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https://telescoper.wordpress.com/2010/11/23/bayes-and-hi-theorem/ | Bayes and his Theorem
My earlier post on Bayesian probability seems to have generated quite a lot of readers, so this lunchtime I thought I’d add a little bit of background. The previous discussion started from the result
$P(B|AC) = K^{-1}P(B|C)P(A|BC) = K^{-1} P(AB|C)$
where
$K=P(A|C).$
Although this is called Bayes’ theorem, the general form of it as stated here was actually first written down, not by Bayes but by Laplace. What Bayes’ did was derive the special case of this formula for “inverting” the binomial distribution. This distribution gives the probability of x successes in n independent “trials” each having the same probability of success, p; each “trial” has only two possible outcomes (“success” or “failure”). Trials like this are usually called Bernoulli trials, after Daniel Bernoulli. If we ask the question “what is the probability of exactly x successes from the possible n?”, the answer is given by the binomial distribution:
$P_n(x|n,p)= C(n,x) p^x (1-p)^{n-x}$
where
$C(n,x)= n!/x!(n-x)!$
is the number of distinct combinations of x objects that can be drawn from a pool of n.
You can probably see immediately how this arises. The probability of x consecutive successes is p multiplied by itself x times, or px. The probability of (n-x) successive failures is similarly (1-p)n-x. The last two terms basically therefore tell us the probability that we have exactly x successes (since there must be n-x failures). The combinatorial factor in front takes account of the fact that the ordering of successes and failures doesn’t matter.
The binomial distribution applies, for example, to repeated tosses of a coin, in which case p is taken to be 0.5 for a fair coin. A biased coin might have a different value of p, but as long as the tosses are independent the formula still applies. The binomial distribution also applies to problems involving drawing balls from urns: it works exactly if the balls are replaced in the urn after each draw, but it also applies approximately without replacement, as long as the number of draws is much smaller than the number of balls in the urn. I leave it as an exercise to calculate the expectation value of the binomial distribution, but the result is not surprising: E(X)=np. If you toss a fair coin ten times the expectation value for the number of heads is 10 times 0.5, which is five. No surprise there. After another bit of maths, the variance of the distribution can also be found. It is np(1-p).
So this gives us the probability of x given a fixed value of p. Bayes was interested in the inverse of this result, the probability of p given x. In other words, Bayes was interested in the answer to the question “If I perform n independent trials and get x successes, what is the probability distribution of p?”. This is a classic example of inverse reasoning. He got the correct answer, eventually, but by very convoluted reasoning. In my opinion it is quite difficult to justify the name Bayes’ theorem based on what he actually did, although Laplace did specifically acknowledge this contribution when he derived the general result later, which is no doubt why the theorem is always named in Bayes’ honour.
This is not the only example in science where the wrong person’s name is attached to a result or discovery. In fact, it is almost a law of Nature that any theorem that has a name has the wrong name. I propose that this observation should henceforth be known as Coles’ Law.
So who was the mysterious mathematician behind this result? Thomas Bayes was born in 1702, son of Joshua Bayes, who was a Fellow of the Royal Society (FRS) and one of the very first nonconformist ministers to be ordained in England. Thomas was himself ordained and for a while worked with his father in the Presbyterian Meeting House in Leather Lane, near Holborn in London. In 1720 he was a minister in Tunbridge Wells, in Kent. He retired from the church in 1752 and died in 1761. Thomas Bayes didn’t publish a single paper on mathematics in his own name during his lifetime but despite this was elected a Fellow of the Royal Society (FRS) in 1742. Presumably he had Friends of the Right Sort. He did however write a paper on fluxions in 1736, which was published anonymously. This was probably the grounds on which he was elected an FRS.
The paper containing the theorem that now bears his name was published posthumously in the Philosophical Transactions of the Royal Society of London in 1764.
P.S. I understand that the authenticity of the picture is open to question. Whoever it actually is, he looks to me a bit like Laurence Olivier…
11 Responses to “Bayes and his Theorem”
1. Bryn Jones Says:
The Royal Society is providing free access to electronic versions of its journals until the end of this month. Readers of this blog might like to look at Thomas Bayes’s two posthumous publications in the Philosophical Transactions.
The first is a short paper about series. The other is the paper about statistics communicated by Richard Price. (The statistics paper may be accessible on a long-term basis because it is one of the Royal Society’s Trailblazing papers the society provides access to as part of its 350th anniversary celebrations.)
Incidentally, both Thomas Bayes and Richard Price were buried in the Bunhill Fields Cemetery in London and their tombs can be seen there today.
2. Steve Warren Says:
You may be remembered in history as the discoverer of coleslaw, but you weren’t the first.
• Anton Garrett Says:
For years I thought it was “cold slaw” because it was served cold. A good job I never asked for warm slaw.
3. telescoper Says:
My surname, in Spanish, means “Cabbages”. So it was probably one of my ancestors who invented the chopped variety.
4. Anton Garrett Says:
Thomas Bayes is now known to have gone to Edinburgh University, where his name appears in the records. He was barred from English universities because his nonconformist family did not have him baptised in the Church of England. (Charles Darwin’s nonconformist family covered their bets by having baby Charles baptised in the CoE, although perhaps they believed it didn’t count as a baptism since Charles had no say in it. Tist is why he was able to go to Christ’s College, Cambridge.)
5. “Cole” is an old English word for cabbage, which survives in “cole slaw”. The German word is “Kohl”. (Somehow, I don’t see PM or President Cabbage being a realistic possibility. 🙂 )
Note that Old King Cole is unrelated (etymologically). Of course, this discussion could cause Peter to post a clip of
Nat “King” Cole
(guess what his real surname is).
To remind people to pay attention to spelling when they hear words, we’ll close with the Quote of the Day:
It’s important to pay close attention in school. For years I thought that
bears masturbated all winter.
—Damon R. Milhem
6. Of course, this discussion could cause Peter to post a clip of
Nat King Cole
(giess what his real surname is).
7. Of course, this discussion could cause Peter to post a clip of
Nat King Cole
(giess what his real surname is).
The first typo was my fault; the extra linebreaks in the second attempt
(tested again here) appear to be a new “feature”.
8. telescoper Says:
The noun “cole” can be found in English dictionaries as a generic name for plants of the cabbage family. It is related to the German kohl and scottish kail or kale. These are all derived from the latin word colis (or caulis) meaning a stem, which is also the root of the word cauliflower.
The surname “Cole” and the variant “Coles” are fairly common in England and Wales, but are not related to the latin word for cabbage. Both are diminutives of the name “Nicholas”.
9. […] I posted a little piece about Bayesian probability. That one and the others that followed it (here and here) proved to be surprisingly popular so I’ve been planning to add a few more posts […]
10. It already has a popular name: Stigler’s law of eponymy. | 2016-09-28 22:12:05 | {"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": 4, "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": 0.6193313002586365, "perplexity": 1089.0032368849284}, "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-2016-40/segments/1474738661768.10/warc/CC-MAIN-20160924173741-00020-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://docs.microej.com/en/latest/ApplicationDeveloperGuide/testsuiteEngine.html | # MicroEJ Test Suite Engine¶
## Introduction¶
The MicroEJ Test Suite Engine is a generic tool made for validating any development project using automatic testing.
This section details advanced configuration for users who wish to integrate custom test suites in their build flow.
The MicroEJ Test Suite Engine allows the user to test any kind of projects within the configuration of a generic Ant file.
The MicroEJ Test Suite Engine is already pre-configured for running test suites on a MicroEJ Platform (either on Simulator or on Device).
## Using the MicroEJ Test Suite Ant Tasks¶
Multiple Ant tasks are available in the testsuite-engine.jar provided in the Build Kit:
• testsuite allows the user to run a given test suite and to retrieve an XML report document in a JUnit format.
• javaTestsuite is a subtask of the testsuite task, used to run a specialized test suite for Java (will only run Java classes).
• htmlReport is a task which will generate an HTML report from a list of JUnit report files.
### The testsuite Task¶
The following attributes are mandatory:
testsuite task mandatory attributes
Attribute Name Description
outputDir The output folder of the test suite. The final report will be generated at [outputDir]/[label]/[reportName].xml, see the testsuiteReportFileProperty and testsuiteReportDirProperty attributes.
harnessScript The harness script must be an Ant script and it is the script which will be called for each test by the test suite engine. It is called with a basedir located at output location of the current test.
The test suite engine provides the following properties to the harness script giving all the informations to start the test:
harnessScript properties
Attribute Name Description
testsuite.test.name The output name of the current test in the report. Default value is the relative path of the test. It can be manually set by the user. More details on the output name are available in the section Specific Custom Properties.
testsuite.test.path The current test absolute path in the filesystem.
testsuite.test.properties The absolute path to the custom properties of the current test (see the property customPropertiesExtension)
testsuite.common.properties The absolute path to the common properties of all the tests (see the property commonProperties)
testsuite.report.dir The absolute path to the directory of the final report.
The following attributes are optional:
testsuite task optional attributes
Attribute Name Description Default value
timeOut The time in seconds before any test is considerated as unknown. Set it to 0 to disable the time-out. 60
verboseLevel The required level to output messages from the test suite. Can be one of those values: error, warning, info, verbose, debug. info
reportName The final report name (without extension). testsuite-report
customPropertiesExtension The extension of the custom properties for each test. For instance, if it is set to .options, a test named xxx/Test1.class will be associated with xxx/Test1.options. If a file exists for a test, the property testsuite.test.properties is set with its absolute path and given to the harnessScript. If the test path references a directory, then the custom properties path is the concatenation of the test path and the customPropertiesExtension value. .properties
commonProperties The properties to apply to every test of the test suite. Those options might be overridden by the custom properties of each test. If this option is set and the file exists, the property testsuite.common.properties is set to the absolute path of the harnessScript file. no common properties
label The build label. timestamp of when the test suite was invoked.
productName The name of the current tested product. TestSuite
jvm The location of your Java VM to start the test suite (the harnessScript is called as is: [jvm] [...] -buildfile [harnessScript]). java.home location if the property is set, java otherwise.
jvmargs The arguments to pass to the Java VM started for each test. None.
testsuiteReportFileProperty The name of the Ant property in which the path of the final report is stored. Path is [outputDir]/[label]/[reportName].xml testsuite.report.file
testsuiteReportDirProperty The name of the Ant property in which is store the path of the directory of the final report. Path is [outputDir]/[label]. testsuite.report.dir
testsuiteResultProperty The name of the Ant property in which you want to have the result of the test suite (true or false), depending if every tests successfully passed the test suite or not. Ignored tests do not affect this result. None
Finally, you have to give as nested element the path containing the tests.
testsuite task nested elements
Element Name Description
testPath Containing all the file of the tests which will be launched by the test suite.
testIgnoredPath (optional) Any test in the intersection between testIgnoredPath and testPath will be executed by the test suite, but will not appear in the JUnit final report. It will still generate a JUnit report for each test, which will allow the HTML report to let them appears as “ignored” if it is generated. Mostly used for known bugs which are not considered as failure but still relevant enough to appears on the HTML report.
Example of test suite task invocation
<!-- Launch the testusite engine -->
<testsuite:testsuite
timeOut="${microej.kf.testsuite.timeout}" outputDir="${target.test.xml}/testkf"
harnessScript="${com.is2t.easyant.plugins#microej-kf-testsuite.microej-kf-testsuite-harness-jpf-emb.xml.file}" commonProperties="${microej.kf.launch.propertyfile}"
testsuiteResultProperty="testkf.result"
testsuiteReportDirProperty="testkf.testsuite.report.dir"
productName="${module.name} testkf" jvmArgs="${microej.kf.testsuite.jvmArgs}"
lockPort="${microej.kf.testsuite.lockPort}" verboseLevel="${testkf.verbose.level}"
>
<testPath refid="target.testkf.path"/>
</testsuite:testsuite>
### The javaTestsuite Task¶
This task extends the testsuite task, specializing the test suite to only start real Java class. This task retrieves the classname of the tests from the classfile and provides new properties to the harness script:
javaTestsuite task properties
Property Name Description
testsuite.test.class The classname of the current test. The value of the property testsuite.test.name is also set to the classname of the current test.
testsuite.test.classpath The classpath of the current test.
<!-- Launch test suite -->
<testsuite:javaTestsuite
verboseLevel="${microej.testsuite.verboseLevel}" timeOut="${microej.testsuite.timeout}"
outputDir="${target.test.xml}/@{prefix}" harnessScript="${harness.file}"
commonProperties="${microej.launch.propertyfile}" testsuiteResultProperty="@{prefix}.result" testsuiteReportDirProperty="@{prefix}.testsuite.report.dir" productName="${module.name} @{prefix}"
jvmArgs="${microej.testsuite.jvmArgs}" lockPort="${microej.testsuite.lockPort}"
retryCount="${microej.testsuite.retry.count}" retryIf="${microej.testsuite.retry.if}"
retryUnless="${microej.testsuite.retry.unless}" > <testPath refid="target.@{prefix}.path"/> <testIgnoredPath refid="tests.@{prefix}.ignored.path" /> </testsuite:javaTestsuite> ### The htmlReport Task¶ This task allow the user to transform a given path containing a sample of JUnit reports to an HTML detailed report. Here is the attributes to fill: • A nested fileset element containing all the JUnit reports of each test. Take care to exclude the final JUnit report generated by the test suite. • A nested element report: • format: The format of the generated HTML report. Must be noframes or frames. When noframes format is choosen, a standalone HTML file is generated. • todir: The output folder of your HTML report. • The report tag accepts the nested tag param with name and expression attributes. These tags can pass XSL parameters to the stylesheet. The built-in stylesheets support the following parameters: • PRODUCT: the product name that is displayed in the title of the HTML report. • TITLE: the comment that is displayed in the title of the HTML report. Note It is advised to set the format to noframes if your test suite is not a Java test suite. If the format is set to frames, with a non-Java MicroEJ Test Suite, the name of the links will not be relevant because of the non-existency of packages. Example of htmlReport task invocation <!-- Generate HTML report --> <testsuite:htmlReport> <fileset dir="${@{prefix}.testsuite.report.dir}">
<include name="**/*.xml"/> <!-- include unary reports -->
<exclude name="**/bin/**/*.xml"/> <!-- exclude test bin files -->
<exclude name="*.xml"/> <!-- exclude global report -->
</fileset>
<report format="noframes" todir="\${target.test.html}/@{prefix}"/>
</testsuite:htmlReport>
## Using the Trace Analyzer¶
This section will shortly explains how to use the Trace Analyzer. The MicroEJ Test Suite comes with an archive containing the Trace Analyzer which can be used to analyze the output trace of an application. It can be used from different forms;
• The FileTraceAnalyzer will analyze a file and research for the given tags, failing if the success tag is not found.
• The SerialTraceAnalyzer will analyze the data from a serial connection.
Here is the common options to all TraceAnalyzer tasks:
• successTag: the regular expression which is synonym of success when found (by default .*PASSED.*).
• failureTag: the regular expression which is synonym of failure when found (by default .*FAILED.*).
• verboseLevel: int value between 0 and 9 to define the verbose level.
• waitingTimeAfterSuccess: waiting time (in s) after success before closing the stream (by default 5).
• noActivityTimeout: timeout (in s) with no activity on the stream before closing the stream. Set it to 0 to disable timeout (default value is 0).
• stopEOFReached: boolean value. Set to true to stop analyzing when input stream EOF is reached. If false, continue until timeout is reached (by default false).
• onlyPrintableCharacters: boolean value. Set to true to only dump ASCII printable characters (by default false).
Here is the specific options of the FileTraceAnalyzer task:
• traceFile: path to the file to analyze.
Here is the specific options of the SerialTraceAnalyzer task:
• port: the comm port to open.
• baudrate: serial baudrate (by default 9600).
• databits: databits (5|6|7|8) (by default 8).
• stopBits: stopbits (0|1|3 for (1_5)) (by default 1).
• parity: none | odd | event (by default none).
## Appendix¶
The goal of this section is to explain some tips and tricks that might be useful in your usage of the test suite engine.
### Specific Custom Properties¶
Some custom properties are specifics and retrieved from the test suite engine in the custom properties file of a test.
• The testsuite.test.name property is the output name of the current test. Here are the steps to compute the output name of a test:
• If the custom properties are enabled and a property named testsuite.test.name is find on the corresponding file, then the output name of the current test will be set to it.
• Otherwise, if the running MicroEJ Test Suite is a Java test suite, the output name is set to the class name of the test.
• Otherwise, from the path containing all the tests, a common prefix will be retrieved. The output name will be set to the relative path of the current test from this common prefix. If the common prefix equals the name of the test, then the output name will be set to the name of the test.
• Finally, if multiples tests have the same output name, then the current name will be followed by _XXX, an underscore and an integer.
• The testsuite.test.timeout property allow the user to redefine the time out for each test. If it is negative or not an integer, then global timeout defined for the MicroEJ Test Suite is used. | 2021-10-16 06:03:46 | {"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": 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": 0.26926159858703613, "perplexity": 3266.692339823195}, "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/1634323583423.96/warc/CC-MAIN-20211016043926-20211016073926-00486.warc.gz"} |
https://www.biostars.org/p/57152/ | T-Test In R On Microarray Data
3
1
Entering edit mode
10.4 years ago
Diana ▴ 900
Hello everyone,
I'm trying to do a simple t-test on my microarray sample in R. My sample looks like this:
gene_id gene sample_1 value_1 sample_2 value_2
XLOC_000001 LOC425783 Renal 20.8152 Heart 14.0945
XLOC_000002 GOLGB1 Renal 10.488 Heart 8.89434
So the t-test is between sample 1 and sample 2 and my code looks like this:
ttestfun = function(x) t.test(x[4], x[6])$p.value p.value = apply(expression_data, 1, ttestfun) It gives me the following error: Error in t.test.default(x[6], x[8]) : not enough 'x' observations In addition: Warning message: In mean.default(x) : argument is not numeric or logical: returning NA What am I doing wrong? Please help. Many thanks. r microarray • 15k views ADD COMMENT 8 Entering edit mode Nag your supervisor to provide some more arrays and allow you to run the experiment again. The arguments to convince him or her are possibly that: • a nonreplicated experiment does not meet the standards of research in the field (does it in any field?) • the data will therefore not be publishable • the money and time invested in the first screen will therefore be wasted ADD REPLY 3 Entering edit mode +1 because I can't give +2 or more. ADD REPLY 9 Entering edit mode 10.4 years ago I think there's some misconceptions operating here from the original questioner. First and foremost, a t-test is not just a way of calculating p-values, it is a statistical test to determine whether two populations have varying means. The p-value that results from the test is a useful indicator for whether or not to support your null hypothesis (that the two populations have the same mean), but is not the purpose of the test. In order to carry out a t-test between two populations, you need to know two things about those populations: 1) the mean of the observations and 2) the variance about that mean. The single value you have for each population could be a proxy for the mean (although it is a particularly bad one - see below), but there is no way that you can know the variance from only one observation. This is why replicates are required for microarray analysis, not a nice optional extra. The reason a single observation on a single microarray is a bad proxy for the population mean is because you have no way of knowing whether the individual tested is typical for the population concerned. Assuming the expression of a given gene is normally distributed among your population (and this is an assumption that you have to make in order for the t-test to be a valid test anyway), your single individual could come from anywhere on the bell curve. Yes, it is most likely that the observation is somewhere near the mean (by definition, ~68% within 1 standard deviation, see the graph), but there is a significant chance that it could have come from either extreme. Finally, I've read what you suggest about the hypergeometric test in relation to RNA-Seq data recently, but again the use of this test is based on a flawed assumption (that the variance of a gene between the 2 populations is equivalent to the population variance). Picking a random statistical test out of the bag, just because it is able to give you a p-value in your particular circumstance is almost universally bad practise. You need to be able to justify it in light of the assumptions you are making in order to apply the test. BTW, your data does not look like it is in log2 scale (if it is, there's an ~32-fold difference between the renal and heart observations for the first gene above) - how have you got the data into R & normalised it? ADD COMMENT 0 Entering edit mode +1 excellent explaination for beginners ADD REPLY 3 Entering edit mode 10.4 years ago It looks like you are trying to do a t-test with one value per group. That is a statistical impossibility (hence, the "not enough 'x' observations" error). Your only real option is to calculate a fold-change between the two samples by calculating a ratio. expression_data$ratio = expression_data[,3]-expression_data[,5] # assumes log scaled data
You can choose 2-fold changed genes by:
expression_data_filtered = expression_data[abs(expression_data$ratio)>2,] After you obtain replicates, you will want to use limma for gene expression analysis. Unmoderated t-tests are probably not the best way to go. ADD COMMENT 0 Entering edit mode Thank you so much Ben and Sean. Actually I'm trying to answer which of the genes are differentially expressed between these two samples and these are the only values I have. I don't have replicate experiments. Basically I want to associate some kind of significance to the differential expression and I thought calculating p-values would do that and hence the t-test. So there's no way I can calculate p-value for each gene with this data? ADD REPLY 3 Entering edit mode Hi, Diana. Unfortunately there is no way a statistical test can be performed without replication. The only option you have to compute p-values is to repeat the experiment. ADD REPLY 0 Entering edit mode Your interpretation is correct--no p-values with the data that you have in hand. ADD REPLY 0 Entering edit mode I don't know if this is a stupid question again, but someone whose working on such data suggested to me that a hypergeometric test can be done with only these values in hand. I wanted to confirm before I embarked on a useless journey. What do you all think? ADD REPLY 0 Entering edit mode How would you apply that test? ADD REPLY 0 Entering edit mode The hypergeometric distribution is used for the analysis of overlaps of gene sets, e.g. given 2 gene sets selected by some arbitrary choice, what is the probability that 100 or more out of the 1000 genes in each set are common to both both. That doesn't fit because you cannot make sensible gene sets yet. ADD REPLY 0 Entering edit mode Another point. The way you are approaching your problem is detrimental to the solution. Instead of responding by picking some random methods which you seemingly don't understand, you should: - respond to our proposal to replicate the experiment (what did your boss say about replication?) - try to understand how tests work ADD REPLY 0 Entering edit mode Thanks. No replicates for now. Maybe in near future. ADD REPLY 2 Entering edit mode 10.4 years ago Ben ★ 2.0k You are applying the t-test to the 4th and 6th value in each row; firstly R doesn't use zero-indexing so you don't seem to have a 6th column and secondly you are comparing two single values each time. For an (unpaired) t-test comparing expression_data$value_1 and expression_data$value_2 try: t.test(expression_data[,3], expression_data[,5])$p.value
edit: of course it's probably more useful to keep the whole returned list than just the p-value
0
Entering edit mode
Thanks a lot. I want to put all pairwise p-values in one object. When I try to use a loop, it gives me the same error again.
for(i in 1:38620)
{
u = t.test(expression_data[i,3], expression_data[i,5])
}
Error in t.test.default(RNA[i, 3], RNA[i, 5]) : not enough 'x' observations
What's wrong with my loop?
3
Entering edit mode
Again, you're trying to perform a t-test on two values... I think you need to look at what a t-test is and think about what you're trying to find from this data. You likely just want to add paired=T to the code I gave you above. See ?t.test in R too.
0
Entering edit mode
I need to do a t-test for each gene and I will be using two values for comparison. My question is: how can I do the pairwise t-test for each of the two values quickly...I was thinking a loop but its giving me an error. I don't want to do a t-test for each gene individually because I have a lot of genes
0
Entering edit mode
As Ben and I point out, you cannot perform a t-test between groups with only 1 member in them. As an aside, using a for-loop like this in R is usually not the best way to go. See the "apply" function for a better approach (can be orders-of-magnitude faster than a for loop). | 2023-04-02 02:31:46 | {"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": 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": 0.3696608543395996, "perplexity": 2290.7514236553943}, "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/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00763.warc.gz"} |
https://www.scootersoftware.com/vbulletin/showthread.php?11869-how-to-compare-multiple-files&p=39716 | # Thread: how to compare multiple files?
1. Visitor
Join Date
Jun 2013
Posts
4
## how to compare multiple files?
hi,
I dont have much exp with BC 3 and scripting, i would like to build script to compare four files for example:
scenario1:
file 1: \\server1\folder1\text.txt
file 2: \\server2\folder1\text.txt
scenario2:
file 3: \\server3\folder1\text.txt
file 4: \\server4\folder1\text.txt
ofc, I would like to have these two comparisons done in the same time.
2. Team Scooter
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Location
Posts
11,375
Hello,
Would you like to generate a report comparing file1 to file2, then generate a 2nd report comparing file3 to file4?
This can be done in scripting using the command line:
bcompare.exe "@c:\bcscript.txt"
Then the script file example could be:
[CODE]
text-report layout:side-by-side output-to:"c:bcreport1.html" output-options:html-color "\\server1\folder1\text.txt" "\\server2\folder1\text.txt"
text-report layout:side-by-side output-to:"c:bcreport2.html" output-options:html-color "\\server3\folder1\text.txt" "\\server4\folder1\text.txt"
Scripting actions follow the general actions you can perform in the graphical interface. Could you provide more details on the steps you are following in the interface and the reports you are generating from there? We can then help with the script to follow similar steps.
3. Visitor
Join Date
Jun 2013
Posts
4
would it be possible to have output in one file instead of multiple files? for example:
bcreport.html
also, where exactly output file bcreport.html will be saved?
4. Visitor
Join Date
Jun 2013
Posts
4
also, would it be possible to note only file differences (if any)?
5. Team Scooter
Join Date
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Location
Posts
11,375
It is not possible to have a single HTML report file for multiple text comparisons unless you open a folder compare, select the multiple files you want to compare, then generate the report. If you pass in pairs of files on the command line, we do not support appended reports together.
Code:
log verbose "c:\bclog.txt"
criteria rules-based
expand all
select diff.files
text-report layout:side-by-side options:display-mismatches output-to:"c:\bcreport.html" output-options:html-color
For a plain text report, you could append them together using a batch file:
Code:
bcompare.exe "@c:\script.txt" "c:\file1" "c:\file2"
type tempReport.txt >> mainreport.txt
bcompare.exe "@c:\script.txt" "c:\file3" "c:\file4"
type tempReport.txt >> mainreport.txt
Where script.txt is
Code:
text-report layout:side-by-side options:display-mismatches output-to:"c:\tempReport.txt" "%1" "%2"
6. Team Scooter
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To show only differences, add the "options:display-mismatches" parameter to the text-report command. Detailed documentation can be found in the Help file -> Scripting Reference, or in the Help file -> Using Beyond Compare -> Automating with Script chapter.
7. Visitor
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thank you, this was very useful! | 2018-02-21 03:33:45 | {"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": 0.21669554710388184, "perplexity": 13077.311894654757}, "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/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00151.warc.gz"} |
http://www.self.gutenberg.org/articles/eng/Sampling_distribution | #jsDisabledContent { display:none; } My Account | Register | Help
# Sampling distribution
Article Id: WHEBN0000520670
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Title: Sampling distribution Author: World Heritage Encyclopedia Language: English Subject: Collection: Statistical Theory Publisher: World Heritage Encyclopedia Publication Date:
### Sampling distribution
In statistics a sampling distribution or finite-sample distribution is the probability distribution of a given statistic based on a random sample. Sampling distributions are important in statistics because they provide a major simplification en route to statistical inference. More specifically, they allow analytical considerations to be based on the sampling distribution of a statistic, rather than on the joint probability distribution of all the individual sample values.
## Contents
• Introduction 1
• Standard error 2
• Examples 3
• Statistical inference 4
• References 5
## Introduction
The sampling distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. It may be considered as the distribution of the statistic for all possible samples from the same population of a given size. The sampling distribution depends on the underlying distribution of the population, the statistic being considered, the sampling procedure employed, and the sample size used. There is often considerable interest in whether the sampling distribution can be approximated by an asymptotic distribution, which corresponds to the limiting case either as the number of random samples of finite size, taken from an infinite population and used to produce the distribution, tends to infinity, or when just one equally-infinite-size "sample" is taken of that same population.
For example, consider a normal population with mean μ and variance σ². Assume we repeatedly take samples of a given size from this population and calculate the arithmetic mean \scriptstyle \bar x for each sample – this statistic is called the sample mean. Each sample has its own average value, and the distribution of these averages is called the "sampling distribution of the sample mean". This distribution is normal \scriptstyle \mathcal{N}(\mu,\, \sigma^2/n) (n is the sample size) since the underlying population is normal, although sampling distributions may also often be close to normal even when the population distribution is not (see central limit theorem). An alternative to the sample mean is the sample median. When calculated from the same population, it has a different sampling distribution to that of the mean and is generally not normal (but it may be close for large sample sizes).
The mean of a sample from a population having a normal distribution is an example of a simple statistic taken from one of the simplest statistical populations. For other statistics and other populations the formulas are more complicated, and often they don't exist in closed-form. In such cases the sampling distributions may be approximated through Monte-Carlo simulations[1][p. 2], bootstrap methods, or asymptotic distribution theory.
## Standard error
The standard deviation of the sampling distribution of a statistic is referred to as the standard error of that quantity. For the case where the statistic is the sample mean, and samples are uncorrelated, the standard error is:
\sigma_{\bar x} = \frac{\sigma}{\sqrt{n}}
where \sigma is the standard deviation of the population distribution of that quantity and n is the sample size (number of items in the sample).
An important implication of this formula is that the sample size must be quadrupled (multiplied by 4) to achieve half (1/2) the measurement error. When designing statistical studies where cost is a factor, this may have a role in understanding cost–benefit tradeoffs.
## Examples
Population Statistic Sampling distribution
Normal: \mathcal{N}(\mu, \sigma^2) Sample mean \bar X from samples of size n \bar X \sim \mathcal{N}\Big(\mu,\, \frac{\sigma^2}{n} \Big)
Bernoulli: \operatorname{Bernoulli}(p) Sample proportion of "successful trials" \bar X n \bar X \sim \operatorname{Binomial}(n, p)
Two independent normal populations:
\mathcal{N}(\mu_1, \sigma_1^2) and \mathcal{N}(\mu_2, \sigma_2^2)
Difference between sample means, \bar X_1 - \bar X_2 \bar X_1 - \bar X_2 \sim \mathcal{N}\! \left(\mu_1 - \mu_2,\, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \right)
Any absolutely continuous distribution F with density ƒ Median X_{(k)} from a sample of size n = 2k − 1, where sample is ordered X_{(1)} to X_{(n)} f_{X_{(k)}}(x) = \frac{(2k-1)!}{(k-1)!^2}f(x)\Big(F(x)(1-F(x))\Big)^{k-1}
Any distribution with distribution function F Maximum M=\max\ X_k from a random sample of size n F_M(x) = P(M\le x) = \prod P(X_k\le x)= \left(F(x)\right)^n
## Statistical inference
In the theory of statistical inference, the idea of a sufficient statistic provides the basis of choosing a statistic (as a function of the sample data points) in such a way that no information is lost by replacing the full probabilistic description of the sample with the sampling distribution of the selected statistic.
In frequentist inference, for example in the development of a statistical hypothesis test or a confidence interval, the availability of the sampling distribution of a statistic (or an approximation to this in the form of an asymptotic distribution) can allow the ready formulation of such procedures, whereas the development of procedures starting from the joint distribution of the sample would be less straightforward.
In Bayesian inference, when the sampling distribution of a statistic is available, one can consider replacing the final outcome of such procedures, specifically the conditional distributions of any unknown quantities given the sample data, by the conditional distributions of any unknown quantities given selected sample statistics. Such a procedure would involve the sampling distribution of the statistics. The results would be identical provided the statistics chosen are jointly sufficient statistics.
## References
1. ^
• Merberg, A. and S.J. Miller (2008). "The Sample Distribution of the Median". Course Notes for Math 162: Mathematical Statistics, on the web at http://web.williams.edu/Mathematics/sjmiller/public_html/BrownClasses/162/Handouts/MedianThm04.pdf pgs 1–9. | 2020-08-10 05:32:56 | {"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": 0.90926194190979, "perplexity": 696.0688358784275}, "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/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00013.warc.gz"} |
https://community.wolfram.com/groups/-/m/t/2373558 | # Why does the DSolve not solve the PDE giving the 'Arbitrary functions'?
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Hello, I have two PDEs (strainDisp11 & strainDisp22) in 2 variables x1 and x2. strainDisp11 is a PDE with the partial differential term in x1 whereas, strainDisp22 is a PDE with the partial differential term in x2 I am trying to solve these two PDEs separately using DSolve (Last two command lines in the attached file), however, the solution is not generated along with the required arbitrary functions C1[1] which should be f1[x2] and C1[1] which should be f2[x1] in the respective solutions of the PDEs. Attached is Notebook for your reference. Appreciate your help.
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Posted 1 month ago
A Tip: Don't use Subscript , because causes problems.
Posted 1 month ago
Thanks! Very much appreciated.
Posted 11 days ago
Hello, I have two PDEs in 2 variables 'r' and 'theta'. I am trying to solve these two PDEs separately using DSolve (The last two command lines in the attached file). The solution is generated as expected for the 1st PDE (Integration with respect to variable 'r'), however, the solution is not generated for the 2nd PDE (Integration with respect to 'theta'). I cannot understand why Mathematica does not solve all the terms and has replaced 'theta' by K[1] in the unsolved integral with limits? Attached is Notebook for your reference. Appreciate your help.
Posted 11 days ago
Maybe: solDispRR = DSolve[strainDispRR == 0, uR, {r, \[Theta]}] // Flatten; solDisp\[Theta]\[Theta] = DSolve[strainDisp\[Theta]\[Theta] == 0, u\[Theta], {r, \[Theta]}] // Flatten; uRFunctionTemp = uR[r, \[Theta]] /. solDispRR[[1]] u\[Theta]FunctionTemp = (u\[Theta][r, \[Theta]] /. solDisp\[Theta]\[Theta][[1]] /. solDispRR[[1]]) // Activate // ExpandAll Looks like MMA can't integrate, a workaround: u\[Theta]FunctionTemp = (Integrate[#, {K[1], 1, \[Theta]}] & /@ (u\[Theta]FunctionTemp[[1, 1]])) + u\[Theta]FunctionTemp[[2]] (*Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}] + (2*P*\[Nu]^2*Log[r]*(Sin[1] - Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + (2*P*\[Nu]*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + (2*P*\[Nu]^2*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + (2*P*Log[r]*(-Sin[1] + Sin[\[Theta]]))/(Pi*\[DoubleStruckCapitalE]) + C[1][r]*) In this line: Integrate[-C[1][K[1]], {K[1], 1, \[Theta]}] what answer do you expect? | 2021-10-28 08:17:36 | {"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": 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": 0.26450031995773315, "perplexity": 7160.811352501623}, "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/1634323588282.80/warc/CC-MAIN-20211028065732-20211028095732-00550.warc.gz"} |
http://mathhelpforum.com/calculus/209791-partial-derivative-notation-question.html | # Math Help - partial derivative notation question
1. ## partial derivative notation question
what does the notation at the bottom mean? the second derivative wrt z over the partial of y times the partial of x. Is that right? and what does that mean procedurally?
2. ## Re: partial derivative notation question
It means to first take the partial derivative of z with respect to y, then take the partial derivative of this result with respect to x.
For a function like this which is continuous and the respective partials exist, the order of differentiation does not matter, i.e.:
$\frac{\delta^2 z}{\delta x\delta y}=\frac{\delta^2 z}{\delta y\delta x}$ | 2014-10-22 04:12:17 | {"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": 0.9681448936462402, "perplexity": 317.01724842933305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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-2014-42/segments/1413507445299.20/warc/CC-MAIN-20141017005725-00146-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://economics.stackexchange.com/questions/14047/why-are-vacancy-rate-and-unemployment-rate-negatively-correlated | # Why are vacancy rate and unemployment rate negatively correlated?
Why is this the case?
Since Vacancy rate is defined as following,
let $A,Q,U$ denote number of vacancies in the economy, labor force, unemployed respectively.
$$\frac{A}{A+Q-U}$$
Here we can see that if unemployed increase vacancy rate would go up?
Why is there a negatively correlation then? Take the beveridge curve as an example :
https://en.wikipedia.org/wiki/Beveridge_curve
• I have no idea what you are asking here. Maybe rephrase the question. – Jamzy Nov 1 '16 at 22:05
• Are you asking 'why is unemployment lower when job vacancies are higher?'. Unemployed people are people are looking for work. When you increase the thing that they are looking for (work), there will be less of them. – Jamzy Nov 1 '16 at 22:08
Adopting your notation, the vacancy rate at any given time is defined as $A/Q$. There is no mechanical relationship between the unemployment rate $U/Q$ and vacancy rate (A/Q). | 2019-10-20 09:38:34 | {"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": 0.7795020341873169, "perplexity": 2063.607101534271}, "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-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00218.warc.gz"} |
https://www.computer.org/csdl/trans/td/2008/08/ttd2008081099-abs.html | Issue No. 08 - August (2008 vol. 19)
ISSN: 1045-9219
pp: 1099-1110
ABSTRACT
Peer-to-peer (P2P) networks often demand scalability, low communication latency among nodes, and low system-wide overhead. For scalability, a node maintains partial states of a P2P network and connects to a few nodes. For fast communication, a P2P network intends to reduce the communication latency between any two nodes as much as possible. With regard to a low system-wide overhead, a P2P network minimizes its traffic in maintaining its performance efficiency and functional correctness. In this paper, we present a novel tree-based P2P network with low communication delay and low system-wide overhead. The merits of our tree-based network include: $(i)$ a tree-shaped P2P network which guarantees that the degree of a node is constant in probability regardless of the system size. The network diameter in our tree-based network increases logarithmically with an increase of the system size. Specially, given a physical network with a power-law latency expansion property, we show that the diameter of our tree network is constant. $(ii)$ Our proposal has the provable performance guarantees. We evaluate our proposal by rigorous performance analysis, and validate by extensive simulations.
INDEX TERMS
Distributed networks, Distributed Systems, Multicast
CITATION
H. Hsiao and C. He, "A Tree-Based Peer-to-Peer Network with Quality Guarantees," in IEEE Transactions on Parallel & Distributed Systems, vol. 19, no. , pp. 1099-1110, 2007.
doi:10.1109/TPDS.2007.70798 | 2018-05-28 10:18:54 | {"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": 0.7336714267730713, "perplexity": 1543.3951099112298}, "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/1526794872766.96/warc/CC-MAIN-20180528091637-20180528111637-00313.warc.gz"} |
https://mathhothouse.me/category/pre-rmo-2/ | ## Category Archives: Pre-RMO
### Rules for Inequalities
If a, b and c are real numbers, then
1. $a < b \Longrightarrow a + c< b + c$
2. $a < b \Longrightarrow a - c < b - c$
3. $a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc$
4. $a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac$ special case: $a < b \Longrightarrow -b < -a$
5. $a > 0 \Longrightarrow \frac{1}{a} > 0$
6. If a and b are both positive or both negative, then $a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}$.
Remarks:
Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.
Regards,
Nalin Pithwa.
### Set Theory, Relations, Functions Preliminaries: II
Relations:
Concept of Order:
Let us say that we create a “table” of two columns in which the first column is the name of the father, and the second column is name of the child. So, it can have entries like (Yogesh, Meera), (Yogesh, Gopal), (Kishor, Nalin), (Kishor, Yogesh), (Kishor, Darshna) etc. It is quite obvious that “first” is the “father”, then “second” is the child. We see that there is a “natural concept of order” in human “relations”. There is one more, slightly crazy, example of “importance of order” in real-life. It is presented below (and some times also appears in basic computer science text as rise and shine algorithm) —-
Rise and Shine algorithm:
When we get up from sleep in the morning, we brush our teeth, finish our morning ablutions; next, we remove our pyjamas and shirt and then (secondly) enter the shower; there is a natural order here; first we cannot enter the shower, and secondly we do not remove the pyjamas and shirt after entering the shower. 🙂
Ordered Pair: Definition and explanation:
A pair $(a,b)$ of numbers, such that the order, in which the numbers appear is important, is called an ordered pair. In general, ordered pairs (a,b) and (b,a) are different. In ordered pair (a,b), ‘a’ is called first component and ‘b’ is called second component.
Two ordered pairs (a,b) and (c,d) are equal, if and only if $a=c$ and $b=d$. Also, $(a,b)=(b,a)$ if and only if $a=b$.
Example 1: Find x and y when $(x+3,2)=(4,y-3)$.
Solution 1: Equating the first components and then equating the second components, we have:
$x+3=4$ and $2=y-3$
$x=1$ and $y=5$
Cartesian products of two sets:
Let A and B be two non-empty sets then the cartesian product of A and B is denoted by A x B (read it as “A cross B”),and is defined as the set of all ordered pairs (a,b) such that $a \in A$, $b \in B$.
Thus, $A \times B = \{ (a,b): a \in A, b \in B\}$
e.g., if $A = \{ 1,2\}$ and $B = \{ a,b,c\}$, tnen $A \times B = \{ (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}$.
If $A = \phi$ or $B=\phi$, we define $A \times B = \phi$.
Number of elements of a cartesian product:
By the following basic counting principle: If a task A can be done in m ways, and a task B can be done in n ways, then the tasks A (first) and task B (later) can be done in mn ways.
So, the cardinality of A x B is given by: $n(A \times B)= n(A) \times n(B)$.
So, in general if a cartesian product of p finite sets, viz, $A_{1}, A_{2}, A_{3}, \ldots, A_{p}$ is given by $n(A_{1} \times A_{2} \times A_{3} \ldots A_{p}) = n(A_{1}) \times n(A_{2}) \times \ldots \times n(A_{p})$
Definitions of relations, arrow diagrams (or pictorial representation), domain, co-domain, and range of a relation:
Consider the following statements:
i) Sunil is a friend of Anil.
ii) 8 is greater than 4.
iii) 5 is a square root of 25.
Here, we can say that Sunil is related to Anil by the relation ‘is a friend of’; 8 and 4 are related by the relation ‘is greater than’; similarly, in the third statement, the relation is ‘is a square root of’.
The word relation implies an association of two objects according to some property which they possess. Now, let us some mathematical aspects of relation;
Definition:
A and B are two non-empty sets then any subset of $A \times B$ is called relation from A to B, and is denoted by capital letters P, Q and R. If R is a relation and $(x,y) \in R$ then it is denoted by $xRy$.
y is called image of x under R and x is called pre-image of y under R.
Let $A=\{ 1,2,3,4,5\}$ and $B=\{ 1,4,5\}$.
Let R be a relation such that $(x,y) \in R$ implies $x < y$. We list the elements of R.
Solution: Here $A = \{ 1,2,3,4,5\}$ and $B=\{ 1,4,5\}$ so that $R = \{ (1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)\}$ Note this is the relation R from A to B, that is, it is a subset of A x B.
Check: Is a relation $R^{'}$ from B to A defined by x<y, with $x \in B$ and $y \in A$ — is this relation $R^{'}$ *same* as R from A to B? Ans: Let us list all the elements of R^{‘} explicitly: $R^{'} = \{ (1,2),(1,3),(1,4),(1,5),(4,5)\}$. Well, we can surely compare the two sets R and $R^{'}$ — the elements “look” different certainly. Even if they “look” same in terms of numbers, the two sets $R$ and $R^{'}$ are fundamentally different because they have different domains and co-domains.
Definition : Domain of a relation R: The set of all the first components of the ordered pairs in a relation R is called the domain of relation R. That is, if $R \subseteq A \times B$, then domain (R) is $\{ a: (a,b) \in R\}$.
Definition: Range: The set of all second components of all ordered pairs in a relation R is called the range of the relation. That is, if $R \subseteq A \times B$, then range (R) = $\{ b: (a,b) \in R\}$.
Definition: Codomain: If R is a relation from A to B, then set B is called co-domain of the relation R. Note: Range is a subset of co-domain.
Type of Relations:
One-one relation: A relation R from a set A to B is said to be one-one if every element of A has at most one image in B and distinct elements in A have distinct images in B. For example, let $A = \{ 1,2,3,4\}$, and let $B=\{ 2,3,4,5,6,7\}$ and let $R_{1}= \{ (1,3),(2,4),(3,5)\}$ Then $R_{1}$ is a one-one relation. Here, domain of $R_{1}= \{ 1,2,3\}$ and range of $R_{1}$ is $\{ 3,4,5\}$.
Many-one relation: A relation R from A to B is called a many-one relation if two or more than two elements in the domain A are associated with a single (unique) element in co-domain B. For example, let $R_{2}=\{ (1,4),(3,7),(4,4)\}$. Then, $R_{2}$ is many-one relation from A to B. (please draw arrow diagram). Note also that domain of $R_{1}=\{ 1,3,4\}$ and range of $R_{1}=\{ 4,7\}$.
Into Relation: A relation R from A to B is said to be into relation if there exists at least one element in B, which has no pre-image in A. Let $A=\{ -2,-1,0,1,2,3\}$ and $B=\{ 0,1,2,3,4\}$. Consider the relation $R_{1}=\{ (-2,4),(-1,1),(0,0),(1,1),(2,4) \}$. So, clearly range is $\{ 0,1,4\}$ and $range \subseteq B$. Thus, $R_{3}$ is a relation from A INTO B.
Onto Relation: A relation R from A to B is said to be ONTO relation if every element of B is the image of some element of A. For example: let set $A= \{ -3,-2,-1,1,3,4\}$ and set $B= \{ 1,4,9\}$. Let $R_{4}=\{ (-3,9),(-2,4), (-1,1), (1,1),(3,9)\}$. So, clearly range of $R_{4}= \{ 1,4,9\}$. Range of $R_{4}$ is co-domain of B. Thus, $R_{4}$ is a relation from A ONTO B.
Binary Relation on a set A:
Let A be a non-empty set then every subset of $A \times A$ is a binary relation on set A.
Illustrative Examples:
E.g.1: Let $A = \{ 1,2,3\}$ and let $A \times A = \{ (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$. Now, if we have a set $R = \{ (1,2),(2,2),(3,1),(3,2)\}$ then we observe that $R \subseteq A \times A$, and hence, R is a binary relation on A.
E.g.2: Let N be the set of natural numbers and $R = \{ (a,b) : a, b \in N and 2a+b=10\}$. Since $R \subseteq N \times N$, R is a binary relation on N. Clearly, $R = \{ (1,8),(2,6),(3,4),(4,2)\}$. Also, for the sake of completeness, we state here the following: Domain of R is $\{ 1,2,3,4\}$ and Range of R is $\{ 2,4,6,8\}$, codomain of R is N.
Note: (i) Since the null set is considered to be a subset of any set X, so also here, $\phi \subset A \times A$, and hence, $\phi$ is a relation on any set A, and is called the empty or void relation on A. (ii) Since $A \times A \subset A \times A$, we say that $A \subset A$ is a relation on A called the universal relation on A.
Note: Let the cardinality of a (finite) set A be $n(A)=p$ and that of another set B be $n(B)=q$, then the cardinality of the cartesian product $n(A \times B)=pq$. So, the number of possible subsets of $A \times B$ is $2^{pq}$ which includes the empty set.
Types of relations:
Let A be a non-empty set. Then, a relation R on A is said to be: (i) Reflexive: if $(a,a) \in R$ for all $a \in A$, that is, aRa for all $a \in A$. (ii) Symmetric: If $(a,b) \in R \Longrightarrow (b,a) \in R$ for all $a,b \in R$ (iii) Transitive: If $(a,b) \in R$, and $(b,c) \in R$, then so also $(a,c) \in R$.
Equivalence Relation:
A (binary) relation on a set A is said to be an equivalence relation if it is reflexive, symmetric and transitive. An equivalence appears in many many areas of math. An equivalence measures “equality up to a property”. For example, in number theory, a congruence modulo is an equivalence relation; in Euclidean geometry, congruence and similarity are equivalence relations.
Also, we mention (without proof) that an equivalence relation on a set partitions the set in to mutually disjoint exhaustive subsets.
Illustrative examples continued:
E.g. Let R be an equivalence relation on $\mathbb{Q}$ defined by $R = \{ (a,b): a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}$. Prove that R is an equivalence relation.
Proof: Given that $R = \{ (a,b) : a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}$. (i) Let $a \in \mathbb{Q}$ then $a-a=0 \in \mathbb{Z}$, hence, $(a,a) \in R$, so relation R is reflexive. (ii) Now, note that $(a,b) \in R \Longrightarrow (a-b) \in \mathbb{Z}$, that is, $(a-b)$ is an integer $\Longrightarrow -(b-a) \in \mathbb{Z} \Longrightarrow (b-a) \in \mathbb{Z} \Longrightarrow (b,a) \in R$. That is, we have proved $(a,b) \in R \Longrightarrow (b,a) \in R$ and so relation R is symmetric also. (iii) Now, let $(a,b) \in R$, and $(b,c) \in R$, which in turn implies that $(a-b) \in \mathbb{Z}$ and $(b-c) \in \mathbb{Z}$ so it $\Longrightarrow (a-b)+(b-c)=a-c \in \mathbb{Z}$ (as integers are closed under addition) which in turn $\Longrightarrow (a,c) \in R$. Thus, $(a,b) \in R$ and $(b,c) \in R$ implies $(a,c) \in R$ also, Hence, given relation R is transitive also. Hence, R is also an equivalence relation on $\mathbb{Q}$.
Illustrative examples continued:
E.g.: If $(x+1,y-2) = (3,4)$, find the values of x and y.
Solution: By definition of an ordered pair, corresponding components are equal. Hence, we get the following two equations: $x+1=3$ and $y-2=4$ so the solution is $x=2,y=6$.
E.g.: If $A = (1,2)$, list the set $A \times A$.
Solution: $A \times A = \{ (1,1),(1,2),(2,1),(2,2)\}$
E.g.: If $A = \{1,3,5 \}$ and $B=\{ 2,3\}$, find $A \times B$, and $B \times A$, check if cartesian product is a commutative operation, that is, check if $A \times B = B \times A$.
Solution: $A \times B = \{ (1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\}$ whereas $B \times A = \{ (2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$ so since $A \times B \neq B \times A$ so cartesian product is not a commutative set operation.
E.g.: If two sets A and B are such that their cartesian product is $A \times B = \{ (3,2),(3,4),(5,2),(5,4)\}$, find the sets A and B.
Solution: Using the definition of cartesian product of two sets, we know that set A contains as elements all the first components and set B contains as elements all the second components. So, we get $A = \{ 3,5\}$ and $B = \{ 2,4\}$.
E.g.: A and B are two sets given in such a way that $A \times B$ contains 6 elements. If three elements of $A \times B$ are $(1,3),(2,5),(3,3)$, find its remaining elements.
Solution: We can first observe that $6 = 3 \times 2 = 2 \times 3$ so that A can contain 2 or 3 elements; B can contain 3 or 2 elements. Using definition of cartesian product of two sets, we get that $A= \{ 1,2,3\}$ and $\{ 3,5\}$ and so we have found the sets A and B completely.
E.g.: Express the set $\{ (x,y) : x^{2}+y^{2}=25, x, y \in \mathbb{W}\}$ as a set of ordered pairs.
Solution: We have $x^{2}+y^{2}=25$ and so
$x=0, y=5 \Longrightarrow x^{2}+y^{2}=0+25=25$
$x=3, y=4 \Longrightarrow x^{2}+y^{2}=9+16=25$
$x=4, y=3 \Longrightarrow x^{2}+y^{2}=16+9=25$
$x=5, y=0 \Longrightarrow x^{2}+y^{2}=25+0=25$
Hence, the given set is $\{ (0,5),(3,4),(4,3),(5,0)\}$
E.g.: Let $A = \{ 1,2,3\}$ and $B = \{ 2,4,6\}$. Show that $R = \{ (1,2),(1,4),(3,2),(3,4)\}$ is a relation from A to B. Find the domain, co-domain and range.
Solution: Here, $A \times B = \{ (1,2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,2),(3,4),(3,6)\}$. Clearly, $R \subseteq A \times B$. So R is a relation from A to B. The domain of R is the set of first components of R (which belong to set A, by definition of cartesian product and ordered pair) and the codomain is set B. So, Domain (R) = $\{ 1,3\}$ and co-domain of R is set B itself; and Range of R is $\{ 2,4\}$.
E.g.: Let $A = \{ 1,2,3,4,5\}$ and $B = \{ 1,4,5\}$. Let R be a relation from A to B such that $(x,y) \in R$ if $x. List all the elements of R. Find the domain, codomain and range of R. (as homework quiz, draw its arrow diagram);
Solution: Let $A = \{ 1,2,3,4,5\}$ and $B = \{ 1,4,5\}$. So, we get R as $(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)$. $domain(R) = \{ 1,2,3,4\}$, $codomain(R) = B$, and $range(R) = \{ 4,5\}$.
E.g. Let $A = \{ 1,2,3,4,5,6\}$. Define a binary relation on A such that $R = \{ (x,y) : y=x+1\}$. Find the domain, codomain and range of R.
Solution: By definition, $R \subseteq A \times A$. Here, we get $R = \{ (1,2),(2,3),(3,4),(4,5),(5,6)\}$. So we get $domain (R) = \{ 1,2,3,4,5\}$, $codomain(R) =A$, $range(R) = \{ 2,3,4,5,6\}$
Tutorial problems:
1. If $(x-1,y+4)=(1,2)$, find the values of x and y.
2. If $(x + \frac{1}{3}, \frac{y}{2}-1)=(\frac{1}{2} , \frac{3}{2} )$
3. If $A=\{ a,b,c\}$ and $B = \{ x,y\}$. Find out the following: $A \times A$, $B \times B$, $A \times B$ and $B \times A$.
4. If $P = \{ 1,2,3\}$ and $Q = \{ 4\}$, find the sets $P \times P$, $Q \times Q$, $P \times Q$, and $Q \times P$.
5. Let $A=\{ 1,2,3,4\}$ and $\{ 4,5,6\}$ and $C = \{ 5,6\}$. Find $A \times (B \bigcap C)$, $A \times (B \bigcup C)$, $(A \times B) \bigcap (A \times C)$, $A \times (B \bigcup C)$, and $(A \times B) \bigcup (A \times C)$.
6. Express $\{ (x,y) : x^{2}+y^{2}=100 , x, y \in \mathbf{W}\}$ as a set of ordered pairs.
7. Write the domain and range of the following relations: (i) $\{ (a,b): a \in \mathbf{N}, a < 6, b=4\}$ (ii) $\{ (a,b): a,b \in \mathbf{N}, a+b=12\}$ (iii) $\{ (2,4),(2,5),(2,6),(2,7)\}$
8. Let $A=\{ 6,8\}$ and $B=\{ 1,3,5\}$. Let $R = \{ (a,b): a \in A, b \in B, a+b \hspace{0.1in} is \hspace{0.1in} an \hspace{0.1in} even \hspace{0.1in} number\}$. Show that R is an empty relation from A to B.
9. Write the following relations in the Roster form and hence, find the domain and range: (i) $R_{1}= \{ (a,a^{2}) : a \hspace{0.1in} is \hspace{0.1in} prime \hspace{0.1in} less \hspace{0.1in} than \hspace{0.1in} 15\}$ (ii) $R_{2} = \{ (a, \frac{1}{a}) : 0 < a \leq 5, a \in N\}$
10. Write the following relations as sets of ordered pairs: (i) $\{ (x,y) : y=3x, x \in \{1,2,3 \}, y \in \{ 3,6,9,12\}\}$ (ii) $\{ (x,y) : y>x+1, x=1,2, y=2,4,6\}$ (iii) $\{ (x,y) : x+y =3, x, y \in \{ 0,1,2,3\}\}$
More later,
Nalin Pithwa
### Set Theory, Relations, Functions Preliminaries: I
In these days of conflict between ancient and modern studies there must surely be something to be said of a study which did not begin with Pythagoras and will not end with Einstein. — G H Hardy (On Set Theory)
In every day life, we generally talk about group or collection of objects. Surely, you must have used the words such as team, bouquet, bunch, flock, family for collection of different objects.
It is very important to determine whether a given object belongs to a given collection or not. Consider the following conditions:
i) Successful persons in your city.
ii) Happy people in your town.
iii) Clever students in your class.
iv) Days in a week.
v) First five natural numbers.
Perhaps, you have already studied in earlier grade(s) —- can you state which of the above mentioned collections are sets? Why? Check whether your answers are as follows:
First three collections are not examples of sets but last two collections represent sets. This is because in first three collections, we are not sure of the objects. The terms ‘successful persons’, ‘happy people’, ‘clever students’ are all relative terms. Here, the objects are not well-defined. In the last two collections, we can determine the objects clearly (meaning, uniquely, or without ambiguity). Thus, we can say that the objects are well-defined.
So what can be the definition of a set ? Here it goes:
A collection of well-defined objects is called a set. (If we continue to “think deep” about this definition, we are led to the famous paradox, which Bertrand Russell had discovered: Let C be a collection of all sets such which are not elements of themselves. If C is allowed to be a set, a contradiction arises when one inquires whether or not C is an element of itself. Now plainly, there is something suspicious about the idea of a set being an element of itself, and we shall take this as evidence that the qualification “well-defined” needs to be taken seriously. Bertrand Russell re-stated this famous paradox in a very interesting way: In the town of Seville lives a barber who shaves everyone who does not shave himself. Does the barber shave himself?…)
The objects in a set are called elements or members of that set.
We denote sets by capital letters : A, B, C etc. The elements of a set are represented by small letters : a, b, c, d, e, f ….etc. If x is an element of a set A, we write $x \in A$. And, we read it as “x belongs to A.” If x is not an element of a set A, we write $x \not\in A$, and read as ‘x does not belong to A.’e.g., 1 is a “whole” number but not a “natural” number.
Hence, $0 \in W$, where W is the set of whole numbers and $0 \not\in N$, where N is a set of natural numbers.
There are two methods of representing a set:
a) Roster or Tabular Method or List Method (b) Set-Builder or Ruler Method
a) Roster or Tabular or List Method:
Let A be the set of all prime numbers less than 20. Can you enumerate all the elements of the set A? Are they as follows?
$A=\{ 2,3,5,7,11,15,17,19\}$
Can you describe the roster method? We can describe it as follows:
In the Roster method, we list all the elements of the set within braces $\{, \}$ and separate the elements by commas.
In the following examples, state the sets using Roster method:
i) B is the set of all days in a week
ii) C is the set of all consonants in English alphabets.
iii) D is the set of first ten natural numbers.
2) Set-Builder Method:
Let P be the set of first five multiples of 10. Using Roster Method, you must have written the set as follows:
$P = \{ 10, 20, 30, 40, 50\}$
Question: What is the common property possessed by all the elements of the set P?
Answer: All the elements are multiples of 10.
Question: How many such elements are in the set?
Answer: There are 5 elements in the set.
Thus, the set P can be described using this common property. In such a case, we say that set-builder method is used to describe the set. So, to summarize:
In the set-builder method, we describe the elements of the set by specifying the property which determines the elements of the set uniquely.
Thus, we can write : $P = \{ x: x =10n, n \in N, n \leq 5\}$
In the following examples, state the sets using set-builder method:
i) Y is the set of all months of a year
ii) M is the set of all natural numbers
iii) B is the set of perfect squares of natural numbers.
Also, if elements of a set are repeated, they are written once only; while listing the elements of a set, the order in which the elements are listed is immaterial. (but this situation changes when we consider sets from the view-point of permutations and combinations. Just be alert in set-theoretic questions.)
Subset: A set A is said to be a subset of a set B if each element of set A is an element of set B. Symbolically, $A \subseteq B$.
Superset: If $A \subset B$, then B is called the superset of set A. Symbolically: $B \supset A$
Proper Subset: A non empty set A is said to be a proper subset of the set B, if and only if all elements of set A are in set B, and at least one element of B is not in A. That is, if $A \subseteq B$, but $A \neq B$ then A is called a proper subset of B and we write $A \subset B$.
Note: the notations of subset and proper subset differ from author to author, text to text or mathematician to mathematician. These notations are not universal conventions in math.
Intervals:
1. Open Interval : given $a < b$, $a, b \in R$, we say $a is an open interval in $\Re^{1}$.
2. Closed Interval : given $a \leq x \leq b = [a,b]$
3. Half-open, half-closed: $a , or $a \leq x
4. The set of all real numbers greater than or equal to a : $x \geq a =[a, \infty)$
5. The set of all real numbers less than or equal to a is $(-\infty, a] = x \leq a$
Types of Sets:
1. Empty Set: A set containing no element is called the empty set or the null set and is denoted by the symbol $\phi$ or $\{ \}$ or void set. e.g., $A= \{ x: x \in N, 1
2. Singleton Set: A set containing only one element is called a singleton set. Example : (i) Let A be a set of all integers which are neither positive nor negative. Then, $A = \{ 0\}$ and example (ii) Let B be a set of capital of India. Then $B= \{ Delhi\}$
We will define the following sets later (after we giving a working definition of a function): finite set, countable set, infinite set, uncountable set.
3. Equal sets: Two sets are said to be equal if they contain the same elements, that is, if $A \subseteq B$ and $B \subseteq A$. For example: Let X be the set of letters in the word ‘ABBA’ and Y be the set of letters in the word ‘BABA’. Then, $X= \{ A,B\}$ and $Y= \{ B,A\}$. Thus, the sets $X=Y$ are equal sets and we denote it by $X=Y$.
How to prove that two sets are equal?
Let us say we are given the task to prove that $A=B$, where A and B are non-empty sets. The following are the steps of the proof : (i) TPT: $A \subset B$, that is, choose any arbitrary element $x \in A$ and show that also $x \in B$ holds true. (ii) TPT: $B \subset A$, that is, choose any arbitrary element $y \in B$, and show that also $y \in A$. (Note: after we learn types of functions, we will see that a fundamental way to prove two sets (finite) are equal is to show/find a bijection between the two sets).
PS: Note that two sets are equal if and only if they contain the same number of elements, and the same elements. (irrespective of order of elements; once again, the order condition is changed for permutation sets; just be alert what type of set theoretic question you are dealing with and if order is important in that set. At least, for our introduction here, order of elements of a set is not important).
PS: Digress: How to prove that in general, $x=y$? The standard way is similar to above approach: (i) TPT: $x < y$ (ii) TPT: $y < x$. Both (i) and (ii) together imply that $x=y$.
4. Equivalent sets: Two finite sets A and B are said to be equivalent if $n(A)=n(B)$. Equal sets are always equivalent but equivalent sets need not be equal. For example, let $A= \{ 1,2,3 \}$ and $B = \{ 4,5,6\}$. Then, $n(A) = n(B)$, so A and B are equivalent. Clearly, $A \neq B$. Thus, A and B are equivalent but not equal.
5. Universal Set: If in a particular discussion all sets under consideration are subsets of a set, say U, then U is called the universal set for that discussion. You know that the set of natural numbers the set of integers are subsets of set of real numbers R. Thus, for this discussion is a universal set. In general, universal set is denoted by or X.
6. Venn Diagram: The pictorial representation of a set is called Venn diagram. Generally, a closed geometrical figures are used to represent the set, like a circle, triangle or a rectangle which are known as Venn diagrams and are named after the English logician John Venn.
In Venn diagram the elements of the sets are shown in their respective figures.
Now, we have these “abstract toys or abstract building-blocks”, how can we get new such “abstract buildings” using these “abstract building blocks”. What I mean is that we know that if we are a set of numbers like 1,2,3, …, we know how to get “new numbers” out of these by “adding”, subtracting”, “multiplying” or “dividing” the given “building blocks like 1, 2…”. So, also what we want to do now is “operations on sets” so that we create new, more interesting or perhaps, more “useful” sets out of given sets. We define the following operations on sets:
1. Complement of a set: If A is a subset of the universal set U then the set of all elements in U which are not in A is called the complement of the set A and is denoted by $A^{'}$ or $A^{c}$ or $\overline{A}$ Some properties of complements: (i) ${A^{'}}^{'}=A$ (ii) $\phi^{'}=U$, where U is universal set (iii) $U^{'}= \phi$
2. Union of Sets: If A and B are two sets then union of set A and set B is the set of all elements which are in set A or set B or both set A and set B. (this is the INCLUSIVE OR in digital logic) and the symbol is : \$latex A \bigcup B
3. Intersection of sets: If A and B are two sets, then the intersection of set A and set B is the set of all elements which are both in A and B. The symbol is $A \bigcap B$.
4. Disjoint Sets: Let there be two sets A and B such that $A \bigcap B=\phi$. We say that the sets A and B are disjoint, meaning that they do not have any elements in common. It is possible that there are more than two sets $A_{1}, A_{2}, \ldots A_{n}$ such that when we take any two distinct sets $A_{i}$ and $A_{j}$ (so that $i \neq j$, then $A_{i}\bigcap A_{j}= \phi$. We call such sets pairwise mutually disjoint. Also, in case if such a collection of sets also has the property that $\bigcup_{i=1}^{i=n}A_{i}=U$, where U is the Universal Set in the given context, We then say that this collection of sets forms a partition of the Universal Set.
5. Difference of Sets: Let us say that given a universal set U and two other sets A and B, $B-A$ denotes the set of elements in B which are not in A; if you notice, this is almost same as $A^{'}=U-A$.
6. Symmetric Difference of Sets: Suppose again that we are two given sets A and B, and a Universal Set U, by symmetric difference of A and B, we mean $(A-B)\bigcup (B-A)$. The symbol is $A \triangle B.$ Try to visualize this (and describe it) using a Venn Diagram. You will like it very much. Remark : The designation “symmetric difference” for the set $A \triangle B$ is not too apt, since $A \triangle B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \triangle B$ the same two statements are joined by “or” used in the ordinary “either…or….” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \triangle B$ if and only if x belongs to either A or B but not both. The set $A \triangle B$ can be regarded as a kind of a “modulo-two-sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).
Let us now present some (easily provable/verifiable) properties of sets:
1. $A \bigcup B = B \bigcup A$ (union of sets is commutative)
2. $(A \bigcup B) \bigcup C = A \bigcup (B \bigcup C)$ (union of sets is associative)
3. $A \bigcup \phi=A$
4. $A \bigcup A = A$
5. $A \bigcup A^{'}=U$ where U is universal set
6. If $A \subseteq B$, then $A \bigcup B=B$
7. $U \bigcup A=U$
8. $A \subseteq (A \bigcup B)$ and also $B \subseteq (A \bigcup B)$
Similarly, some easily verifiable properties of set intersection are:
1. $A \bigcap B = B \bigcap A$ (set intersection is commutative)
2. $(A \bigcap B) \bigcap C = A \bigcap (B \bigcap C)$ (set intersection is associative)
3. $A \bigcap \phi = \phi \bigcap A= \phi$ (this matches intuition: there is nothing common in between a non empty set and an empty set :-))
4. $A \bigcap A =A$ (Idempotent law): this definition carries over to square matrices: if a square matrix is such that $A^{2}=A$, then A is called an Idempotent matrix.
5. $A \bigcap A^{'}=\phi$ (this matches intuition: there is nothing in common between a set and another set which does not contain any element of it (the former set))
6. If $A \subseteq B$, then $A \bigcap B =A$
7. $U \bigcap A=A$, where U is universal set
8. $(A \bigcap B) \subseteq A$ and $(A \bigcap B) \subseteq B$
9. i: $A \bigcap (B \bigcap )C = (A \bigcap B)\bigcup (A \bigcap C)$ (intersection distributes over union) ; (9ii) $A \bigcup (B \bigcap C)=(A \bigcup B) \bigcap (A \bigcup C)$ (union distributes over intersection). These are the two famous distributive laws.
The famous De Morgan’s Laws for two sets are as follows: (it can be easily verified by Venn Diagram):
For any two sets A and B, the following holds:
i) $(A \bigcup B)^{'}=A^{'}\bigcap B^{'}$. In words, it can be captured beautifully: the complement of union is intersection of complements.
ii) $(A \bigcap B)^{'}=A^{'} \bigcup B^{'}$. In words, it can be captured beautifully: the complement of intersection is union of complements.
Cardinality of a set: (Finite Set) : (Again, we will define the term ‘finite set’ rigorously later) The cardinality of a set is the number of distinct elements contained in a finite set A and we will denote it as $n(A)$.
Inclusion Exclusion Principle:
For two sets A and B, given a universal set U: $n(A \bigcup B) = n(A) + n(B) - n(A \bigcap B)$.
For three sets A, B and C, given a universal set U: $n(A \bigcup B \bigcup C)=n(A) + n(B) + n(C) -n(A \bigcap B) -n(B \bigcap C) -n(C \bigcup A) + n(A \bigcap B \bigcap C)$.
Homework Quiz: Verify the above using Venn Diagrams.
Power Set of a Set:
Let us consider a set A (given a Universal Set U). Then, the power set of A is the set consisting of all possible subsets of set A. (Note that an empty is also a subset of A and that set A is a subset of A itself). It can be easily seen (using basic definition of combinations) that if $n(A)=p$, then $n(power set A) = 2^{p}$. Symbol: $P(A)$.
Homework Tutorial I:
1. Describe the following sets in Roster form: (i) $\{ x: x \hspace{0.1in} is \hspace{0.1in} a \hspace{0.1in} letter \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} word \hspace{0.1in} PULCHRITUDE\}$ (II) $\{ x: x \hspace{0.1in } is \hspace{0.1in} an \hspace{0.1in} integer \hspace{0.1in} with \hspace{0.1in} \frac{-1}{2} < x < \frac{1}{2} \}$ (iii) $\{x: x=2n, n \in N\}$
2. Describe the following sets in Set Builder form: (i) $\{ 0\}$ (ii) $\{ 0, \pm 1, \pm 2, \pm 3\}$ (iii) $\{ \}$
3. If $A= \{ x: 6x^{2}+x-15=0\}$ and $B= \{ x: 2x^{2}-5x-3=0\}$, and $x: 2x^{2}-x-3=0$, then find (i) $A \bigcup B \bigcup C$ (ii) $A \bigcap B \bigcap C$
4. If A, B, C are the sets of the letters in the words, ‘college’, ‘marriage’, and ‘luggage’ respectively, then verify that $\{ A-(B \bigcup C)\}= \{ (A-B) \bigcap (A-C)\}$
5. If $A= \{ 1,2,3,4\}$, $B= \{ 3,4,5, 6\}$, $C= \{ 4,5,6,7,8\}$ and universal set $X= \{ 1,2,3,4,5,6,7,8,9,10\}$, then verify the following:
5i) $A\bigcup (B \bigcap C) = (A\bigcup B) \bigcap (A \bigcup C)$
5ii) $A \bigcap (B \bigcup C)= (A \bigcap B) \bigcup (A \bigcap C)$
5iii) $A= (A \bigcap B)\bigcup (A \bigcap B^{'})$
5iv) $B=(A \bigcap B)\bigcup (A^{'} \bigcap B)$
5v) $n(A \bigcup B)= n(A)+n(B)-n(A \bigcap B)$
6. If A and B are subsets of the universal set is X, $n(X)=50$, $n(A)=35$, $n(B)=20$, $n(A^{'} \bigcap B^{'})=5$, find (i) $n(A \bigcup B)$ (ii) $n(A \bigcap B)$ (iii) $n(A^{'} \bigcap B)$ (iv) $n(A \bigcap B^{'})$
7. In a class of 200 students who appeared certain examinations, 35 students failed in MHTCET, 40 in AIEEE, and 40 in IITJEE entrance, 20 failed in MHTCET and AIEEE, 17 in AIEEE and IITJEE entrance, 15 in MHTCET and IITJEE entrance exam and 5 failed in all three examinations. Find how many students (a) did not flunk in any examination (b) failed in AIEEE or IITJEE entrance.
8. From amongst 2000 literate and illiterate individuals of a town, 70 percent read Marathi newspaper, 50 percent read English newspapers, and 32.5 percent read both Marathi and English newspapers. Find the number of individuals who read
8i) at least one of the newspapers
8ii) neither Marathi and English newspaper
8iii) only one of the newspapers
9) In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take the tea and milk both and everyone takes at least one beverage, find the number of students in the hostel.
10) There are 260 persons with a skin disorder. If 150 had been exposed to chemical A, 74 to chemical B, and 36 to both chemicals A and B, find the number of persons exposed to (a) Chemical A but not Chemical B (b) Chemical B but not Chemical A (c) Chemical A or Chemical B.
11) If $A = \{ 1,2,3\}$ write down the power set of A.
12) Write the following intervals in Set Builder Form: (a) $(-3,0)$ (b) $[6,12]$ (c) $(6,12]$ (d) $[-23,5)$
13) Using Venn Diagrams, represent (a) $(A \bigcup B)^{'}$ (b) $A^{'} \bigcup B^{'}$ (c) $A^{'} \bigcap B$ (d) $A \bigcap B^{'}$
Regards,
Nalin Pithwa.
### References for IITJEE Foundation Mathematics and Pre-RMO (Homi Bhabha Foundation/TIFR)
1. Algebra for Beginners (with Numerous Examples): Isaac Todhunter (classic text): Amazon India link: https://www.amazon.in/Algebra-Beginners-Isaac-Todhunter/dp/1357345259/ref=sr_1_2?s=books&ie=UTF8&qid=1547448200&sr=1-2&keywords=algebra+for+beginners+todhunter
2. Algebra for Beginners (including easy graphs): Metric Edition: Hall and Knight Amazon India link: https://www.amazon.in/s/ref=nb_sb_noss?url=search-alias%3Dstripbooks&field-keywords=algebra+for+beginners+hall+and+knight
3. Elementary Algebra for School: Metric Edition: https://www.amazon.in/Elementary-Algebra-School-H-Hall/dp/8185386854/ref=sr_1_5?s=books&ie=UTF8&qid=1547448497&sr=1-5&keywords=elementary+algebra+for+schools
4. Higher Algebra: Hall and Knight: Amazon India link: https://www.amazon.in/Higher-Algebra-Knight-ORIGINAL-MASPTERPIECE/dp/9385966677/ref=sr_1_6?s=books&ie=UTF8&qid=1547448392&sr=1-6&keywords=algebra+for+beginners+hall+and+knight
5. Plane Trigonometry: Part I: S L Loney: https://www.amazon.in/Plane-Trigonometry-Part-1-S-L-Loney/dp/938592348X/ref=sr_1_16?s=books&ie=UTF8&qid=1547448802&sr=1-16&keywords=plane+trigonometry+part+1+by+s.l.+loney
The above references are a must. Best time to start is from standard VII or standard VIII.
-Nalin Pithwa.
### Pre RMO Practice question: 2018: How long does it take for a news to go viral in a city? And, a cyclist vs horseman
Problem 1:
Some one arrives in a city with very interesting news and within 10 minutes tells it to two others. Each of these tells the news within 10 minutes to two others(who have not heard it yet), and so on. How long will it take before everyone in the city has heard the news if the city has three million inhabitants?
Problem 2:
A cyclist and a horseman have a race in a stadium. The course is five laps long. They spend the same time on the first lap. The cyclist travels each succeeding lap 1.1 times more slowly than he does the preceding one. On each lap the horseman spends d minutes more than he spent on the preceding lap. They each arrive at the finish line at the same time. Which of them spends the greater amount of time on the fifth lap and how much greater is this amount of time?
I hope you enjoy “mathematizing” every where you see…
Good luck for the Pre RMO in Aug 2018!
Nalin Pithwa.
### How to solve equations: Dr. Vicky Neale: useful for Pre-RMO or even RMO training
Dr. Neale simply beautifully nudges, gently encourages mathematics olympiad students to learn to think further on their own…
### A nice dose of practice problems for IITJEE Foundation math and PreRMO
It is said that “practice makes man perfect”.
Problem 1:
Six boxes are numbered 1 through 6. How many ways are there to put 20 identical balls into these boxes so that none of them is empty?
Problem 2:
How many ways are there to distribute n identical balls in m numbered boxes so that none of the boxes is empty?
Problem 3:
Six boxes are numbered 1 through 6. How many ways are there to distribute 20 identical balls between the boxes (this time some of the boxes can be empty)?
Finish this triad of problems now!
Nalin Pithwa.
### IITJEE Foundation Math and PRMO (preRMO) practice: another random collection of questions
Problem 1: Find the value of $\frac{x+2a}{2b--x} + \frac{x-2a}{2a+x} + \frac{4ab}{x^{2}-4b^{2}}$ when $x=\frac{ab}{a+b}$
Problem 2: Reduce the following fraction to its lowest terms:
$(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \div (\frac{x+y+z}{x^{2}+y^{2}+z^{2}-xy-yz-zx} - \frac{1}{x+y+z})+1$
Problem 3: Simplify: $\sqrt[4]{97-56\sqrt{3}}$
Problem 4: If $a+b+c+d=2s$, prove that $4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}=16(s-a)(s-b)(s-c)(s-d)$
Problem 5: If a, b, c are in HP, show that $(\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+\frac{9}{b^{2}}=\frac{25}{ac}$.
May u discover the joy of Math! 🙂 🙂 🙂
Nalin Pithwa.
### Pre-RMO (PRMO) Practice Problems
Pre-RMO days are back again. Here is a list of some of my random thoughts:
Problem 1:
There are five different teacups, three saucers, and four teaspoons in the “Tea Party” store. How many ways are there to buy two items with different names?
Problem 2:
We call a natural number “odd-looking” if all of its digits are odd. How many four-digit odd-looking numbers are there?
Problem 3:
We toss a coin three times. How many different sequences of heads and tails can we obtain?
Problem 4:
Each box in a 2 x 2 table can be coloured black or white. How many different colourings of the table are there?
Problem 5:
How many ways are there to fill in a Special Sport Lotto card? In this lotto, you must predict the results of 13 hockey games, indicating either a victory for one of two teams, or a draw.
Problem 6:
The Hermetian alphabet consists of only three letters: A, B and C. A word in this language is an arbitrary sequence of no more than four letters. How many words does the Hermetian language contain?
Problem 7:
A captain and a deputy captain must be elected in a soccer team with 11 players. How many ways are there to do this?
Problem 8:
How many ways are there to sew one three-coloured flag with three horizontal strips of equal height if we have pieces of fabric of six colours? We can distinguish the top of the flag from the bottom.
Problem 9:
How many ways are there to put one white and one black rook on a chessboard so that they do not attack each other?
Problem 10:
How many ways are there to put one white and one black king on a chessboard so that they do not attack each other?
I will post the answers in a couple of days.
Nalin Pithwa.
### Three in a row !!!
If my first were a 4,
And, my second were a 3,
What I am would be double,
The number you’d see.
For I’m only three digits,
Just three in a row,
So what must I be?
Don’t say you don’t know!
Cheers,
Nalin Pithwa. | 2019-09-22 12:42:00 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 362, "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": 0.9647656083106995, "perplexity": 602.2883204159482}, "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/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00541.warc.gz"} |
http://math.stackexchange.com/questions/222974/probability-of-getting-2-aces-2-kings-and-1-queen-in-a-five-card-poker-hand-pa | # Probability of getting 2 Aces, 2 Kings and 1 Queen in a five card poker hand (Part II)
So I reworked my formula in method 1 after getting help with my original question - Probability of getting 2 Aces, 2 Kings and 1 Queen in a five card poker hand. But I am still getting results that differ...although they are much much closer than before, but I must still be making a mistake somewhere in method 1. Anyone know what it is?
Method 1
$P(2A \cap 2K \cap 1Q) = P(Q|2A \cap 2K)P(2A|2K)P(2K)$
$$= \frac{1}{12}\frac{{4 \choose 2}{46 \choose 1}}{50 \choose 3}\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}$$
$$= \frac{(6)(17296)(6)(46)}{(2598960)(19600)(12)}$$
$$= 4.685642 * 10^{-5}$$
Method 2
$$\frac{{4 \choose 2} {4 \choose 2}{4 \choose 1}}{52 \choose 5} = \frac{3}{54145}$$
$$5.540678 * 10^{-5}$$
-
Please make an effort to make the question self-contained and provide a link to your earlier question. – Sasha Oct 28 '12 at 19:56
I think we would rather ahve you edit your initial question by adding your new progress. This avoids having loss of answer and keeps track of progress – Jean-Sébastien Oct 28 '12 at 19:56
But there already answers to my original question so those answers would not make sense now that I am using a new formula for method 1. – sonicboom Oct 28 '12 at 20:03
Conditional probability arguments can be delicate. Given that there are exactly two Kings, what's the $46$ doing? That allows the possibility of more Kings. – André Nicolas Oct 28 '12 at 20:26
The $46$ is because have already taken two kings from the pack leaving us with 50. And now we have chosen 2 aces and we have to pick the other 1 card from the 50 remaining cards less the 4 aces? – sonicboom Oct 28 '12 at 20:42
show 1 more comment
$$\frac{1}{11}\frac{{4 \choose 2}{44 \choose 1}}{48 \choose 3}\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}$$
If you wrote this as $$\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}\frac{{4 \choose 2}{44 \choose 1}}{48 \choose 3}\frac{{4 \choose 1}{40 \choose 0}}{44 \choose 1}$$ it might be more obvious why they are the same. | 2014-03-07 11:01:44 | {"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": 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": 0.7363124489784241, "perplexity": 334.5089065648005}, "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/1393999642170/warc/CC-MAIN-20140305060722-00084-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/35899/relation-between-independent-increments-and-markov-property | # Relation between independent increments and Markov property
Independent increments and Markov property.do not imply each other. I was wondering
• if being one makes a process closer to being the other?
• if there are cases where one implies the other?
Thanks and regards!
-
To see this, assume that $(X_n)_{n\ge0}$ has independent increments, that is, $X_0=0$ and $X_n=Y_1+\cdots+Y_n$ for every $n\ge1$, where $(Y_n)_{n\ge1}$ is a sequence of independent random variables. The filtration of $(X_n)_{n\ge0}$ is $(\mathcal{F}^X_n)_{n\ge0}$ with $\mathcal{F}^X_n=\sigma(X_k;0\le k\le n)$. Note that $$\mathcal{F}^X_n=\sigma(Y_k;1\le k\le n),$$ hence $X_{n+1}=X_n+Y_{n+1}$ where $X_n$ is $\mathcal{F}^X_n$ measurable and $Y_{n+1}$ is independent on $\mathcal{F}^X_n$. This shows that the conditional distribution of $X_{n+1}$ conditionally on $\mathcal{F}^X_n$ is $$\mathbb{P}(X_{n+1}\in\mathrm{d}y|\mathcal{F}^X_n)=Q_n(X_n,\mathrm{d}y), \quad \mbox{where}\quad Q_n(x,\mathrm{d}y)=\mathbb{P}(x+Y_{n+1}\in\mathrm{d}y).$$ Hence $(X_n)_{n\ge0}$ is a Markov chain with transition kernels $(Q_n)_{n\ge0}$.
@Didier: Thanks! But I think it doesn't because of the following. First $P(X(t_3) | X(t_2), X(t_1)) = P(X(t_3)-X(t_2)|X(t_2), X(t_2)-X(t_1))$. Next $P(X(t_3)-X(t_2)|X(t_2), X(t_2)-X(t_1)) = P(X(t_3)-X(t_2)|X(t_2))$, if and only if $X(t_3)-X(t_2)$ and $X(t_2)-X(t_1))$ are conditionally independent given $X(t_2)$, which can not be implied by $X(t_3)-X(t_2)$ and $X(t_2)-X(t_1))$ are independent. Any mistake? – Tim Apr 29 '11 at 20:54
What is $P(W|U,V)$ for three random variables $U$, $V$, $W$? – Did Apr 29 '11 at 22:43
Why should "independent increments" require that $Y_j$ are independent of $X_0$? $X_0$ is not an increment. – Robert Israel Apr 29 '11 at 23:08
@Didier: Thanks! 1) I still have no clue how to explain and correct (2) in my last comment. Would you point me where in what texts/materials? 2) Generally when saying increments of a stochastic process, is $X_0$ an increment? Does the definition of an independent-increment process require $X_0=0$? – Tim May 3 '11 at 12:29
Invoking "smartness" here is a way to avoid following the explicit suggestions I made, which would lead you to understand the problem. It is also a cheap shot at my advice, considering the time and work I spent on your questions. // Since once again you are stopped by matters of definitions I suggest to come back to definitions: consider random variables $\xi$ and $\eta$ and a sigma-algebra $G$ such that $\xi$ is independent on $H=\sigma(\eta)\vee G$. Why is $E(u(\xi+\eta)\mid H)=E(u(\xi+\eta)\mid\eta)$ for every bounded $u$? Why is this related to your question? .../... – Did Nov 6 '11 at 8:48 | 2016-06-27 02:45:07 | {"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": 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": 0.9062880873680115, "perplexity": 229.43696224200997}, "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-2016-26/segments/1466783395620.56/warc/CC-MAIN-20160624154955-00053-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-is-the-derivative-of-f-x-e-2x-ln-x | What is the derivative of f(x)=(e^(2x))(ln(x))?
Mar 3, 2017
$f ' \left(x\right) = {e}^{2 x} \left(2 \ln x + \frac{1}{x}\right)$
Explanation:
The derivative of $\ln x$ is $\frac{1}{x}$
The derivative of ${e}^{g} \left(x\right)$is ${e}^{g} \left(x\right) \cdot g ' \left(x\right)$
The derivative of $h \left(x\right) \cdot l \left(x\right)$ is $h ' \left(x\right) \cdot l \left(x\right) + h \left(x\right) \cdot l ' \left(x\right)$
Then
$f ' \left(x\right) = {e}^{2 x} \cdot 2 \cdot \ln x + {e}^{2 x} \cdot \frac{1}{x}$
$= {e}^{2 x} \left(2 \ln x + \frac{1}{x}\right)$ | 2019-10-15 07:23:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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": 0.8896889090538025, "perplexity": 561.6444014124368}, "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-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00023.warc.gz"} |
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1187
str and repr | Pydon't
Python's str and repr built-in methods are similar, but not the same. Use str to print nice-looking strings for end users and use repr for debugging purposes. Similarly, in your classes you should implement the __str__ and __repr__ dunder methods with these two use cases in mind.
415
What all promotions actually mean
Nowadays stores come up with all sorts of funky promotions to catch your eye... But how much money do you actually save with each type of promotion?
2318
Assignment expressions and the walrus operator := | Pydon't
The walrus operator := can be really helpful, but if you use it in convoluted ways it will make your code worse instead of better. Use := to flatten a sequence of nested ifs or to reuse partial computations.
712
Problem #028 - hidden key 🗝️
There is a key hidden in one of three boxes and each box has a coin on top of it. Can you use the coins to let your friend know where the key is hiding?
992
EAFP and LBYL coding styles | Pydon't
In Python, if you are doing something that may throw an error, there are many cases in which it is better to "apologise than to ask for permission". This means you should prefer using a try block to catch the error, instead of an if statement to prevent the error.
1145
Unpacking with starred assignments | Pydon't
How should you unpack a list or a tuple into the first element and then the rest? Or into the last element and everything else? Pydon't unpack with slices, prefer starred assignment instead.
400
Problem #027 - pile of coconuts 🥥
Five sailors and their monkey were washed ashore on a desert island. They decide to go get coconuts that they pile up. During the night, each of the sailors, suspicious the others wouldn't behave fairly, went to the pile of coconuts take their fair share. How many coconuts were there in the beginning..?
1769
Pydon't disrespect the Zen of Python
The "Zen of Python" is the set of guidelines that show up in your screen if you import this. If you have never read them before, read them now and again from time to time. If you are looking to write Pythonic code, write code that abides by the Zen of Python.
860
Problem #026 - counting squares
I bet you have seen one of those Facebook publications where you have a grid and you have to count the number of squares the grid contains, and then you jump to the comment section and virtually no one agrees on what the correct answer should be... Let's settle this once and for all!
542
Problem #025 - knight's tour
Alice and Bob sit down, face to face, with a chessboard in front of them. They are going to play a little game, but this game only has a single knight... Who will win?
3605
Pydon't Manifesto
"Pydon'ts" are short, to-the-point, meaningful Python programming tips. A Pydon't is something you should not do when programming in Python. In general, following a Pydon't will make you write more Pythonic code.
752
Problem #024 - hats in a line
Some people are standing quiet in a line, each person with a hat that has one of two colours. How many people can guess their colour correctly?
430
Filling your Pokédex - a probabilistic outlook
Join me in this blog post for Pokéfans and mathematicians alike. Together we'll find out how long it would take to fill your complete Pokédex by only performing random trades.
457
Implementing an interpreter in 14 lines of Python.
In this blog post I'll show you how you can write a full interpreter for the brainf*ck programming language in just 14 lines of Python. Be prepared, however, to see some unconventional Python code!
498
Problem #023 - guess the polynomial
In this problem you have to devise a strategy to beat the computer in a "guess the polynomial" game.
310
Twitter proof: consecutive integers are coprime
Let's prove that if $$k$$ is an integer, then $$\gcd(k, k+1) = 1$$. That is, any two consecutive integers are coprime.
295
Twitter proof: maximising the product with a fixed sum
Let's prove that if you want to maximise $$ab$$ with $$a + b$$ equal to a constant value $$k$$, then you want $$a = b = \frac{k}{2}$$.
442
Problem #022 - coprimes in the crowd
This simple problem is an example of a very interesting phenomenon: if you have a large enough "universe" to consider, even randomly picked parts exhibit structured properties.
725
Problem #021 - predicting coin tosses
Alice and Bob are going to be locked away separately and their faith depends on their guessing random coin tosses!
409
Let's build a simple interpreter for APL - part 3 - the array model
In this blog post we will go over some significant changes, from implementing APL's array model to introducing dyadic operators! | 2021-04-22 22:41:04 | {"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": 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": 0.2345961183309555, "perplexity": 1445.706594397109}, "config": {"markdown_headings": false, "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-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00371.warc.gz"} |
https://mathematica.stackexchange.com/questions/194590/integers-to-soundnote-for-midi-generation | Integers to soundnote for midi generation
To generate a midi file given two lists of integers of equal length, one for the note pitch and one for the corresponding note duration (tempo), I'd like to use the built in Midi file generator, but am not sure how to proceed to map the integers to the soundnote "C#" etc, I would like to use an 88 note mapping like a piano, and perhaps 5 discrete note duration values. Thanks.
I saw this but it takes a sound note and gives a number, whereas I'd like to generate 88 soundnotes scaled linearly from my list of integers.
Getting MIDI SoundNote Pitches as Numeric Values
This is what I have so far, a 0.25second fixed note duration, and a list of values which I am not sure about for the range of soundnotes they generate:
Sound[SoundNote[#, 0.25, "Piano"] & /@ {0, -7, -50, 7, 12, 50, 0, -10,
50, -50, 0, 0, 10, 60, 65, 67}]
Export["sequence.mid", %]
Thanks.
cheers, Jamie
If you want to use 2 integer lists, try this
pitch = {0, 2, 4, 5, 7, 9, 11, 12};
tempo = {.5, 1, .5, 1, .3, .2, .1, .1};
Sound[SoundNote[#, #2, "Piano"] & @@@ Transpose@{pitch, tempo}]
As for the mapping the 88 keys are the Range[-39,48]
-39 is A-1, -38 is A#-1 ,-37 is B-1 , -36 is C0 ,-35 is C#0 etc
If Mod[Tone,12]=0 then you have a C
so -36 is C0, -24 is C1, -12 is C2 , 0 is C3, 12 is C4 ... 48 is C7
using Mod[#,12] you can easily find the tones
0 is C, 1 is C#, 2 is D, 3 is D#, 4 is E, 5 is F, 6 is F#, 7 is G, 8 is G#, 9 is A, 10 is A# and 11 is B
Mod[#,12] actually is the reminder of the division #/12,
so it can take values from 0 to 11 which are the 12 notes
But if you don't want to use integers you can use the builtin notation:
pitch = {"C3", "D3", "E3", "F3", "G3", "A3", "B3", "C4"};
tempo = {.5, 1, .5, 1, .3, .2, .1, .1};
Sound[SoundNote[#, #2, "Piano"] & @@@ Transpose@{pitch, tempo}] | 2020-02-24 12:40:19 | {"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": 0.5054821372032166, "perplexity": 2888.0448661706596}, "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-2020-10/segments/1581875145941.55/warc/CC-MAIN-20200224102135-20200224132135-00184.warc.gz"} |
https://www.transtutors.com/questions/set-up-a-definite-integral-that-represents-the-length-of-the-curve-y-x-cos-x--5375990.htm | # Set up a definite integral that represents the length of the curve y = x + cos x...
Set up a definite integral that represents the length of the curve y = x + cos x for 0 5 x 5 it. Then use your calculator to find the length rounded off to four decimal places. Note: x is given in radians.
Attachments:
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Free Plagiarism Checker | 2020-05-25 13:54:58 | {"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": 0.8035172820091248, "perplexity": 433.085334932267}, "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/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00470.warc.gz"} |
http://www.drumtom.com/q/what-is-the-domain-range-of-y-10-x-compare-to-the-equation-y-1-x | FIND THE ANSWERS
# What is the domain range of y= -10/x. compare to the equation y = 1/x?
### Answer this question
• What is the domain range of y= -10/x. compare to the equation y = 1/x?
## Answers
Discusses the domain and range of a function, ... I'll just list the x-values for the domain and the y-values for the range: domain: {–3, –2, –1 ...
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Positive: 54 %
Compute domain and range for functions of several variables. HOME ABOUT PRODUCTS BUSINESS RESOURCES ... domain of f(x,y) = log(1-(x^2+y^2)) Related Wolfram ...
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Positive: 51 %
### More resources
... For function y=1/x-2 Give the y values for x =-1,0,1,2,3,4 ... Start with the given equation Plug in Calculate the y value by following the order of ...
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Positive: 54 %
{x x 1, x R} and a range of {y ... y 8. How many roots does the equation 2 81 ... of the function 2 2 4 x x y. b) Identify the domain, range, ...
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Positive: 49 %
everything maths & science. ... Functions of the form y = 1 x. ... Domain and range. For y = a x + q, the function is undefined for x = 0.
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Positive: 35 %
Discovering the characteristics. For functions of the general form $$f(x) = y = a(x + p)^2 + q$$: Domain and range. The domain is \(\left\{x:x\in ℝ\right ...
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Positive: 12 %
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Discover Questions | 2017-01-24 03:28:46 | {"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": 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": 0.31406155228614807, "perplexity": 2512.0758548780245}, "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-2017-04/segments/1484560283689.98/warc/CC-MAIN-20170116095123-00537-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://eurekamathanswers.com/use-of-integers/ | An integer is a number that includes 0, positive numbers, and negative numbers. It can never be a fraction, decimal, or percent. Integers are mainly used in our day-to-day lives in mathematical terms. Get to know the definition, operations, and use of integers in the below-mentioned sections. Also, get the example questions with solutions for the convenience of grade 6 math students.
Also, Check: Use of Integers as Directed Numbers
## What is an Integer?
Integers are a set of counting numbers (positive and negative) along with zero. Some of the examples of integers are -4, -3, -2, -1, 0, 1, 2, 3. The integers are represented by Z. The types of integers are positive integers, negative integers, and zero.
### Integer Rules
The rules depend on the operations performed on integers given below:
• If the sign of both integers is the same, then the sum will have the same sign.
• If the sign of one integer is positive, the sign of another integer is negative, then the sign of the result is the sign of the larger number.
• Subtraction Rule:
• Convert the operation to addition by changing the sign of the subtrahend.
• Multiplication Rule:
• Multiply the sign of integers to get the result sign.
• Division Rule:
• Divide the signs of two operands and get the resultant sign.
### Real Life Examples of Integers
The examples on using integers are along the lines:
• If profit is represented by positive integers, losses are by negative integers.
• The rise in the price of a product is represented by positive integers and fill in price by negative integers.
• If heights above the sea level are represented by positive integers, then depths below sea level by negative integers, and so on.
### Integers as Directed Numbers
If a number represents direction, then the number is called a directed number. The below-given examples explain it in detail.
Example:
If +4 represents 4 m towards East, then -5 represents 5 m towards its opposite direction i.e towards west.
If a positive integer shows a particular direction, then the negative integer shows the opposite direction.
### Example Questions on Use of Integers
Question 1:
Write an integer to describe a situation
(i) Losing Rs 100
(ii) Owing Rs 1500
(iii) Depositing $500 in a bank Solution: (i) An integer representing a loss of Rs 100 is -100. (ii) An integer representing an owing of Rs 1500 is -1500 (iii) An integer representing depositing$500 in a bank is +500
Question 2:
Write an appropriate integer for each of the following:
(i) earned Rs 800 interest
(ii) a decrease of 5 members
(iii) an increase of 3 inches
Solution:
(i) Earned Rs 800 interest is represented by +800
(ii) Decrease of 5 members is represented by -5
(iii) An increase of 3 inches is represented by +3
### Frequently Asked Question’s on Using Integers
1. What are the applications of integers?
Integers are used to signify two contradicting situations. The positive and negative integers have different applications. Integers can compare and measure changes in temperature, credits, debits calculation by the bank.
2. What are the integer rules?
The integer rules are the sum of two integers is an integer, a difference of two integers is an integer. Multiplication of two or more integers is an integer. The division of integers may or may not be an integer.
3. What are the integer properties?
The main properties of integers are closure property, commutative property, associative property, identity property, and distributive property.
4. What are the 5 integer operations?
The operations with integers are addition, subtraction, multiplication, and division. | 2022-01-22 09:07:35 | {"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": 0.48777544498443604, "perplexity": 961.7523615772751}, "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/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00002.warc.gz"} |
http://www.maths.usyd.edu.au/u/UG/JM/MATH1111/Quizzes/quiz32.html | ## MATH1111 Quizzes
Local Linearity and the Differential Quiz
Web resources available Questions
This quiz tests the work covered in lecture on local linearity and the differential and corresponds to Section 14.3 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There is a useful applet at http://www.slu.edu/classes/maymk/banchoff/TangentPlane.html - take some time to read the instructions and add your own functions.
There are more web quizzes at Wiley, select Section 3. This quiz has 10 questions.
Suppose $f\left(3,2\right)=4\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{f}_{x}\left(3,2\right)=-2$ and ${f}_{y}\left(3,2\right)=3$ for some surface $z=f\left(x,y\right)\phantom{\rule{0.3em}{0ex}}.$
Which of the following is the tangent plane to the surface at $\left(3,2,4\right)\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
a) $4z=-2\left(x-3\right)+3\left(y-2\right)$ b) $z=4-2\left(x-3\right)+3\left(y-2\right)$ c) $z+4=2\left(x+3\right)+3\left(y+2\right)$ d) $4z=3\left(x+3\right)-2\left(y+2\right)$
Choice (a) is incorrect
Try again, check the formula for the tangent plane.
Choice (b) is correct!
The tangent at the point $\left(a,b\right)$ on the surface is $z=f\left(a,b\right)+{f}_{x}\left(a,b\right)\left(x-a\right)+{f}_{\left(}a,b\right)\left(y-b\right)$ so the above equation is correct.
Choice (c) is incorrect
Try again, check the formula for the tangent plane.
Choice (d) is incorrect
Try again, check the formula for the tangent plane.
Is the plane $z=12+8\left(x-1\right)+7\left(y-2\right)$ the tangent plane to the surface, $f\left(x,y\right)={x}^{2}+3xy+{y}^{2}-1$ at $\left(1,2\right)\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
a) Yes. b) No
Choice (a) is incorrect
$f\left(1,2\right)=10$ and $z=12$ at $\left(1,2\right)$ so the plane does not touch the surface.
Choice (b) is correct!
${f}_{x}\left(x,y\right)=2x+3y$ so ${f}_{x}\left(1,2\right)=2+6=8$
${f}_{y}\left(x,y\right)=3x+2y$ so ${f}_{y}\left(1,2\right)=3+4=7$
$f\left(1,2\right)=10$ so the tangent plane is $z=10+8\left(x-1\right)+7\left(y-2\right)\phantom{\rule{0.3em}{0ex}}.$
Which of the following is the tangent plane to the surface $f\left(x,y\right)={x}^{2}-2xy-3{y}^{2}$ at the point $\left(-2,1,5\right)\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
a) $z+6x+2y+15=0$ b) $z-6x-2y+5=0$ c) $z+6x+2y+5=0\phantom{\rule{0.3em}{0ex}}.$ d) None of the above, since $\left(-2,1,5\right)$ is not on the surface.
Choice (a) is incorrect
Try again, look carefully at the signs of the constant terms.
Choice (b) is incorrect
Try again, carefully rearrange your equation.
Choice (c) is correct!
${f}_{x}\left(x,y\right)=2x-2y$ so ${f}_{x}\left(-2,1\right)=-4-2=-6$
${f}_{y}\left(x,y\right)=-2x-6y$ so ${f}_{y}\left(-2,1\right)=4-6=-2$
$f\left(-2,1\right)=5$ so the tangent plane is $z=5-6\left(x+2\right)-2\left(y-1\right)⇒z+6x+2y+5=0$ as required.
Choice (d) is incorrect
Try again, the point is on the surface.
Which of the following is the differential of $f\left(x,y\right)=sinxy\phantom{\rule{0.3em}{0ex}}{e}^{xy}\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
a) $df=cosxy\phantom{\rule{0.3em}{0ex}}{e}^{xy}\left({y}^{2}\phantom{\rule{0.3em}{0ex}}dx+{x}^{2}\phantom{\rule{0.3em}{0ex}}dy\right)$ b) $df={e}^{xy}\left(cosxy+sinxy\right)\left(x\phantom{\rule{0.3em}{0ex}}dx+y\phantom{\rule{0.3em}{0ex}}dy\right)$ c) $df={e}^{xy}\left(-cosxy+sinxy\right)\left(y\phantom{\rule{0.3em}{0ex}}dx+x\phantom{\rule{0.3em}{0ex}}dy\right)$ d) $df={e}^{xy}\left(cosxy+sinxy\right)\left(y\phantom{\rule{0.3em}{0ex}}dx+x\phantom{\rule{0.3em}{0ex}}dy\right)$
Choice (a) is incorrect
Try again, you must use the product rule to differentiate $f\left(x,y\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, you have not differentiated $f\left(x,y\right)$ correctly.
Choice (c) is incorrect
Try again, you have not differentiated $sinxy$ correctly.
Choice (d) is correct!
${f}_{x}\left(x,y\right)=ycosxy\phantom{\rule{0.3em}{0ex}}{e}^{xy}+sinxy\left(y{e}^{xy}\right)=y{e}^{xy}\left(cosxy+sinxy\right)$ using the product rule, and
${f}_{y}\left(x,y\right)=xcosxy\phantom{\rule{0.3em}{0ex}}{e}^{xy}+sinxy\left(x{e}^{xy}\right)=x{e}^{xy}\left(cosxy+sinxy\right)$ using the product rule, so
$df=y{e}^{xy}\left(cosxy+sinxy\right)\phantom{\rule{0.3em}{0ex}}dx+x{e}^{xy}\left(cosxy+sinxy\right)\phantom{\rule{0.3em}{0ex}}dy={e}^{xy}\left(cosxy+sinxy\right)\left(y\phantom{\rule{0.3em}{0ex}}dx+x\phantom{\rule{0.3em}{0ex}}dy\right)\phantom{\rule{0.3em}{0ex}}.$ | 2018-03-20 11:58:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 45, "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": 0.794314444065094, "perplexity": 679.7104532559423}, "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/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00660.warc.gz"} |
https://mathematicsgre.com/viewtopic.php?f=1&t=382 | ## FR0568 #41
Forum for the GRE subject test in mathematics.
thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am
### FR0568 #41
Hey can any solve this for me?
I know how to solve the line intergral, but what are the upper and lower limits?
origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm
### Re: FR0568 #41
When you make new threads like this, please post the problem to make it easier for everyone.
The question is
Let C be the circle $$x^2 + y^2 = 1$$ oriented counterclockwise in the xy-plane. What is the value of the line integral
$$\oint_C (2x-y) dx + (x+3y)dy$$
A) 0
B) 1
C) pi/2
D) pi
E) 2pi
The limits you need for the integral will depend on the parametrization of the circle you use. You could use the parametrization $$y = \sqrt{1-x^2}$$ for the top half of the circle, and then your x would go from 1 to -1. You'll also need to compute the integral on the bottom half.
However, I think actually attempting to compute that line integral would be excessively difficult and miss the point of the question, use the other techniques at your disposal.
thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am
### Re: FR0568 #41
God it s hard integrating this line integral, have you got another method in mind origin? Coz I could sure it rite now.
origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm
### Re: FR0568 #41
And spoil all the fun of it?
Alright, Green's Theorem.
Basically anytime you have a surface integral, you should be checking if its easier to integrate the boundary, and anytime you have a closed line integral, you should be checking if its easier to integrate the surface. The GRE guys are tricky like that.
mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am
### Re: FR0568 #41
line integral was pretty straigt forward here.
and the ans I calculated is 2pi
thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am
### Re: FR0568 #41
hey greens theorem did the trick! Hardly took any time!! thanks!!! | 2023-02-05 11:10:53 | {"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": 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": 0.8142852187156677, "perplexity": 2269.68555150114}, "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/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00373.warc.gz"} |
https://www.transtutors.com/questions/consider-the-following-multilayer-perceptron-network-the-transfer-function-of-the-hi-2011189.htm | Consider the following multilayer perceptron network. (The transfer function of the hidden layer...
Consider the following multilayer perceptron network. (The transfer function of the hidden layer is
The initial weights and biases are:
Perform one iteration of the standard steepest descent backpropagation (use matrix operations) with learning rate a = 0.5 for the following input/ target pair:
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Free Plagiarism Checker | 2021-04-21 06:04:06 | {"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": 0.8393312692642212, "perplexity": 3066.8290442374864}, "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/1618039508673.81/warc/CC-MAIN-20210421035139-20210421065139-00105.warc.gz"} |
https://www.tensorflow.org/graphics/api_docs/python/tfg/geometry/transformation/axis_angle/from_euler_with_small_angles_approximation | # tfg.geometry.transformation.axis_angle.from_euler_with_small_angles_approximation
Converts small Euler angles to an axis-angle representation.
Under the small angle assumption,
$$\sin(x)$$
and
$$\cos(x)$$
can be approximated by their second order Taylor expansions, where
$$\sin(x) \approx x$$
and
$$\cos(x) \approx 1 - \frac{x^2}{2}$$
. In the current implementation, the smallness of the angles is not verified.
#### Note:
The conversion is performed by first converting to a quaternion representation, and then by converting the quaternion to an axis-angle.
#### Note:
In the following, A1 to An are optional batch dimensions.
angles A tensor of shape [A1, ..., An, 3], where the last dimension represents the three small Euler angles. [A1, ..., An, 0] is the angle about x in radians [A1, ..., An, 1] is the angle about y in radians and [A1, ..., An, 2] is the angle about z in radians.
name A name for this op that defaults to "axis_angle_from_euler_with_small_angles_approximation".
A tuple of two tensors, respectively of shape [A1, ..., An, 3] and [A1, ..., An, 1], where the first tensor represents the axis, and the second represents the angle. The resulting axis is a normalized vector. | 2020-05-26 18:06:14 | {"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": 0.9281443357467651, "perplexity": 1994.3965365572162}, "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/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00328.warc.gz"} |
https://www.effortlessmath.com/math-puzzles/algebra-puzzle-challenge-50/ | # Algebra Puzzle – Challenge 50
This is another math puzzle and brain teaser that is interactive, challenging, and entertaining for those who love Math challenges!
## Challenge:
If 11 workers can build 11 cars in 11 days, then how many days would it take 7 workers to build 7 cars?
A- 7
B- 9
C- 11
D- 14
E- 18
### The Absolute Best Book to challenge your Smart Student!
If, 11 workers can build one car per day, then, one worker can make a car in 11 days. (Each worker can build $$\frac{1}{11}$$ of a car per day. So, it takes 11 days for a worker to make a car) | 2022-01-20 08:06:06 | {"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": 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": 0.1920827031135559, "perplexity": 1741.780503342366}, "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/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00023.warc.gz"} |
http://clay6.com/qa/48042/a-charge-of-8-mc-is-located-at-the-origin-calculate-the-work-done-in-taking | Want to ask us a question? Click here
Browse Questions
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0 votes
# A charge of $8 \;mC$ is located at the origin. Calculate the work done in taking a small charge of $−2 \times 10^{−9} C$ from a point $P (0, 0, 3 cm)$ to a point $Q (0, 4 cm, 0),$ via a point $R (0, 6 cm, 9 cm).$
Can you answer this question?
## 1 Answer
0 votes
$(B)1.27 J$
Hence B is the correct answer.
answered Jun 23, 2014 by | 2016-12-07 14:24:36 | {"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": 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": 0.7803388237953186, "perplexity": 644.9861641090696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00280-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1644647/how-can-we-prove-a-statement-is-provable?noredirect=1 | # How can we prove a statement is provable?
Given a concrete mathematical statement, such as BSD conjecture(https://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture), do we know if it is provable?
• I suspect the answer, in the vast majority of specific cases, is going to be, quite simply, "We don't." I've never heard of a non-independence result that doesn't itself discern whether the statement is true or false. I would be interested in finding out if such a thing exists - by the consistency theorem, you could start with two models, one of the statement and one of its negation, and try and derive a contradiction. – Dustan Levenstein Feb 7 '16 at 16:21
• What do you mean by provable? If you mean does a proof exist - then it is just as hard as to prove conjecture. Only provably correct conjecture provably exist a proof. If you mean if it is possible to have a proof, however, then it is easy. The only thing that you cannot write a proof are "non-statements". For example, one cannot write a proof to "Good Morning", or "How are you" – Andrew Au Feb 7 '16 at 16:27
• @AndrewAu I was thinking people are trying to prove BSD conjecture, but is it possible that the conjecture is not provable by Godel's incompleteness theorem? – Qixiao Feb 7 '16 at 17:46
• A statement is not "provable" in and of itself. It is only provable relative to a particular axiom system. The most common way to show an axiom system doesn't prove a statement is to build a model of the system that doesn't satisfy the statement. For BSD there seems to be no specific reason to suspect it is unprovable from ZFC set theory. – Carl Mummert Feb 13 '16 at 13:34
• In general, however, there is no algorithm that can decide whether arbitrary statements are provable from ZFC. They have to be considered on a case by case basis. – Carl Mummert Feb 13 '16 at 13:35
You're using the wrong term. You mean to ask whether we can tell if a conjecture is decidable, meaning that it is either provable or disprovable. But no we cannot tell whether a statement is decidable if the quantifier complexity is too high. Furthermore, it may be possible that even the decidability of a statement is itself undecidable! (See below for an example.)
First read https://math.stackexchange.com/a/1643073/21820, to ensure that you fully understand the import of Godel's incompleteness theorem. After that, consider the following. $\def\imp{\rightarrow}$
[We work in a meta-system and assume that $PA$ is omega-consistent.]
Let $φ = \square_{PA} Con(PA) \lor \square_{PA} \neg Con(PA)$. [So $φ$ expresses "Con(PA) is decidable over $PA$".]
If $PA \vdash φ$:
Within $PA$:
$\square Con(PA) \lor \square \neg Con(PA)$.
If $\square Con(PA)$:
$\neg Con(PA)$. [by the internal incompleteness theorem]
$\square \bot$.
$\square \neg Con(PA)$. [by (D1),(D2)]
$\square \neg Con(PA)$. [by basic logic]
$\neg Con(PA)$. [because $PA$ is omega-consistent]
Contradiction. [with the external incompleteness theorem]
Therefore $PA \nvdash φ$.
If $PA \vdash \neg φ$:
Within $PA$:
$\neg \square Con(PA)$. [by basic logic]
If $\square \bot$:
$\square Con(PA)$. [by (D1),(D2)]
$\neg \square \bot$.
$Con(PA)$.
Therefore $PA \nvdash \neg φ$.
Thus $φ$ is independent of $PA$. | 2019-05-21 09:37:56 | {"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": 0.7555591464042664, "perplexity": 467.9351876700904}, "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-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00470.warc.gz"} |
https://socratic.org/questions/how-do-you-use-the-product-rule-to-differentiate-g-x-x-2-1-x-2-2x | # How do you use the product rule to differentiate g(x)=(x^2+1)(x^2-2x)?
Jan 14, 2017
$g ' \left(x\right) = 4 {x}^{3} - 6 {x}^{2} + 2 x - 2$
#### Explanation:
$\text{Given " g(x)=f(x).h(x)" then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{g ' \left(x\right) = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$
$\text{here } f \left(x\right) = {x}^{2} + 1 \Rightarrow f ' \left(x\right) = 2 x$
$\text{and } h \left(x\right) = {x}^{2} - 2 x \Rightarrow h ' \left(x\right) = 2 x - 2$
$\Rightarrow g ' \left(x\right) = \left({x}^{2} + 1\right) \left(2 x - 2\right) + \left({x}^{2} - 2 x\right) .2 x$
$= 2 {x}^{3} - 2 {x}^{2} + 2 x - 2 + 2 {x}^{3} - 4 {x}^{2}$
$= 4 {x}^{3} - 6 {x}^{2} + 2 x - 2$ | 2019-02-17 10:30:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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": 0.9969086647033691, "perplexity": 2172.637939934715}, "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/1550247481832.13/warc/CC-MAIN-20190217091542-20190217113542-00265.warc.gz"} |
https://math.stackexchange.com/questions/1403926/expected-value-of-die-rolls-roll-n-keep-1 | # Expected value of die rolls - roll $n$, keep $1$
I know how to calculate expected value for a single roll, and I read several other answers about expected value with rerolls, but how does the calculation change if you can make your reroll before choosing which die to keep?
For instance, what is the expected value of rolling $2$ fair $6$-sided dice and keeping the higher value? And can you please generalize to $n$ $x$-sided dice?
• If you wish to find distribution of $\max$ of $n$ i.i.d. random variables, then $P(\max\{X_1,..,X_n\}<x) = P(X_1<x,...,X_n<x)=$/*they are independent*/$=\prod_{i=1}^n P(X_i<x)=$/*probabilities are equal*/$=(P(X_1<x))^n$ – Slowpoke Aug 20 '15 at 14:29
• The expectation of the sum without rerolling is (for $k$ $n-sided$ dices) : $\frac{k(n+1)}{2}$ – Peter Aug 20 '15 at 14:34
• @hcl14, thank you for the response. Could you please show an example? I don't understand some of the notation you are using. – Catherine Aug 20 '15 at 14:39
• @Catherine The notation means, that if you have maximum of a few variables be less or equal than some value, then every variable is less or equal than that value. That allows you to easily write the distribution function $F_{max}(x)=P(\max\{...\}\leq x)$ of the maximum as a product of distribution functions of the variables. Then expectation can be easily computed: as long as for one m-sided dice $P(X_1\leq x) = x/m$, then $F_{max}(x)=(x/m)^n$ and $P(\max \{..\}=x) = F_{max}(x)-F_{max}(x-1)$. $E[\max]=\sum_{x=1}^m x*P(\max \{..\}=x)$ which will lead to the result in Jason Carr's answer. – Slowpoke Aug 20 '15 at 15:00
So to calculate this one in particular isn't all that difficult, but it's a special case of order statistics, that is, generating statistics about multiple events when they're ordered. You'd need to use that for the middle.
In this case where we take the one highest die, consider that to be less than or equal any given value, we must have both dice no greater than that value. So, it's the intersection of the probabilities that each individual die is no greater than than the value. If we have a cumulative distribution function for a die, then it describes the probability that the die will roll at most some value.
In the case of s-sided dice, we have that $P(X \le a) = \{\frac{a}{s}, a \in [1,s]\}$.
To find out what the intersection of multiple dice is, we take the intersection of their probabilities, so noting that that intersection of a number of distinct events is $\prod(P)$ we can get that our new $P(X \le a)$ is $\frac{a^n}{s^n}$ or similar equivalent expressions for the intersection of n s-sided dice
Now in order to get the expected value, we need to get the probability distribution function, that is $P(X = a)$. To do this we'll take the discrete difference. We can't really simplify this, so we'll just take $P(X = a) = \frac{(a)^n}{s^n} - \frac{(a - 1)^n}{s^n}$ Then we can take the summation of each of these for all $a \in [1,s]$
Then the expected value is the sum $\sum_{a=1}^s{a({\frac{(a)^n}{s^n} - \frac{(a - 1)^n}{s^n})}}$
To start with two dices. You can make a table and insert the maximum values of the two dices.
$\begin{array}{|c|c|c|c|c|c|} \hline \text{dice 1 / dice 2 } & 1 &2 &3 &4 &5 &6 \\ \hline\hline \hline 1 & &2 & &4 &5 & \\ \hline 2 & 2 &2 & &4 &&6 \\ \hline 3&3 &3 &3 &4 &5&6 \\ \hline 4 & & &&4&5&6 \\ \hline 5 &5 &5&5&5&5& \\ \hline 6 &6&&&6&6&6 \\ \hline \end{array}$
I left out some values to leave some work to do for you.
The probability for each combination is $p_{ij}=\frac{1}{36}$
The expected value then is $E(x)=\sum_{i=1}^6\sum_{j=1}^6 p_{ij} \cdot max(x_i,x_j)$
A start: Let $X$ be the "larger" number when you roll two fair $n$-sided dice. Then $$\Pr(X=k)=\Pr(X\le k)-\Pr(X\le k-1).$$ But the probability that $X$ is $\le a$ is the probability both dice are $\le a$. This is $\frac{a^2}{n^2}$.
Remark: There are easier (and arguably better) ways to show that $\Pr(X\le k)=\frac{2k-1}{n^2}$. But the above trick is a useful one.
The same idea works for tossing $d$ $n$-sided dice. The probability that the maximum $X$ is $k$ is $\frac{k^d}{n^d}-\frac{(k-)^d}{n^d}$. | 2019-10-20 02:50:11 | {"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": 0.8720001578330994, "perplexity": 223.04828978898834}, "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-2019-43/segments/1570986702077.71/warc/CC-MAIN-20191020024805-20191020052305-00414.warc.gz"} |
https://tex.stackexchange.com/questions/361159/how-to-draw-a-pretty-tree-diagram-i-already-did-one-but-it-is-very-ugly | # How to draw a pretty tree-diagram (I already did one, but it is very ugly)
\documentclass{standalone}
\usepackage{forest}
\usepackage{rotating}
\begin{document}
\centering
\begin{forest}
for tree={
if level=0{
align=center,
l sep=20mm,
}{%
align={@{}C{1.5em}@{}},
edge path={
\noexpand\path [draw, \forestoption{edge}] (!u.parent anchor) -- +(0,-5mm) -| (.child anchor)\forestoption{edge label};
},
},
draw,
font=\sffamily\bfseries,
parent anchor=south,
child anchor=north,
l sep=10mm,
edge={thick, rounded corners=1pt},
thick,
inner color=gray!5,
outer color=gray!20,
rounded corners=2pt,
fzr/.style={
alias=fzr,
align=center,
child anchor=west,
fill=green!25,
edge path={
\noexpand\path[\forestoption{edge}]
([yshift=-1em]!u.parent anchor) -- (.child anchor)\forestoption{edge label};
},
},
}
[manager, alias=master, align=center
% [expatriate,fzr]
[\rotatebox{90}{bureau}]
[\rotatebox{90}{production}
[\rotatebox{90}{line}]
]
[\rotatebox{90}{finance}]
[\rotatebox{90}{quality},align=center
%[quality supervisor,fzr]
[\rotatebox{90}{laboratory}]
[\rotatebox{90}{review}]
]
[\rotatebox{90}{supply}
[\rotatebox{90}{material}]
[\rotatebox{90}{\parbox{2.5cm}{Semi-finished\\ products}}]
[\rotatebox{90}{\parbox{2.5cm}{Finished\\ product}}]
]
]
\node [draw,font=\sffamily\bfseries,thick,fill=green!25,rounded corners=2pt,xshift=25mm,] (fuzeren) [yshift=-1.3em,] {expatriate};
\path [draw,thick] ([yshift=-.5em]!master.south) -- (fuzeren);
\end{forest}
\end{document}
It appears to be something like this:
In fact I modified the code from someone and I don't really understand everything inside the code. The diagram I desire is:
1. The 'manager' aligns with 'quality' and the 'supply' aligns with 'semi finished products' (I already let the 'quality' align = center. I don't know why it doesn't align with manager).
2. The turnings of the connecting line should not have fillet, to let the intersection to be straight right.
3. It should spare some place for the 'expatriat' for that it doesn't touch the horizontal line.
4. The frame of each level should be aligned by the upper border.
Anyway, would someone give a solution to help achieve my desired tree-diagram?
• Who's someone? Please attribute code and provide a link to the original. – cfr Mar 30 '17 at 12:27
• See the Forest manual for better ways to rotate nodes. You don't want to use \rotatebox here, I don't think. – cfr Mar 30 '17 at 13:47
• Don't you get a compilation error? If we had the original source, we might at least be able to compare them. – cfr Mar 30 '17 at 23:34
• You didn't test this before uploading, did you? How is the C column type defined? – cfr Mar 30 '17 at 23:35
• The original code uses ctex, a chinese language package and all contents are in chinese. The original code, if you want, is here : paste.ubuntu.com/24288316 Thanks a lot. – jiexiang wen Mar 31 '17 at 14:53
Without working code, with no clue where the original source with missing stuff might be found, it is easier to just start from scratch.
Use the edges library for forked edges. Have Forest do the rotation. Have Forest place the expatriate. Then the code is much simpler, cleaner and more straightforward.
If you don't understand something in the code, look it up in the manual. If you don't understand the explanation, ask. If you use code from somebody else, attribute it. People have names. They are not anonymous some-bodies.
\documentclass[border=10pt]{standalone}
\usepackage[edges]{forest}
\begin{document}
\begin{forest}
forked edges,
for tree={
edge+={thick},
inner color=gray!5,
outer color=gray!20,
rounded corners=2pt,
draw,
thick,
tier/.option=level,
align=center,
font=\sffamily,
},
where level<=1{}{
rotate=90,
anchor=east,
},
[manager
[, coordinate, calign with current
[bureau]
[production
[line]
]
[finance]
[quality, calign with current
[laboratory]
[review]
]
[supply
[material]
[Semi-finished\\products, calign with current]
[Finished\\product]
]
]
[expatriate, inner color=green!10, outer color=green!25, child anchor=west, edge path'={(!u.parent anchor) -| (.child anchor)}, before drawing tree={y'+=7.5pt} ]
]
\end{forest}
\end{document}
• Thank you. I have provided the original code in comment. – jiexiang wen Mar 31 '17 at 14:57
• And the source is a friend, he wanted to do this and his code does't run neither, I modified into english and post here for an answer. – jiexiang wen Mar 31 '17 at 15:05 | 2019-08-20 05:02:40 | {"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": 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": 0.5196300745010376, "perplexity": 4913.866983804146}, "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-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00411.warc.gz"} |
https://aitopics.org/mlt?cdid=arxivorg%3A0D86D0F0&dimension=pagetext | to
### Quality Evaluation of GANs Using Cross Local Intrinsic Dimensionality
Generative Adversarial Networks (GANs) are an elegant mechanism for data generation. However, a key challenge when using GANs is how to best measure their ability to generate realistic data. In this paper, we demonstrate that an intrinsic dimensional characterization of the data space learned by a GAN model leads to an effective evaluation metric for GAN quality. In particular, we propose a new evaluation measure, CrossLID, that assesses the local intrinsic dimensionality (LID) of real-world data with respect to neighborhoods found in GAN-generated samples. Intuitively, CrossLID measures the degree to which manifolds of two data distributions coincide with each other. In experiments on 4 benchmark image datasets, we compare our proposed measure to several state-of-the-art evaluation metrics. Our experiments show that CrossLID is strongly correlated with the progress of GAN training, is sensitive to mode collapse, is robust to small-scale noise and image transformations, and robust to sample size. Furthermore, we show how CrossLID can be used within the GAN training process to improve generation quality.
### Manifold regularization with GANs for semi-supervised learning
Generative Adversarial Networks are powerful generative models that are able to model the manifold of natural images. We leverage this property to perform manifold regularization by approximating a variant of the Laplacian norm using a Monte Carlo approximation that is easily computed with the GAN. When incorporated into the semi-supervised feature-matching GAN we achieve state-of-the-art results for GAN-based semi-supervised learning on CIFAR-10 and SVHN benchmarks, with a method that is significantly easier to implement than competing methods. We also find that manifold regularization improves the quality of generated images, and is affected by the quality of the GAN used to approximate the regularizer.
### Variational Approaches for Auto-Encoding Generative Adversarial Networks
Auto-encoding generative adversarial networks (GANs) combine the standard GAN algorithm, which discriminates between real and model-generated data, with a reconstruction loss given by an auto-encoder. Such models aim to prevent mode collapse in the learned generative model by ensuring that it is grounded in all the available training data. In this paper, we develop a principle upon which auto-encoders can be combined with generative adversarial networks by exploiting the hierarchical structure of the generative model. The underlying principle shows that variational inference can be used a basic tool for learning, but with the in- tractable likelihood replaced by a synthetic likelihood, and the unknown posterior distribution replaced by an implicit distribution; both synthetic likelihoods and implicit posterior distributions can be learned using discriminators. This allows us to develop a natural fusion of variational auto-encoders and generative adversarial networks, combining the best of both these methods. We describe a unified objective for optimization, discuss the constraints needed to guide learning, connect to the wide range of existing work, and use a battery of tests to systematically and quantitatively assess the performance of our method.
### Perturbative GAN: GAN with Perturbation Layers
Perturbative GAN, which replaces convolution layers of existing convolutional GANs (DCGAN, WGAN-GP, BIGGAN, etc.) with perturbation layers that adds a fixed noise mask, is proposed. Compared with the convolu-tional GANs, the number of parameters to be trained is smaller, the convergence of training is faster, the incep-tion score of generated images is higher, and the overall training cost is reduced. Algorithmic generation of the noise masks is also proposed, with which the training, as well as the generation, can be boosted with hardware acceleration. Perturbative GAN is evaluated using con-ventional datasets (CIFAR10, LSUN, ImageNet), both in the cases when a perturbation layer is adopted only for Generators and when it is introduced to both Generator and Discriminator.
### ChainGAN: A sequential approach to GANs
We propose a new architecture and training methodology for generative adversarial networks. Current approaches attempt to learn the transformation from a noise sample to a generated data sample in one shot. Our proposed generator architecture, called $\textit{ChainGAN}$, uses a two-step process. It first attempts to transform a noise vector into a crude sample, similar to a traditional generator. Next, a chain of networks, called $\textit{editors}$, attempt to sequentially enhance this sample. We train each of these units independently, instead of with end-to-end backpropagation on the entire chain. Our model is robust, efficient, and flexible as we can apply it to various network architectures. We provide rationale for our choices and experimentally evaluate our model, achieving competitive results on several datasets. | 2022-01-20 05:48:18 | {"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": 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": 0.537478506565094, "perplexity": 1010.6213934574956}, "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-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00120.warc.gz"} |
http://es.mathworks.com/help/robust/ref/psinfo.html?nocookie=true | # psinfo
Inquire about polytopic or parameter-dependent systems created with `psys`
## Syntax
```psinfo(ps)
[type,k,ns,ni,no] = psinfo(ps)
pv = psinfo(ps,'par')
sk = psinfo(ps,'sys',k)
sys = psinfo(ps,'eval',p)
```
## Description
`psinfo ` is a multi-usage function for queries about a polytopic or parameter-dependent system `ps` created with `psys`. It performs the following operations depending on the calling sequence:
• `psinfo(ps)` displays the type of system (affine or polytopic); the number `k` of `SYSTEM` matrices involved in its definition; and the numbers of `ns`, `ni`, `no` of states, inputs, and outputs of the system. This information can be optionally stored in MATLAB® variables by providing output arguments.
• `pv = psinfo(ps,'par')` returns the parameter vector description (for parameter-dependent systems only).
• `sk = psinfo(ps,'sys',k)` returns the k-th `SYSTEM` matrix involved in the definition of `ps`. The ranking k is relative to the list of systems `syslist` used in `psys`.
• `sys = psinfo(ps,'eval',p)` instantiates the system for a given vector p of parameter values or polytopic coordinates.
For affine parameter-dependent systems defined by the `SYSTEM` matrices S0, S1, . . ., Sn, the entries of `p` should be real parameter values p1, . . ., pn and the result is the LTI system of `SYSTEM` matrix
S(p) = S0 + p1S1 + . . .+ pnSn
For polytopic systems with `SYSTEM` matrix ranging in
Co{S1, . . ., Sn},
the entries of `p` should be polytopic coordinates p1, . . ., pn satisfying pj ≥ 0 and the result is the interpolated LTI system of `SYSTEM` matrix
$S=\frac{{p}_{1}{S}_{1}+\cdots +{p}_{n}{S}_{n}}{{p}_{1}+\cdots +{p}_{n}}$ | 2015-05-06 17:44:39 | {"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": 0.8108884692192078, "perplexity": 2428.035381164915}, "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-2015-18/segments/1430458969644.56/warc/CC-MAIN-20150501054249-00022-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-10-exponents-and-radicals-review-exercises-chapter-10-page-693/1 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
By the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the given statement is $\text{ TRUE .}$ | 2020-04-08 16:43:35 | {"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": 0.8470041751861572, "perplexity": 537.4119410260097}, "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/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00367.warc.gz"} |
End of preview.
The dataset was derived from OpenWebMath after cleaning about 1.65m garbled and non-mathematical documents.
You can find the detailed cleaning process on my blog.
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