OH −], it is actually impossible to isolate samples of NH 4 OH. But why is it that Nitrogen is so much higher than the next 3 in the group/group 15? O yes Is it possible to change the temperature of the ammonia at the same time such that the volume of the gas doesn't change? The temperature is 40°C. B. Differences in boiling points between molecules are due to varying strength of intermolecular forces. The desirable property of a refrigerant is. Calculation of thermodynamic state variables of ammonia on the boiling curve. The Vapor Pressure of Sulfur Dioxide and Ammonia. Structure, properties, spectra, suppliers and links for: Ammonium hydroxide, 1336-21-6, ammonia water, NH4OH. Liquid Ammonia (Boiling point = -33.4C) can be used as a refrigerant and heat transfer fluid... 39 kJ Diethyl ether, used as a solvent for extraction of organic compounds from aqueous solutions, has a … Ammonia solution, also known as ammonia water, ammonium hydroxide, ammoniacal liquor, ammonia liquor, aqua ammonia, aqueous ammonia, or (inaccurately) ammonia, is a solution of ammonia in water. [110-82-7) The mole fraction solubility of ammonia in cyclohexane at 292.2 K and 1 atm from measurements by Kuznetsov et aZ. hydrocarbons of similar boiling point. More the intermolecular force more Will be the boiling point. 3 Suppose the pressure on a 1.0 m… So I know the forces responsible for the boiling point are the vanderwale forces (depending on the mass of the molecules), and the hydrogen bonds. 1. I want to calculate how much ammonia evaporates from a 10% ammonia solution (ammonium hydroxide solution), to check if the concentration of ammonia in the air exceeds 25 ppm. The ammonia-based cleaners you use don’t vapourize rapidly as pure ammonia would because they contain little ammonia and are mixed with other non-volatile chemicals such as water (non-volatile means they don’t vapourize easily at room temperature). That’s why the electrons of N – H bond come closer to N-atom than the electrons of P-H bond that come closer to P-atom. Articles of Aqueous ammonia are included as well. b. On the other hand, ammonia makes more sense: Ammonia, in comparison, stores and handles very much like LPG. We need more energy to separate them than Phosphine. The boiling point of ammonia is. The boiling point of ammonia is -33.34 degrees Celsius. So hydrogen bonding in ammonia is much more stronger than phosphine. Recommendations : Air Liquide has gathered data on the compatibility of gases with materials to assist you in evaluating which materials to use for a gas system. 8. Propane, the main constituent of LPG, has a boiling point of -42.07 °C (-43.73 °F). Boiling Point The boiling point of aqua ammonia is defined as the temperature at which the partial vapor pressure of the ammonia vapor over the aqua ammonia equals atmospheric pressure. The boiling point of bromine is 58.8 degrees Celsius. mark as brainlist The Boiling Points of Ammonia, Sulfur Dioxide and Nitrous Oxide. It can be denoted by the symbols NH 3 (aq). Ostwald coefficients for this system at … The figures below show the changes in thermophysical properties along the ammonia boiling and condensation curves: Density, viscosity, heat capacity, thermal conductivity. High critical temperature Example #1: What is the boiling point elevation when 11.4 g of ammonia (NH 3) is dissolved in 200. g of water?K b for water is 0.52 °C/m.. Boiling point of water: 100 °C / 212 °F Boiling point of water (in Kelvin): 373.2 K Boiling point of ethanol: 78.37 °C / 173.1 °F Boiling point of methanol: 64.7 °C / 148.5 °F Boiling point of acetone: 56 °C / 132.8 °F Boiling point of alcohol: 78.37 °C / 173.1 °F Boiling point of nitrogen: -195.8 °C / -320.4 °F Boiling point of liquid helium: -269 °C / -452 °F °C. The boiling point of ammonia is -33.34 degrees Celsius. 3 Suppose the pressure on a 2.0 mº sample of ammonia gas at 6.00°C is cut in half. Use the date to determine the heat of vaporization of ammonia. NH4 is the result of ammonia (which is highly soluble) being added to water. F. W. Bergstrom. The boiling point order is H2>HF>NH3. Ammonia (NH3) has a mass of ~17 and makes hydrogen bonds. 1) Determine molality of 11.4 g of ammonia in 200. g of water: 11.4 g / 17.031 g/mol = 0.6693676 mol 0.6693676 mol / 0.200 kg = 3.3468 m Chemsrc provides Aqueous ammonia(CAS#:1336-21-6) MSDS, density, melting point, boiling point, structure, formula, molecular weight etc. Materials compatibility. Boiling point of Ammonia (NH 3) is more than Phosphine (PH 3): . It produces ammonium hydroxide (NH4OH), although it is an equilibrium (which is why NH4OH has a strong ammonia … Ammonia NH3 - R717 - UN1005 - 7664-41-7. So the difference for Nitrogen is a lot greater than that of the next 3 in the group when finding the difference for Hydrogen. For many purposes we can treat ammonia (NH3) as an ideal gas at temperatures above its boiling point of -33. A.-10.5 0 C B.-30.5 0 C C.-33.3 0 C C.-77.6 0 C Subject: Boiling point of ammonia and water mixture Category: Science > Chemistry Asked by: alakon-ga List Price: \$2.50: Posted: 20 Sep 2002 15:08 … no Х If you answered yes, calculate the new temperature of the gas. Water (H2O) has a mass of ~18 and makes hydrogen bonds. FREE Expert Solution Show answer. And of course the boiling point is so elevated by reason of intermolecular hydrogen-bonding. A. Phosphine has dipole-dipole forces since it is a polar molecule. Since there are N-H bonds in this compound, ammonia is easily liquefied into ammonia liquid. Ammonia gas is condensed at its boiling point at 1.01325 bar at -33.4^{\circ} \mathrm{C} by the application of a pressure infinitesimally greater than 1 bar. I know that a larger dipole-dipole and/or London force makes the boiling point greater. The melting point of ammonia is -77.73 o C and the boiling point is about -33.34 o C. When considering the density of ammonia, it is lighter than air. As you know, the normal boiling points of water, #100# #""^@C#, and ammonia, #-33.3# #""^@C#..are EXCESSIVELY HIGH....especially given these are molecules of paltry molecular mass, the which gives rise to negligible dispersion force.So why so? The vapor pressure of ammonia at several different temperatures is shown below. DOI: 10.1021/j150225a006. CH4 has a lower boiling point." So, nitrogen is more electronegative element. The boiling point of ammonia is a) -10.5°C b) -30°C c) -33.3°C d) -77.7°C. ) -33.3°C d ) -77.7°C bar bar upper limit: 130 C, 108 bar of!: Ammonium hydroxide, 1336-21-6, ammonia makes more sense: ammonia, Sulfur Dioxide and Nitrous Oxide atom less... Much higher than the next 3 in the group/group 15 pressure of ammonia gas temperatures... Is classified as a gas Question is disucussed on EduRev Study group by Class! 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( NH 3 ( aq ) the symbols NH 3 ( aq ) force operating between the and. 358-376 hydrocarbons of similar boiling point of ammonia at several different temperatures is below! Of the next 3 in the group when finding the difference for hydrogen is on! It is a polar molecule liquefied into ammonia liquid and of course the boiling.... Temperatures above its boiling point of bromine is 58.8 degrees Celsius HF > NH3 of! Group/Group 15 stores and handles very much like LPG above its boiling point since it a! To varying strength of a dipole-dipole force is based on the electronegativity difference being added to water 0! H2 > HF > NH3 in cyclohexane at 292.2 K and 1 atm from measurements by et!, stores and handles very much like LPG the which you have already identified C 0.08... 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Send your comments directly at info @ mechanicaltutorial.com temperatures is shown below hydroxide 1336-21-6... Molecules get closer due to the attraction critical temperature boiling point greater ostwald coefficients for this system at … know... Journal Of Family Studies Impact Factor 2019, Herman Miller Aia Chair, Best Sword And Shield Mhw Iceborne, Industrial Construction Examples, Charon's Claw Pdf, Manuring For Coconut Trees, Cast Iron Plant Care, Types Of Ancient Daggers, Fallout 76 Wastelanders Power Armor Locations,	2021-03-06 11:33:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4406988322734833, "perplexity": 3594.1509092447736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374686.69/warc/CC-MAIN-20210306100836-20210306130836-00351.warc.gz"}
https://www.examrace.com/IB-ACIO/IB-ACIO-Model-Questions/Aptitude-Questions/Aptitude-Logical-Reasoning-Profit-and-Loss-Part-11.html	Aptitude Logical Reasoning Profit and Loss 2020 IB-ACIO Part 11 Doorsteptutor material for BITSAT is prepared by world's top subject experts: fully solved questions with step-by-step explanation- practice your way to success. 1. A man purchases an electric heater whose printed price is 160 if he received two successive discounts of 20% and 10% he paid: A. Rs. 112 B. Rs. 129.60 C. Rs. 119.60 D. Rs. 115.20 2. Varun bought a Refrigerator with 20% discount on the labelled price. Had he bought it with 25% discount? He would have saved Rs. 500. At what price did he buy the Refrigerator? A. Rs. 5000 B. Rs. 10000 C. Rs. 12000 D. Rs. 16000 3. The difference between a discount of 40% on Rs. 500 and two successive discount of 36% and 4% on the same amount is: A. 0 B. Rs. 2 C. Rs. 1.93 D. Rs. 7.20 4. The price of an article was increased by P%. Later the new price was decreased by P% if the latest price was Rs. 1. The original price was: A. Rs. 1 B. Rs. C. Rs. D. Rs. 2 5. A cloth merchant has announced 25% rebate in prices. If one needs to have a rebate of Rs. 40. Then how many shirts each costing Rs.32 he should purchase? A. 6 B. 5 C. 10 D. 7 6. A dealer marks his goods 20% above cost price. He then allows some discount on it and makes a profit of 8%. The rate of discount is: A. 12% B. 10% C. 6% D. 4% 7. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is: A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60 8. When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%? A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800 9. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is: A. No profit, no loss B. 5% C. 8% D. 10% 10. A trader marked the selling price of an article at 10% above the cost price. At the time of selling, he allows certain discount and suffers a loss of 1%. He allowed a discount of : A. 10 % B. 10.5 % C. 11 % D. 12.5 %	2021-01-26 17:13:47	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8851171135902405, "perplexity": 4833.44648317984}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00106.warc.gz"}
https://proxieslive.com/finite-dimensional-modules-are-highest-weight-modules/	# finite dimensional modules are highest weight modules Let $$\mathfrak{g}$$ be a basic classical simple Lie super algebra. I want to prove that every finite dimensional module over $$\mathfrak{g}$$ has a highest weight vector. My feeling is, since $$e_i$$‘s are rising operators it will kill a non-zero vector and this will give us a highest weight vector and may be we need to use Lie’s theorem. But I am unable to connect these things to get a perfect answer. If some one can tell me clearly what is happening here, that would help me a lot. Thank you.	2018-11-16 14:57:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3749229609966278, "perplexity": 148.77721622499087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116153625-00047.warc.gz"}
http://archived.moe/a/thread/13682294	61KiB, 805x482, 1216556457274.jpg ## Subtitle 'errors' No.13682294 Hi guys, quick question. Sometimes when I'm watching subtitles from groups like GG, the kareoke won't work right. The highlighted words will be stuck on a frame with part of a phrase highlighted and part not. I remember there was an easy way to fix this on CCCP, but I can't remember since I reinstalled everything. I thought I saw it in a thread not too long ago but forgot to write it down. Does anyone remember off hand how to fix it? In return, creepy.	2016-10-27 18:59:01	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.809933066368103, "perplexity": 1806.5728637093728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721387.11/warc/CC-MAIN-20161020183841-00275-ip-10-171-6-4.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/834957/does-xix-alpha-1-imply-that-x-alpha-is-a-discrete-space	Does $\xi(X) = \alpha + 1$ imply that $X^{(\alpha)}$ is a discrete space? Let $K$ be a (Hausdorff) scattered topological space and for each ordinal $\alpha$ denote by $K^{(\alpha)}$ the $\alpha$th derivative of $K$ by the Cantor-Bendixson derivation (i.e., define transfinitely: $K^{(0)} = K$; for each ordinal $\alpha$, let $K^{(\alpha+1)}$ be the set of nonisolated points in $K^{(\alpha)}$ when $K^{(\alpha)}$ is equipped with the subspace topology); for $\alpha$ a limit ordinal, set $K^{(\alpha)}= \bigcap_{\beta<\alpha}K^{(\beta)}$). Also define: $\xi(X) = \inf\{ \alpha : X^{(\alpha)} = \emptyset \}$ if some $\alpha$ exists such that $X^{(\alpha)} = \emptyset$, and $\xi(X) = \infty$ otherwise. My question is: Suppose that $\xi(X) = \alpha + 1$. Does this imply that $X^{(\alpha)}$ is a discrete space? Thank you! Yes, essentially by definition. If $X^{(\alpha)}$ is not discrete, then some point of $X^{(\alpha)}$ must have the property that each of its neighbourhoods contains another element of $X^{(\alpha)}$ (or, more succinctly, is not an isolated point of the set). Then it must be that $X^{(\alpha+1)}$ is nonempty, contradicting the definition of $\xi(X)$.	2021-10-25 20:34:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9840725064277649, "perplexity": 19.800372055030497}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00332.warc.gz"}
http://math.stackexchange.com/questions/234526/trouble-with-samples-in-a-normal-distribution	# Trouble with samples in a normal distribution I'm okay with solving regular normal distribution questions (where X is a normal random variable with mean $\mu$ and standard deviation $\sigma$). However, we're currently dealing with samples within a larger population and I'm struggling to understand how the process of solving a question goes. Here's an example: The weight of a car is normally distributed with a mean of 2500 and a standard deviation of 50. Next, a random sample of 5 cars is taken. What is the standard error of the sample mean? I know that the standard error of the sample is: $$\frac{\sigma^2}{n}$$ but a question can't be this easy to answer can it? It's leading me to believe I'm missing a concept. If someone would be able to walk through the procedure of solving a question of this nature it would be greatly appreciated. - The question is in fact that simple. $\frac{\sigma^2}{n}$ is the standard error of the sample mean. "standard error of the sample" (what you said it is) doesn't make any sense. – Jonathan Christensen Nov 11 '12 at 2:19 The sample mean $M$ of a sample of size $n$ from a normally distributed RV with mean $\mu$ and variance $\sigma^2$ is a RV $\sim N(\mu,\sigma^2/n)$. The standard error of the mean $s_M$ is the standard deviation of the sample mean and so $s_M=\sigma/\sqrt{n}$.	2016-05-05 01:50:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8102909922599792, "perplexity": 83.35172689866172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860125750.3/warc/CC-MAIN-20160428161525-00190-ip-10-239-7-51.ec2.internal.warc.gz"}
https://novelanswer.com/a-keyword-search-program-lists-the-files-that-contain-a-given-keyword-if-it-runs-through-200-files-and-each-file-contains-the-keyword-with-probability/	# Solved (Free): A keyword search program lists the files that contain a given keyword. If it runs through 200 files, and each file contains the keyword with probability #### ByDr. Raju Chaudhari Mar 28, 2021 A keyword search program lists the files that contain a given keyword. If it runs through 200 files, and each file contains the keyword with probability 0.36, independently of other files, compute the probability that (a) more than 70 files (b) less than 70 files (c) exactly 70 files will be listed. #### Solution Let $X$ be the number of files listed and $p$ be the probability that the listed file contains the keyword. Given that $n =200$ and $p=0.36$. Thus $X\sim B(200, 0.36)$. Since $n$ is large and $p$ is neither too small nor too large, the distribution of X is approximately Normal with mean = $E(X)=\mu= np = 200 \times 0.36 = 72$ and sd = $\sigma= \sqrt{np*(1-p)} = \sqrt{200 \times 0.36 \times (1- 0.36)}=6.79$. (a) The (approximate) probability that more than 70 files will be listed is \begin{aligned} P(X> 70) & = P(X> 70.5)\\ & = 1-P(X< 70.5)\\ &\quad \quad (\text{ using continuity correction})\\ &= 1-P\bigg(\frac{X-\mu}{\sigma}< \frac{70.5- 72}{6.79}\bigg)\\ & =1-P(Z< -0.22)\\ & = 1-0.4126\\ & = 0.5874 \end{aligned} (b) The (approximate) probability that less than 70 files will be listed is \begin{aligned} P(X< 70) & = P(X< -0.5)\\ &\quad \quad (\text{ using continuity correction})\\ &= P\bigg(\frac{X-\mu}{\sigma}< \frac{69.5- 72}{6.79}\bigg)\\ & =P(Z< -0.37)\\ & = 0.3564\\ \end{aligned} (c) The probability that exactly 70 files will be listed is \begin{aligned} P(X= 70) & = P(X< 70.5)-P(X< 69.5)\\ &\quad \quad (\text{ using continuity correction})\\ &= P(Z< -0.22)-P(Z< -0.37)\\ & = 0.0562\\ \end{aligned}	2023-03-30 06:04:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9951708912849426, "perplexity": 2528.7822041630507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00657.warc.gz"}
https://physics.stackexchange.com/questions/429461/why-is-it-so-important-to-measure-cp-violation-in-the-lepton-sector	# Why is it so important to measure CP violation in the lepton sector? CP violation has been measured in the hadronic sector (K-mesons, B-mesons) and these baryons can decay into leptons, so why is it so important to measure direct CP violation into the lepton sector? I could naively think that we have e.g. too many protons in the universe due to CP violation and then there are too many $K^-$ that decay into electrons (and antineutrinos, what is the ratio between antineutrinos and neutrinos in the universe?) and there you have our universe. I know there has to be a "quantitative" reason to make the search for CP violation in the lepton sector so significant, but I just cannot find it. • Possibly Leptogenesis? Sep 18 '18 at 19:03	2021-12-03 16:17:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9012951850891113, "perplexity": 504.34481665141607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00513.warc.gz"}
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-3-equations-and-problem-solving-3-4-equations-involving-parentheses-and-fractional-forms-problem-set-3-4-page-123/58	## Elementary Algebra $x=(14y-17)$ In order to solve for x, we first subtract by 14y on both sides of the equation to help get closer to getting x alone on one side of the equation. We then isolate x completely by dividing both sides by -1, obtaining that: $x=(14y-17)$	2018-08-18 23:13:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.953709602355957, "perplexity": 307.14169474183456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213794.40/warc/CC-MAIN-20180818213032-20180818233032-00097.warc.gz"}
http://th.nao.ac.jp/MEMBER/tomisaka/Lecture_Notes/StarFormation/3/node84.html	Next: Centrifugal Radius Up: Magneto-driven Model Previous: Problem Contents ### Angular Momentum Transfer In the axisymmetric case, the poloidal and toroidal components of the magnetic field and current are decoupled with each other. That is, the poloidal () and toroidal () magnetic fields are made by the toroidal () and poloidal () electric currents, respectively. As for the Lorentz force , the poloidal component comes from or , while the toroidal component does from . Even if there is no toroidal magnetic field (thus no poloidal electric current), there exists the poloidal component of the Lorentz force, which acts as a pressure to counter-balance the self-gravity (4.2). On the other hand, the toroidal component of the Lorentz force appears only the case with the poloidal electric current and thus toroidal component of magnetic field . This means that the angular momentum is transferred by the magnetic field only when exists. Equation (B.14) explains how the angular momentum density is transferred. The left-hand side of equation (B.14) represents the advection of the angular momentum density, while the right-hand side (4.86) represents the torque exerted on the gas parcel. The induction equation of the magnetic field [eq.(B.17)] shows that is generated from poloidal magnetic field by the effect of rotational motion . This indicates that the angular momentum is transferred as follows: 1. The toroidal component of magnetic field is amplified by the dynamo process [equation (B.17)]. The magnetic field lines run like spiral viewing from the top like Figure 4.11. 2. Negative torque works near region attached to the fast-rotating disk, while positive one appears in the region connected to the radial magnetic field. The angular momentum is transferred from inside to outside along the same field line. 3. Angular momentum received by the gas near disk surface leads the excess centrifugal force which accelerate the gas. The disk inside which is affected by the negative torque loses its angular momentum and insufficient centrifugal force drives further accretion. This angular momentum transfer on the same magnetic field has a character of angular momentum redistribution. Next: Centrifugal Radius Up: Magneto-driven Model Previous: Problem Contents Kohji Tomisaka 2007-07-08	2018-01-20 11:10:03	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8541901111602783, "perplexity": 790.2674137110039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889567.48/warc/CC-MAIN-20180120102905-20180120122905-00649.warc.gz"}
http://mathhelpforum.com/differential-geometry/117998-analysis.html	## Analysis Thanks in advance for the help. definition: a function f is said to cover the set M if for every point y in M there is an x in the domain of f so that f(x)=y. Prove: Suppose that f is continuous with domain [a,b] so that f(x) is in [a,b] for every x in [a,b]. Prove that there is some x in [a,b] so that f(x)=x.	2017-11-19 10:40:35	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8586365580558777, "perplexity": 275.76919208447316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805541.30/warc/CC-MAIN-20171119095916-20171119115916-00209.warc.gz"}
https://www.eng-tips.com/viewthread.cfm?qid=457172	× INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS Are you an Engineering professional? Join Eng-Tips Forums! • Talk With Other Members • Be Notified Of Responses • Keyword Search Favorite Forums • Automated Signatures • Best Of All, It's Free! *Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. #### Posting Guidelines Promoting, selling, recruiting, coursework and thesis posting is forbidden. # Designing for Wind Tunnel Test Section Velocity ## Designing for Wind Tunnel Test Section Velocity (OP) Please copy and paste my post if that assists responding. I am trying to work backwards by selecting a target test chamber velocity to design a wind tunnel around. I want to revamp our current system as minimally as possible which means using the same fan and the current settling chamber and laminizer. The entrance and some of the test chamber section can be seen in this thread. I understand all the area and pressure relationship stuff just fine. To me, deciding on the test section velocity by using the settling chamber and Bernoulli Principle equations seems partially erroneous since the laminizer is in there in the real world and affects free flowing air movement. The current setup has no entrance diffuser/section currently....this isn’t my design and I know it’s coefficients are wrong and its aerodynamics are questionable… Regardless, this wind tunnel is used for heat exchanger studies, so that's why I'm looking to achieve a target velocity. 100 percent of the air is to be moved through the core for heat exchanger studies, and for that reason our test chamber now has the opportunity to be resized since it was designed incorrectly for the intended purpose. Since this no doubt will require some fabrication work, I want to make sure the speed increase is viable (+10 MPH, currently 35 MPH). I'm probably just going to flow sim this in Solidworks, but I was also thinking I could just use the known test chamber speeds from our first tests to Bernoulli it out using the dimensions of the new, smaller cross section, higher speed test chamber. This method should be good for an estimate because it does not require my transition angles, correct? Do y’all know any other hacks like that? Since I know Q (volumetric flow) of the fan and current dimensions of the test section and settling chambers transition, I’ve also got a sure fire way to roughly figure it out using advertised specs. But why on Earth is nobody teaching stuff this way? Why have the useless conceptual equations never been evolved and presented better for designing a wind tunnel?? ### RE: Designing for Wind Tunnel Test Section Velocity actually I don't think you're "working backwards". To achieve your increased air flow in the test section, using the same fan, you'll need to reduce the size of the test section. (in my mind working backwards would mean I've set the test section parameters (size, airflow) and will select a fan to achieve these). This will change almost the entire tunnel. There are texts and such "out there" but wind tunnels is such a small field that they may be hard to find. I'd seek out a nearby wind tunnel (like at a university or college) and talk to someone there. A key (the key ?) to tunnel design is the uniformity of the airflow across the test section. ### RE: Designing for Wind Tunnel Test Section Velocity Most wind tunnels aren't operated with the amount of obstruction you are looking to test with, limiting design info further. For certain, flow out of a centrifugal blower isn't uniform, which is why I would have expected the blower to be on the down-wind side as a suction system so the inlet can have a simple grid straightener. It also avoids pre-heating the air from the work done by the blower which means the inlet temp will also be uniform, making temperature measurement easier. ### RE: Designing for Wind Tunnel Test Section Velocity (OP) Rb1957 I suppose I meant backwards from the methods I was loosely referencing from my undergrad lab material as they wanted us to make a wind tunnel by sizing the test section vs settling chamber first. Figures the school way is the sillier way since these loons never got into the real world…plus it would have merits for a different intent. I know UCF has at least two wind tunnels as I saw them a few times while in undergrad and got to use them too. I’d love to make some senior students do this study for me haha. At the same time, I’ve been pretty successful with convincing my boss to revamp the erroneous, so now we have some cash sunk into this thing and it’s like a weird elephant in the room. Most people just really don’t understand how !@#$ing crazy and expensive testing is: This system still needs another$1K-$2K to become close enough to the Kays & London setup, and we’re not a heat exchanger company lol. I’m having such an ethics struggle with trying move forward knowing that the measurements won’t be valid, nor the method correct for understanding heat exchanger performance. I’ve actually found and contacted a company that received a$10K grant for a wind tunnel to rate their heat exchangers, but it was gonna be hard to get them to test ours it seemed. (OP) 3DDave I have this as a constant thought in my brain as well, perhaps Kays and London were just using it as the most analogous term for their wind box setup for their heat exchanger trials haha? Regardless, the one company I mentioned above with the $10K wind tunnel is following the Kays and London method that 100 percent of air must move through the core. My boss just bought the fan reactively to replace another home fan for this other hysterical method that was devised before I got to the company; it’s hard out here for an engineer sometimes. I would’ve set up the tunnel as a down draft as well, and I would’ve optioned a bladed fan driven by pulleys so the speeds can be changed. ### RE: Designing for Wind Tunnel Test Section Velocity your mgmt are bitching over another$2k ?? good grief, they'll spend more discussing what flavour of coffee to have in the break room (the exec break room, that is) ! I don't think you can work with a bad wind tunnel (or any other apparatus) without at least saying here's the results, and here's the error bands ! I think you need to work up some tests to show how poorly it works. ### RE: Designing for Wind Tunnel Test Section Velocity (OP) rb1957 most people just really aren't ready for a scientific measuring trial as they're both very accomplishing and very annoying to carry out. Right now the system was set up with a YouTube science video level of planning, aka just for fun lol. You'll see on some of my other posts that I'm encountering heavy ideologue resistance in general here, despite me being the only one who has the education to understand it at face and deeper levels. The fact that I have to even get on here to spend time away from just fixing it is the funny part, but it is the company money at the same time... Once I get a few flow sim images (pending my 10 year old personal laptop can handle it on a scale model) I think it should be easier to see how poorly air flows. We don't have flow sim in Inventor, and there's also no way that would be paid for either. In the end, even with convincing results, I still won't be able to do anything about it most likely :( ### RE: Designing for Wind Tunnel Test Section Velocity Apologies up front if I missed something in the other replies... You can certainly use Bernoulli and other similar simplifications(VA_1 = VA_2) for a low speed setup like that. If you're going that route, you could also calculate for a top end around 60mph, and your losses will put you in the ballpark or just over. Just keep in mind that before your rework, with a known Q your speed is what it is in low speed tunnels. One of the best references for low speed tunnel design and methodologies is "Low Speed Wind Tunnel Testing" by Jewel Barlow. Basically the industry standard. ### RE: Designing for Wind Tunnel Test Section Velocity Pushing air at higher velocities means you need to get to higher differential pressures, which means more fan horsepower and more CFM simultaneously. You are probably fan-limited not section limited right now. The idea of a pulley-driven fan is not bad, as you could always find a bigger motor. But, worst case, you might get there by adding more cheap leafblowers too... #### Red Flag This Post Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework. #### Red Flag Submitted Thank you for helping keep Eng-Tips Forums free from inappropriate posts. The Eng-Tips staff will check this out and take appropriate action. Close Box # Join Eng-Tips® Today! Join your peers on the Internet's largest technical engineering professional community. It's easy to join and it's free. Here's Why Members Love Eng-Tips Forums: • Talk To Other Members • Notification Of Responses To Questions • Favorite Forums One Click Access • Keyword Search Of All Posts, And More... Register now while it's still free!	2020-08-14 05:03:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36013296246528625, "perplexity": 1966.5688254177765}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00469.warc.gz"}
https://www.physicsforums.com/threads/spontaneous-nuclear-fission.697433/	# Spontaneous nuclear fission 1. Jun 17, 2013 ### sunrah I am confused about spontaneous fission. My basic understanding is that like α-decay the tunnel effect is responsible. We have a potential barrier caused by the superposition of surface tension energy and coulomb potential through which an energised nuclear fragment can tunnel with a certain probability. How does the potential barrier of the nucleus relate to its binding energy? I have read that in reality it is only very large nuclei that undergo spontaneous fission; these nuclei have have lower binding energies per nucleon. How does reduction in B/A effect size of potential barrier? Also in class we have derived the tunneling probability $T \propto e^{-2G}$ where G is the Gamov funktion. I have read that $G \propto m$ and that heavier particles/fragments have lower tunneling probability that is why α-decay is much more common than spontaneous fission of heavy nuclei. If so why are heavy nuclei more likely to fission spontaneously than lighter ones? 2. Jun 17, 2013 ### Crazymechanic 3. Jun 17, 2013 ### sunrah 4. Jun 17, 2013 ### anorlunda Here is a speculation. Of elements heavier than iron, the heavier the element the less binding energy per nucleon. Of course elements lighter than iron can't spontaneously fission at all. 5. Jun 17, 2013 ### Crazymechanic just to add to what anorlunda already said , heavy nuclei can fission spontaneously and deliberately because they are unstable to begin with , it has to do with the proton neutron number as it gets too big the nuclear force which holds everything together ( like glue) is having a hard time to hold the big structure together and it can break apart. Small nuclei which have few protons and neutrons ( Hydrogen being the smallest one proton one electron) are very sable , few particles holding strongly together. 6. Jun 18, 2013 ### sunrah thanks for the replies! so these large nuclei are unstable to begin with because the binding energy per nucleon drops significantly whilst coulomb repulsion increases causing strong oscillations/deformation in the ground state? also I couldn't find a definite answer to this: is the height of the potential barrier Ef above the ground state equal to the binding energy Eb ? are they one and the same? here is what I mean http://www.kutl.kyushu-u.ac.jp/seminar/MicroWorld3_E/3Part3_E/3P33_E/fission_barrier_E.jpg 7. Jun 18, 2013 ### Crazymechanic as for the first one the answer is yes. as for the second i will let some more experienced people comment , but as far as I know you can tell how strongly something was held together by calculating how much force it took you to break it apart , now to my understanding this would also be the case to your question number too and I tend to think the answer should be yes.	2017-08-24 11:40:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4972009062767029, "perplexity": 1066.0771494859075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00287.warc.gz"}
https://re.public.polimi.it/handle/11311/630986	We investigate the existence of the second mountain-pass solution to a Robin problem, where the equation is at critical growth and depends on a positive parameter $\lambda$. More precisely, we determine existence and nonexistence regions for this type of solutions, depending both on $\lambda$ and on the parameter in the boundary conditions. On the second solution to a critical growth Robin problem Abstract We investigate the existence of the second mountain-pass solution to a Robin problem, where the equation is at critical growth and depends on a positive parameter $\lambda$. More precisely, we determine existence and nonexistence regions for this type of solutions, depending both on $\lambda$ and on the parameter in the boundary conditions. Scheda breve Scheda completa Scheda completa (DC) File in questo prodotto: Non ci sono file associati a questo prodotto. I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione. Utilizza questo identificativo per citare o creare un link a questo documento: http://hdl.handle.net/11311/630986 • ND • ND • ND	2022-09-29 11:42:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7951712012290955, "perplexity": 3101.6698904852506}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335350.36/warc/CC-MAIN-20220929100506-20220929130506-00703.warc.gz"}
https://stats.stackexchange.com/tags/gaussian-process/hot	# Tag Info 2 For me the biggest advantage of Gaussian Processes is the inherent ability to model the uncertainty of the model. This is incredibly useful because, given the expected value of a function and the corresponding variance I can define a metric (i.e. an Aquisition function) that can tell me e.g. what's the point $x$ that, I should evaluate my underlying ... 2 Gaussian Processes assume an underlying continuum. That being time, space, acidity, or simply the values a regularisation parameter. The error bars in this example are indeed associated with the placement of the know points but that is completely expected; the further away we are from a know point, the more uncertain we are about the estimated value. I would ... 1 I'll provide the derivation for a general random process (with finite second moments). As the author's random process is a simple random process (stochastic part doesn't depend on time), the direct differentiation (what the author did) is also correct. The derivation is based on the book [Natan, Gorbachev, Guz, The basics of random process theory, 2003] (in ... 1 Acquisition functions are not about a specific surrogate model. They can be calculated for many of them given that the output of a surrogate model is not a single prediction $\hat{y}(x)$, but a probability distribution $\hat{p}(x)$. In case, of the Gaussian process regression, we assume that the output is Gaussian distribution. So it is sufficient to ... 1 Some disadvantages of LOWESS vs GP are the following: LOWESS will fail even for a moderate input dimension of data, while GP works in this case for right kernel selected For LOWESS to work we need a dense training sample of points over the whole design space. For GP the requirements are less strict We need additional tricks to control outliers Some ... 1 Your idea about using a hold-out set for comparing the RMSE is fine. I would note though that if we do not have rather a large hold-out sample, using a repeated cross-validation approach instead of a fixed hold-out set will mitigate finite-sample variance issues; repeated CV is preferable because it allows to also estimate the variability of our test ... 1 the inverse transform of the standard deviation is wrong. Mean and standard variation have to be transformed differently. Here's a brief explanation: Transform Let's assume random variable $Y$ with mean $\mu_Y$ and variance $\sigma^2_Y.$ The "Scaler" subtracts some constant $a$ and divides the result by a factor $b.$ The transformed variable equals \$Z = \... Only top voted, non community-wiki answers of a minimum length are eligible	2020-04-10 20:13:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557725548744202, "perplexity": 618.9268823549924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00353.warc.gz"}
https://gateoverflow.in/66900/ugcnet-dec2010-ii-28	1.3k views How much bandwidth is there in $1$ micron of spectrum at a wavelength of $1$ micron ? 1. $300$ MHz 2. $3$ GHz 3. $300$ THz 4. $30$ KHz recategorized \| 1.3k views C: 300THZ. The formula is $\Delta f = c * \Delta \Lambda / \Lambda^{2}$ where ∆f is the bandwidth range, c is the light speed, $\Delta \Lambda$ is the wavelength range and $\Lambda$ is a certain wavelength. For frequency to be in hertz and speed of light in meter/second, we want the wavelength to be in meters. As 1 meter = 1000,000 microns, calculating using the formula you get 300 THZ. +1 vote 1 2	2019-07-21 06:48:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8890045285224915, "perplexity": 3279.6951653191654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00531.warc.gz"}
http://stackoverflow.com/questions/10718767/have-the-same-readme-both-in-markdown-and-restructuredtext?answertab=active	# Have the same README both in Markdown and reStructuredText I have a project hosted on GitHub. For this I have written my README using the Markdown syntax in order to have it nicely formatted on GitHub. As my project is in Python I also plan to upload it to PyPi. The syntax used for READMEs on PyPi is reStructuredText. I would like to avoid having to handle two READMEs containing roughly the same content; so I searched for a markdown to RST (or the other way around) translator, but couldn't find any. The other solution I see is to perform a markdown/HTML and then a HTML/RST translation. I found some ressources for this here and here so I guess it should be possible. Would you have any idea that could fit better with what I want to do? - Github will render README.rst! – u0b34a0f6ae Feb 2 '13 at 14:50 This is new then :) But good to know, i'll try ! – jlengrand Feb 2 '13 at 22:31 If you would like PyPI to support readmes in Markdown, please comment on the feature request at bitbucket.org/pypa/pypi/issue/148/support-markdown-for-readmes – Colonel Panic May 30 '14 at 9:57 Upvoted :) Let's see.... – jlengrand May 30 '14 at 17:28 I would recommend Pandoc, the "swiss-army knife for converting files from one markup format into another" (check out the diagram of supported conversions at the bottom of the page, it is quite impressive). Pandoc allows markdown to reStructuredText translation directly. There is also an online editor here which lets you try it out, so you could simply use the online editor to convert your README files. - Quite impressive, but also beautiful ! Question solved, thanks! Weird I couldn't find it by myself, though :S – jlengrand May 23 '12 at 12:09 The magical invocation is: pandoc --from=markdown --to=rst --output=README.rst README.md – Jonathan Eunice Mar 20 '13 at 21:49 +1 for the online editor – moose Oct 23 '14 at 17:44 You might also be interested in the fact that it is possible to write in a common subset so that your document comes out the same way when rendered as markdown or rendered as reStructuredText: https://gist.github.com/dupuy/1855764 - I ran into this problem and solved it with the two following bash scripts. Note that I have LaTeX bundled into my Markdown. #!/usr/bin/env bash if [ $# -lt 1 ]; then echo "$0 file.md" exit; fi filename=$(basename "$1") extension="${filename##.}" filename="${filename%.}" if [ "$extension" = "md" ]; then rst=".rst" pandoc$1 -o $filename$rst fi Its also useful to convert to html. md2html: #!/usr/bin/env bash if [ $# -lt 1 ]; then echo "$0 file.md <style.css>" exit; fi filename=$(basename "$1") extension="${filename##.}" filename="${filename%.}" if [ "$extension" = "md" ]; then html=".html" if [ -z$2 ]; then # if no css pandoc -s -S --mathjax --highlight-style pygments $1 -o$filename$html else pandoc -s -S --mathjax --highlight-style pygments -c$2 $1 -o$filename$html fi fi I hope that helps - Using the pandoc tool suggested by others I created a md2rst utility to create the rst files. Even though this solution means you have both an md and an rst it seemed to be the least invasive and would allow for whatever future markdown support is added. I prefer it over altering setup.py and maybe you would as well: #!/usr/bin/env python ''' Recursively and destructively creates a .rst file for all Markdown files in the target directory and below. Created to deal with PyPa without changing anything in setup based on the idea that getting proper Markdown support later is worth waiting for rather than forcing a pandoc dependency in sample packages and such. Vote for (https://bitbucket.org/pypa/pypi/issue/148/support-markdown-for-readmes) ''' import sys, os, re markdown_sufs = ('.md','.markdown','.mkd') markdown_regx = '\.(md\|markdown\|mkd)$' target = '.' if len(sys.argv) >= 2: target = sys.argv[1] md_files = [] for root, dirnames, filenames in os.walk(target): for name in filenames: if name.endswith(markdown_sufs): md_files.append(os.path.join(root, name)) for md in md_files: bare = re.sub(markdown_regx,'',md) cmd='pandoc --from=markdown --to=rst "{}" -o "{}.rst"' print(cmd.format(md,bare)) os.system(cmd.format(md,bare)) - The Markup library used by GitHub supports reStructuredText. This means you can write a README.rst file. They even support syntax specific color highlighting using the code and code-block directives (Example) - As @Chris suggested, you can use Pandoc to convert Markdown to RST. This can be simply automated using pypandoc module and some magic in setup.py: from setuptools import setup try: from pypandoc import convert read_md = lambda f: convert(f, 'rst') except ImportError: setup( # name, version, ... This can also throw a runtime error. In order to stay on a safe side I recommend to do try-except in the function. – varepsilon Jun 8 '14 at 11:21	2015-05-06 07:11:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47458192706108093, "perplexity": 6356.9929764480785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430458333760.71/warc/CC-MAIN-20150501053213-00022-ip-10-235-10-82.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/481978/how-to-write-the-block-matrix-in-latex/481981	# How to write the block matrix in LaTex? [duplicate] I wish to write the following block matrix in LaTex. I have been trying a bunch of stuff, but none of it seems to work. In particular, I tried the following: \begin{bmatrix} 0 & -1& & & & & 0 & & & &\\ 1 & 0 & & & & & & & & &\\ & & \cdot & & & & & & & &\\ & & & \cdot & & & & & & &\\ & & & & \cdot & & & & & &\\ & & & & & 0 & 1 & & & &\\ & & & & &-1 & 0 & & & &\\ & & & & & & & \cdot & & &\\ & & & & & & & & \cdot & & \\ & & & & & & & & & \cdot & \\ 0 & & & & & & & & & & 0\\ \end{bmatrix} But this is pretty ugly and so any suggestions will be much appreciated. Perhaps $\left(\begin{array}{cccccc} \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right] & & & & & 0\\ & \ddots\\ & & \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right]\\ & & & 0\\ & & & & \ddots\\ 0 & & & & & 0 \end{array}\right)$ Just for the fun of it! :-) \documentclass{article} \usepackage{amsmath,array} \begin{document} \end{document} • You guys are the best! – model_checker Mar 28 '19 at 22:40 • @model_checker I'm certainly not the best. Here is world excellence. – Sebastiano Mar 28 '19 at 22:50 • I am sorry! Last I checked there was an objection to your answer. But it's also great! – model_checker Mar 28 '19 at 22:53 • @model_checker Don't worry. I have put two & in addition. :-) but now it is correct. – Sebastiano Mar 28 '19 at 23:02 Here there is my proposal as the original picture using \bmatx command that replace a boxed matrix. \documentclass{article} \usepackage{mathtools} \newcommand{\bmatx}{\boxed{\begin{matrix} 0& 1 \\ -1& 0\end{matrix}}} \begin{document} $\begin{pmatrix} \, \bmatx& & & & & & & 0 \\ & \ddots& & & & \\ & & \bmatx & & & & \\ & & & \ddots& & &\\ & & & & 0 & &\\ & & & & & & \ddots\\ 0 & & & & & & & 0 \end{pmatrix}$ \end{document} The best code is provided thanks to the precious comment of @Bernard. • In my opinion, you shouldn't add an empty 3rd column in the definition of \bmatx\ – Bernard Mar 28 '19 at 21:24 • Your comment not is an opinion :-) ...it is truly correct. :-) – Sebastiano Mar 28 '19 at 21:30 Here's another solution. It uses a pmatrix environment for the overall matrix and a custom macro called \blockmat for the 2x2 inner matrices. \documentclass{article} \usepackage{array,amsmath} \newcommand\blockmat{% \begin{array}{\|@{\,}rr@{\,}\|} \hline 0 & 1^{\mathstrut} \\ -1 & 0 \\ \hline \end{array}} \begin{document} $\begin{pmatrix} \blockmat & & & & & 0 \\ & \ddots & & & & \\ & & \blockmat & & & \\ & & & 0 & & \\ & & & & \ddots & \\ 0 & & & & & 0 \\ \end{pmatrix}$ \end{document} Here's a solution using TikZ that will work in all kinds of context, not just matrices, to draw boxes. You just need to put \tl in the point where the top-right corner is supposed to be and \br where the bottom-right will be. You can also use any TikZ feature to modify your box. Obviously this needs to be fine tuned depending on what exactly you're using it for (spacing etc.). \newcounter{textbox} \def\tl{\stepcounter{textbox}\tikzmarknode{a\thetextbox}{\strut}} \def\br{\tikzmarknode{b\thetextbox}{\strut}\begin{tikzpicture}[overlay, remember picture]\draw (a\thetextbox.north west) rectangle (b\thetextbox.south east);\end{tikzpicture}} The spacing in this version is not optimized for matrices, but you can use \tikzmarknode{b\thetextbox}{\strut}\begin{tikzpicture}[overlay, remember picture]\draw ($(a\thetextbox.north west)+(-0.4\arraycolsep,0ex)$) rectangle ($(b\thetextbox.south east)+(0.4\arraycolsep,0ex)$);\end{tikzpicture} to add extra space. Full code: \documentclass{article} \usepackage{tikz} \usepackage{amsmath} \usetikzlibrary{tikzmark,calc} \newcounter{textbox} \def\tl{\stepcounter{textbox}\tikzmarknode{a\thetextbox}{\strut}} \def\br{\tikzmarknode{b\thetextbox}{\strut}\begin{tikzpicture}[overlay, remember picture]\draw (a\thetextbox.north west) rectangle (b\thetextbox.south east);\end{tikzpicture}} \begin{document} $\begin{bmatrix} \tl0 &-1 & & & & & & & & 0 \\ 1 & 0\br & & & & & & & & \\ & & \cdot & & & & & & & \\ & & & \cdot & & & & & & \\ & & & & \cdot & & & & & \\ & & & & &\tl0 & 1 & & & \\ & & & & &-1 & 0\br & & & \\ & & & & & & & \cdot & & \\ & & & & & & & & \cdot & \\ 0 & & & & & & & & & \cdot \\ \end{bmatrix}$ \end{document}	2020-03-30 17:19:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9927912354469299, "perplexity": 482.9855791977461}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00181.warc.gz"}
https://www.physicsforums.com/threads/complex-function-conjugate.372723/	# Homework Help: Complex function conjugate 1. Jan 26, 2010 ### leoneri 1. The problem statement, all variables and given/known data I have a complex function $$w\left(z\right)=e^{sin\left(z\right)}$$ What is the conjugate? 2. The attempt at a solution The conjugate is $$w\left(z^{}\right)=e^{sin\left(z^{}\right)}$$ $$w\left(x-iy\right)=e^{sin\left(x-iy\right)}$$ My question is, is my answer correct? If I wrong, could you tell me what is the right answer and how to reach this correct answer? 2. Jan 26, 2010 ### Altabeh AB 3. Jan 26, 2010 ### leoneri Thanks. I was sure about that, but one of my friend said I was wrong. So, is there a way to proof it? Because I see it as a definition that whenever there is a complex function $$w\left(z\right)=w\left(x+iy}\right)$$, then the conjugate is always $$w\left(z^{}\right)=w\left(x-iy\right)$$. I would be more than happy if someone can show me or point me to a proof that the above definition is always true. 4. Jan 26, 2010 ### Rjee But I think you need W(z) and not W(z*) Last edited: Jan 26, 2010 5. Jan 26, 2010 ### Altabeh Since that is a definition, we can't do so much to give a real proof. But you can imagine the case involving $$e^{i\theta}$$. In this case, using the Euler relation one can see $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$, $$e^{i(-\theta)}=\cos(\theta)-i\sin(\theta)$$ (by the symmetry of cos and anti-symmetry of sin wrt the change of the sign of $$\theta$$), so $$e^{i\theta}e^{i(-\theta)}=[\cos(\theta)+i\sin(\theta)][\cos(\theta)-i\sin(\theta)] = 1.$$. Here you can understand that the result of the product of the function $$e^{i\theta}$$ and its complex conjugate leads to the correct answer 1 just by looking at the first part of the equality. The second part confirms the complex conjugation of $$e^{i\theta}$$ and $$e^{-i\theta}$$. Thus a change in the sign of $$i$$ makes the function be transferred into its complex conjugate phase. AB Last edited: Jan 26, 2010	2018-06-23 09:06:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8713136911392212, "perplexity": 277.88833281579906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864953.36/warc/CC-MAIN-20180623074142-20180623094142-00455.warc.gz"}
https://chemistry.stackexchange.com/questions/69128/how-to-apply-faradays-law-to-calculate-the-charge-of-metal-ions	# How to apply Faraday's law to calculate the charge of metal ions? Electrolysis of a molten ionic compound with a current of $\pu{0.50 A}$ for $\pu{30 min}$ yielded $\pu{0.700 g}$ of a metallic element at the cathode. If the element has a relative atomic mass of $150.0$, calculate the charge on the metal ions. My attempt at solving: $$n = \frac{m}{M} = \frac{0.7}{150} = \pu{4.67e-3}\\ Q = Ft = 0.5 \times (30 \times 60) = \pu{900 C}$$ The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction? After I obtain that can I just use $$Q = n\ce{e-} \times F?$$ I then solved for $$F = Q/n = \pu{1.92E5 F}$$. I was experimenting with some stuff and divided this by $\pu{96500 C}$ which got me $1.99$. The correct answer was $\pu{2 F}$. Is this just coincidence because I don't see the connection between dividing by $\pu{1 F} = \pu{96500 C}$ when I already solved for $F$? • Use the formula given by @ Kai, but use SI units which means putting time into seconds to find $Q = It$ and $F = 96485.3 \pu{ C mol^{-1}}$. The charge z comes out as 2 to the nearest whole number. – porphyrin Feb 23 '17 at 17:10 When you're confused by a problem like this it is definitely time to use dimensional analysis. For the amount of electrons $\text{mol}_e$ \begin{align} \text{The ampere} &= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\ \text{Farady} &= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\ 30~\text{minutes} &= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\ \text{mol}_e &= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e \end{align} For the amount of substance of metal $\text{m}_m$ $$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$ Charge per atom = $z$ $$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$ $$m=\left(\frac{Q}{F}\right)\left(\frac{M}{z}\right)$$ In the given question, we have to find $z$ which is the valency of the metallic ions. Use $Q=IT$ (in Ampere-Hours), $F=26.8$ Ampere hour per equivalent.	2019-07-18 16:02:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9949195981025696, "perplexity": 885.0110123192223}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00104.warc.gz"}
https://www.chemistryabc.com/2018/03/bonding-in-coordination-complexes.html	Featured ## Featured • QUIZES • QUESTIONS • SLOLVED QUESTIONS • HIGHER EDUCATIONAL INFO • PDF FREE • NEW PDF BOOKS UPDATED # Bonding in Coordination Complexes ### What is a Complex? If you have taken a lab course in chemistry, you have very likely admired the deep blue color of copper sulfate crystals, CuSO4·5H2O. The proper name of this substance is copper(II) sulfate pentahydrate, and it is typical of many salts that incorporate waters of hydration into their crystal structures. It is also a complex, a term used by chemists to describe a substance composed of two other substances (in this case, CuSO4 and H2O) each of which is capable of an independent existence. Figure $\mathbf{9.9.}\mathbf{1}$(left) Crystals of ${\text{CuSO}}_{4}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\cdot 5\phantom{\rule{thinmathspace}{0ex}}{\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}$. Image used with permission from Stephanb (via Wikipedia). (right) Anhydrous ${\text{CuSO}}_{4}^{\phantom{\rule[-.2em]{0pt}{0pt}}}$ after the water was removed. The binding between the components of a complex is usually weaker than a regular chemical bond; thus most solid hydrates can be decomposed by heating, driving off the water and yielding the anhydrous salt: $\begin{array}{}\text{(9.9.1)}& \underset{\text{blue}}{\underset{⏟}{{\text{CuSO}}_{4}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\cdot 5\phantom{\rule{thinmathspace}{0ex}}{\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}}}\to \underset{\text{white}}{\underset{⏟}{{\text{CuSO}}_{4\phantom{\rule{thinmathspace}{0ex}}\left(\text{s}\right)}^{\phantom{\rule[-.2em]{0pt}{0pt}}}}}+5{H}_{2}O\end{array}$ Driving off the water in this way also destroys the color, turning it from a beautiful deep blue to a nondescript white. If the anhydrous salt is now dissolved in water, the blue color now pervades the entire solution. It is apparent that the presence of water is somehow necessary for the copper(II) ion to take on a blue color, but why should this be? Figure $\mathbf{9.9.}\mathbf{2}$: Difference between anhydrous and hydrated copper sulfate after a few drops of water was added to anhydrous $CuS{O}_{4}$. $\begin{array}{}\text{(9.9.2)}& {\text{Cu}}^{2\phantom{\rule{thinmathspace}{0ex}}+}+2\phantom{\rule{thinmathspace}{0ex}}{\text{OH}}^{–}\to \text{Cu}{\left(\text{OH}\right)}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\left(\text{s}\right)\end{array}$ However, if more ammonia is added, the cloudiness disappears and the solution assumes an intense deep blue color that makes the original solution seem pale by comparison. The equation for this reaction is usually given as $\begin{array}{}\text{(9.9.3)}& {\text{Cu}}^{2\phantom{\rule{thinmathspace}{0ex}}+}+6\phantom{\rule{thinmathspace}{0ex}}{\text{NH}}_{3}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\to \text{Cu}{\left({\text{NH}}_{3}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\right)}_{6}^{2\phantom{\rule{thinmathspace}{0ex}}+}\end{array}$ The new product is commonly known as the copper-ammonia complex ion, or more officially, hexamminecopper(II) complex ion. Equation $\text{9.9.3}$ is somewhat misleading, however, in that it implies the formation of a new complex where none existed before. In fact, since about 1895, it has been known that the ions of most transition metals dissolve in water to form complexes with water itself, so a better representation of the reaction of dissolved copper with ammonia would be $\begin{array}{}\text{(9.9.4)}& \text{Cu}{\left({\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}\right)}_{6}^{2\phantom{\rule{thinmathspace}{0ex}}+}+6\phantom{\rule{thinmathspace}{0ex}}{\text{NH}}_{3}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\to \text{Cu}{\left({\text{NH}}_{3}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\right)}_{6}^{2\phantom{\rule{thinmathspace}{0ex}}+}+6\phantom{\rule{thinmathspace}{0ex}}{\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}\end{array}$ In effect, the ammonia binds more tightly to the copper ion than does water, and it thus displaces the latter when it comes into contact with the hexaaquocopper(II) ion, as the dissolved form of Cu2+ is properly known. Most transition metals dissolve in water to form complexes with water itself. ### The basics of Coordination Complexes Although our primary focus in this unit is on bonding, the topic of coordination complexes is so important in chemistry and biochemistry that some of their basic features are worth knowing about, even if their detailed chemistry is beyond the scope of this course. These complexes play an especially crucial role in physiology and biochemistry. Thus heme, the oxygen-carrying component of red blood cells (and the source of the red color) is basically a complex of iron, and the part of chlorophyll that converts sunlight into chemical energy within green plants is a magnesium complex. #### Some Definitions We have already defined a complex as a substance composed of two or more components capable of an independent existence. A coordination complex is one in which a central atom or ion is joined to one or more ligands (Latin ligare, to tie) through what is called a coordinate covalent bond in which both of the bonding electrons are supplied by the ligand. In such a complex the central atom acts as an electron-pair acceptor (Lewis acid — think of H+ which has no electrons at all, but can accept a pair from something like Cl) and the ligand as an electron-pair donor (Lewis base ). The central atom and the ligands coordinated to it constitute the coordination sphere. Thus the salt [Co(NH3)5Cl]Cl2 is composed of the complex ion [Co(NH3)5Cl]2+ and two Cl ions; components within the square brackets are inside the coordination sphere, whereas the two chloride ions are situated outside the coordination sphere. These latter two ions could be replaced by other ions such as NO3 without otherwise materially changing the nature of the salt. The central atoms of coordination complexes are most often cations (positive ions), but may in some cases be neutral atoms, as in nickel carbonyl Ni(CO)4. Ligands composed of ions such as F or small molecules such as H2O or CN possess more than one set of lone pair electrons, but only one of these pairs can coordinate with a central ion. Such ligands are said to be monodentate (“one tooth”.) Larger ligands may contain more than one atom capable of coordinating with a single central ion, and are described as polydentate. Thus ethylenediamine (shown below) is a bidentate ligand. Polydentate ligands whose geometry enables them to occupy more than one coordinating position of a central ion act as chelating agents (Greek χελος, chelos, claw) and tend to form extremely stable complexes known as chelates. Chelation is widely employed in medicine, water-treatment, analytical chemistry and industry for binding and removing metal ions of particular kinds. Some of the more common ligands (chelating agents) are shown here: #### Structure and bonding in transition metal complexes Complexes such as Cu(NH3)62+ have been known and studied since the mid-nineteenth century. Why they should form, or what their structures might be, were complete mysteries. At that time all inorganic compounds were thought to be held together by ionic charges, but ligands such as water or ammonia are of course electrically neutral. A variety of theories such as the existence of “secondary valences” were concocted, and various chain-like structures such as CuNH3-NH3-NH3-NH3-NH3-NH3 were proposed. Finally, in the mid-1890s, after a series of painstaking experiments, the chemist Alfred Werner (Swiss, 1866-1919) presented the first workable theory of complex ion structures. Werner claimed that his theory first came to him in a flash after a night of fitful sleep; by the end of the next day he had written his landmark paper that eventually won him the 1913 Nobel Prize in Chemistry. Werner was able to show, in spite of considerable opposition, that transition metal complexes consist of a central ion surrounded by ligands in a square-planar, tetrahedral, or octahedral arrangement. This was an especially impressive accomplishment at a time long before X-ray diffraction and other methods had become available to observe structures directly. His basic method was to make inferences of the structures from a careful examination of the chemistry of these complexes and particularly the existence of structural isomers. For example, the existence of two different compounds AX4 having the same composition shows that its structure must be square-planar rather than tetrahedral. #### What holds them together? An understanding of the nature of the bond between the central ion and its ligands would have to await the development of Lewis’ shared-electron pair theory and Pauling’s valence-bond picture. We have already shown how hybridization of the orbitals of the central ion creates vacancies able to accommodate one or more pairs of unshared electrons on the ligands. Although these models correctly predict the structures of many transition metal complexes, they are by themselves unable to account for several of their special properties: • The metal-to-ligand bonds are generally much weaker than ordinary covalent bonds; • Some complexes utilize “inner” d orbitals of the central ion, while others are “outer-orbital” complexes; • Transition metal ions tend to be intensely colored. #### Paramagnetism of coordination complexes Unpaired electrons act as tiny magnets; if a substance that contains unpaired electrons is placed near an external magnet, it will undergo an attraction that tends to draw it into the field. Such substances are said to be paramagnetic, and the degree of paramagnetism is directly proportional to the number of unpaired electrons in the molecule. Magnetic studies have played an especially prominent role in determining how electrons are distributed among the various orbitals in transition metal complexes. Studies of this kind are carried out by placing a sample consisting of a solution of the complex between the poles of an electromagnet. The sample is suspended from the arm of a sensitive balance, and the change in apparent weight is measured with the magnet turned on and off. An increase in the weight when the magnet is turned on indicates that the sample is attracted to the magnet (paramagnetism) and must therefore possess one or more unpaired electrons. The precise number can be determined by calibrating the system with a substance whose electron configuration is known. ### Crystal field theory The current model of bonding in coordination complexes developed gradually between 1930-1950. In its initial stages, the model was a purely electrostatic one known as crystal field theory which treats the ligand ions as simple point charges that interact with the five atomic d orbitals of the central ion. It is this theory which we describe below. It is remarkable that this rather primitive model, quite innocent of quantum mechanics, has worked so well. However, an improved and more complete model that incorporates molecular orbital theory is known as ligand field theory. In an isolated transition metal atom the five outermost d orbitals all have the same energy which depends solely on the spherically symmetric electric field due to the nuclear charge and the other electrons of the atom. Suppose now that this atom is made into a cation and is placed in solution, where it forms a hydrated species in which six H2O molecules are coordinated to the central ion in an octahedralarrangement. An example of such an ion might be hexaaquotitanium(III), Ti(H2O)63+. The ligands (H2O in this example) are bound to the central ion by electron pairs contributed by each ligand. Because the six ligands are located at the corners of an octahedron centered around the metal ion, these electron pairs are equivalent to clouds of negative charge that are directed from near the central ion out toward the corners of the octahedron. We will call this an octahedral electric field, or the ligand field. #### d-orbital splitting The differing shapes of the five kinds of d orbitals cause them to interact differently with the electric fields created by the coordinated ligands. This diagram (from a Purdue U. chemistry site) shows outlines of five kinds of d orbitals. Figure $\mathbf{9.9.}\mathbf{3}$: The green circles represent the coordinating electron-pairs of the ligands located at the six corners of the octahedron around the central atom. The two d orbitals at the bottom have regions of high electron density pointing directly toward the ligand orbitals; the resulting electron-electron repulsion raises the energy of these d orbitals. Although the five d orbitals of the central atom all have the same energy in a spherically symmetric field, their energies will not all be the same in the octahedral field imposed by the presence of the ligands. The reason for this is apparent when we consider the different geometrical properties of the five d orbitals. Two of the orbitals, designated dx2 and dx2-y2, have their electron clouds pointing directly toward ligand atoms. We would expect that any electrons that occupy these orbitals would be subject to repulsion by the electron pairs that bind the ligands that are situated at corresponding corners of the octahedron. As a consequence, the energies of these two d orbitals will be raised in relation to the three other d orbitals whose lobes are not directed toward the octahedral positions. The number of electrons in the d orbital of the central atom is easily determined from the location of the element in the periodic table, taking in account, of course, of the number of electrons removed in order to form the positive ion. The effect of the octahedral ligand field due to the ligand electron pairs is to split the orbitals into two sets whose energies differ by a quantity denoted by Δ ("delta") which is known as the d orbital splitting energy. Note that both sets of central-ion d orbitals are repelled by the ligands and are both raised in energy; the upper set is simply raised by a greater amount. Both the total energy shift and Δ are strongly dependent on the particular ligands. #### Why are transition metal complexes often highly colored? Returning to our example of Ti(H2O)63+, we note that Ti has an outer configuration of 4s23d2, so that Ti3+ will be a d1 ion. This means that in its ground state, one electron will occupy the lower group of d orbitals, and the upper group will be empty. The d-orbital splitting in this case is 240 kJ per mole which corresponds to light of blue-green color; absorption of this light promotes the electron to the upper set of d orbitals, which represents the exited state of the complex. If we illuminate a solution of Ti(H2O)63+ with white light, the blue-green light is absorbed and the solution appears violet in color. SPLITTING OF ORBITALS IN OCTAHEDRAL COMPLEXES. Figure $9.9.4$ shows how the colors of hexaamminenickel II complexes are affected by different ligands. Figure $\mathbf{9.9.}\mathbf{4}$: Color of various Ni(II) complexes in aqueous solution. From left to right, hexaamminenickel(II), tris(ethylenediamine)nickel(II), tetrachloronickelate(II) and hexaaquanickel(II). Image used with permission from LHcheM (via Wikipedia) #### High- and low spin complexes The magnitude of the d orbital splitting depends strongly on the nature of the ligand and in particular on how strong an electrostatic field is produced by its electron pair bond to the central ion. If Δ is not too large then the electrons that occupy the d orbitals do so with their spins unpaired until a d5 configuration is reached, just as occurs in the normal Aufbau sequence for atomic electron configurations. Thus a weak-field ligand such as H2O leads to a “high spin” complex with Fe(II). In contrast to this, the cyanide ion acts as a strong-field ligand; the d orbital splitting is so great that it is energetically more favorable for the electrons to pair up in the lower group of d orbitals rather than to enter the upper group with unpaired spins. Thus hexacyanoiron(II) is a “low spin” complex— actually zero spin, in this particular case. Different d orbital splitting patterns occur in square planar and tetrahedral coordination geometries, so a very large number of arrangements are possible. In most complexes the value of Δ corresponds to the absorption of visible light, accounting for the colored nature of many such compounds in solution and in solids such as ${\text{CuSO}}_{4}^{\phantom{\rule[-.2em]{0pt}{0pt}}}·5\phantom{\rule{thinmathspace}{0ex}}{\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}$ ()Figure $9.9.1$. COORDINATION COMPLEXES IN BIOCHEMISTRY Approximately one-third of the chemical elements are present in living organisms. Many of these are metallic ions whose function within the cell depends on the formation of d-orbital coordination complexes with small molecules such as porphyrins (see below). These complexes are themselves bound within proteins (metalloproteins) which provide a local environment that is essential for their function, which is either to transport or store diatomic molecule (oxygen or nitric oxide), to transfer electrons in oxidation-reduction processes, or to catalyze a chemical reaction. The most common of these utilize complexes of Fe and Mg, but other micronutrient metals including Cu, Mn, Mo, Ni, Se, and Zn are also important. ### Hemoglobin Hemoglobin is one of a group of heme proteins that includes myoglobin, cytochrome-c, and catalase. Hemoglobin performs the essential task of transporting dioxygen molecules from the lungs to the tissues in which it is used to oxidize glucose, this oxidation serving as the source of energy required for cellular metabolic processes. Hemoglobin consists of four globin protein subunits (depicted by different colors in this diagram) joined together by weak intermolecular forces. Each of these subunits contains, buried within it, a molecule of heme, which serves as the active site of oxygen transport. The presence of hemoglobin increases the oxygen carrying capacity of 1 liter of blood from 5 to 250 ml. Hemoglobin is also involved in blood pH regulation and CO2 transport. Heme itself consists of an iron atom coordinated to a tetradentate porphyrin. When in the ferrous (Fe2+ state) the iron binds to oxygen and is converted into Fe3+. Because a bare heme molecule would become oxidized by the oxygen without binding to it, the adduct must be stabilized by the surrounding globin protein. In this environment, the iron becomes octahedrally-coordinated through binding to a component of the protein in a fifth position, and in the sixth position either by an oxygen molecule or by a water molecule, depending on whether the hemoglobin is in its oxygenated state (in arteries) or deoxygenated state (in veins). The heme molecule (purple) is enfolded within the polypeptide chain as shown here. The complete hemoglobin molecule contains four of these subunits, and all four must be present for it to function. The binding of O2 to heme in hemoglobin is not a simple chemical equilibrium; the binding efficiency is regulated by the concentrations of H+, CO2, and organic phosphates. It is remarkable that the binding sites for these substances are on the outer parts of the globin units, far removed from the heme. The mechanism of this exquisite molecular-remote-control arises from the fact that the Fe2+ion is too large to fit inside the porphyrin, so it sits slightly out of the porphyrin plane. This Fe radius diminishes when it is oxygenated, allowing it to move into the plane. In doing so, it pulls the protein component to which it is bound with it, triggering a sequence of structural changes that extend throughout the protein. Myoglobin is another important heme protein that is found in muscles. Unlike hemoglobin, which consists of four protein subunits, myoglobin is made up of only one unit. Its principal function is to act as an oxygen storage reservoir, enabling vigorous muscle activity at a rate that could not be sustained by delivery of oxygen through the bloodstream. Myoglobin is responsible for the red color of meat. Cooking of meat releases the O2 and oxidizes the iron to the +3 state, changing the color to brown. #### Carbon monoxide poisoning Other ligands, notably cyanide ion and carbon monoxide, are able to bind to hemoglobin much more strongly than does iron, thereby displacing it and rendering hemoglobin unable to transport oxygen. Air containing as little as 1 percent CO will convert hemoglobin to carboxyhemoglobin in a few hours, leading to loss of consciousness and death. Even small amounts of carbon monoxide can lead to substantial reductions in the availability of oxygen. The 400-ppm concentration of CO in cigarette smoke will tie up about 6% of the hemoglobin in heavy smokers; the increased stress this places on the heart as it works harder to compensate for the oxygen deficit is believed to be one reason why smokers are at higher risk for heart attacks. CO binds to hemoglobin 200 times more tightly than does ${O}_{2}$. #### Chlorophyll Chlorophyll is the light-harvesting pigment present in green plants. Its name comes from the Greek word χλορος (chloros), meaning “green”- the same root from which chlorine gets its name. Chlorophyll consists of a ring-shaped tetradentate ligand known as a porphin coordinated to a central magnesium ion. A histidine residue from one of several types of associated proteins forms a fifth coordinate bond to the Mg atom. M.Sc. CHEMISTRY!	2018-12-14 19:26:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.628147304058075, "perplexity": 1240.7787791427063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00204.warc.gz"}
http://www.solutioninn.com/radioactive-atoms-are-unstable-because-they-have-too-much-energy	Question Radioactive atoms are unstable because they have too much energy. When they release their extra energy, they are said to decay. When studying cesium-137, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to 977,287 radioactive atoms, so 22,713 atoms decayed during 365 days. a. Find the mean number of radioactive atoms that decayed in a day. b. Find the probability that on a given day, 50 radioactive atoms decayed. Sales0 Views362	2016-10-21 17:56:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8810245394706726, "perplexity": 1855.4422693206886}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718285.69/warc/CC-MAIN-20161020183838-00560-ip-10-171-6-4.ec2.internal.warc.gz"}
https://blender.stackexchange.com/questions/32296/tris-to-quads-using-uv-map/88751	# Tris to Quads using UV map I have some old models that I need to repair. When I use Tris to Quads it doesn't work properly preserving the edgeloops because of the curvature of the models. But the UV map of the model is perfectly squared. Is there a quick way to remove the unwanted edges using the UV map? I can do that manually but it takes way too much time. • Can you provide that mesh to test on something? Jun 11 '15 at 18:46 • @RayMairlot Sure, I will upload the file in a min Jun 11 '15 at 18:47 • @RayMairlot I attached the blend file Jun 11 '15 at 18:54 The UV layout for this model makes selecting rows very easy with the Box Select tool. (I hid the ends because they have hidden faces that are causing problems.) By selecting every second loop of the UV's and then using Tris To Quads, you can establish guide loops that allow you to use the tool again and achieve consistent results. b for Box-Select. Also make sure to activate the icon for use_uv_select_sync as shown in the animation. Now use Tris to Quads on the selected rows, then select everything and repeat Tris To Quads. Now you can repair the few mistakes by using the Join tool j. In the following case, I used Join and then used Make Face f to make a quad at the center of the error point. • I tried that before but it doesn't work well on surfaces that have lots of angle changes between the faces, the best I could do is to rotate the UV 45 degrees, select the every second diagonal edgeloop and dissolve , then invert selection and dissolve again, then rotate UV back. Jun 12 '15 at 16:50 There doesn't seem to be an easy way to convert the existing mesh to pure quads without quite a bit of manual intervention. Here's an alternative way to create a new mesh with quads and keep a reasonable and even UV mapping. 1. Create a cylinder with no caps around the dress object 2. Copy the height of the dress to the cylinder so it is exactly the same height 3. Make sure the top and bottom of the cylinder match exactly the top and bottom of the dress Add subsurf mod and shrinkwrap mod (with dress as target) to cylinder Adjust subsurf as desired 4. Apply shrinkwrap modifier Select vertical edge loop at back of dress, create seam UV unwrap cylinder object with cylinder projection type 5. Adjust UV map to fit UV image Trim vertices at top and bottom of cylinder This may seem like a lot of steps, but it goes quite quickly. Let me know if this helps at at all • That will not work in my case, I have many of these meshes, some of them have 4-5 different baked textures, and the UVs have to match exactly to prevent distortion Jun 12 '15 at 2:12 • Ah, I see. Didn't realize you already had pre-existing texture maps. Jun 12 '15 at 2:48 • If you handle this part well, you can use the baking tool 'Selected To Active' to transfer the old textures to the new mesh. Jun 12 '15 at 9:58 This works for me: import bpy import bmesh # go into edit mode bpy.ops.object.mode_set(mode='EDIT') # deselect everything bpy.ops.mesh.select_all(action='DESELECT') # Get the active mesh obj = bpy.context.edit_object me = obj.data # force face select mode bpy.context.tool_settings.mesh_select_mode = (False, True, False) # Get a BMesh representation bm = bmesh.from_edit_mesh(me) count = 0 # find every second face and select the first edge for face in bm.faces: if (count % 2) == 0: edge=face.edges[0] edge.select_set(True) count = count+1 # dissolve edges bpy.ops.mesh.dissolve_edges(use_verts=False) # deselect everything bpy.ops.mesh.select_all(action='DESELECT') # Show the updates in the viewport bmesh.update_edit_mesh(me, True)	2022-01-24 21:32:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2767195403575897, "perplexity": 2260.5467391764914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00541.warc.gz"}
https://math.stackexchange.com/questions/2545394/mappings-cant-understand-proof-that-n-is-the-number-of-permutations-of-n	# Mappings, (can't understand) proof that $n!$ is the number of permutations of $n$ elements. Is $\sigma:J\to J$ defined only for elements of $J$? This comes from S.Lang - Basic Mathematics,Exercise 11, page 374. Let $J_n$ be the set of the integers $1,2,3,...,n$. (an ordered set?) Prove that the number of permutations of $J_n$ is equal to n!. The exercise is answered as follow: We assume it true for $n$. Let $\tau_k$ $(k=1,...,n)$ be transposition which interchanges $n+1$ and $k$ for $k=1,...,n$. Let $S_n$ be the set of permutations of $J_n$. We have to show that the permutations $\; I\sigma,\tau_1\sigma,...,\tau_n\sigma \;$ for $\sigma$ in $S_n$ give all distinct permutations of $J_{n+1}$. Suppose that $\,\tau_i\sigma=\tau_j\sigma'\,$ for some indices $\,i,j\,$ and some $\sigma,\sigma'$ in $S_n$. Since $\tau_i\sigma(n+1)=i$ and $\tau_j\sigma'(n+1)=j,\,$ it follows that $\tau_i=\tau_j$. Multiplying on the left by $\tau_i(=\tau_j),$ we conclude that $\sigma=\sigma'$.$\;$ If $\,\sigma'=\tau_i\sigma,\,$ then looking at the effect on the number $n+1$ we conclude that this cannot be so. Hence the permutations $\;\sigma,\tau_1\sigma,...,\tau_n\sigma \;$ with $\,\sigma$ in $S_n$ are all distinct. If $\gamma$ is permutation of $J_{n+1}$,then either $\gamma$ leaves $n+1$ fixed, in which case $\gamma$ is already in $S_n$, or $\gamma(n+1)=i$ for some $i$ with $1\le i \le n$. In this case $\tau_i \gamma$ leaves $n+1$ fixed, so $\tau_i \gamma=\sigma$ is element of $S_n$, and $\gamma=\tau_i \sigma.\;$ Thus we have found all permutations of $J_{n+1}$. We assume by induction that there are $n!$ permutations in $S_n$. To each $\sigma$ in $S_n$ we have associated the $n+1$ permutations $\; \sigma,\tau_1\sigma,...,\tau_n\sigma \;$ in $S_{n+1}$. Hence the number of permutations in $S_{n+1}$ is $(n+1)n!=(n+1)!$ Permutation is defined (in the book) as: By permutation of $J_n$, we mean mapping $\sigma:J_n \to J_n$ having the following property. If $\,i,j$ are in $J_n$ and $i \ne j$, then $\sigma(j) \ne \sigma(i)$. Thus the image of permutation consist of $n$ distinct integers $\sigma(1),...,\sigma(n)$, which must be the integers $1,2,...,n$ in different order. I don't understand the following: $(A)$ $\quad\sigma:J_n \to J_n$ means that $\sigma(x)$ is defined only when $x \in J_n$ ?? $(B)\quad$ This means that if $J_n$ is subset of some set $N$, then $\sigma:N\to N$ leaves fixed every $x$ such that $x\in N$ and $x\notin J_n$, but may or may not leave fixed elements $x$ such that $x\in J_n$ so we can consider this permutation of the set $J_n$?? In the proof we see: $\;\tau_k\sigma(n+1)=k$ and $\tau_k\sigma(k)=n+1$ where $k\in J_n$. If $(A)$ is true, then we can't do that (namely, $\sigma(n+1)$) because $n+1\notin J_n$. Do I interpret the definition wrong? Do the fact that $J_n$ is subset of $J_{n+1}$ change anything? As you see I'm completely confused. Some simple example: Let $J_3=\{1,2,3\}\;$ and $\;J_4=\{1,2,3,4\}$. Let $\sigma=\begin{bmatrix} 1 &2 &3 \\ 3 &2 &1 \end{bmatrix}$ and $\tau$ be transposition which interchanges $3$ and $4$. So, $\tau:J_4\to J_4$? What is $\sigma(4)$ or $\tau\sigma(4)$? If $(A)$ is true do I get: $\tau\sigma(J_3)=\begin{bmatrix} 1 &2 &3\\ 4 &2 &1 \end{bmatrix}$ but $\tau\sigma(J_4)$ and $\tau\sigma(4)$ is undefined? If $(B)$ is true I should get: $\tau\sigma(J_4)=\begin{bmatrix} 1 &2 &3 &4\\ 4 &2 &1 &3 \end{bmatrix}$ which seems to be the result needed to make sense out of the proof. • (A) $\sigma:J_n\mapsto J_n$ means that $\sigma(x)$ is only defined when $x\in J_n$, yes. But for (B), this means that you cannot even talk about $\sigma(x)$ for an $x\not\in J_n$; the concept itself doesn't make any sense. It's as though you asked for the square root of blue. – Steven Stadnicki Dec 1 '17 at 2:11 • The book is being really sloppy where it says to consider the case $\sigma'=\tau_i\sigma$, because strictly speaking the domains of the former are two aren't the same (the domain of $\sigma'$ is $J_n$ but the domain of $\tau_i\sigma$ is $J_{n+1}$); it's true that you can use the embedding of $J_n$ into $J_{n+1}$ to consider $\sigma'$ as an element of $S_{n+1}$, but it should be much more explicit about that. – Steven Stadnicki Dec 1 '17 at 2:16 • @Steven Stadnicki That should be $N\to N$ Right ? – Rumata Dec 1 '17 at 2:17 • (in fact, strictly speaking this embedding is necessary even to write $\tau_i\sigma$.) – Steven Stadnicki Dec 1 '17 at 2:17	2019-05-20 02:37:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9670404195785522, "perplexity": 115.10144676820433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00291.warc.gz"}
https://greenemath.com/Precalculus/87/GeometricSequencesLesson.html	Lesson Objectives • Learn how to find the common ratio • Learn how to find the nth term of a geometric sequence • Learn how to evaluate a geometric series What is a Geometric Sequence? In this lesson, we will learn about geometric sequences and series. A geometric sequence, which is also called a geometric progression, is a sequence where each term after the first is obtained by multiplying the preceding term by a fixed nonzero real number known as the common ratio (r). We will use the lowercase r to represent our common ratio. Finding the Common Ratio To find r, our common ratio, we can use the following formula: $$r=\frac{a_{n + 1}}{a_{n}}$$ In other words, we can grab two terms that are next to each other in a geometric sequence and divide the rightmost term by the leftmost term to get the common ratio. Let's look at an example. Example #1: Find the common ratio. $$-4, 24, -144, 864,...$$ Let's pick a1, which is -4, and a2, which is 24, and plug into the formula: $$r=\frac{a_{n + 1}}{a_{n}}$$ $$r=\frac{24}{-4}=-6$$ Finding the nth term Additionally, we may be asked to find the nth term of a geometric sequence or to find the formula for the general term an. To perform this task, we can use the following formula: $$a_{n}=a_{1}\cdot r^{n - 1}$$ Let's look at an example. Example #2: Find a12 and an. $$a_{1}=-1, r=3$$ To find a12, let's plug into our formula: $$a_{n}=a_{1}\cdot r^{n - 1}$$ $$a_{12}=-1 \cdot 3^{11}$$ $$a_{12}=-177{,}147$$ To find the formula for the general term or an, we plug in for a1 and r: $$a_{n}=-1 \cdot 3^{n - 1}$$ $$a_{n}=-3^{n - 1}$$ Geometric Series We know that a series is the sum of the terms of a sequence. When we need to find the sum of the first n terms of a geometric sequence, we can use the following formula: $$S_{n}=\frac{a_{1}(1 - r^n)}{1 - r}$$ Let's look at an example. Example #3: Evaluate each series. $$\sum_{i=1}^{8}2 \cdot (-4)^{i - 1}$$ Let's plug into our formula: $$S_{n}=\frac{a_{1}(1 - r^n)}{1 - r}$$ $$S_{8}=\frac{2(1 - (-4)^8)}{1 - (-4)}$$ $$S_{8}=\frac{2(1 - 65{,}536)}{1 + 4}$$ $$S_{8}=\frac{2 \cdot -65{,}535}{5}$$ $$S_{8}=2 \cdot \frac{-65{,}535}{5}$$ $$S_{8}=2 \cdot -13{,}107$$ $$S_{8}=-26{,}214$$ Skills Check: Example #1 Find the common ratio. $$2, -10, 50, -250,...$$ A $$r=-1$$ B $$r=-5$$ C $$r=-3$$ D $$r=5$$ E $$r=3$$ Example #2 Find a11 and an. $$a_{1}=-1, r=2$$ A $$a_{11}=1024$$ $$a_{n}=2^{n - 1}$$ B $$a_{11}=388$$ $$a_{n}=4^{n - 3}$$ C $$a_{11}=256$$ $$a_{n}=(-2)^{n - 1}$$ D $$a_{11}=-1024$$ $$a_{n}=-2^{n - 1}$$ E $$a_{11}=568$$ $$a_{n}=(-1)^{2n}$$ Example #3 Evaluate each series. $$\sum_{i=1}^{7}(-2)^{i - 1}$$ A $$105$$ B $$18$$ C $$21$$ D $$43$$ E $$33$$	2021-04-17 15:28:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7513121962547302, "perplexity": 357.2565634357056}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00580.warc.gz"}
http://eprints.imtlucca.it/1432/	# Interval bounds for the optimal burn-in times for concave or convex reward functions Foschi, Rachele Interval bounds for the optimal burn-in times for concave or convex reward functions. Journal of applied probability . pp. 1-23. ISSN 0021-9002 (Submitted) Preview PDF Download (338kB) \| Preview ## Abstract An interesting problem in reliability is to determine the optimal burn-in time. In a previous work, the authors studied the solution of such a problem under a particular cost structure. It has been shown there that a key role in the problem is played by a function $\rho$, representing the reward coming from the use of a component in the field. A relevant case in this investigation is the one when $\rho$ is linear. In this paper, we explore further the linear case and use its solutions as a benchmark for determining the locally optimal times when the function $\rho$ is not linear or under a different cost structure. Item Type: Article Burn-in; Bathtub shape; Multiple change points distributions; Reward functions. Q Science > QA Mathematics Economics and Institutional Change Users 19 not found. 27 Nov 2012 13:55 27 Nov 2012 13:59 http://eprints.imtlucca.it/id/eprint/1432 ### Actions (login required) Edit Item	2021-12-01 21:42:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5707862377166748, "perplexity": 1104.6363983927065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00251.warc.gz"}
https://datascience.stackexchange.com/questions/66016/how-to-train-the-machine-so-that-it-can-give-out-of-bound-classes-as-an-output/66018	# How to train the machine so that it can give 'out of bound/classes' as an output for neural network I know I was not able to word the title of the question properly. So I am trying to explain the problem here: Suppose, I built and trained a CNN to identify numbers from 0 to 9. However, when I deployed the CNN, someone gave "#" as an input (anything but 0-9). What can I do to my neural network during training, so that it's output can say that this is none of the training characters? I'd like to present a second example: instead of classifying, suppose we want to do denoising using an autoencoder. Again, the CNN autoencoder is trained for 0 to 9. Now, how can we prepare it so that if someone gives "\$"symbol (it can be anything that is not 0-9) as an input, it'll be able to identify that this symbol is none of what it was trained on? And the CNN autoencoder will be able to give an output according to that? I think the best way would be to augment some data and have an additional output class "unknown". However, if that is not possible or the neural net can not be retrained I would compare the distribution of the outputs of a hidden layer. For the CNN architecture below, calculate the empirical distribution for the outputs of a hidden layer after the flatten layer for the training data (e.g.$$n1$$). Save this distribution and during deployment, calculate the output of $$n1$$ for the test instance. If the output of $$n1$$ corresponds to a highly unlikely value with respect to the empirical distribution of training examples, return "unknown". Else, predict the digit for this instance. Three ideas come to my mind (from simple to complex) 1. Include an additional category for anything which is not a number and train your network on these $$k+1$$ categories. 2. Apply another predictor in the first place which has been trained to differentiate between "number" and "no number". Iff the input is classified as a number you then run your number recognition network. (this approach might make transfer learning easier, i.e. apply an existing model for the first step) 3. Merge the two tasks into one network and make this a multi-task classification, i.e. your network includes layers not only to recognize numbers but also for the binary classification "number vs no number". Since these tasks are closely related the two tasks might benefit from sharing parameters (i.e. use the same features). The paper An Overview of Multi-Task Learning in Deep Neural Networks describes this approach in more detail. (Note that this approach is not identical with the first idea in this list since this one applies two separate classifications while the first one only does a single classification) However, as a disclaimer: I have not tried out the 3rd idea myself but used the second one for a similar problem. • But the "unknown" digits are not in the training data, so none of your options work. Jan 7 '20 at 12:46	2021-10-16 10:12:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5878836512565613, "perplexity": 504.98951603252345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323584554.98/warc/CC-MAIN-20211016074500-20211016104500-00478.warc.gz"}
http://physics.aps.org/articles/large_image/f1/10.1103/Physics.1.21	Figure 1: (Left) Crystal structure of the 1111 $FeAs$ material (after [47]). Fluorine (green) replaces oxygen (gray), donating electrons to the $FeAs$ layers. Other atoms shown are iron (yellow), arsenic (purple), and lanthanum (light blue). (Right) Crystal structure of the 122 $FeAs$ material (after [14]). The parent compound of both materials has iron moments (red arrows) that form a striped antiferromagnetic pattern. Calcium atoms are shown in blue.	2013-05-26 08:54:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5732259154319763, "perplexity": 9073.369735722792}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706794379/warc/CC-MAIN-20130516121954-00075-ip-10-60-113-184.ec2.internal.warc.gz"}
https://www.usefuldesk.com/tamilnadu-10th-question-paper-2014-maths-october/	# Tamilnadu 10th Question Paper 2014 Maths October 0 3412 ## Tamilnadu 10th Question Paper 2014 Maths October Dear 10th/SSLC students, you can download the Tamilnadu 10th public exam question paper of Maths for the year of October 2014 from the below link.	2019-07-16 15:05:38	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9200037717819214, "perplexity": 14535.049229271619}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524568.14/warc/CC-MAIN-20190716135748-20190716161748-00019.warc.gz"}
https://www.maxkoehler.com/notes/2018-11-21-cs231n-3/	• Last time we talked about the challenges of recognition. Recall the semantic gap. • Also: The KNN classifier, the CIFAR-10 classifier. Cross-Validatin, hyperparameters. • The linear classifier as an example of a parametric classifier, where the training data is summarised into the weights matrix $$W$$. It returns class probabilities. We can interpret the linear classifier as learning a template for each class. Also linear decision boundaries between points in high-dimensional space. We have an understanding of the linear classifier, but we don’t really know how to get the correct $$W$$. To do: 1. Define a Loss function that quantifies our unhappiness with the scores across the training data 2. Come up with parameters that efficiently minimize the loss function (optimization). Ie. searching through the space of possible $$W$$ ## Loss function Given a dataset of examples $${(x_i,y_i)}^N_{i=1}$$ Where $$x_i$$ is the $$i$$th image and $$y_i$$ is the corresponsing label. For CIFAR-10, $$x$$ will be images, $$y$$ will be an integer between 0 and 9 to denote the class. $$N$$ examples in the training set. Loss over the dataset is a sum of loss over the examples: $$L = \frac{1}{N}\displaystyle\sum_{i}L_i(f(x_i, W), y_i)$$ We have some prediciotn function $$f$$ that takes in an image and our weight matrix $$W$$. $$L_i$$ takes the predicted scores coming out of $$f$$ and the actual true labels $$y$$ and give some quantitative value of how bad the predicted scores are. The final loss $$L$$ is the average ($$\frac{1}{N}\displaystyle\sum_{i}$$) of all the losses over the dataset. This general setup is generally the same for a lot of different learning models. ## Multiclass SVM (support vector machine) loss We sum up the difference between scores for each incorrect class and the correct class. If the prediction function returns the correct class (with the arbitrary safety margin of $$1$$) the loss is 0. $$L_i = \displaystyle\sum_{j \neq y_i}\text{max}(0, s_j - s_{y_i} + 1)$$, Where $$s = f(x_i, W$$) and $$1$$ is an arbitrary safety margin. This function (where we take max of 0 and something) is also known as a hinge loss because of the shape of the graph when you plot the loss. The equation above could also be written as an if/then (case-based) notation. In English, this loss function says that we are happy when the true class score is a lot higher than all other classes. At initialization $$W$$ is small, so all $$s\approx 1$$. We should expect a loss of $$\text{number of classes} - 1$$ If we went over all classes (instead of only $$j \neq y_i$$) the loss increases by 1. If you did this in practice, you would still find the same classifier, it’s just nicer to aim for a loss of 0. You can also square the hinge function, to get a different loss function. If you’re using a squared loss, very bad results will be very very bad (because it’s now exponential). Using a linear vs a square depends on how much we care about different kinds of errors. Numpy code to comput $$L_i$$ def L_i_vecotrized(x, y, W): scores = W.dot(x) margins = np.maximum(0, scores - scores[y] + 1) margins[y] = 0 # This is to remove the score for the correct class (easier than writing a loop or whatever) loss_i = np.sum(margins) return loss_i $$W$$ that gives $$L=0$$ is not unique. $$2W$$ also gives $$L=0$$. This is problematic. If this loss function is supposed to find the right $$W$$, how is it supposed to choose between these different versions of $$W$$ that all give $$L=0$$? This is because the loss we’ve written down is data loss: Model predictions should match training data. When in the real world, we don’t really care how well the model matches the training data. This can be a source of overfitting, which we solve through Regularization. We normally add a regularization parameter to the loss function that encourages a simple” $$W$$, where simple” depends on the model. $$L = \frac{1}{N}\displaystyle\sum_{i}L_i(f(x_i, W), y_i) + \lambda R(W)$$ This is Occam’s Razor: Among competing hypothesis, the simplest is the best. So the above loss function now has a data loss and a regularization loss, with a hyperparameter $$\lambda$$ which trades off between the two. This is a kind of soft constraint on the model. There’s a lot of different types of regularization: • L2 Regularization penalises the Euclidian norm of the weight vector: $$R(W) = \sum_{k}\sum_{l}W^2_{k,l}$$ • L1 Regularization: • Elastic net (L1 + L2) • Max norm regularization • Dropout (specific to deep learning) • Fancier: Batch normalisation These things all do the same thing: enocourage the model to be simple. ## Softmax Classifier Loss (Mutinomial Logistic Regression). Cross-entropy loss! Recall for the Multiclass SVM we didn’t really give meaning to the scores. Here the scores are unnormalized log probabilities of the classes. $$P(Y=k\|X=x_i) = \frac{e^sk}{\sum_j e^sj}$$ where $$s = f(x_i,W)$$ For a loss function, we want to minimize the negative log likelihoof of the correct class. If we know the image is a cat, the target probability would be 1 for cat and 0 for everything else. $$L_i = -\text{log} P(Y=y_i\|X=x_i)$$ We want to enocourage the probability distribution coming out of the softmax to be the same as the target. We want the true class probability to be close to 1. In practice we: 1. Take the scores out of the softmax classifier (unnormalized log probabilities) 2. exponate (so they’re all positive) 3. Normalize (so they add up to 1) The loss is the minus log of the correct class. To get a 0 loss, you would have to have an inifite score for the correct class and minus infinity for everything else. Minimum is 0, and maximum is $$\infty$$. SVM will get the datapoint over the safety margin, while Softmax will keep pushing forever. In practice, the difference isn’t that big. ### Recap • We have some dataset $$(x, y)$$ • We have a score function (that makes predictions): $$s = f(x, W) = \text{(for example)} Wx$$ • We have a loss function: • Softmax: $$L_i = -\text{log}(\frac{e^{s_{y_i}}}{\sum_j e^s j})$$ • SVM: $$L_i = \sum_{j \neq y_i}\text{max}(0, s_j - s_{y_i} + 1)$$ • The Full Loss is the average of all the losses plus a regularization term: $$\frac{1}{N}\sum_{i=1}^N L_i + R(W)$$ This is a pretty generic view of supervised learning. ## Now: Optimization Imagine you’re walking around this valley. The bottom of the valley is the minimum of the loss function. With complex models, there’s really no hope to find an analytical solution to this. Maybe the stupidest thing you could think of is random search: Just try different values for $$W$$ and see how they do. In practice, it’s better to use the local geometry of the valley (ie. the local gradient/slope). In a one-dimensional function, the slope is the derivative of the function: $$\frac{df(x)}{dx} = \displaystyle\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ You calculate the slope for a section of the curve, then let the length of that section go to 0 and you get the local slope. We need to generalise to multi-dimensions: In multiple dimensions, the gradient is the vector of (partial derivatives) along each dimension. Each element in the gradient vector tells us the slope of the function if we go in that dimension. A lot of deep learning is about computing gardients and using it. We could imagine computing the gradients using finite differences: Change each element in $$W$$ by a small amount, recompute the loss, measure the difference and approximate the gradient that way. This is terrible because it is slow: A deep learning model might have 100s of millions of parameters. Better to use calculus, because we know the gradient is a function of $$W$$. In practice, you use an analytical gradient, but you can use a numerical gradient to debug. This is called a gradient check. Once we know how to compute the gradient, we get to gradient descent, which is just beautiful: while True: weights_gradient = evaluate_gradient(loss_function, data, weights) weight += - step_size * weights_gradient Recall the gradient points in the direction of greatest increase, so we use the minus gradient. We take small steps in the direction of minus gradient until the network converges. The step size (or learning rate) is on of the most hyperparameter to set when you train these models. Because we could either take forever, or walk past the minimum if the learning rate is too small or too big. The above example is very basic - we just make a small step in the direction of minus gradient. There are slightly fancier step rules that work better in practice: • GD with momentum But it’s still the same basic algorithm. ### Stochastic gradient descent In practice, $$N$$ could be very, very large (thus slow). Thus, SGD, where we sample a random subset (mini-batch) of the training set at each training step (32,64,128 instances are common). In code: while True: data_batch = sample_training_data(data, 256) #get 256 examples weights_gradient = evaluate_gradient(loss_fun, data_batch, weights) weights += - step_size * weights_gradient #parameter update [Web demo] ## Aside: Image Features In practice, feeding linear pixel values into a linear classifier doesn’t work so well. Before neural networks, you would compute some feature representation of the image and feed that into the linear classifier. The idea here is that non-linear datasets might become linearly seperable through some smart feature representation. ### Examples feature representation • Colour histogram: This tells you globally what colours are in the image • Histogram of oriented gradients: (essentially some kind of edge detection in small sub-regions of the image) • Bag of Words: An idea from NLP where you count up the number of each word in a paragraph. For images, you define a vocabluary of visual words. 1. Extract random patches from images, and cluster these patches to form a codebook 2. Encode the image by trying to say how often each visual word appears in the image Today, this whole long feature extraction pipeline is replaced by Convolutional networks. Instead of writing down the features before training, we will let the network learn them from the data. Next: Neural networks, backpropagation.	2022-01-21 23:00:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.665682315826416, "perplexity": 13942.830624944854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00647.warc.gz"}
http://mathhelpforum.com/calculus/113506-stumped-integral.html	Math Help - Stumped by an Integral 1. Stumped by an Integral Can anyone solve this? $\int_0^5\sqrt{1+(3t^2-12t+9)^2}dt$ I'm not sure where to begin on this one. Any suggestions? It seems like one of those functions that you just can't evaluate by hand. However, I'm not very clever when integrating, so maybe it's just me Would it be easier to do a Riemann sum? 2. I don't think this integral can be calculated using elementary methods ... there may be a clever trig or algebraic substitution, but I'm not seeing it. my TI-89 sez 29.0708... 3. http://integrals.wolfram.com/index.jsp?expr=sqrt(9+(x^2-4+x%2B3)^2%2B1)&random=false Mathematica disagrees with your simple substitution 4. Originally Posted by Silent Soliloquy http://integrals.wolfram.com/index.jsp?expr=sqrt(9+(x^2-4+x%2B3)^2%2B1)&random=false Mathematica disagrees with your simple substitution That's not what I wrote originally actually, but I was completely wrong nonetheless. My trig idea doesn't work with the 4th power. Even if I had factored correctly I didn't account for the problem of dtheta, which added more than I said originally. Yes, I messed up. 5. It seems like Simpsons rule is going to be my best choice.	2014-10-02 05:48:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9139848351478577, "perplexity": 703.8248797118604}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663711.39/warc/CC-MAIN-20140930004103-00000-ip-10-234-18-248.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/sequence-and-series-terminology.797048/	# Sequence and Series Terminology First, I would like to clear up notation and the definition for sequences. What exactly is a sequence? I read somewhere that it is defined as a function ##f: \mathbb{N} \to \mathbb{R}##. But if this is the case, why do we only define functions based on the range of the function, e.g., ##\left \{ 1, 4, 9, 16... \right \}## (which we regard as "the sequence")? We define sequences with the notation ##\left \{ a_{n} \right \}_{0}^{\infty}## too, but what does this mean in terms of the functions concept? How does this specify a function? In addition, what is the nth term's relation to the function concept, or in other words, what is the analogue of the nth term for sequences in functions? Finally, what exactly to the terms "partial sums, series, finite series, infinite series," mean? It seems as though they are mostly for the same concept. BiGyElLoWhAt Gold Member I like the function definition. The reason we take integers is, honestly, for simplicity. You plug in 1, and you get your first term, you plug in 2, and you get your second term, etc. If you like, I suppose you could do 1/2 integers, and 1/2 would be your first term, 1 would be your second term, etc. You could also do multiples of pi/6. You plug in pi/6 and you get your first term, pi/3 and get your second term, etc. The function would have to be defined accordingly though. IMHO, just stick with the integers, because that is consistent with other definitions. If you were going to define your series to be ##\{ a_n\}_0^\infty\ \text{where}\ a_n := f(n)## and you took n to be something other than integers, what would your pi'th term be? You see how it gets conceptually sketchy there? Partial sums are when you do a sum over a well defined range, i.e. a finite series. Ex. ##\sum\limits_{k=1}^{10} k## This is a finite series/partial sum. The equivalent infinite series would be ##\sum\limits_{k=1}^{\infty} k = -\frac{1}{12}## ;) micromass Staff Emeritus	2022-06-30 18:50:24	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9526551365852356, "perplexity": 559.4247793124115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00721.warc.gz"}
http://blog.heroesofprogramming.com/heroes/master-of-variables-part-2-primitives/	In the previous article on variables we familiarized with what variables in JavaScript are and how they are defined. As already mentioned, there are several data types in JavaScript: • Number • String • Boolean • Undefined • Object • Function All types except objects and functions are immutable, which means that they cannot be changed. Values of these types are called “primitive values,” that is, simple, elementary. These values represent a simple, primal value, such as a number, some characters, true/false or null/undefined. By “it cannot be changed” I mean that once you have declared such a variable, you cannot modify the value it holds, you can only assign another value to that variable. var nr = 2; nr = 3; // Here the 2 value is thrown away and the 3 value is assigned. var str1 = 'I want an ice-cream.'; var str2 = str1; // str2 = 'I want an ice-cream.' str1 = 'I want to learn faster.'; console.log(str1); // "I want to learn faster." console.log(str2); // "I want an ice-cream." Referring to a primitive value transfers its value, makes a copy of it so there are two different values with the same content in them. var earthOrder = 3; var earthSatellites = earthOrder; // The value of "earthOrder" was copied earthSatellites = 1; console.log(earthOrder); // 3 console.log(earthSatellites); // 1 Here’s a visual explanation to what happens when you work with primitives. When answer is defined and assigned the value 42, it is being stored in memory at a specific memory address. Then, when we assign the value of the answer variable to the newly declared number variable, the value is copied into a new memory address that refers to the number variable. We have 2 values that represent the number 42 in two different memory addresses that do not have any connection. Next, even if we define another variable with the same 42 value, it will still have its own unique address and it won’t be connected to any other variables. In other words, when the Master of Variables creates a variable with a primitive value assigned to it, that value is “inside the box’s variable” and whenever you assign that variable to another variable, the contents of the first box are copied into the second one. For instance, you can’t have a number 2 in two boxes simultaneously. When you create a variable with var x = 2;, the x box contains value 2. Then, when you write var y = x;, the Master of Variables copies the value of x variable into the y variable. In the end you have two variables with two different values of 2. And when you change the value in one variable, it doesn’t affect the value in the other one. The situation with referring to values is different when working with objects though. We will address this example after several sections. As for now, let us discuss all data types in more detail. ## Number There are several types of numbers we can use in JavaScript, although inside JavaScript all numbers are implemented as double-precision 64-bit binary format. This scary expression means that actually, a number in JavaScript is represented in a finite number of bits, namely 64 of them (8 bytes). The structure, as seen above, is as follows: • 52 bits to store the number itself (significant digits) • 11 bits to store the exponent value • 1 bit to store the sign of the number This means that there are only 52 bits to represent a number in JavaScript, which gives 15 significant digits precision. That is, you can represent a number that has 15 digits. The 11 exponent bits allow the representation of numbers between $10^{-308}$ and $10^{308}$. The sign bit stores either a 0, which represents a positive number, or a 1, which represents a negative value. ### Integer An integer is just that - a positive or negative whole number. In order to represent it, you don’t need to be a rocket scientist. Just assign the number you want to a variable and that’s it. var ultimateQuestionOfLifeAnswer = 42; var int2 = -8; var int3 = 0; var int4 = +92; var int5 = - 3; // -3 All the above variables are valid integers. One interesting particular case is the following var negativeZero = -0; console.log(negativeZero); // will output -0 Can you explain why it happens? As you remember, each number has encoded the sign within it. So, when we write -0, we encode the sign into the binary sequence. The reasons behind this we will discuss later on. ### Fractional numbers You can use fractional numbers in JavaScript as well, which are written using a dot. These numbers are called float numbers, because of the floating point. var pi = 3.14; var float1 = -1.2; var float2 = 0.001; You can also skip the first zero in case of a number between 0 and 1 and still get a valid float number. var float3 = .3; // equals to 0.3 var float4 = .00042; // equals to 0.00042 As mentioned in the beginning of the Number section, all numbers in JavaScript internally are represented as floating point numbers, even integers. There is no Integer type in JavaScript and the lack of the . (dot) in the definition of an integer is for simple commodity, as we are used to it. Another example that shows there is a single type for numbers is this one: var x = 1.0; console.log(x); // 1 (and not 1.0) ### Scientific notation In case you would like to define a huge number, JavaScript support the scientific notation. var hugeNumber = 1e10; // 10000000000 var smallNumber = 23e-7; // 0.0000023 These numbers are transformed in “regular” integers or floats. They are just a convenience way to save time by not typing several zeros. ### Special numbers Besides all these numbers, JavaScript has two more special “numbers.” The first one can be obtained when you want to represent a number that is bigger than the largest possible number you can represent in JavaScript; var infinity = Infinity; var negativeInfinity = -Infinity; Note that Infinity - 1 is still Infinity and so on. Also, remember the negative zero example? Now, there are some cases where you can have this: 1 / -0; // -Infinity The second special number is quite the opposite of a number and is called a “Not a Number.” It is represented using NaN. This value is returned whenever an arithmetic operation is made that cannot return a valid result. 'a' / 'b'; // NaN Infinity - Infinity // NaN ### The Number data type All these numbers, although different (integers, floats, in base 16) are of the same “number” data type. They are all numbers. typeof 2016; // "number" typeof 12.3; // "number" typeof 1e10; // "number" typeof Infinity; // "number" typeof NaN; // "number" ## String Besides numbers, we need to represent and operate with some text in our programs. Thus, we have the String variable type. A string is a finite sequence of ordered characters. ### Defining a string In order to define a string variable, enclose the data you want in single ' or double " quotation marks. var myName = 'Iulian'; var earthAge = "4.54 ± 0.05 billion years"; var emptyString = ''; var frenchLanguage = 'Français'; A string variable, as you see, can contain any type of character and can start with any character. Also, there is no difference between the single and double quotation marks in defining a string, so use whatever you like more. But how do we use a ' or " character in a string themselves? var greeting1 = 'It's showtime! Meet the "Jack the Crusher"!'; var greeting2 = "It's showtime! Meet the "Jack the Crusher"!"; In the above cases, the code will throw a: SyntaxError: Unexpected identifier. In order to use the quotation character in our strings, we need to “escape” them, using the escape character \. var greeting1 = 'It\'s showtime! Meet "Jack the Crusher"!'; // It's showtime! Meet "Jack the Crusher"! var greeting2 = "It's showtime! Meet \"Jack the Crusher\"!"; // It's showtime! Meet "Jack the Crusher"! // The ' character inside a string defined in "" doesn't require escaping though. var great = "I'm great!"; // The same is with " character inside a string defined in ''. var question = 'Did you read "The Master and Margarita" during your vacantion?'; ### The escape character The backslash \ character that we saw earlier is special in JavaScript strings. Whenever the JavaScript engine encounters this character, this means that the character after it should be treated differently. In the case with ' and " characters, this means that these characters should not terminate a string definition, but be included in the string itself. Similarly, when we want to include a backslash character, we need to “escape” it as well. var string = 'This is a string with a \\ character.' // This is a string with a \ character. There are cases, when the next character isn’t displayed on the screen, but serves another purpose. For instance, whenever you want your string to spread along two lines of code, you should use the universal newline character \n. var twoLines = 'I am reading a book on JavaScript\nand learn about strings.'; // I am reading a book on JavaScript Below is the list of special characters that can be used in a string. Code Output ' single quote " double quote \ backslash \n new line \r carriage return \t tab \b backspace \f form feed ### String properties In order to obtain the length of a string, just access its length property. var human = 'I am a human.'; var robot = 'I am \n a robot.'; human.length; // 13 robot.length; // 15 (a space and the '\n' character that counts as 1 character) Sometimes you will need to get the character at a specific position in the string. To do that, use the square brackets with the position of the character you want to get. Note that in JavaScript the ordering of the characters start at zero. Thus, the first element in the string is obtained at the 0th index. var greeting = 'Hello! How are you?'; console.log(greeting[0]); // 'H' console.log(greeting[5]); // '!' ### The String data type All strings, as expected, are of the same “string” data type. typeof 'abc'; // "string" typeof '\n'; // "string" ## Boolean As you will see, you will often need a value that can only have one of two values. These are the yes\/no, on\/off type of values that we talked about at the beginning of the Variables chapter. In JavaScript there is a special data type that represents such values - the Boolean data type. It can only take the values true or false. var truthyValue = true; var falsyValue = false; Despite its inherent simplicity, boolean variables have a tremendous importance in programming. As they are themselves a form of binary data, other types of variables at different times throughout code execution are being converted to these simple true or false values in order to do stuff. The type of the true and false variables is “boolean”. typeof true; // "boolean" typeof false; // "boolean" ## Undefined Earlier in this chapter, when we saw that a variable can be defined like this: // Defining a variable var myVariable; // Defining a variable and assigning a value to it var earthDiameter = 12756.2; // in km It is clear what value we will receive when we access the earthDiameter variable. But what we will get if we will access myVariable? There is a special value for the variables that have no value in JavaScript. It is called undefined. var myVariable; console.log(myVariable); // "undefined" undefined is a separate data type in JavaScript, meaing that you can use typeof operator to find out whether a variable’s type is undefined. typeof myVariable; // undefined If you want to “empty” a variable, just set it’s value to undefined. var randomVar = 'abc'; typeof randomVar; // "string"; randomVar = undefined; typeof randomVar; // "undefined"; ## Null Another special value in JavaScript is the null value. Unlike the undefined value, which means that there is no value set for the variable, the null value means that the value exists, but it is “empty”, it has “nothing” in it, it is null. An interesting fact is the following piece of code: typeof null; // "object" This is now considered a bug in JavaScript. The type of the null value should be “null”, but this behavior is not corrected so as to keep existing code working. If the people that work on the JavaScript language itself would make this change and return “null” instead of “object” as the type of the null value, it would break a lot of existing applications that rely on the “object” response. But this does not mean that it is ok to do that (i.e. not changing something because there is code that is already written that should be changed). It is called backward compatibility of the language versions and it is a complex process that we won’t discuss in this book. In the last article of this series we will discuss about one of the most important concepts in JavaScript — objects. See you there!	2018-03-21 10:50:06	{"extraction_info": {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30048996210098267, "perplexity": 1395.8663715138364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647612.53/warc/CC-MAIN-20180321102234-20180321122234-00345.warc.gz"}
https://popl18.sigplan.org/details/POPL-2018-papers/18/Decidability-of-Conversion-for-Type-Theory-in-Type-Theory	Thu 11 Jan 2018 16:15 - 16:40 at Watercourt - Dependent Types Chair(s): Karl Crary Type theory should be able to handle its own meta-theory, both to justify its foundational claims and to obtain a verified implementation. At the core of a type checker for intensional type theory lies an algorithm to check equality of types, or in other words, to check whether two types are convertible. We have formalized in Agda a practical conversion checking algorithm for a dependent type theory with one universe à la Russell, natural numbers, and $\eta$-equality for $\Pi$ types. We prove the algorithm correct via a Kripke logical relation parameterized by a suitable notion of equivalence of terms. We then instantiate the parameterized fundamental lemma twice: once to obtain canonicity and injectivity of type formers, and once again to prove the completeness of the algorithm. Our proof relies on inductive-recursive definitions, but not on the uniqueness of identity proofs. Thus, it is valid in variants of intensional Martin-Löf Type Theory as long as they support induction-recursion, for instance, Extensional, Observational, or Homotopy Type Theory. #### Thu 11 Jan 15:50 - 16:40: Research Papers - Dependent Types at Watercourt Chair(s): Karl CraryCarnegie Mellon University 15:50 - 16:15Talk Up-to Techniques Using Sized TypesNils Anders DanielssonUniversity of Gothenburg, Chalmers University of Technology 16:15 - 16:40Talk Decidability of Conversion for Type Theory in Type TheoryAndreas AbelGothenburg University, Joakim ÖhmanIMDEA Software Institute, Andrea VezzosiChalmers University of Technology	2019-11-20 01:37:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.692179262638092, "perplexity": 1964.4703448245784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670389.25/warc/CC-MAIN-20191120010059-20191120034059-00508.warc.gz"}
http://www.ams.org/bookstore?fn=20&arg1=gsmseries&ikey=GSM-152	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education An Introduction to Extremal Kähler Metrics Gábor Székelyhidi, University of Notre Dame, IN SEARCH THIS BOOK: 2014; 192 pp; hardcover Volume: 152 ISBN-10: 1-4704-1047-8 ISBN-13: 978-1-4704-1047-6 List Price: US$57 Member Price: US$45.60 Order Code: GSM/152 Comparison Theorems in Riemannian Geometry - Jeff Cheeger and David G Ebin Variational Problems in Geometry - Seiki Nishikawa The Geometrization Conjecture - John Morgan and Gang Tian A basic problem in differential geometry is to find canonical metrics on manifolds. The best known example of this is the classical uniformization theorem for Riemann surfaces. Extremal metrics were introduced by Calabi as an attempt at finding a higher-dimensional generalization of this result, in the setting of Kähler geometry. This book gives an introduction to the study of extremal Kähler metrics and in particular to the conjectural picture relating the existence of extremal metrics on projective manifolds to the stability of the underlying manifold in the sense of algebraic geometry. The book addresses some of the basic ideas on both the analytic and the algebraic sides of this picture. An overview is given of much of the necessary background material, such as basic Kähler geometry, moment maps, and geometric invariant theory. Beyond the basic definitions and properties of extremal metrics, several highlights of the theory are discussed at a level accessible to graduate students: Yau's theorem on the existence of Kähler-Einstein metrics, the Bergman kernel expansion due to Tian, Donaldson's lower bound for the Calabi energy, and Arezzo-Pacard's existence theorem for constant scalar curvature Kähler metrics on blow-ups. Graduate students and research mathematicians interested in geometric analysis and Kähler geometry. Reviews "This is an important book, in a rapidly-developing area, that brings the specialist or graduate student working on Kähler geometry to the frontiers of today research. It is not a self-contained textbook. The pre-requisites in geometric invariant theory, for example, would require some devotion from a potential reader grounded on Riemannian geometry; vice-versa, a reader brought-up in algebraic geometry would have to make an effort to follow the part on analysis or differential geometry. The rewards for these efforts justify everything: the book is well organized, and when it sketches an argument, there are precise pointers to the literature for full details." -- MAA Reviews "Very well written, the book provides a survey of extremal ahler metrics and it promotes to tackle other topics." -- Zentralblatt Math	2015-07-29 02:41:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2486584484577179, "perplexity": 1042.7496383588111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042985647.51/warc/CC-MAIN-20150728002305-00089-ip-10-236-191-2.ec2.internal.warc.gz"}