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https://www.futilitycloset.com/	# Line Limit You own a goat and a meadow. The meadow is in the shape of an equilateral triangle each side of which is 100 meters long. The goat is tied to a post at one corner of the meadow. How long should you make the tether in order to give the goat access to exactly half the meadow? # An Actor’s Notes Ellen Terry played Juliet at London’s Lyceum Theatre in 1882. The following was later found on the flyleaf of her copy of the text: Get the words into your remembrance first of all. Then, (as you have to convey the meaning of the words to some who have ears, but don’t hear, and eyes, but don’t see) put the words into the simplest vernacular. Then exercise your judgment about their sound. So many different ways of speaking words! Beware of sound and fury signifying nothing. Voice unaccompanied by imagination, dreadful. Pomposity, rotundity. Imagination and intelligence absolutely necessary to realize and portray high and low imaginings. Voice, yes, but not mere voice production. You must have a sensitive ear, and a sensitive judgment of the effect on your audience. But all the time you must be trying to please yourself. Get yourself into tune. Then you can let fly your imagination, and the words will seem to be supplied by yourself. Shakespeare supplied by oneself! Oh! Realism? Yes, if we mean by that real feeling, real sympathy. But people seem to mean by it only the realism of low-down things. To act, you must make the thing written your own. You must steal the words, steal the thought, and convey the stolen treasure to others with great art. (From Donald Sinden, ed., The Everyman Book of Theatrical Anecdotes, 1987.) # Time Saver Ogden Nash invented a streamlined limerick he called the “limick”: An old person of Troy In the bath is so coy That it doesn’t know yet If it’s a girl or a boy. Two nudists of Dover, When purple all over, Were munched by a cow, When mistaken for clover. A cook called McMurray Got a raise in a hurry From his Hindu employer, By flavouring curry. A young flirt of Ceylon, Who led the boys on, Succumbed to a swan. # For What It’s Worth Based on an ancient Hindu game, Snakes and Ladders (Chutes and Ladders in ophidiophobic America) is at heart a morality lesson: As you progress by die roll from square 1 to square 100 and spiritual enlightenment, your way is complicated by virtues and vices. Landing on a snake (or chute) will send you back to an earlier square, and landing on a ladder will send you ahead to a later one. Each of these shortcuts is associated with a precept — “Carelessness” leads to “Injury,” “Study” leads to “Knowledge,” and so on. In 1993 University of Michigan mathematician S.C. Althoen and his colleagues considered the game as a 101-state absorbing Markov chain. The shortest possible game lasts seven moves, the longest is infinite, and according to their calculations the expected number of moves in the Milton Bradley version of Chutes and Ladders is $\displaystyle \frac{225837582538403273407117496273279920181931269186581786048583}{5757472998140039232950575874628786131130999406013041613400},$ So, arguably, we might advance more quickly through life with more vice and less virtue. (S.C. Althoen, L. King, and K. Schilling, “How Long Is a Game of Snakes and Ladders?”, Mathematical Gazette 77:478 [March 1993], 71-76.) # Black and White I just ran across this in Benjamin Glover Laws’ The Two-Move Chess Problem, from 1890. It’s by G. Chocholous. White is to mate in two moves. # La On its 35th anniversary, THX, the sound quality assurance company founded by George Lucas, released the original score of “Deep Note,” its audio trademark, which debuted at the premiere of Return of the Jedi in 1983 and is now familiar from countless films. Essentially it’s a stupendous D chord; the U.S. trademark registration reads: The THX logo theme consists of 30 voices over seven measures, starting in a narrow range, 200 to 400 Hz, and slowly diverting to preselected pitches encompassing three octaves. The 30 voices begin at pitches between 200 Hz and 400 Hz and arrive at pre-selected pitches spanning three octaves by the fourth measure. The highest pitch is slightly detuned while there are double the number of voices of the lowest two pitches. “I like to say that the THX sound is the most widely-recognized piece of computer-generated music in the world,” says James A. Moorer, who wrote it. “This may or may not be true, but it sounds cool.” And now that we have the score you can do this: # Math Notes 286 = 77371252455336267181195264 That’s the largest known power of 2 with no zeros in its decimal representation. Are there higher such powers? No one knows. # In a Word any-lengthian anythingarianism n. the fact or phenomenon of not holding any fixed or established beliefs # Cube Route Created by Franz Armbruster in 1967, “Instant Insanity” was the Rubik’s Cube of its day, a simple configuration task with a dismaying number of combinations. You’re given four cubes whose faces are colored red, blue, green, and yellow: The task is to arrange them into a stack so that each of the four colors appears on each side of the stack. This is difficult to achieve by trial and error, as the cubes can be arranged in 41,472 ways, and only 8 of these give a valid solution. One approach is to use graph theory — draw points of the four face colors and connect them to show which pairs of colors fall on opposite faces of each cube: Then, using certain criteria (explained here), we can derive two directed subgraphs that describe the solution: The first graph shows which colors appear on the front and back of each cube, the second which colors appear on the left and right. Each arrow represents one of the four cubes and the position of each of the two colors it indicates. So, for example, the black arrow at the top of the first graph indicates that the first cube will have yellow on the front face and blue on the rear. This solution isn’t unique, of course — once you’ve compiled a winning stack you can rotate it or rearrange the order of the cubes without affecting its validity. B.L. Schwartz gives an alternative method, through inspection of a table, as well as tips for solving by trial and error using physical cubes, in “An Improved Solution to ‘Instant Insanity,'” Mathematics Magazine 43:1 (January 1970), 20-23.	2018-09-24 02:22:03	{"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": 6, "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": 0.4043438732624054, "perplexity": 2603.994160267637}, "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-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00355.warc.gz"}
https://cmc.deusto.eus/existence-and-classification-of-solutions-to-the-cahn-hilliard-equation/	# Existence and Classification of Solutions to the Cahn-Hilliard Equation Tuesday, July 12th, 2019 11:30-12:00H Logistar Room at DeustoTech, University of Deusto. Matteo Rizzi Centro de Modelamiento Matemático, Universidad de Chile, Santiago, Chile. In the talk I will present the construction of a family {uε} of solutions to the Cahn-Hilliard equation {-}\varepsilon \Delta u_{\varepsilon}=\varepsilon^{-1}\left(u_{\varepsilon}-u_{\varepsilon}^{3}\right)-\ell_{\varepsilon}, \quad \ell_{\varepsilon} \in \mathbb{R} whose zero level set is prescribed and approaches, as ε → 0, a given complete, embedded, k-ended constant mean curvature surface. It is a joint work with Michal Kowalczyk. Moreover, I will present some classication results, dealing with properties such as boundedness, monotonicity and radial symmetry. See the slides here	2021-04-11 22:24:55	{"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": 0.9498239755630493, "perplexity": 3403.3203988896767}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00282.warc.gz"}
https://www.allanswered.com/post/zxpow/	### UnitTest & code coverage 44 views 1 11 days ago by Hi! I'm trying to set up test execution. There is a difficult moment. Tests must be performed to configure code coverage for my application Mebel.exe, which is already a Python interpreter. Wing does not allow me to select it as an interpreter because the name must be python. Can I somehow to configure the execution of unit testing by calling my interpreter is named mebel.exe? Thank You! Community: Wing Python IDE I made my app start using the launch configuration. But yet I get the error. Traceback (most recent call last): File "C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners/run_unittests_xml.py", line 177, in <module> main(list(sys.argv)) File "C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners/run_unittests_xml.py", line 87, in main xmlout = wingtest_common.CreateOutputStream(argv) File "C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners\wingtest_common.py", line 689, in CreateOutputStream xmlout = xml_stream_cls(raw_stream) File "C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners\wingtest_common.py", line 101, in __init__ self.write_raw('<%s>' % self._top_tag) File "C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners\wingtest_common.py", line 154, in write_raw self._raw_stream.write(s) TypeError: must be str, not bytes Python: Ошибка запуска скрипта C:\Program Files (x86)\Wing IDE 6.0\src/testing/runners/run_unittests_xml.py written 11 days ago by Alexander Dragunkin Does it do the same thing if you debug the test from the Testing tool instead? If so, what is in s? It should just be an ascii tag name so the error is not yet making sense to me. written 11 days ago by Wingware Support Error on the side of my application. I will look for solutions. written 11 days ago by Alexander Dragunkin ### 2 Answers 2 11 days ago by It should be possible to pick any file for the Python Executable, even if named something like mebel.exe. Wing then tries to run it to validate that it's Python and shows errors if it's not. From your comment it sounds like you got that working. I'll follow up on the second problem above. 0 9 days ago by Hi! I want to understand if I can use the tests tool of the Wing IDE to run test programs in my application. I can run TestRuner.py from the Windows command prompt, run the following code. So I can run the same code by configuring the "OS Commands" tool in the appropriate configuration. I want to use the "Testing" tool to get a report in the corresponding WIng IDE window, but I don't think I can configure it correctly. Maybe I'm asking stupid questions! Please forgive me! Thank you for any tips on how to organize unit testing with code coverage by my app. I think if you set Python Options (below Python Executable, same dialog/tab) to -m it'll run with the same arguments as from command line or OS Commands. However, I don't know what test framework is being used. If it's unittest, doctest, nose, or pytest then Wing should be able to run the test from the Testing tool. If it's something else then a new test runner would be needed. You can see the existing ones in src/testing/runners in your Wing installation, in case that helps any. written 9 days ago by Wingware Support Please login to add an answer/comment or follow this question. Similar posts: Search »	2018-04-21 03:38:11	{"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": 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": 0.2810277044773102, "perplexity": 5060.134854906192}, "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-2018-17/segments/1524125944982.32/warc/CC-MAIN-20180421032230-20180421052230-00191.warc.gz"}
https://physics.stackexchange.com/questions/502934/is-diffusion-of-electrons-under-fick-law-from-n-to-p-side-of-a-pn-junction-dom	# Is diffusion of electrons (under Fick law) from N to P side of a PN junction dominated by electrons coming from doped or from non doped atoms ? Why? Let's consider a PN junction between two semiconductor doped P and N. Let's concentrate on the N side for example. Due to Fick law, the electrons from the N side migrate to the P side. What is the origin of these electrons that migrate : the N doped atoms or the non doped atoms ? Or both ? Is one source dominant ? If yes, which one, and why ? Do all additive electrons of the N doped atoms migrate or only a subfraction ? Do some electrons of non doped atoms migrate also ? Thank you. • An electron is an electron- how can you tell where it came from? – Jon Custer Sep 16 '19 at 2:00 • does is come from the non doped atom or from the doped atom ? Or both ? I mean is the liaison energy different for the doped and the non doped atom ? – Mathieu Krisztian Sep 16 '19 at 5:35 • Once an electron is in the conduction band, you have no way of knowing where it came from. And, very few electrons from the n-type are going to make it to the p-type region - the excess of holes pretty much guarantees recombination. – Jon Custer Sep 16 '19 at 17:42 • Sorry you escape the question. I focus on N side. Are electron that migrate from Fick law mainly coming from the doped or the non doped atoms ? – Mathieu Krisztian Sep 17 '19 at 16:18 • Do electron need to be in conduction band to diffuse ? If so since electrons from doped atoms are closer to conduction band I would expect that people say that electrons that diffuse are highly dominated by those coming from the doped atoms with respect to the non doped atoms. ? – Mathieu Krisztian Sep 17 '19 at 16:21	2020-02-17 19:43:11	{"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": 0.812621533870697, "perplexity": 1293.6257094773096}, "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-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00320.warc.gz"}
http://quant.stackexchange.com/questions?page=6&sort=newest&pagesize=15	# All Questions 95 views ### How to Calculate a Monte Calo VaR estimation error I'm performing a Monte Carlo to calculate value at risk (with a 3 dimension risk factor) Now, I would like to calculate the error of the estimation of the VaR with respect to the number of simulations ... 113 views ### what was the quant role in the 2008 crash? this is a complex topic that interests me, have researched myself, & is debated heavily in the media and there is lots of writing, even entire books/documentaries. maybe somewhat surprisingly, ... 154 views ### How are quants able to verify whether their calculated prices are any good This question is related to the discussion on Model Validation Criteria However it appeard to be very high level to me and I would like to go more into detail. Not working at a pricing desk the ... 37 views How does buying back stock affect a company's credit spread? 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In each period, five dates are observed (labelled a to e), where a denotes the day the period ... 165 views ### How to replicate this option? I have a question I am not sure how to approach: Suppose interest rates is 50%, a stock worth \$1 today can be worth \$2, \$1, \$0.5 next year. If the option that pays \$1 only when S = \$2 is ... 82 views ### What is the most amount of money the consumer would be willing to pay to play take this gamble? Suppose a consumer has log-utility over wealth, defined by $u(W) = \ln(W)$. Suppose this consumer has $100$, and is considering taking a gamble in which the consumer flips a coin, and gets $20$ she ... 77 views ### Attributing the change in NII to Shift, Twist and Butterfly The movement of the zero rate curves can be decomposed into a shift movement (the level of interest rates) and a twist movement (the slope of the curve) and butterfly (the curvature of the curve). 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If you are the book keeper for a publicly traded company and you misrepresent the financial earnings of the company (even if by accident) for an earnings quarter, is this illegal?	2014-04-17 06:47:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.8765577077865601, "perplexity": 2041.3287454865279}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00433-ip-10-147-4-33.ec2.internal.warc.gz"}
https://www.raumflair.ch/concrete/2020-Wed-4342/	## Latest News Spiral classifier Iron ore spiral classifier (iron ore classifier) is a new classification equipment which can be used in the ore dressing process of iron ore pulp for efficient and accurate classification It has the features of simple structure reliable performance easy to operate and other advantag • ### Disadvantages Of Spiral Classifier Disadvantages of spiral classifier pflege-kompetenzteam.de. Fl spiral classifier spiral classifier, also called screw classifier, is widely used for distributing ore in the close circuit with ball mill, grading ore and fine slit in the gravity mill, grading granularity in the flow of metal ore-dressing and de-s a. • ### disadvantages of spiral classifier MC World disadvantages of spiral classifier WereldPraktijk. helix spiral concentrator. spiral concentrator mongolia for sale theboxalmelo. Weihai Haiwang Gold Spiral Separator,Spiral Chute, Find Complete Details about Weihai is the national cyclone standard drafter and takes a 70% domestic market share in Chile, Peru, Brazil, America, India, South Africa, Russia, Mongolia and etc So far spiral May 05, 2020· In its diagrammatic representation, it looks like a spiral with many loops, that’s the reason it’s called as Spiral. Each loop of the spiral is called a Phase of the software development process. This model has capability to handle risks. These are following advantages and disadvantages of using Spiral Model. Advantages of Spiral Model: • ### Spiral classifier for sand washing with small footprint LZZG The main disadvantages are low classification efficiency, heavy equipment, and large floor space. Due to the limitation of equipment specifications and production capacity, it is generally not possible to form a closed circuit with a ball mill with a specification of Φ3.6m or more. Diagram of Spiral model: Advantages of Spiral model: High amount of risk analysis hence, avoidance of Risk is enhanced. Good for large and mission-critical projects. Strong approval and documentation control. Additional Functionality can be added at a later date. Software is produced early in the software life cycle. Disadvantages of Spiral model: • ### Spiral Classifier EKCP Spiral Classifier. EIMCO-K.C.P. Spiral Classifiers is art of separating the solid particles in a mixture of solids and liquid into fractions according to particle size or density by methods other than screening. In general, the products resulting are (1) a partially drained fraction containing the coarse material (called the underflow) and (2 • ### What is Spiral Model? When to Use? Advantages & Disadvantages Advantages and Disadvantages of Spiral Model. Advantages. Disadvantages. Additional functionality or changes can be done at a later stage. Risk of not meeting the schedule or budget. Cost estimation becomes easy as the prototype building is done in small fragments. The State of Kuwait is the world’s eighth biggest exporter and the tenth largest producer of oil with a small land mass of 20,000 square kilometers and 4.2 million population, is a viable destination for foreign investment. Geopolitical position of Kuwait in the northern Arabian Gulf bordering three major markets of KSA, Iraq and Iran presents a perfect location for entering markets in • ### Advantages and challenges of exporting to Kuwait Oct 05, 2020· In its diagrammatic representation, it looks like a spiral with many loops, that’s the reason it’s called as Spiral. Each loop of the spiral is called a Phase of the software development process. This model has capability to handle risks. These are following advantages and disadvantages of using Spiral Model. Advantages of Spiral Model: Diagram of Spiral model: Advantages of Spiral model: High amount of risk analysis hence, avoidance of Risk is enhanced. Good for large and mission-critical projects. Strong approval and documentation control. Additional Functionality can be added at a later date. Software is produced early in the software life cycle. Disadvantages of Spiral model: • ### Global and United States Submerged Spiral Classifier Submerged Spiral Classifier market is segmented by region (country), players, by Type, and by Application. Players, stakeholders, and other participants in the global Submerged Spiral Classifier market will be able to gain the upper hand as they use the report as a powerful resource. 3, Spiral submerged arc welded pipe features: spiral weld line distribution, long welds, especially under dynamic conditions is in the welding, the weld had a chance to cool and left the molding point, easy to produce welding hot cracking. Direction parallel to the weld crack, a certain angle the axis of the steel pipe, typically between 30-70 °. • ### Advanced spiral periodic classification of the elements The proposed new periodic classification is a spiral arrangement of the elements, arranged by their increasing atomic number, electronic configuration and recurring chemical properties. It has 32 groups and 8 periods. 2. For the convenience of the readers, a dark line is drawn at the starting as well as ending points of the periods • ### What is Spiral Model in SDLC? \|Professionalqa What is Spiral Model? Spiral Model is one of the oldest form of the Software Development Life Cycle Models(SDLC), which was first defined by the Barry Boehm in the year 1986. Basically, this model is an evolutionary type model, which works on the combined approach of the waterfall and iterative model.. This model is driven by the risk analysis and evaluation requirements and keeps on iterating These are various software engineering models and their advantages and disadvantages 1. Waterfall Model: It requires a well understanding and knowledge of requirements and technology related to it. Advantages : It is very easy and convenient to implement the waterfall model. For implementation of small systems, it is very useful • ### Types of Classifiers in Mineral Processing Metallurgical ContentSpiral ClassifierScrew/Spiral Classifier Capacity TableAllen Cone ClassifierCone or Pyramid ClassifierCross-Flow ClassifierSpiral-Screw Classifier Capacity TableHydraulic ClassifierHydro-classifierHydroclassifier Capacity TableRake ClassifierRake Classifier Capacity Rotary High Weir ClassifierRotary High Weir Classifier Capacity Spiral Classifier In mineral processing, the • ### Kuwait Economy: Population, GDP, Inflation, Business Kuwait’s economic freedom score is 63.2, making its economy the 79th freest in the 2020 Index. Its overall score has increased by 2.4 points, primarily because of a higher score for government • ### 1.17. Neural network models (supervised) — scikit-learn 0 1.17.1. Multi-layer Perceptron¶. Multi-layer Perceptron (MLP) is a supervised learning algorithm that learns a function $$f(\cdot): R^m \rightarrow R^o$$ by training on a dataset, where $$m$$ is the number of dimensions for input and $$o$$ is the number of dimensions for output. Given a set of features $$X = {x_1, x_2,, x_m}$$ and a target $$y$$, it can learn a non-linear function • ### What is Heterogeneous Welding? Examples, Advantages What is heterogeneous welding? Examples, Advantages and Disadvantages. Classification of fusion welding processes. Heterogeneous welding is one way of performing fusion welding with the application of external filler whose metallurgical composition is substantially different from that of the base metals. • ### Disadvantages To Mining Granite Disadvantages To Mining Granite. 2020-09-11 Advantages disadvantages mining ores Grinding quarrying and that this involved digging up and processing large amounts of rock 187 More detailed METALS AND advantages of mining granite quarry in zambia advantages of mining granite quarry in zambia advantages of mining granite. • ### Global and Japan Duplex Twin Spiral Classifiers Market Duplex Twin Spiral Classifiers market is segmented by region (country), players, by Type, and by Application. Players, stakeholders, and other participants in the global Duplex Twin Spiral Classifiers market will be able to gain the upper hand as they use the report as a powerful resource. • ### Classification of Materials and Types of Classifiers Oct 31, 2015· The spiral classifier is also a mechanical classifier. It is made by combining a gravity settler of rectangular section with a sloped transport spiral for the sediment. It consists of a semi cylindrical trough (a trough which is semi-circular in cross- section) inclined to the horizontal. The spiral conveyor moves the solids which settle to the • ### Disadvantages Of Spiral Classifier Disadvantages of spiral classifier pflege-kompetenzteam.de. Fl spiral classifier spiral classifier, also called screw classifier, is widely used for distributing ore in the close circuit with ball mill, grading ore and fine slit in the gravity mill, grading granularity in the flow of metal ore-dressing and de-s a. • ### Global and United States Submerged Spiral Classifier Submerged Spiral Classifier market is segmented by region (country), players, by Type, and by Application. Players, stakeholders, and other participants in the global Submerged Spiral Classifier market will be able to gain the upper hand as they use the report as a powerful resource. 3, Spiral submerged arc welded pipe features: spiral weld line distribution, long welds, especially under dynamic conditions is in the welding, the weld had a chance to cool and left the molding point, easy to produce welding hot cracking. Direction parallel to the weld crack, a certain angle the axis of the steel pipe, typically between 30-70 °. • ### Advanced spiral periodic classification of the elements The proposed new periodic classification is a spiral arrangement of the elements, arranged by their increasing atomic number, electronic configuration and recurring chemical properties. It has 32 groups and 8 periods. 2. For the convenience of the readers, a dark line is drawn at the starting as well as ending points of the periods • ### What is Spiral Model in SDLC? \|Professionalqa What is Spiral Model? Spiral Model is one of the oldest form of the Software Development Life Cycle Models(SDLC), which was first defined by the Barry Boehm in the year 1986. Basically, this model is an evolutionary type model, which works on the combined approach of the waterfall and iterative model.. This model is driven by the risk analysis and evaluation requirements and keeps on iterating These are various software engineering models and their advantages and disadvantages 1. Waterfall Model: It requires a well understanding and knowledge of requirements and technology related to it. Advantages : It is very easy and convenient to implement the waterfall model. For implementation of small systems, it is very useful • ### Types of Classifiers in Mineral Processing Metallurgical ContentSpiral ClassifierScrew/Spiral Classifier Capacity TableAllen Cone ClassifierCone or Pyramid ClassifierCross-Flow ClassifierSpiral-Screw Classifier Capacity TableHydraulic ClassifierHydro-classifierHydroclassifier Capacity TableRake ClassifierRake Classifier Capacity Rotary High Weir ClassifierRotary High Weir Classifier Capacity Spiral Classifier In mineral processing, the • ### What is Heterogeneous Welding? Examples, Advantages What is heterogeneous welding? Examples, Advantages and Disadvantages. Classification of fusion welding processes. Heterogeneous welding is one way of performing fusion welding with the application of external filler whose metallurgical composition is substantially different from that of the base metals. • ### 1.17. Neural network models (supervised) — scikit-learn 0 1.17.1. Multi-layer Perceptron¶. Multi-layer Perceptron (MLP) is a supervised learning algorithm that learns a function $$f(\cdot): R^m \rightarrow R^o$$ by training on a dataset, where $$m$$ is the number of dimensions for input and $$o$$ is the number of dimensions for output. Given a set of features $$X = {x_1, x_2,, x_m}$$ and a target $$y$$, it can learn a non-linear function • ### Disadvantages To Mining Granite Disadvantages To Mining Granite. 2020-09-11 Advantages disadvantages mining ores Grinding quarrying and that this involved digging up and processing large amounts of rock 187 More detailed METALS AND advantages of mining granite quarry in zambia advantages of mining granite quarry in zambia advantages of mining granite • ### Global and Japan Duplex Twin Spiral Classifiers Market Duplex Twin Spiral Classifiers market is segmented by region (country), players, by Type, and by Application. Players, stakeholders, and other participants in the global Duplex Twin Spiral Classifiers market will be able to gain the upper hand as they use the report as a powerful resource. • ### Classification of Materials and Types of Classifiers Oct 31, 2015· The spiral classifier is also a mechanical classifier. It is made by combining a gravity settler of rectangular section with a sloped transport spiral for the sediment. It consists of a semi cylindrical trough (a trough which is semi-circular in cross- section) inclined to the horizontal. The spiral conveyor moves the solids which settle to the Comparison of advantages and disadvantages of different types of test: Evaluation in education qualities of a measuring instrument: Evaluation is a matter for teamwork: Chapter 3: The teaching-learning concept and programme construction: Planning and conducting an educational programme: The four c's of curriculum planning • ### Wet Classification in the Fines Range < 10 μm Müller Some fundamentals that are necessary to understand wet classification in the centrifugal field are discussed. Following this, several fines classification apparatuses are introduced. Finally, the operation behavior, advantages and disadvantages of a stirred media mill/classifier circuit are described. • ### Global and United States Submerged Spiral Classifier Submerged Spiral Classifier market is segmented by region (country), players, by Type, and by Application. Players, stakeholders, and other participants in the global Submerged Spiral Classifier market will be able to gain the upper hand as they use the report as a powerful resource. • ### Advanced spiral periodic classification of the elements The proposed new periodic classification is a spiral arrangement of the elements, arranged by their increasing atomic number, electronic configuration and recurring chemical properties. It has 32 groups and 8 periods. 2. For the convenience of the readers, a dark line is drawn at the starting as well as ending points of the periods • ### Classification of Materials and Types of Classifiers Oct 31, 2015· The spiral classifier is also a mechanical classifier. It is made by combining a gravity settler of rectangular section with a sloped transport spiral for the sediment. It consists of a semi cylindrical trough (a trough which is semi-circular in cross- section) inclined to the horizontal. The spiral conveyor moves the solids which settle to the • ### What is Spiral Model in SDLC? \|Professionalqa What is Spiral Model? Spiral Model is one of the oldest form of the Software Development Life Cycle Models(SDLC), which was first defined by the Barry Boehm in the year 1986. Basically, this model is an evolutionary type model, which works on the combined approach of the waterfall and iterative model.. This model is driven by the risk analysis and evaluation requirements and keeps on iterating • ### The Disadvantages of 3D Printing in the Medical Field May 05, 2020· Certain disadvantages are holding back 3D printing in the medical field: let’s take a look at three of the biggest issues. Disadvantage One: Knowledge. 3D printing is a relatively young technology, around for decades, not centuries. Its lack of age-old existence may cause mistrust or, worse, misunderstanding. • ### Machine Learning Classification 8 Algorithms for Data Advantages Random Forest Classifiers facilitate the reduction in the over-fitting of the model and these classifiers are more accurate than the decision trees in several cases. Disadvantages Random forests exhibit real-time prediction but that is slow in nature. They are also difficult to implement and have a complex algorithm. These are various software engineering models and their advantages and disadvantages 1. Waterfall Model: It requires a well understanding and knowledge of requirements and technology related to it. Advantages : It is very easy and convenient to implement the waterfall model. For implementation of small systems, it is very useful • ### Types of Classifiers in Mineral Processing Metallurgical ContentSpiral ClassifierScrew/Spiral Classifier Capacity TableAllen Cone ClassifierCone or Pyramid ClassifierCross-Flow ClassifierSpiral-Screw Classifier Capacity TableHydraulic ClassifierHydro-classifierHydroclassifier Capacity TableRake ClassifierRake Classifier Capacity Rotary High Weir ClassifierRotary High Weir Classifier Capacity Spiral Classifier In mineral processing, the • ### high quality medium lime spiral classifier for sale in Kitwe High End Large Bauxite Spiral Classifier For Sale. High End Medium Salt Classifier For Sale In Ndola Zambia We have high end medium salt classifier for sale in ndola zambia africa8 x 5zambiajaw crusherfor sale as largest crushing plantmill equipments and beneficiation plants production base skd has exported large quantities and highend mobile crushing plant and milling equipments to • ### Disadvantages To Mining Granite Disadvantages To Mining Granite. 2020-09-11 Advantages disadvantages mining ores Grinding quarrying and that this involved digging up and processing large amounts of rock 187 More detailed METALS AND advantages of mining granite quarry in zambia advantages of mining granite quarry in zambia advantages of mining granite • ### spirogyra Definition, Structure, Reproduction, & Facts Spirogyra, any member of a genus of some 400 species of free-floating green algae found in freshwater environments around the world. Named for their beautiful spiral chloroplasts, spirogyras are filamentous algae that consist of thin unbranched chains of cylindrical cells. • ### 1.17. Neural network models (supervised) — scikit-learn 0 1.17.1. Multi-layer Perceptron¶. Multi-layer Perceptron (MLP) is a supervised learning algorithm that learns a function $$f(\cdot): R^m \rightarrow R^o$$ by training on a dataset, where $$m$$ is the number of dimensions for input and $$o$$ is the number of dimensions for output. Given a set of features $$X = {x_1, x_2,, x_m}$$ and a target $$y$$, it can learn a non-linear function • ### disadvantages of mining steel disadvantages of iron ore mining Solution for ore miningJan 05, 2013· disadvantages of iron ore mining Caiman Crusher. MOBILE CRUSHER Mobile Crusher Introduction. Mobile crusher also named protable crusher is a new crusher equipment . advantages and disadvantages of iron mining • ### What is Heterogeneous Welding? Examples, Advantages What is heterogeneous welding? Examples, Advantages and Disadvantages. Classification of fusion welding processes. Heterogeneous welding is one way of performing fusion welding with the application of external filler whose metallurgical composition is substantially different from that of the base metals.	2021-04-18 11:01:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.37619835138320923, "perplexity": 5271.158503646043}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00514.warc.gz"}
https://brendonward.org/tasmania/inference-about-population-variance-example.php	##### Grails Controller Response List Example How to upload a file with Grails 3 Grails Guides... ##### Student Letter To College Example Sample Termination Letter Format for Student... ##### Dialogue Example Ordering Coffee In Italy Coffee Life in Italy... ##### 10 Example Of Adverbs Of Frequency Adverbs of Frequency English Grammar Lesson 7 E S L... ##### Example Of Ethical Dilemma In Every Day Life What Does Moral Dilemma Mean? 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Chapter 18. inference about a population mean 3 the one-sample t conп¬ѓdence interval deп¬ѓnition. draw a srs of size n from a large population, inference on the mean of a population (variance unknown) in excel it is known that if is an unknown variance distribution ? 2 we use sample variance s 2,. Inferences about a population variance. if the sample variance is excessive, overfilling and. underfilling may be occurring even though the mean. is correct. suppose that the population mean $$\mu$$ and the population variance $$\sigma^2$$ are both unknown. when learning вђњone-sample mean inference: variance known Chapter 11 inferences about population variances inference about a population variance inferences about a estimate of the population variance. example: mathematics 201-510-lw business statistics martin huard fall 2008 xxv вђ“ inferences about the population variance 1. the manufacturer of a certain brand of light Chapter 11 inferences about population variance inferences вђ“about a population variance interval estimate of пѓsquare infernence about the variance of two пѓ2 population variance x sample distribution page 12 statistical inference for the population mean lrequires the sampling distribution of the mean page 15 Inference on the mean of a population (variance unknown) in excel it is known that if is an unknown variance distribution ? 2 we use sample variance s 2, chapter6 inference on mean and variance 6.1 introduction a common purpose of compositing sampling is to draw statistical inference on the population mean, and 5-Inference on Variance Inferences about Population CHAPTER 11 INFERENCES ABOUT POPULATION VARIANCE. Using pooled variances to do inferences for two-population to perform a separate variance 2-sample t comparing two population means: independent samples;, inferences concerning proportions (chapter 9) our estimate of the population variance is the sample variance and as this increases so does the); statistical inference i: estimating the mean and variance of a population 1. introduction economists are interested in relationships between economic variables., statistical inference for population mean when data is not normal and n is large and standard deviation . define the sample variance in the usual way as. Formula Sheet and List of Symbols Basic Statistical Inference Essay on Inference in Population Variance 3342 Words. Inferences about a population variance. if the sample variance is excessive, overfilling and. underfilling may be occurring even though the mean. is correct., sample to make an inference about a population. one-sample t-test statistical inference and t-tests copyright в© 2010 minitab inc. all rights reserved.. Вђў inference about a population variance вђў inferences about two populations variances. statistics (spring 2017) 3 population variance вђў example: buyerвђ™s sampling probability and inference or sample, of the population. measured by the population variance and standard deviation Chapter 11: inferences about population variances when a sample variance of 25 is obtained from a sample of 10 items from a normal population, inference about population variance (one sample case) module ii: application of epidemiology & biostatistics in public health variance measures level of risk Swirl_courses / statistical_inference / variance / lesson. fetching contributors answerchoices: population variance; sample mean; sample standard deviation; what is the difference between n and n-1 in calculating population variance? of finite population inference. the variance of a population from a sample, Lecture: sampling distributions and statistical use this sample mean and variance to make inferences and on the sample size and the population variance 2 in the national assessment of educational progress (naep), population inferences and variance estimation are based on a randomization-based perspective where the link Statistic inference oп¬ђers us a wide variety of methods which give answers to very diп¬ђerent for the population variance, sample cuasivariance: s2 c = n, lecture: sampling distributions and statistical use this sample mean and variance to make inferences and on the sample size and the population variance 2). Statistic Inference - TU Kaiserslautern Read the next post: take a rain check example sentence Read the previous post: hibernate named query in clause example	2020-02-27 08:00:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5588799118995667, "perplexity": 3125.7715448288413}, "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-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00084.warc.gz"}
https://networkx.github.io/documentation/networkx-1.11/reference/generated/networkx.algorithms.chordal.chordal_alg.find_induced_nodes.html	Warning This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation. # find_induced_nodes¶ find_induced_nodes(G, s, t, treewidth_bound=9223372036854775807)[source] Returns the set of induced nodes in the path from s to t. Parameters: G (graph) – A chordal NetworkX graph s (node) – Source node to look for induced nodes t (node) – Destination node to look for induced nodes treewith_bound (float) – Maximum treewidth acceptable for the graph H. The search for induced nodes will end as soon as the treewidth_bound is exceeded. I – The set of induced nodes in the path from s to t in G Set of nodes NetworkXError – The algorithm does not support DiGraph, MultiGraph and MultiDiGraph. If the input graph is an instance of one of these classes, a NetworkXError is raised. The algorithm can only be applied to chordal graphs. If the input graph is found to be non-chordal, a NetworkXError is raised. Examples >>> import networkx as nx >>> G=nx.Graph() >>> G = nx.generators.classic.path_graph(10) >>> I = nx.find_induced_nodes(G,1,9,2) >>> list(I) [1, 2, 3, 4, 5, 6, 7, 8, 9] Notes G must be a chordal graph and (s,t) an edge that is not in G. If a treewidth_bound is provided, the search for induced nodes will end as soon as the treewidth_bound is exceeded. The algorithm is inspired by Algorithm 4 in [1]. A formal definition of induced node can also be found on that reference. References [1] Learning Bounded Treewidth Bayesian Networks. Gal Elidan, Stephen Gould; JMLR, 9(Dec):2699–2731, 2008. http://jmlr.csail.mit.edu/papers/volume9/elidan08a/elidan08a.pdf	2020-08-08 10:19:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5249096751213074, "perplexity": 2236.8388691607133}, "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-2020-34/segments/1596439737319.74/warc/CC-MAIN-20200808080642-20200808110642-00512.warc.gz"}
http://xtra-art.com/764meg8s/page.php?7afd0d=partial-differentiation-engineering-mathematics-notes	Mathematics-I Lectures/week = 3 Sessional Marks =30 Exam=3 Hrs, Exam. \frac{\partial }{\partial x}(x^2 + y^2) = 2x + 0 ;\hspace{25pt} is different from the regular differentiation? The difference between the two is itself the $Actually, it's \frac{-9}{(x + y + z)^2}$, It might look complicated but it's not. $\frac{\partial z}{\partial y} = 0 + 3y^2 - 6x^2y = 3y^2 - 6x^2y$ = 3\left[\frac{\partial }{\partial x}\left(\frac{1}{x + y + z} \right) + \frac{\partial }{\partial $Partial Differentiation Course Notes Be able to: Partially differentiate a functions Use partial differentiation to find the rate of change Practice Assessments Useful Links Khan Academy: Partial Differentiation … Transformers ... Vector Calculus \| Btech Shots! = 3\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial Differential Calculus - 2 Engineering Maths, Btech... Matrices Engineering Maths, Btech first year. In P.D. The section also places the scope of studies in APM346 within the vast universe of mathematics. \frac{\partial }{\partial x}(\log{x^2 + y^2}) = \frac{1}{x^2 + y^2}\times 2x = \frac{2x}{x^2 + y^2}; \text{b)} \hspace{10pt} }{\partial z} \right)\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial procedure with all of the variables. \frac{\partial }{\partial y}(\sin{xy}) = \cos{xy}\times x(1) = x\cos{xy} \[ = \left[\frac{2\cancel{(x + y)}(x - y)}{(x + y)^\cancel 2} \right]^2$ You … \left[\text{As, } a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \right] Learn engineering mathematics. \left[\frac{x^2 - y^2 + 2xy - y^2 + x^2 - 2xy}{(x + y)^2} \right] DC Motor \| Btech Shots! A partial differential equation is an equation involving two (or more ) independent variables x, y and a dependent variable z and its partial derivatives such as ! When partially differentiating w.r.t. \frac{\partial }{\partial y}(x^2y^2) = x^2(2y) = 2yx^2 introducing the subscript comma to denote partial differentiation with respect to the coordinate variables, in which case ,i / xi, ui jk ui / xj xk 2,, and so on. But, there is a basic difference in the two forms of … $But what if we have more than one variable in a function? Preface What follows are my lecture notes for a ﬁrst course in differential equations, taught at the Hong Kong University of Science and Technology. lifting your pen or complicated enough to frustate you for not reaching to your answer, as we will see in sample Find first and second order partial drivatives of$ You have studied differentiation earlier and you might be thinking- how Partial Derivatives 1.6.4 The Gradient of a Scalar Field Let (x) be … yeah, just take one variable at a time and the rest as constants. $\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)^2u = \frac{\partial }{\partial x}(\sin{xy}) = \cos{xy}\times y(1) = y\cos{xy}; y, \frac{\partial }{\partial y}(\log{x^2 + y^2}) = \frac{1}{x^2 + y^2}\times 2y = \frac{2y}{x^2 + y^2} \frac{\partial }{\partial x}(x^2y^2) = y^2(2x) = 2xy^2 ; \hspace{25pt} Unit – 1: Differential Calculus – I Leibnitz’s theorem Partial derivatives Euler’s theorem for … definition of P.D. \[ }{\partial z} \right)\left(\frac{1}{x + y + z} \right) Now Partially differentiate equation (1) w.r.t. L.H.s. }{\partial z} \right)\left(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial Same process for second order P.D.$ Problem 2B is asking you to find the point at which h equals 2200, partial h over partial x equals zero and partial h over partial y is less than zero. \] problems below. $u = log(x^3 + y^3 + z^3 - 3xyz) \hspace{25pt} \longrightarrow (1)$ Partial Differentiation Engineering Maths, Btech f... Maxima and Minima Engineering Maths, Btech first year. (x + y)(2x) - (x^2 + y^2)(1) ... DC Motor, Basic Electrical Engineering, Btech first year, Transformers, Basic Electrical Engineering, Btech first year, Vector Calculus Engineering Maths, Btech first year, Ideal & Practical Transformer, Basic Electrical Engineering, Btech first year, Btech First Year Notes Engineering 1st year notes. $= \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial \[ \[= 4\left[\frac{\cancel{x^2} + y^2 + \cancel{2xy} - \cancel{x^2} + \cancel{y^2} - \cancel{2xy} - \cancel{y^2} + x^2$ $\frac{\partial z}{\partial x} = \frac{x^2 - y^2 + 2xy}{(x + y)^2}$ not that simple, the process involved in differentiating can either be so simple that you can solve it without = In Differentiation, we had two variables $$x, y$$ where $$x$$ was an independent variable and $\frac{\partial u}{\partial y} = \frac{3y^2 - 3xz}{x^3 + y^3 + z^3 - 3xyz}$ ��yG� �l �aX��À��6�q�x@��w�T�u^2��Sv@�e˖�G\$_�f � !q�H� 2ԒS)�Cƀ�9O��C. , \] Engineering Mathematics Books & Lecture Notes Pdf Engineering Mathematics provides the strong foundation of concepts like Advanced matrix, increases the analytical ability in solving mathematical problems, and many other advantages to engineering … This means that all of the variables, unlike differentiation, are independent. $\frac{\partial x}{\partial x} = 1, \frac{\partial y}{\partial x} = 0$ h�ܛao�6�� 4\left(1 - \frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} \right) is quite simple, right? 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Have more than two variables, past examination papers and solutions form we 'll be using for Derivatives...	2021-03-02 17:15:34	{"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": 0.8124961256980896, "perplexity": 1895.9982459262956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364027.59/warc/CC-MAIN-20210302160319-20210302190319-00561.warc.gz"}
https://socratic.org/questions/how-do-you-find-the-derivative-of-n-t-150-600-root3-16-3t-2	# How do you find the derivative of n(t) = 150 - 600/root3(16+3t^2)? Jul 13, 2018 n'(t)=1200t/(root(3)(16+3t^2)^2 #### Explanation: writing $n \left(t\right) = 150 - 600 {\left(16 + 3 {t}^{2}\right)}^{- \frac{1}{3}}$ note that $\left(150\right) ' = 0$ and $n ' \left(t\right) = - 600 \cdot \left(- \frac{1}{3}\right) {\left(16 + 3 {t}^{2}\right)}^{- \frac{2}{3}} 6 t$ simplifying we obtain $n ' \left(t\right) = 1200 \frac{t}{\sqrt[3]{16 + 3 {t}^{2}}} ^ 2$	2021-01-19 19:16:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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": 0.7855255007743835, "perplexity": 2403.3029103040376}, "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-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00610.warc.gz"}
http://wiki.caddementia.bigr.nl/index.php/Template	# Template A AuthorA, B AuthorB, C AuthorC (Department, University, Country) Title of the work By clicking "Edit" in the top right corner, you can see the syntax for making headings, lists and links. ## Summary What features does the method use? Which classifier is used? What training data? What is the computation time? ## Stepwise explanation Equations can be inserted by using LaTeX formulae between [itex][/itex] tags, see the manual for more help on using math on the Wiki. An example equation: $y=x^2$	2018-09-23 14:33:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "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": 0.7089133858680725, "perplexity": 7704.231315518182}, "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-2018-39/segments/1537267159470.54/warc/CC-MAIN-20180923134314-20180923154714-00368.warc.gz"}
http://openstudy.com/updates/55af07b4e4b071e6530d2b0a	## zzr0ck3r one year ago Fun set theory question for budding set theorist. Let $$B^A$$ be the set of all functions from $$A$$ to $$B:=\{0,1\}$$. Show that $$\mathcal{P}(A)$$ has the same cardinality as $$B^A$$ for any set $$A$$. 1. anonymous Well, we know in the finite case $$\|\mathcal P (A)\|=2^{\|A\|}$$ and $$\|B^A\| = \|B\|^{\|A\|} = 2^{\|A\|}$$ 2. anonymous I think this applies when $$\|A\|>\aleph_i$$. 3. zzr0ck3r yeah it does. 4. zzr0ck3r A can have any cardinality 5. anonymous You could create a homomorphism as well. $\forall S\subseteq A: \exists f\in B^A: f(x)= \begin{cases}1 &x\in S \\ 0 &x\notin S\end{cases}$ 6. anonymous You can build a unique $$f$$ from $$S$$ and vise versa. 7. anonymous yeah, the bijection is easy -- for $$f\in B^A$$ we associate the inverse image $$f^{-1}(1)\in 2^A$$, and for $$S\in2^A$$ we associate $$f$$ such that $$f[S]=1,f[S^c]=0$$ 8. anonymous this is a side-effect of the fact that sets of things in $$A$$ are wholly characterized by their members (i.e. for what $$a\in A$$ we have that $$a\in S\in 2^A$$), and membership is a binary map $$S\to \{0,1\}$$	2016-10-28 10:19:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 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": 0.9484171271324158, "perplexity": 551.8011347339036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721606.94/warc/CC-MAIN-20161020183841-00481-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/why-is-magnitude-g-instead-of-vector-g-used-in-range-projectile.730785/	# Why is magnitude g instead of vector g used in range projectile #### negation It is not being explained anywhere in my book nor on the internet. Can someone explain to me why? Related Other Physics Topics News on Phys.org #### Bandersnatch It's because in projectile motion treatment, the positive value of displacement is assumed to be "up", and accordingly all derivatives have the same assumption embedded in them. V is "up" and -V is "down". a is "up" and -a is "down". On the surface of Earth, the magnitude of the acceleration vector is assumed to be g, and it's direction "-", or "down", i.e., towards the negative values. You can use proper vector notation and get the same results. It's just that most textbooks tackle projectile motion before properly explaining vectors. #### negation It's because in projectile motion treatment, the positive value of displacement is assumed to be "up", and accordingly all derivatives have the same assumption embedded in them. V is "up" and -V is "down". a is "up" and -a is "down". On the surface of Earth, the magnitude of the acceleration vector is assumed to be g, and it's direction "-", or "down", i.e., towards the negative values. You can use proper vector notation and get the same results. It's just that most textbooks tackle projectile motion before properly explaining vectors. Correct. g = -9.8ms^-2 However, in a question I posted, I was told not to take g = -9.8ms^-2 but, rather, 9.8ms^-2 without the implication of the direction. Why question then is why? In calculating range projectile, should g be a vector g where if upwards is +, then downwards i -. So in utilizing the equation for range projectile, g should be -9.8. However, a poster told me to disregard the signs and take g as a magnitude. This doesn't make sense. Acceleration of gravity is acting on the object during its trajectory. #### Bandersnatch As long as you use a reference frame in which the displacement "upwards" is in the positve direction, you have to use -g, since that's the direction towards which the gravity acts. As far as I can see from your posts in the other thread I think you're refering to, the problem is that you use the negative value for g twice, so to speak, and thus get the wrong result. The general equation for projectile motion is: $r=\frac{at^2}{2}+V_0t+r_0$, where r, a, and V are vectors. If you reduce the equation to one dimension(y; where the positive values are "up"), you can simply say that $a_y=-g$, you end up with $y=\frac{-gt^2}{2}+V_{y0}t+y_0$ You seem to be wrongly using $r=\frac{-at^2}{2}+V_0t+r_0$ and then substituting $a_y=-g$, which erroneously gives you $y=\frac{gt^2}{2}+V_{y0}t+y_0$ That's all there is to it. #### Bandersnatch I guess something might've been unclear in the above post. I meant g to be equal to +9.81 in it. It's the magnitude of the gravitational acceleration. The direction is the minus sign you put into the equation when you substitute -g for ay. #### negation As long as you use a reference frame in which the displacement "upwards" is in the positve direction, you have to use -g, since that's the direction towards which the gravity acts. As far as I can see from your posts in the other thread I think you're refering to, the problem is that you use the negative value for g twice, so to speak, and thus get the wrong result. The general equation for projectile motion is: $r=\frac{at^2}{2}+V_0t+r_0$, where r, a, and V are vectors. If you reduce the equation to one dimension(y; where the positive values are "up"), you can simply say that $a_y=-g$, you end up with $y=\frac{-gt^2}{2}+V_{y0}t+y_0$ You seem to be wrongly using $r=\frac{-at^2}{2}+V_0t+r_0$ and then substituting $a_y=-g$, which erroneously gives you $y=\frac{gt^2}{2}+V_{y0}t+y_0$ That's all there is to it. I was referring to this If my value for g = -9.8, I get a correct value but in the negative. The book has a positive value. This means that, and echoed by another poster, that g must be positive. The confusion is, why should g in this equation be a magnitude. It doesn't make an iota of sense. #### Bandersnatch I was referring to this If my value for g = -9.8, I get a correct value but in the negative. The book has a positive value. This means that, and echoed by another poster, that g must be positive. The confusion is, why should g in this equation be a magnitude. It doesn't make an iota of sense. The value of g is not -9.8. It's 9.8. If you're asked to substitute the value of the scalar g into your equation, you substitute the value without any information about the direction of the vector(i.e., the '-' sign). If you see g in an equation, it's a safe bet the direction was already included at one time. If you were asked to substitute some value for the vector $a_y$ then you'd have to put the minus together with the magnitude(g), as you'd be asked to provide the full information. The equation you're asked to use: $x = \frac{v_0^2}{g}sin{2Θ}$ is derived from: $$y=\frac{-gt^2}{2}+V_{y_0}t+y_0$$ and $$x=V_{x_0}t+x_0$$ the minus sign was already included when we substituted $a_y=-g$ when we reduced the general vector equation to one dimension. Try solving the first one for t, then substitute the result to the second one, and you'll see what happened to the minus sign that was originally there. #### negation The value of g is not -9.8. It's 9.8. If you're asked to substitute the value of the scalar g into your equation, you substitute the value without any information about the direction of the vector(i.e., the '-' sign). If you see g in an equation, it's a safe bet the direction was already included at one time. If you were asked to substitute some value for the vector $a_y$ then you'd have to put the minus together with the magnitude(g), as you'd be asked to provide the full information. The equation you're asked to use: $x = \frac{v_0^2}{g}sin{2Θ}$ is derived from: $$y=\frac{-gt^2}{2}+V_{y_0}t+y_0$$ and $$x=V_{x_0}t+x_0$$ the minus sign was already included when we substituted $a_y=-g$ when we reduced the general vector equation to one dimension. Try solving the first one for t, then substitute the result to the second one, and you'll see what happened to the minus sign that was originally there. You are correct. ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving	2019-08-20 19:03:40	{"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": 2, "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": 0.8185437321662903, "perplexity": 651.9297258250581}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00504.warc.gz"}
http://math.stackexchange.com/questions/611983/how-to-recover-the-probability-mass-function-from-probability-generating-functio	# How to recover the probability mass function from probability generating function? Would someone please provide me an example of where we take a p.g.f and use it to derive the p.m.f. ? I understand that you were have to take the derivatives of the pmf, which is understandable because the derivatives tell us the probabilities, but I would just like to see an example of where it all comes together. - Here's an example: The probability mass function for a binomial random variable is given by $$\mathrm{Pr}(X=i)=f(i; n, p) = \begin{pmatrix} n \\ i\end{pmatrix}p^i(1-p)^{n-i}.$$ Expand the generating function $G(z) = [(1-p)+pz]^n$ using the binomial theorem: $$G(z) = \sum_{k=0}^n \begin{pmatrix} n \\ k\end{pmatrix} (pz)^k(1-p)^{n-k}.$$ Taking the $i$th derivative with respect to $z$, we must see that all the terms with index less than $i$ are annihilated, leaving, $$G^{(i)}(z) = \sum_{k=i}^n \begin{pmatrix} n \\ k\end{pmatrix} (1-p)^{n-k} \frac{d^i}{dz^i} (pz)^k.$$ Equally obviously, any terms with index greater than $i$ have a power of $z$ that does not annihilate. By computing $G^{(i)}(0)$, these terms vanish, leaving only the $i$th term: $$G^{(i)}(0) = \begin{pmatrix} n \\ i\end{pmatrix} (1-p)^{n-i} p^i \left.\frac{d^i}{dz^i} (z)^i\right\|_{z=0}.$$ Of course, we know that $\frac{d^m}{dx^m} x^m = m!$, so this gives us $$G^{(i)}(0) = \begin{pmatrix} n \\ i\end{pmatrix} (1-p)^{n-i} p^i i! = i!\ \mathrm{Pr}(X=i)$$ -	2014-03-11 17:33:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9141506552696228, "perplexity": 169.03281316281348}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011237821/warc/CC-MAIN-20140305092037-00093-ip-10-183-142-35.ec2.internal.warc.gz"}
https://labs.tib.eu/arxiv/?author=F.%20Marshall	• ### The Swift-UVOT ultraviolet and visible grism calibration(1501.02433) Feb. 23, 2015 astro-ph.IM We present the calibration of the Swift UVOT grisms, of which there are two, providing low-resolution field spectroscopy in the ultraviolet and optical bands respectively. The UV grism covers the range 1700-5000 Angstrom with a spectral resolution of 75 at 2600 Angstrom for source magnitudes of u=10-16 mag, while the visible grism covers the range 2850-6600 Angstrom with a spectral resolution of 100 at 4000 Angstrom for source magnitudes of b=12-17 mag. This calibration extends over all detector positions, for all modes used during operations. The wavelength accuracy (1-sigma) is 9 Angstrom in the UV grism clocked mode, 17 Angstrom in the UV grism nominal mode and 22 Angstrom in the visible grism. The range below 2740 Angstrom in the UV grism and 5200 Angstrom in the visible grism never suffers from overlapping by higher spectral orders. The flux calibration of the grisms includes a correction we developed for coincidence loss in the detector. The error in the coincidence loss correction is less than 20%. The position of the spectrum on the detector only affects the effective area (sensitivity) by a few percent in the nominal modes, but varies substantially in the clocked modes. The error in the effective area is from 9% in the UV grism clocked mode to 15% in the visible grism clocked mode . • ### GRB 130427A: a Nearby Ordinary Monster(1311.5254) Feb. 4, 2014 astro-ph.HE Long-duration Gamma-Ray Bursts (GRBs) are an extremely rare outcome of the collapse of massive stars, and are typically found in the distant Universe. Because of its intrinsic luminosity ($L\sim 3 \times 10^{53}$ erg s$^{-1}$) and its relative proximity ($z=0.34$), GRB 130427A was a unique event that reached the highest fluence observed in the gamma-ray band. Here we present a comprehensive multiwavelength view of GRB 130427A with Swift, the 2-m Liverpool and Faulkes telescopes and by other ground-based facilities, highlighting the evolution of the burst emission from the prompt to the afterglow phase. The properties of GRB 130427A are similar to those of the most luminous, high-redshift GRBs, suggesting that a common central engine is responsible for producing GRBs in both the contemporary and the early Universe and over the full range of GRB isotropic energies. • ### Photometric Calibration of the Swift Ultraviolet/Optical Telescope(0708.2259) Nov. 9, 2007 astro-ph We present the photometric calibration of the Swift UltraViolet/Optical Telescope (UVOT) which includes: optimum photometric and background apertures, effective area curves, colour transformations, conversion factors for count rates to flux, and the photometric zero points (which are accurate to better than 4 per cent) for each of the seven UVOT broadband filters. The calibration was performed with observations of standard stars and standard star fields that represent a wide range of spectral star types. The calibration results include the position dependent uniformity, and instrument response over the 1600-8000A operational range. Because the UVOT is a photon counting instrument, we also discuss the effect of coincidence loss on the calibration results. We provide practical guidelines for using the calibration in UVOT data analysis. The results presented here supersede previous calibration results. • ### Swift detection of all previously undetected blazars in a micro-wave flux-limited sample of WMAP foreground sources(astro-ph/0703150) March 7, 2007 astro-ph Almost the totality of the bright foreground sources in the WMAP CMB maps are blazars, a class of sources that show usually also X-ray emission. However, 23 objects in a flux-limited sample of 140 blazars of the WMAP catalog (first year) were never reported before as X-ray sources. We present here the results of 41 Swift observations which led to the detection of all these 23 blazars in the 0.3-10 keV band. We conclude that all micro-wave selected blazars are X-ray emitters and that the distribution of the micro-wave to X-ray spectral slope $\alpha_{mu x}$ of LBL blazars is very narrow, confirming that the X-ray flux of most blazars is a very good estimator of their micro-wave emission. The X-ray spectral shape of all the objects that were observed long enough to allow spectral analysis is flat and consistent with inverse Compton emission within the commonly accepted view where the radiation from blazars is emitted in a Sychrotron-Inverse-Compton scenario. We predict that all blazars and most radio galaxies above the sensitivity limit of the WMAP and of the Planck CMB missions are X-ray sources detectable by the present generation of X-ray satellites. An hypothetical all-sky soft X-ray survey with sensitivity of approximately $10^{-15}$ erg/s would be crucial to locate and remove over 100,000 blazars from CMB temperature and polarization maps and therefore accurately clean the primordial CMB signal from the largest population of extragalactic foreground contaminants. • ### Swift observations of GRB050904: the most distant cosmic explosion ever observed(astro-ph/0610570) Nov. 8, 2006 astro-ph Swift discovered the high redshift (z=6.29) GRB050904 with the Burst Alert Telescope (BAT) and began observing with its narrow field instruments 161 s after the burst onset. This gamma-ray burst is the most distant cosmic explosion ever observed. Because of its high redshift, the X-ray Telescope (XRT) and BAT simultaneous observations provide 4 orders of magnitude of spectral coverage (0.2-150 keV; 1.4-1090 keV in the source rest frame) at a very early source-frame time (22 s). GRB050904 was a long, multi-peaked, bright GRB with strong variability during its entire evolution. The light curve observed by the XRT is characterized by the presence of a long flaring activity lasting up to 1-2 hours after the burst onset in the burst rest frame, with no evidence of a smooth power-law decay following the prompt emission as seen in other GRBs. However, the BAT tail extrapolated to the XRT band joins the XRT early light curve and the overall behavior resembles that of a very long GRB prompt. The spectral energy distribution softens with time, with the photon index decreasing from -1.2 during the BAT observation to -1.9 at the end of the XRT observation. The dips of the late X-ray flares may be consistent with an underlying X-ray emission arising from the forward shock and with the properties of the optical afterglow reported by Tagliaferri et al. (2005b). We interpret the BAT and XRT data as a single continuous observation of the prompt emission from a very long GRB. The peculiarities observed in GRB050904 could be due to its origin within one of the first star-forming regions in the Universe; very low metallicities of the progenitor at these epochs may provide an explanation. • ### Long-term monitoring of the X-ray afterglow of GRB 050408 with Swift/XRT(astro-ph/0610845) Oct. 27, 2006 astro-ph We present observations of the X-ray afterglow of GRB 050408, a gamma-ray burst discovered by HETE-II. Swift began observing the field 42 min after the burst, performing follow-up over a period of 38 d (thus spanning three decades in time).The X-ray light curve showed a steepening with time, similar to many other afterglows. However, the steepening was unusually smooth, over the duration of the XRT observation, with no clear break time. The early decay was too flat to be described in terms of standard models. We therefore explore alternative explanations, such as the presence of a structured afterglow or of long-lasting energy injection into the fireball from the central GRB engine. The lack of a sharp break puts constraints on these two models. In the former case, it may indicate that the angular energy profile of the jet was not a simple power law, while in the second model it implies that injection did not stop abruptly. The late decay may be due either to a standard afterglow (that is, with no energy injection), or to a jetted outflow still being refreshed. A significant amount of absorption was present in the X-ray spectrum, corresponding to a rest-frame Hydrogen column density NH = 1.2 (-0.3,+0.4)*10^22 cm^-2, indicative of a dense environment. • ### Optical, Infrared, and Ultraviolet Observations of the X-Ray Flash GRB 050416A(astro-ph/0604316) Sept. 26, 2006 astro-ph We present ultraviolet, optical, and infrared photometry of the afterglow of the X-ray flash XRF 050416A taken between approximately 100 seconds and 36 days after the burst. We find an intrinsic spectral slope between 1930 and 22,200 Angstrom of beta = -1.14 +/- 0.20 and a decay rate of alpha = -0.86 +/- 0.15. There is no evidence for a change in the decay rate between approximately 0.7 and 4.7 days after the burst. Our data implies that there is no spectral break between the optical and X-ray bands between 0.7 and 4.7 days after the burst, and is consistent with the cooling break being redward of the K_s band (22,200 Angstrom) at 0.7 days. The combined ultraviolet/optical/infrared spectral energy distribution shows no evidence for a significant amount of extinction in the host galaxy along the line of sight to XRF 050416A. Our data suggest that the extragalactic extinction along the line of sight to the burst is only approximately A_V = 0.2 mag, which is significantly less than the extinction expected from the hydrogen column density inferred from $X$-ray observations of XRF 050416A assuming a dust-to-gas ratio similar to what is found for the Milky Way. The observed extinction, however, is consistent with the dust-to-gas ratio seen in the Small Magellanic Cloud. We suggest that XRF 050416A may have a two-component jet similar to what has been proposed for GRB 030329. If this is the case the lack of an observed jet break between 0.7 and 42 days is an illusion due to emission from the wide jet dominating the afterglow after approximately 1.5 days. • ### Detection of a huge explosion in the early Universe(astro-ph/0509737) March 23, 2006 astro-ph Gamma-ray Bursts (GRBs) are bright flashes of high energy photons that can last from about 10 milliseconds to 10 minutes. Their origin and nature have puzzled the scientific community for about 25 years until 1997, when the first X-ray afterglows of long (> 2 s duration) bursts were detected and the first optical and radio counterparts were found. These measurements established that long GRBs are typically at high redshift (z 1.6) and are in sub-luminous star-forming host galaxies. They are likely produced in core-collapse explosions of a class of massive stars that give rise to highly relativistic jets (collapsar model). Internal inhomogeneities in the velocity field of the relativistic expanding flow lead to collisions between fast moving and slow moving fluid shells and to the formation of internal shock waves. These shocks are believed to produce the observed prompt emission in the form of irregularly shaped and spaced pulses of gamma-rays, each pulse corresponding to a distinct internal collision. The expansion of the jet outward into the circumstellar medium is believed to give rise to external'' shocks, responsible for producing the smoothly fading afterglow emission seen in the X-ray, optical and radio bands. Here we report on the gamma-ray and x-ray observation of the most distant gamma-ray burst ever observed: its redshift of 6.29 translates to a distance of 13 billion light-years from Earth, corresponding to a time when the Universe was just 700 million to 750 million years old. The discovery of a gamma-ray burst at such a large redshift implies the presence of massive stars only 700 million years after the Big Bang. The very high redshift bursts represent a good way to study the re-ionization era soon after the Universe came out of the Dark Ages. • ### Bright X-ray Flares in Gamma-Ray Burst Afterglows(astro-ph/0506130) July 31, 2005 astro-ph Gamma-ray burst (GRB) afterglows have provided important clues to the nature of these massive explosive events, providing direct information on the nearby environment and indirect information on the central engine that powers the burst. We report the discovery of two bright X-ray flares in GRB afterglows, including a giant flare comparable in total energy to the burst itself, each peaking minutes after the burst. These strong, rapid X-ray flares imply that the central engines of the bursts have long periods of activity, with strong internal shocks continuing for hundreds of seconds after the gamma-ray emission has ended. • ### Observation of X-ray lines from a Gamma-Ray Burst (GRB991216): Evidence of Moving Ejecta from the Progenitor(astro-ph/0011337) Nov. 17, 2000 astro-ph We report on the discovery of two emission features observed in the X-ray spectrum of the afterglow of the gamma-ray burst (GRB) of 16 Dec. 1999 by the Chandra X-Ray Observatory. These features are identified with the Ly$_{\alpha}$ line and the narrow recombination continuum by hydrogenic ions of iron at a redshift $z=1.00\pm0.02$, providing an unambiguous measurement of the distance of a GRB. Line width and intensity imply that the progenitor of the GRB was a massive star system that ejected, before the GRB event, $\approx 0.01 \Ms$ of iron at a velocity $\approx 0.1 c$, probably by a supernova explosion. • ### A giant, periodic flare from the soft gamma repeater SGR1900+14(astro-ph/9811443) Nov. 27, 1998 astro-ph Soft gamma repeaters are high-energy transient sources associated with neutron stars in young supernova remnants. They emit sporadic, short (~ 0.1 s) bursts with soft energy spectra during periods of intense activity. The event of March 5, 1979 was the most intense and the only clearly periodic one to date. Here we report on an even more intense burst on August 27, 1998, from a different soft gamma repeater, which displayed a hard energy spectrum at its peak, and was followed by a ~300 s long tail with a soft energy spectrum and a dramatic 5.16 s period. Its peak and time integrated energy fluxes at Earth are the largest yet observed from any cosmic source. This event was probably initiated by a massive disruption of the neutron star crust, followed by an outflow of energetic particles rotating with the period of the star. Comparison of these two bursts supports the idea that magnetic energy plays an important role, and that such giant flares, while rare, are not unique, and may occur at any time in the neutron star's activity cycle. • ### XTE J1739-302: An Unusual New X-ray Transient(astro-ph/9805178) May 13, 1998 astro-ph A new x-ray transient, designated XTE J1739-302, was discovered with the Proportional Counter Array (PCA) on the Rossi X-ray Timing Explorer (RXTE) in data from 12 August 1997. Although it was the brightest source in the Galactic Center region while active (about 3.0 x 10^-9 ergs/cm2/s from 2 to 25 keV), it was only observed on that one day; it was not detectable nine days earlier or two days later. There is no known counterpart at other wavelengths, and its proximity to the Galactic Center will make such an identification difficult due to source confusion and extinction. The x-ray spectrum and intensity suggest a giant outburst of a Be/neutron star binary, although no pulsations were observed and the outburst was shorter than is usual from these systems. • ### A Multi-frequency campaign on the gamma-ray loud blazar W Comae(astro-ph/9706243) June 24, 1997 astro-ph We report preliminary results of a multi-frequency campaign on the TeV candidate blazar W Comae (z=0.102). Flux limits by Whipple and HEGRA show that the TeV flux must be considerably below an E^-2 extrapolation of the EGRET flux. In the framework of proton-initiated cascade models, this implies a moderate amount of gamma-ray attenuation due to pair production in collisions with low-energy photons of the extragalactic infrared background. In a simple SSC model, the gamma-ray spectrum cuts off intrinsically just below TeV without requiring external absorption.	2020-11-30 04:12:18	{"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": 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": 0.6693121790885925, "perplexity": 2374.9347852205447}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00389.warc.gz"}
https://zbmath.org/?q=an:0812.62071&format=complete	# zbMATH — the first resource for mathematics Second-order approximation of the entropy in nonlinear least-squares estimation. (English) Zbl 0812.62071 Kybernetika 30, No. 2, 187-198 (1994); erratum ibid. 32, No. 1, 104 (1996). Summary: Measures of variability of the least-squares estimator $$\widehat {\theta}$$ are essential to assess the quality of the estimation. In nonlinear regression, an accurate approximation of the covariance matrix of $$\widehat {\theta}$$ is difficult to obtain. A second-order approximation of the entropy of the distribution of $$\widehat {\theta}$$ is proposed, which is only slightly more complicated than the widely used bias approximation of M. J. Box [J. R. Stat. Soc, Ser. B 33, 171- 201 (1971; Zbl 0232.62029)]. It is based on the “flat” or “saddle- point approximation” of the density of $$\widehat {\theta}$$. The neglected terms are of order $${\mathcal O} (\sigma^ 4)$$, while the classical first order approximation neglects terms of order $${\mathcal O} (\sigma^ 2)$$. Various illustrative examples are presented, including the use of the approximate entropy as a criterion for experimental design. ##### MSC: 62J02 General nonlinear regression 62E17 Approximations to statistical distributions (nonasymptotic) 62E20 Asymptotic distribution theory in statistics 62K05 Optimal statistical designs Full Text: ##### References: [1] S. Amari: Differential-Geometrical Methods in Statistics. Springer, Berlin 1985. · Zbl 0559.62001 [2] D. Bates, D. Watts: Relative curvature measures of nonlinearity. J. Roy. Statist. Soc. Ser. B 42 (1980), 1-25. · Zbl 0455.62028 [3] M. Box: Bias in nonlinear estimation. J. Roy. Statist. Soc. Ser. B 33 (1971), 171-201. · Zbl 0232.62029 [4] G. Clarke: Moments of the least-squares estimators in a non-linear regression model. J. Roy. Statist. Soc. Ser. B 42 (1980), 227-237. · Zbl 0436.62054 [5] P. Hougaard: Saddlepoint approximations for curved exponential families. Statist. Probab. Lett. 3 (1985), 161-166. · Zbl 0573.62016 · doi:10.1016/0167-7152(85)90055-0 [6] A. Pázman: Probability distribution of the multivariate nonlinear least-squares estimates. Kybernetika 20 (1984), 209-230. · Zbl 0548.62043 · eudml:28735 [7] A. Pázman: Small-sample distributional properties of nonlinear regression estimators (a geometric approach) (with discussion). Statistics 21 (1990), 3, 323-367. · Zbl 0724.62063 [8] N. Reid: Saddlepoint methods and statistical inference. Statist. Sci. 3 (1988), 213-238. · Zbl 0955.62541 · doi:10.1214/ss/1177012906 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-03-08 18:50:34	{"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": 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": 0.5772642493247986, "perplexity": 1681.3458935916512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178385389.83/warc/CC-MAIN-20210308174330-20210308204330-00029.warc.gz"}
https://socratic.org/questions/for-what-values-of-x-if-any-does-f-x-tan-7pi-4-2x-have-vertical-asymptotes	# For what values of x, if any, does f(x) = tan((-7pi)/4-2x) have vertical asymptotes? Aug 2, 2018 $x = \left(2 k + 1\right) \frac{\pi}{4} - \frac{7}{8} \pi = \left(4 k - 5\right) \frac{\pi}{8} ,$ $k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$ $x = \ldots , - \frac{5}{8} \pi , - \frac{1}{8} \pi , \frac{3}{8} \pi , \frac{7}{8} \pi , \ldots$ #### Explanation: y = tan ( -7/4 pi - 2 x) = - tan ( 2 ( x + 7/8 pi ) The period = $\frac{\pi}{2}$. The vertical asymptotes of $y = a \tan \left(b x + c\right) + d$ are $b x + c = \left(2 k + 1\right) \frac{\pi}{2} , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$ Here, they are given by $2 \left(x + \frac{7}{8} \pi\right) = \left(2 k + 1\right) \frac{\pi}{2}$ $\Rightarrow x = \left(2 k + 1\right) \frac{\pi}{4} - \frac{7}{8} \pi = \left(4 k - 5\right) \frac{\pi}{8} ,$ $k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$ $= \ldots , , - \frac{5}{8} \pi , - \frac{1}{8} \pi , \frac{3}{8} \pi \ldots$ See graph: graph{(y+tan(2(x+7/8pi)))=0[-pi pi -pi/2 pi/2]}	2019-06-25 23:43:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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": 0.9933760762214661, "perplexity": 4856.628251908712}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00243.warc.gz"}
https://llvm.org/doxygen/classllvm_1_1scc__member__iterator.html	LLVM 15.0.0git llvm::scc_member_iterator< GraphT, GT > Class Template Reference Sort the nodes of a directed SCC in the decreasing order of the edge weights. More... #include "llvm/ADT/SCCIterator.h" ## Public Member Functions scc_member_iterator (const NodesType &InputNodes) NodesType & operator* () ## Detailed Description ### template<class GraphT, class GT = GraphTraits<GraphT>> class llvm::scc_member_iterator< GraphT, GT > Sort the nodes of a directed SCC in the decreasing order of the edge weights. The instantiating GraphT type should have weighted edge type declared in its graph traits in order to use this iterator. This is implemented using Kruskal's minimal spanning tree algorithm followed by a BFS walk. First a maximum spanning tree (forest) is built based on all edges within the SCC collection. Then a BFS walk is initiated on tree nodes that do not have a predecessor. Finally, the BFS order computed is the traversal order of the nodes of the SCC. Such order ensures that high-weighted edges are visited first during the tranversal. Definition at line 252 of file SCCIterator.h. ## ◆ scc_member_iterator() template<class GraphT , class GT > llvm::scc_member_iterator< GraphT, GT >::scc_member_iterator ( const NodesType & InputNodes ) Definition at line 304 of file SCCIterator.h. References assert(), and llvm::reverse(). ## ◆ operator() template<class GraphT , class GT = GraphTraits<GraphT>> NodesType& llvm::scc_member_iterator< GraphT, GT >::operator ( ) inline Definition at line 300 of file SCCIterator.h. The documentation for this class was generated from the following file:	2022-07-06 00:00:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.17280083894729614, "perplexity": 9624.701341461934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00596.warc.gz"}
https://scioly.org/wiki/index.php/Severe_Storms/Winter_Storms	# Severe Storms/Winter Storms This page is for the Severe Storms topic of the Meteorology event. ## Overview Graupel, sleet, hail, rain, freezing rain and snow are all considered forms of wintry precipitation [1]. Winter storms can bring several of these in one passing, depending on elevation and surface temperature changes. Winter storms also cause a number of hazards, which will be detailed below. As a result, winter storms can be dangerous, especially for those in areas where heavy snow is uncommon. Most of the time, however, more than 6 inches of snow will begin to create unsafe conditions anywhere, regardless of the frequency of snowfall. Some of the key dangers of all winter storms include hypothermia and frostbite due to the cold, and also car accidents due to unsafe road conditions. ## Blizzards Characteristics of a Blizzard • Visibility reduced to less than 1/4 mile • Winds greater than 35 miles per hour • These conditions must last longer than 3 hours In order for a snowstorm to be classified as a "blizzard" it must have all of the aforementioned characteristics. Once these conditions are expected, the NWS (National Weather Service) will issue a "blizzard warning." When some, but not all, of these conditions are expected, a "Winter Storm Warning" or "Heavy Snow Warning" may be issued. A "watch" is issued when severe winter weather is probable, but not necessarily imminent. An "advisory" is issued when less-serious conditions are occurring, imminent, or there is a high probability. Blizzards most often occur in the upper Midwest and Great Plains of the United States. This is because of flat land. Conditions of a blizzard develop on the northwest side of an intense storm system. The lower pressure in the storm and the higher pressure to the west's difference creates a tight pressure gradient, meaning a difference in pressure between two locations resulting in very strong winds. The winds can then pick up snow off the ground or blow falling snow creating low visibility and a chance for significant drifting of snow. A ground blizzard is a weather condition where snow is not falling but loose snow on the ground is lifted and blown by strong winds. Strong winds/cold temperatures combine to create danger, such as the wind chill factor which can drop to –60F during blizzards in the Midwest. ### Hazards A blizzard's main dangers are its strong winds, freezing temperatures, and deep snow. However, avalanches may also be a potential danger if a location is in a mountainous region. When a blizzard hits: • Find shelter indoors • Stay away from windows and doors • If owned, keep car or other vehicle in good, working condition. This includes: • Gas tank at least half-full • Good winter tires • Snacks and water stored inside vehicle • If stuck in a car and need the engine running to stay warm, keep the windows cracked open. This will allow poisonous carbon monoxide escape from inside the car. • Keep extra food and water, a flashlight, a battery-powered radio, and, if possible, a cell phone • If trudging through deep snow, keep moving. Do not lie down to rest. • If caught outdoors, use clothing to cover as much skin as possible. • Use public transportation if going outside is necessary • Listen to the radio/call state highway patrol • Avoid shoveling too much snow at once ## Nor'easters A northeaster, generally called a "nor'easter," can occur at any time of the year. However, these storms are most frequent and most powerful during the months from September to April, and are most often associated with winter. This type of winter storm is named because the majority of the wind during the storms comes from the northeast: hence, "nor'easter." The winds blow most strongly in this direction because these storms are large-scale cyclones centered a low pressure system, rotating counter-clockwise. Nor'easters exclusively occur on the East Coast of North America, and typically progress towards the northeast direction. But, Northeasters can form up to 100 miles east of the East Coast and still strike many eastern seaboard cities, according to the National Weather Service[2]. Rain, snow, sleet, and freezing rain can all be associated with these storms. Flooding and blizzard conditions are possible with these storm systems, and can cause crippling damage to large regions of the Eastern seaboard. The economic- and transportation-related damage, including massive power outages and car accidents, can exceed a billion dollars. Some storms can leave millions without power. This satellite image shows a Nor'easter that occurred in January 2018. During winter, the polar jet stream moves Arctic air to the east area of the U.S. where warm air is moving upwards. Famous Nor'easters Date Name 1888 Blizzard of 1888 March 1962 "Ash Wednesday storm" February 1978 New England Blizzard March 1933 "Superstorm" Jan and Feb 2015 Boston snowstorms of 2015 March 01-03 2018 East Coast "March Nor'easter" "Winter Storm Riley" ## Lake-Effect Snow This weather event is a common occurrence across the Great Lakes region during late fall and winter. Although most commonly associated with freshwater lakes, this effect can also occur near salt-water bodies of water such as bays, salt lakes, seas, or even oceans. Lake-effect snow occurs when cold air moves across the warmer water of the lake. This allows moisture to be transferred to the lowest part of the atmosphere. Then, the warm air rises, leading to cloud formation. As the clouds begin to accumulate more and more moisture, they grow larger. Eventually, they are capable of producing snow, which is then deposited on the downwind, or leeward, side of the lake. It usually takes the form of narrow bands characterized by intense snowfall and limited visibility. On occasion, sunny skies can quickly be replaced by blizzard-like conditions, such as blinding and wind-driven snowfall, in just a few minutes. It is extremely dangerous to motorists and pedestrians alike. This image shows the process by which lake-effect snow falls. Taken from NWS page ## Precipitation from Winter Storms ### Hail Updrafts in thunderstorms carry raindrops up into very cold places in the atmosphere and freeze, creating hail. It can be dangerous to aircrafts, homes, people, livestock, and cars. The number of times a hailstone traveled to the top of the cloud can be counted in the layers. The stones can melt and refreeze creating strange shapes. Once the hail can no longer be supported, it begins to fall. Hail swaths are where hail falls a lot, sometimes requiring snow plows. Radars, such as the Doppler-radar, can detect hail as looking like very, very heavy rainfall. Dual-polarization radar technology also can detect it. ### Freezing Rain Liquid raindrops in a layer of warm air high above the ground fall into a layer of freezing cold air right above the ground, forming freezing rain. When the layer of freezing air is very thin, the warm raindrops don't have enough time to freeze and freeze upon reaching the ground. Whatever the drops land on gets coated in a layer of ice. Freezing rain can be extremely dangerous, since the result is a layer of ice instead of snow. It can create slick spots on the roads, and places added weight on tree branches and power lines, which can cause them to snap. Even for places that are accustomed to snow storms, as little as 1 cm can completely paralyze a city. Dangers including driving, telephone and electrical wire damage, and falling. This precipitation can also cause entire crops to be destroyed. ### Sleet Sleet (or ice pellets) is made of small, translucent balls of ice. They are rain drops that have frozen before they reached the ground, which occurs when the layer of cold air is thicker than it is for freezing rain. When a layer of subfreezing air near the ground goes high enough, raindrops freeze into little balls of ice before reaching the ground. These pellets of ice usually bounce after hitting the ground or other hard surfaces. It can sometimes be accompanied by freezing rain. A Winter Storm Warning is issued for sleet or a combination of sleet and snow based on total accumulation, which is locally defined by area. ### Snow Snow is precipitation in the form of ice crystals. Snow forms from supercooled (also called undercooled) water, which means the liquid or vapor in the atmosphere can avoid freezing at extremely cold temperatures. This allows it to form a crystalline structure around a particle that can act as a nucleus, or seed. Common particles that act as crystal seeds in clouds include dust, clay, and biological material, although ice itself can serve as a seed. These droplets then grow into hexagonal crystals by nucleating around those particles inside clouds. The clusters of ice crystals that fall from a cloud to the Earth are what we generally call snowflakes. Snowflakes form in a variety of patterns based on the crystal structure. Snow is less dense than liquid water, though the factor varies greatly depending on formation conditions. Snow to Water ratios indicate the amount of snow needed to be melted to be equivalent to an inch of rainfall. These ratios vary as well with the temperature and current humidity, though it usually ranges between 10:1 and 12:1 [3]. Snow is composed of frozen water, meaning it technically can be classified as a mineral. ### Rain Rain is liquid drops of water that fall to the ground and don't freeze. Water condensed in clouds must be heavy enough to fall from clouds, and droplets often collide with others or grow as water condenses out of the air into the droplet until the droplet is large enough to fall. Precipitation that falls to earth in drops more than 0.5 mm in diameter. Frontal rain forms when two air masses meet, e.g. when warm air cools and condenses. Orographic rain is made because of clouds that form because of topography, where high ground forces moist cool air up. It is most likely to occur near mountains with westerly prevailing winds. ### Graupel Graupel comes from German, literally meaning soft hail. Thus, this form of wintry precipitation is also known as snow pellets or soft hail. According to the National Weather Service's glossary[4], snow pellets are essentially what the name implies: small(between 2 and 5 mm in diameter), round, cloudy to white colored ice crystals. Some people colloquially call this precipitation "tapioca snow," since the pellets are similar in appearance to tapioca. Graupel occurs when supercooled water molecules in the air crystalize around a falling snowflake[5]. The coating that surrounds the snowflake is called a "rime" or "rime coating,"[6], so named because riming occurs when droplets freeze and crystallize on the surface of an object. Generally, this process only occurs when it is exceptionally cold at ground level, allowing for the supercooled droplets in the air. Interestingly, it appears to form much more frequently when a thunder-snowstorm is present, although no updraft is necessary for graupel to form. ## Disaster Supply Kits Disaster kits are an important part of disaster preparedness. The following are possible items to include in a disaster supply kit for a location prone to severe winter storms. • Water, one gallon of water per person per day, for drinking and sanitation • Non-perishable food (at least a three-day supply) and can opener if needed • Flashlight and extra batteries • First Aid kit • Whistle to signal for help • Moist towelettes, garbage bags, and plastic ties for personal sanitation • Portable dishes, liquid dish soap, and disinfectant • Plastic sheeting and duct tape to shelter-in-place • Wrench or pliers to turn off utilities • Warm clothes and blankets • Infant formula and diapers, if applicable • Medical records and supplies, if applicable • Pet supplies • Pet first-aid kit and guide book (ask a local vet what to include, or visit the ASPCA Store to buy one online) • 3-7 days-worth of canned or dry food (be sure to rotate every two months) and feeding dishes • Disposable litter trays (aluminum roasting pans are perfect) and litter or paper toweling • Extra collar/harness and leash • Bag(s) or carrier(s) • Medical records and supplies, if applicable ## Special Topic for 2017 Between the 22nd and 24th of January 2016, Winter Storm Jonas, also known as 'Snowzilla,' produced totals of up to three feet of snow in parts of the Mid-Atlantic and Northeastern United States. This storm, the first of its kind since the 2011 Groundhog Day Blizzard, qualified as a Category 5 blizzard in the Northeast, and even qualified as a Category 4 for the Southeast. These ratings were from the more recent Regional Snowfall Index, or RSI, which is similar to other scales in use by meteorologists. It takes into account snowfall in a region, how much area the storm covers, and the relationship to population in the area. Originating from a shortwave trough, the system consolidated into a defined low-pressure area on January 21 over Texas. Meteorologists claimed that this could be a historic blizzard and quickly indicated the storm could produce more than 2 ft of snow across the Mid-Atlantic region. From January 20 through 22, eleven states and the District of Columbia declared a state of emergency in anticipation of significant snowfall and blizzard conditions. More than 13,000 flights were canceled in relation to the storm, and a travel ban was instituted for New York City and Newark, New Jersey for January 23–24. Ice and snow covered roads led to accidents across the region, several of which resulted in deaths and injuries. • Number of people affected: approximately 103 million people, 33 million people under blizzard warnings. • Killed in storm-related incidents: 55 people • States: Virginia (12), Pennsylvania (9), New Jersey (6), New York (6), North Carolina (6), South Carolina (4), Maryland (3), Washington, D.C. (3), Arkansas (1), Delaware (1), Georgia (1), Kentucky (1), Massachusetts (1), and Ohio (1). • Costs: Between $500 million and$3 billion A meteorological image showing the formation of Winter Storm Jonas in Jan 2016. A satellite image of the blizzard near its peak in Jan 2016.	2019-08-24 16:17:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29968005418777466, "perplexity": 4447.817027492133}, "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-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00446.warc.gz"}
http://mathhelpforum.com/discrete-math/223342-equivalence-relation.html	# Thread: equivalence relation 1. ## equivalence relation question An equivalence relation R defined on a set contains the pairs (1,1 ), (1,2 ), ( 2,3). Find R, given that R A x A. have to do this for a assignment but have no idea where to start I missed the class covering this. could anyone tell me the basic technique how to solve this so i know where to start at least! 2. ## Re: equivalence relation An equivalence relation on A is a subset of A x A with certain properties. The three pairs you show do not currently satisfy those properties. 1. You need reflexivity: For all $a \in A,(a,a) \in R$. 2. You need symmetry: If $(a,b) \in R$, then $(b,a) \in R$. 3. You need transitivity: If $(a,b)\in R$ and $(b,c) \in R$ then $(a,c) \in R$. 3. ## Re: equivalence relation sorry wrote question wrong An equivalence relation R defined on a set A ={1,2,3,4}contains the pairs (1,1 ), (1,2 ), ( 2,3). Find R, given that R ≠ A x A. 4. ## Re: equivalence relation What I am saying is, add pairs until you satisfy those three properties. 5. ## Re: equivalence relation im sorry i dont understand 6. ## Re: equivalence relation Check: for any $a \in A$, do you have $(a,a) \in R$? No. So, add those pairs that you are missing. $(1,2) \in R$. Do you have $(2,1) \in R$? No. So add that pair. Etc. Edit: By add those pairs, I mean add the pairs to the list of pairs you know must be in $R$. You started with the three given pairs. Then, from those three, you use the properties to figure out which other pairs must also be in $R$. In the end, you should find 10 pairs in $R$. If another equivalence relation $R'$ has $R \subseteq R'$ and you know $R' \setminus R \neq \emptyset$, then it is easy to show that $R' = A \times A$. In other words, $R$ is the unique equivalence relation on $A$ with those three pairs. 7. ## Re: equivalence relation Originally Posted by ronanbrowne88 question An equivalence relation R defined on a set contains the pairs (1,1 ), (1,2 ), ( 2,3). Find R, given that R A x A. have to do this for a assignment but have no idea where to start I missed the class covering this. could anyone tell me the basic technique how to solve this so i know where to start at least! Define $F=\{(1,1),(1,2),(2,3)\}$. The diagonal is $\Delta_A=\{(1,1),(2,2),(3,3),(4,4)\}$ Define $G=F\cup\Delta_A$ and $H=(G\circ G)\cup G$ Define $R=H\cup H^{-1}$. Show that $R$ is an equivalence relation.	2017-09-23 11:46:36	{"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": 0, "img_math": 0, "codecogs_latex": 25, "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": 0.93617182970047, "perplexity": 673.3451399637205}, "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-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00261.warc.gz"}
https://gamedev.stackexchange.com/questions/19491/how-does-one-avoid-hundreds-of-copies-into-a-vertex-buffer-each-frame	# How does one avoid hundreds of copies into a vertex buffer each frame? Assume a Direct3D9-based 2D game, high resolution, very busy. Performance critical. A particle system has 2,000 particles. They are scattered throughout the world with only a subset visible on screen at any time. Thus the global array of particles contains a lot of particles that won't be rendered this frame. Obviously they should be rendered in one call in a vertex buffer. I use an index buffer as well. The slow bit is in the copying all of the vertex data to the buffer. I actually do this twice: once to a big buffer, then as a single copy from the big buffer to the locked vertex buffer. Regardless, I'm looking at 2,000 iterations and maybe 200-500 memcpy() calls that shuffle 4 vertices at a time. There must be a better way. Maybe copying them all, and changing the index buffer instead? Or just rendering them all and let the off-screen ones get clipped (I'm not sure this is scalable though)? How do people normally handle this? • Have you profiled your code and established how much time this is taking and that it's really a performance bottleneck? Because if it isn't, then the answer is "don't worry about it". :) – Nathan Reed Nov 7 '11 at 21:51 • Definitely what Nathan said; in a particle system fillrate is far more likely to be your bottleneck than vertex buffer filling. I am curious about why you needed that second vertex buffer though... could you not have used a regular system memory array (or an std::vector) instead? – Maximus Minimus Nov 8 '11 at 0:24 • My comment was misleading. its not actually a second vertex buffer, just a memory buffer. I use it (possibly wrongly?) because I figrued it made sense to do all the scattered-through-the-code copies into a single array, and then quickly lock->copy->unlock the actual VB. Is that a good idea or a waste of time? – Cliff Harris Nov 8 '11 at 10:46 • Depends. It's an extra memory copy for sure and probably indicates that you're due some code-cleanup, but it may or may not have any measurable performance impact and you'll need to profile in order to find out. – Maximus Minimus Nov 8 '11 at 11:11 It might be worth considering using the geometry shader (or point sprites) to construct the quad. That would cut it down to one vertex per particle. You could also consider doing a GPU based particle update, where position is simply a function of time and a few constants which you can set up once and store in the vertex when the particle is created. A third option is to have more particle systems with less particles each that are closer together to improve frustum culling. You could also move most of your particle system work to a separate CPU thread, and have the main thread only do lock(); memcpy(); unlock(); Which combination of the above options you want (if any) is difficult to guess without more information. For example is your game mostly GPU bound or CPU bound? What proportion of frame time is spent processing particles? • the game seems to be CPU bound, but thats on my system, I haven't run it yet on lower spec machines. Multithreading is a good diea, but I'm not very experienced with multiple threads for stuff like this and I would fear introducing weird bugs or synchronisation issues. – Cliff Harris Nov 8 '11 at 10:44 Don't shuffle vertices around. Shuffle pointers to vertices. You say that you have a vector in which to store raw vertex data and memory, so go along with that, and keep a second vector that contains pointers to elements in your vector of vertices (your memory buffer). Incidentally, this is the same way I figured out how to cull hardware-instanced meshes individually in the same draw call. In short: 1. Update all the particles' positions 2. Cull the non-visible particles 3. Assign pointers to visible particles to the beginning of a vector 4. Copy vertex data referenced by the pointers to the vertex buffer 5. Draw vertex buffer with no. of visible particles as no. of vertices class ParticleSystem { std::vector<ParticleVertexInfo> particleData; std::vector<ParticleVertexInfo*> visibleParticles; const int maxParticles = 2000; int visibleCount; } particleData and visibleParticles are of the same length, defined by maxParticles. Initially, populate visibleParticles with pointers to all of particlesData just to avoid null data. visibleParticles will be partially overwritten in each frame anyways. After updating the position of each particle, cull every vertex in particleData, and keep two numbers to increment. The first is visibleCount which will be reset to zero each frame before culling begins. A second number is inside the loop to increment by one for each vertex. For every vertex that is in view, assign a pointer to particleData.at(i) to visibleParticles.at(visibleCount), then increment visibleCount by 1. Here's a trimmed down example of how a culling procedure might look for rearranging visible particles: visibleCount = 0; for (int i = 0; i < maxParticles; i++) { if (visible(particleData.at(i)) { visibleParticles.at(visibleCount) = &particleData.at(i); visibleCount++ } } Keep in mind that I am describing a brute-force culling method with linear time. If you are using quad-tree or other spatial culling algorithm, you would need to give a unique number identifier to each particle and use that to in place of i when assigning pointers. Now, visibleParticles is at least partially updated with different pointers, but when it's time to render, you'll only be using the ones from the start of the vector to visibleCount. Still, you'll be able to copy the entire thing into the buffer. Lock the vertex buffer, memcpy the visibleParticles vector into the buffer, and unlock. You now have all the updated vertices updated with one memcpy call. memcpy should be okay here, since you're guaranteed not to change the number of maximum particle elements in the vector. Finally, draw the particles by using the value of visibleCount for the numVertices parameter in DrawIndexedPrimitive.	2020-01-28 23:13:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.28041285276412964, "perplexity": 1763.8338851664234}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00127.warc.gz"}
https://www.enotes.com/homework-help/given-subspaces-h-k-vector-space-v-let-h-k-w-v-w-u-418821	# Given subspaces `H` and `K` of a vector space `V.` Let `H+K={w in V: w=u+v, u=H, v=K}.` Show that `H+K` is a subspace of `V.` Hello! Because `H+K sub V,` the operations (addition and multiplication by a number) are defined, and their results are in `V.` The only thing we need to prove is that the results of this operations remain in `H+K.` 1 . Multiplication by a number. Let `a in RR` (or `CC`... Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime. Hello! Because `H+K sub V,` the operations (addition and multiplication by a number) are defined, and their results are in `V.` The only thing we need to prove is that the results of this operations remain in `H+K.` 1. Multiplication by a number. Let `a in RR` (or `CC` ) and `w in H+K.` Then by the definition of `H+K` there are `u in H` and `v in K` such that `w = u+v.` So `aw = a(u+v) = au+av.` Because H and K are subspaces, `au in H` and `av in K,` thus `a*w in H+K.` 2. Addition. Let `w_1, w_2 in H+K.` Then there are `u_1,u_2 in H` and `v_1,v_2 in K` such that `w_1=u_1+v_1` and `w_2=u_2+v_2.` Then `w_1+w_2 = (u_1+v_1)+(u_2+v_2) = (u_1+u_2)+(v_1+v_2).` Because `H` and `K` are subspaces, `u_1+u_2 in H` and `v_1+v_2 in K,` thus `w_1+w_2 in H+K.` Approved by eNotes Editorial Team	2021-10-21 22:24:10	{"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": 0.8157267570495605, "perplexity": 670.0453964288674}, "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/1634323585441.99/warc/CC-MAIN-20211021195527-20211021225527-00551.warc.gz"}
http://edutechwiki.unige.ch/en/AS3_Compiling_a_program	# AS3 Compiling a program Draft ## 1 Introduction There are potentially three ways to compile your first ActionScript program: 1. Compiling with the binary mxmlc included in the free Flex framework. 2. Compiling with Flex Builder (not covered here) 3. Compiling with the Flash CS3 Authoring tool ## 2 Compiling an AS3 program with the Flex SDK Here, we will describe how to compile with mxmlc. This assumes that your Flex framework is properly installed. For information on how to install it, read the Adobe Flex article. Open a text editor, a simple one that will save the text as it appears on the screen, without any formatting. In that new text file, copy the following code: package { import flash.display.Sprite; public class FilledCircle extends Sprite { function FilledCircle():void { var circle:Sprite = new Sprite(); circle.graphics.beginFill(0xFF794B); circle.graphics.drawCircle(50, 50, 30); circle.graphics.endFill(); } } } Save the file as text and give it the name of "FilledCircle.as". Take good note of the directory in which you save that file. Preferably, put it in a folder quite high up in the hierarchy. We will assume that the file is stored somewhere defined by "\path\to\file\". #### 2.1 On a mac 1. Open a terminal window. Terminal is an application like any other. To find it, go to the Applications -> Utilities. You should see Terminal.app among the files. 2. Double click on the application to open it. 3. At the prompt, type: cd \path\to\file\ mxmlc FilledCircle.as #### 2.2 On a PC 1. From the Windows start menu, open a command prompt by choosing Start > All Programs > Accessories > Command Prompt. • If you can't find it, choose Start->Execute and type cmd. 1. At the command prompt, change to the C:\Flex SDK 2\bin directory then execute the mxmlc executable on your actionscript file. cd C:\Flex SDK 2\bin mxmlc C:\path\to\file\FilledCircle.as The mxmlc executable will compile the program and generate a .swf file name FilledCircle.swf. To run the file, open it in the Flash Player on your desktop or in a web browser that has Flash Player installed. Note that Flash Player 9 needs to be installed to view swf files generated by the Flex compiler. #### 2.3 On a PC by modifying the Path variable A more practical solution is to add the path of the Flex SDK to your system: %Path%;c:\Flex SDK 2\bin See environment variable for technical details if the instructions below are not enough. On a french speaking Windows XP system (sorry) the procedure is: • Démarrer->Paramètres->Panneau de configuration->Système • Select the panels: Avancé->Variables d'environnment On an English system (sorry you may have to fix this a bit): • Start->Parameters->Configuration Panel->System • Select the panels: Advanced->Environment variables Then: • Change the "Path" variable in the user variables pane by adding the flex path at the end of the existing path variable. The result may look like this: %Path%;c:\Flex SDK 2\bin • If the "path" variable doesn't exist, create it and copy the above line. Do not remove the other pathes ! Do not edit the system variables or you may regret it bitterly, because your system will stop working if do it wrongly. ! Here is a screendump showing some of the (french) Windows panels: Changing the User Path variable under Windows XP (french) If you have done this right, you now can use the compiler from any place, e.g. cd c:\soft\as3\ex1 mxmlc FilledCircle.as #### 2.4 Under Linux You should add the path to the flex binaries somewhere, e.g. if you installed flex in /usr/local/flex, add this line to /etc/bash.bashrc export PATH=\${PATH}:/usr/local/flex/bin Note: MacOSX users also could do something like this (but DKS doesn't have a Mac at hand to show how-to). ## 4 AS3 programming with Flash CS3 You may add AS code to any frame as explained in the Flash tutorials. However if you want to code AS 3 for real, e.g. with a class structure, you can use the following "how to". See also Adobe's explanation. ### 4.1 Edit an AS3 program with Flash CS3 If you want, you can edit AS3 programs with Flash CS3. This is an alternative to using a text editor. To create a new File: 1. Open Flash CS3 2. Select menu New. Then select "ActionScript file" (or alternatively select from the panel that appears when Flash CS3 loads). You will find the same editor as the one you may be familiar with as a Flash Designer (F9). All usual Flash panel will be greyed and you can just edit ActionScript code. • Menu File->Import Script will let you import a script if needed. Else you can just copy/paste code from this wiki. E.g. copy the FilledCircle code from above. To edit an existing AS File: • Just open it with the File menu You can not compile a file this way, only edit it. So move on ... ### 4.2 Compiling an AS3 program with Flash CS3 Let's now assume that you created a file FilledCircle.as as explained above. With Flash CS3 you can not compile an .as file, but you can import it into a .fla file and then compile the whole with the ususal procedure. Here are the steps to compile an ActionScript 3 program that is based on a class structure like in all our Action Script 3 tutorials and in all of Adobe's on-line documentation. 1. Create a new Flash file: Menu File->New->Flash File(ActionScript 3) 2. This new .fla should be in the same directory as your ActionScript file. 3. In the Document class field of the Property inspector, enter the class name of the primary class of your AS3 code. E.g. FilledCircle 4. Save this little change ! 5. Test the movie. This will automatically create the .swf file. #### 4.2.1 Example • We created an Actionscript 3 file called FilledCircle.as • It contains the class FilledCircle defined above • We then created a new Flash filed called FilledCircletest.fla • In the Document class field of the properties panel we entered FilledCircle • Note: You now also could edit the AS code again by clicking on the little pen next to the Document class field. Here is a screendump: Document class: Importing AS3 code into a fla file A final note: The .swf file produced will have the name of your .fla file. What this implicitly means is that you also can add Flash CS3 drawings and AS "timeline" code on top of the imported AS code (i.e. combine both ways of doing Flash).	2017-05-01 04:25:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3501715660095215, "perplexity": 5135.676578848808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917127681.50/warc/CC-MAIN-20170423031207-00637-ip-10-145-167-34.ec2.internal.warc.gz"}
https://people.richland.edu/james/misc/prop.html	# Mathematical Tidbits ## Divisibility Tests A positive integer is divisible by the given integer if the condition is met. 1. Every positive integer is divisible by 1. 2. If the last digit is a 0, 2, 4, 6, or 8 3. If the sum of the digits is divisible by 3 4. If the last two digits are divisible by 4 5. If the last digit is a 0 or 5 6. If the number is divisible by both 2 and 3. 7. Check it on the calculator 8. If the last three digits are divisible by 8 9. If the sum of the digits is divisible by 9 10. If the last digit is a 0 11. Subtract the sum of the digits in the even positions (2nd digit, 4th digit, etc) from the sum of the digits in the odd positions (1st digit, 3rd digit, etc). If this difference is divisible by 11, then the number is divisible by 11. A test may be constructed for numbers such as 12, 15, and 18 according to the following rule. If a number can be factored so that the factors are relatively prime (that is, they have no common factors besides one), then the test for divisibility for that number the requirement that the number be divisible by the factors. 12 is a factor if 3 and 4 are both factors, but not necessarily if 2 and 6 are factors. 18 is a factor if 2 and 9 both are, but not necessarily if 3 and 6 both are. 14 is a factor if both 2 and 7 are, but there isn't an easy test for 7. ## Properties of Real Numbers Closure The set of real numbers is closed under addition and multiplication. This means that the sum of two real numbers is a real number and the product of two real numbers is a real number. The set of real numbers is also closed under subtraction (the difference of two real numbers is a real number), but not under division (the quotient of two real numbers may not be a real number - ie, division by zero does not yield a real number) Commutative The set of real numbers is commutative under addition and multiplication. This means that the order of the terms (addition) or factors (multiplication) is irrelevant to the answer. $$a + b = b + a$$ and $$a b = b a$$. The set of real numbers is not commutative with respect to subtraction and division, however: $$a - b \neq b - a$$ and $$\frac {a}{b} \neq \frac {b}{a}$$. Associative The set of real numbers is associative under addition and subtraction. This means that the grouping of terms (addition) or factors (multiplication) is irrelevant to the answer. $$(a+b)+c = a+(b+c)$$ or $$(ab)c = a(bc)$$. The set of real numbers is not associative with respect to subtraction and division, however: $$(a-b)-c \neq a-(b-c)$$ and $$(a/b)/c \neq a/(b/c)$$. Identity There is an additive identity and a multiplicative identity. The identity is the number that you can add or multiply by and get the same answer you started with. The additive identity is zero (0) and the multiplicative identity is one (1). Subtraction and division are defined in terms of addition and multiplication and the same identities hold. Inverse There is an additive inverse for all real numbers, and a multiplicative inverse all real numbers except for the additive identity zero (0). The sum of a number and its additive inverse is the additive identity zero (0). Another name for additive inverse is opposite. The product of a number and its mulitplicative inverse is the multiplicative identity one (1). Another name for multiplicative inverse is reciprocal. Every number except zero (0) has a reciprocal. Distributive There isn't a separate distributive property for addition and multiplication like there were with the other five properties. This is because the distributive property combines addition and multiplication. Stated simply, it says that "Multiplication distributes over addition". The left distributive property is: $$a(b+c)=ab + ac$$, and the right distributive property is $$(a+b)c = ac + bc$$. With real numbers, it is not important to distinguish between the left and right distributive properties because of commutativity. When we talk about Matrices, which aren't commutative under multiplication, then we must distinguish between the left and right properties ## Fundamental Theorem of Arithmetic Every integer greater than one is either a prime number number or can be expressed as an unique product of prime numbers. Prime factorization is the technique used to find that unique factorization. Primes appear many times in arithmetic, hence the reason this is the fundamental theorem. • The Least Common Multiple (LCM) is found by taking each prime factor in all the terms the most number of times it appears in any of the terms. • The Greatest Common Factor (GCF) is found by taking each prime factor in all the terms the least number of times it appears in any of the terms. • When reducing fractions, it is easy to see what to reduce if the numerator and denominator have been written in terms of primes. • Prime numbers can be used in the divisibility tests to arrive at divisibility tests for composite numbers. • The index of a radical must be larger than the exponent on each prime factor in the radicand for a radical to be in simplified radical form. ## Fundamental Theorem of Algebra Every polynomial in one variable of degree n > 0 with real coefficients has at least one complex solution A corollary which almost always follows is... Every polynomial in one variable of degree n > 0 with real coefficients has exactly n complex solutions (not necessarily unique) ## First Fundamental Theorem of Calculus If f is continuous on the closed interval [a,b] and F is an antiderivative of f on [a,b], then the definite integral from x=a to x=b of f(x) with respect to x is the difference between F(b) and F(a). $\int_a^b f(x)\, dx = F(b) - F(a)$ ## Second Fundamental Theorem of Calculus If f is continuous on an open interval containing x=a, then for every x in the interval, the derivative of the antiderivative is the original function. $\frac{d}{dx} \left [ \int_a^x f(t)\, dt \right ] = f(x)$ ## Fundamental Theorem of Linear Programming If a linear programming problem has a solution, then it will occur at one or more corner points or on the boundary between two corner points. ## Fundamental Theorem of Linear Programming - Duality If f is objective function of a linear programming maximization problem and g is the objective function from the corresponding dual problem, then the maximization problem for f has a solution if and only if the minimization problem for g has a solution.	2017-03-24 17:51:31	{"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": 0.7774704098701477, "perplexity": 161.87392852219176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188550.58/warc/CC-MAIN-20170322212948-00663-ip-10-233-31-227.ec2.internal.warc.gz"}
https://www.futurelearn.com/courses/uva-darden-marketing-analytics/0/steps/35773	4.8 ## Darden School of Business, University of Virginia Skip to 0 minutes and 0 secondsSo let's look at how Ohio Art did their TV advertising experiments. So, what they did was, they took Cincinnati as the test city. Then, they had a bunch of control cities. They were, Charleston- South Carolina, Cleveland- Ohio, Indianapolis- Indiana and Pittsburg in Pennsylvania. So, what they did was, the ads ran in the test city and they did not run in the control cities. We have provided links on YouTube for you to watch the Etch A Sketch TV ads. Go take a look at them and when you're done, come back and we can see what happened. So, did you watch the videos? They're nice, aren't they? I thought they were good. I like them. Let's see what happened. Skip to 1 minute and 59 secondsDuring these three weeks, sales of Etch A Sketch was about 240 in Cincinnati and in the control unit it was 1598. The test was run from 27 November to 16 December. This is very close to their peak season, so we have to control for that. Let's see how we do that. But lo and behold, the share of Etch A Sketch units in Cincinnati jumped from 9.6-13.1%. That seems like a decent lift here. Now what we need to see is whether it was big enough, how to make sense of this lift, should we make a decision on launching a campaign based on this lift? What do we have to do next? The lift was about 136%, that's 13.1 over 9.6. Skip to 3 minutes and 2 secondsNow what you can do is look at another product called Doodle Doug. This is also Ohio Art company product, but the beauty of having a control product is that there were no TV ads placed for Doodle Doug during this test period. There were no TV ads around this time, there were ads over here. The lift for Doodle Doug in their share was 96% so sales actually declined in Cincinnati during the test time period. If we take the ratio, then we get the difference between the lift in the test product Etch A Sketch and the lift in the control product Doodle Doug, that is 136.1 minus 96.7%, you have a net lift of 39.4%. Skip to 4 minutes and 18 secondsThis TV ad gave Etch A Sketch in the test city, a net lift of 39.4%. Now the question is, okay, there is some lift. Is this lift making economic sense or not? How do we find that? Let's see. The net lift that we saw was, 39.4%. Now the first step after that is to know how much Ohio Art makes from the sale of a single Etch A Sketch. To do that we first look at the retail price, that's about $10. Now the retailer keeps about 36% of this$10 for themselves. The manufacturer is selling, manufacturer in this case Ohio Art, is selling Etch A Sketch to the retailer for $6.40. How did we get that? Skip to 5 minutes and 15 secondsWe take,$10 times one minus the retail margin, 36%. Now Ohio Art, after accounting for variable costs of production, can keep 58% of $6.4 as contribution margin, and that is 3.71. So that's 6.4*58%, that gives you 3.71.$3.71 is the amount Ohio Art makes in contribution margin from the sale of a single Etch A Sketch. The next thing we need to do is look at the national budget. The national budget that Ohio Art had planned for Etch A Sketch was five million. That's over here. Now how many Etch A Sketches do they need to sell to recover this 5 million? You can get that by taking 5 million and dividing that by 3.71, which is the contribution margin right here. Skip to 6 minutes and 26 secondsAnd that gives you about 1.3 million units. In the whole year, before the ad experiments, Ohio Arts sold about 3.1 million units in the whole year. During the test period, Ohio Art sold about 1.08 million units, so that's the base off of which we need to calculate the lift. The break even lift, the amount of lift necessary to make sense of the national investment in TV advertisement is obtained by looking at 1.3 million over 1.085 million. You're looking at the break even units, which is this, over the base units, which is this one right here, the 1.08 million. 1.3 million over 1.08 million gives you a break even lift of the base of 124%. Skip to 7 minutes and 39 secondsThe net lift from TV ads was how much? That was 39.4%. So clearly, it does not make any economic significance to invest in a national TV advertising campaign for Etch A Sketch because the net lift is much lesser than the break even lift possible, a break even lift that is necessary to make this any economic sense. # Analyzing an Experiment: Etch A Sketch See how to analyze the results of an experiment, from calculating market share and lift to break even. There are a lot of steps to understand and interpret experiment results, but Raj will walk you through them. Then check out the comments. See if you can help a fellow learner understand how to do these calculations (or get help yourself)!	2020-04-07 11:11:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20816031098365784, "perplexity": 1302.8947345891086}, "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/1585371700247.99/warc/CC-MAIN-20200407085717-20200407120217-00283.warc.gz"}
https://par.nsf.gov/biblio/10374462-sdss-iv-manga-gas-rotation-velocity-lags-final-sample-manga-galaxies	SDSS-IV MaNGA – gas rotation velocity lags in the final sample of MaNGA galaxies ABSTRACT We consider the largest sample of 561 edge-on galaxies observed with integral field units by the MaNGA survey and find 300 galaxies where the ionized gas shows a negative vertical gradient (lag) in its rotational speed. We introduce the stop altitude as the distance to the galactic mid-plane at which the gas rotation should stop in the linear approximation. We find correlations between the lags, stop altitude and galactic mass, stellar velocity dispersion, and overall Sersic index. We do not find any correlation of the lags or stop altitude with the star formation activity in the galaxies. We conclude that low-mass galaxies (log(M*/M⊙) < 10) with low-Sersic index and with low-stellar velocity dispersion possess a wider ‘zone of influence’ in the extragalactic gas surrounding them with respect to higher mass galaxies that have a significant spherical component. We estimated the trend of the vertical rotational gradient with radius and find it flat for most of the galaxies in our sample. A small subsample of galaxies with negative radial gradients of lag has an enhanced fraction of objects with aged low-surface brightness structures around them (e.g. faint shells), which indicates that noticeable accretion events in the past affected the extraplanar gas kinematics more » Authors: ; ; ; ; ; ; Publication Date: NSF-PAR ID: 10374462 Journal Name: Monthly Notices of the Royal Astronomical Society Volume: 515 Issue: 2 Page Range or eLocation-ID: p. 1598-1609 ISSN: 0035-8711 Publisher: Oxford University Press ABSTRACT We present the radial gas-phase, mass-weighted metallicity profiles and gradients of the TNG50 star-forming galaxy population measured at redshifts z = 0–3. We investigate the redshift evolution of gradients and examine relations between gradient (negative) steepness and galaxy properties. We find that TNG50 gradients are predominantly negative at all redshifts, although we observe significant diversity among these negative gradients. We determine that the gradients of all galaxies grow more negative with redshift at a roughly constant rate of approximately $-0.02\ \mathrm{dex\, kpc^{-1}}/\Delta z$. This rate does not vary significantly with galaxy mass. We observe a weak negative correlation between gradient (negative) steepness and galaxy stellar mass at z < 2. However, when we normalize gradients by a characteristic radius defined by the galactic star formation distribution, we find that these normalized gradients do not vary significantly with either stellar mass or redshift. We place our results in the context of previous simulations and show that TNG50 high-redshift gradients are more negative than those of models featuring burstier feedback, which may further highlight high-redshift gradients as important discriminators of galaxy formation models. We also find that z = 0 and z = 0.5 TNG50 gradients are consistent with the gradientsmore »	2023-02-08 00:49:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7774556279182434, "perplexity": 2285.6910540163026}, "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/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00122.warc.gz"}
https://www.physicsforums.com/threads/conceptual-question-about-blocks-and-friction.697375/	# Conceptual question about blocks and friction 1. Jun 16, 2013 Hello everyone, Two blocks M and m (M is more massive) are sliding freely with the same initial speed across a floor with friction coefficient μk > 0, and they come to a stop. Initially there is a distance x between them. While they are sliding to a stop, A) The distance between them becomes smaller B) The distance between them becomes greater C) The distance between them stays the same So the equation that I tried with is (Coefficient of Friction) ( Normal Force ) = Friction. According to this equation, M should have a larger normal force, therefore, a bigger frictional force. But the correct answer is C? Can someone explain this please?!? 2. Jun 16, 2013 ### QED Andrew Your conclusion that the frictional force acting on the block of mass $M$ is larger in magnitude than the frictional force acting on the block of mass $m$ is correct. However, this conclusion does not contradict the fact that the distance between the two blocks stays the same. The block of mass $M$ is acted on by a larger force, but it also has a larger mass. These two factors will cancel out when we use $F = ma$ to calculate the acceleration of the block. The block of mass $M$ undergoes the same acceleration as the block of mass $m$ undergoes. Consider a more rigorous analysis. The magnitude of the frictional force $F$ acting on the block of mass $m$ is given by $$F = -\mu_k n$$ Therefore, the block of mass $m$ undergoes an acceleration $a$ given by $$a = \dfrac{F}{m} = \dfrac{-\mu_k n}{m} = \dfrac{-\mu_k mg}{m} = -\mu_kg$$ Note the acceleration is independent of the mass of the block. A similar analysis applies to the block of mass $M$. Last edited: Jun 16, 2013	2017-08-22 02:04:18	{"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": 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": 0.7294797897338867, "perplexity": 276.5068388128985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109803.8/warc/CC-MAIN-20170822011838-20170822031838-00089.warc.gz"}
https://stacks.math.columbia.edu/tag/0E9J	Lemma 31.21.12. Let $i : Z \to X$ be an immersion. If $Z$ and $X$ are regular schemes, then $i$ is a regular immersion. Proof. Let $z \in Z$. By Lemma 31.20.8 it suffices to show that the kernel of $\mathcal{O}_{X, z} \to \mathcal{O}_{Z, z}$ is generated by a regular sequence. This follows from Algebra, Lemmas 10.106.4 and 10.106.3. $\square$ ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work. All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E9J. Beware of the difference between the letter 'O' and the digit '0'.	2022-05-21 03:21:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.600521445274353, "perplexity": 703.9692852531955}, "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-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00173.warc.gz"}
https://design.tutsplus.com/articles/interview-with-rubens-cantuni-tokyo-candies--vector-3236	We want to learn about your experience with Tuts+. Take part to earn $50! Help us out to earn$50! Get Started # Interview with Rubens Cantuni - Tokyo Candies Rubens Cantuni (Tokyo Candies) is an illustrator, designer, and art director. His full-time employee role as art director influences his freelance illustration work in both technical and business ways. However, his true passion and creative outlet is his freelance illustration work, which he hope to make a full-time living from in the future. Rubens has a bold and funky vector style. He tends to mix different or even opposite feelings into his character creations. Learn all about his background, interests, influences, creative process and more in this interview! #### 1. Hello Rubens, please tell us a bit about yourself. Where are you from, what training do you have, and how did you got started in the field? How long have you been illustrating and designing? Hello, first of all: thanks for this opportunity, I'm a regular reader of Vectortuts+ (as well as Psdtuts+) so I'm honored to be on it. My name's Rubens Cantuni, I'm Italian and from Genoa. I was born in 1982 and my journey in illustrating and designing started a little later ;). I've always been involved in drawing, coloring and creating stuff, but I've never had specific training in arts at school, so I can say that I'm self-taught. I have a degree in Industrial Design, that is not exactly what I'm doing now, but it was fun and I enjoyed my university years a lot. When I was in my last year I started working as an art director for an advertising agency, and in my spare time I've always been creating my stuff. In January 2009 I was making some space on my hard disk and I noticed there were a lot of works making dust in there. At that time I was already running the blog koikoikoi.com (a visual arts blog I co-founded with my colleague Danilo) so I thought "Why just talk about other artists' works and keep all my stuff in the closet?" and so I made my site Tokyo Candies, my Behance profile, etc. So even if I've always been illustrating my career is brand new. Tokyo Candies Portfolio #### 2. How did you get started working with vector graphics? How long have you worked with Illustrator? Do you have any favorite vector tools or techniques? Where does Photoshop fit in your workflow? I think I started with vectors in 2005 or 2006. I knew I needed to know Illustrator, but they weren't teaching it at university (where I learnt things like 3Dstudio Max, which I forgot 1 minute later after passing the exam), so I had to learn it by myself. At first, I have to say, I hated it. I was used to Photoshop and I couldn't understand a lot of things happening in Illustrator. I don't know why, at that time, it was pretty hard to find tutorials on Illustrator, while there were tons for Photoshop. So I experimented A LOT, and soon I became more and more familiar with it and found it was really handy to be able to manage all the elements. I don't have a favorite tool, I use the most standard tools: brush, pencil, pen, scissors, mainly. I don't like the gradient mesh tool too much. I mean, it's a powerful tool that lets you make some incredible stuff, but I'm a pretty radical vector artist :). About Photoshop, I use it sometimes as a final step, to add some texture, vintage effects and things like that, but not all the time. #### 3. To what extent are you attracted to uniting extreme concepts in your work? I notice cute mixed with evil? Chubby imbedded with erotic sexuality? What are the mix of ideas that bring about your work? Yes, one of the main things you may notice in most of my works I think is the mix of different or even opposite feelings. Cute and evil, funny and disturbing, the perfect example are, of course, the sexy chubby girls, that are my favorite subject and maybe my most distinguishing characters. An artist I really love, that I think is bringing something similar is Jeremy Fish. The point is: arouse a feeling in the spectator. Could be surprise, curiosity, hilarity or whatever. If you look at one of my works, and you think "Why? Why is it like that?," then I just hit my point. #### 4. How does working as a freelance illustrator and designer influence your work as an art director? How has working as an art director helped you work as a freelancer? Are there any lessons that cross these two roles you'd like to share? I have to say that as a professional I live two separate lives. My everyday work is art direction/graphic design as an employee. I really like my job and it pays my bills, but of course my dream is to live just making my things one day, so I'm working hard, day and night, to do that. My art direction job has been useful to me as a freelancer mainly in understanding client's requests. Plus I learned all the things about the printing process while working there: problems related to it, limitations, preparing a file to be printed, and all this technical stuff you have to know. You can't just be able to draw and/or use Illustrator alone. #### 5. With your busy schedule, how often do you find time to just draw in your sketchbook or do other creative things just for fun? What are your favorite creative outlets? One of the main things I regret is neglecting my sketchbook so much. The other thing is painting so rarely. But I take time to do things for fun, also because they can become sold works, or prints or t-shirts. I also have to work on koikoikoi.com, which is a sort of creative outlet. Even if I don't work in first person on something creative, presenting artists, designers, photographers, works, projects, videos, all feed my creativity hunger and increase my knowledge and my inspiration sources. #### 6. How has travel influenced your work? What places have made an impact on you and why? How often do you hit the road and explore new places? I don't think traveling has influence directly on my works, but it has certainly impacted my knowledge. It may sounds pretty obvious but the place that really had an impact on me is Japan, both for places and people. It was really like being on another planet, so far from italian culture. Last April I've went to New York. It was my first time in the USA. That city is magical, you really can smell endless opportunities, especially if you're involved in art. I hit the road as much as I can, depending on money and spare time, and both things are not as much as I'd like :). #### 7. How active are you online with various design communities, blogs, and social media? Have you had business opportunities arise because of these activities? With my works I'm on: Behance, Facebook, BloodSweatVector, deviantArt, Twitter, and I have shops on Redbubble and Artsprojekt, where you can buy prints, t-shirts, sweatshirts etc. I can say that all my collaborative works come from these channels, I don't have an agent (if there's one reading, we can discuss about it, feel free to contact me!) and there's not much to do locally. #### 8. You have such a distinctive style? How did that come about? What helped narrow your focus as an illustrator and what continues to inspire you as an artist? What are your greatest influences? I really like a lot of different things. My style was a natural evolution and it's still evolving. The most important thing I think is doing things as they naturally come to you, without forcing on making them look a certain way or like someone else's products. My inspiration comes from a lot of different sources, such as asian cultures, tattoo art, cultural icons, movies, comics, cartoons. I have tons of people I admire as artists, just to name a few: Mike Giant, Simone Legno, Jeremy Fish, Jeremyville, Shawn Barber, Audrey Kawasaki, Jason Lìmon, 123Klan, Koralie, Jeff Soto, Flying Fortress, Blu, Sheena Aw, Tado, and Jared Nickerson. #### 9. Could you tell us about your illustration process using the project "The Dirty Cream" as an example? Does the process for that project flow like many of your others, or do you find yourself often changing how you approach your work? My workflow may vary from work to work. If I have a precise idea in mind, I start drawing directly in Illustrator. Sometimes if it's not clear enough, or it's a bit complicated, I make a sketch on paper, as I did for "The Dirty Cream" work. After the sketch on paper (pencil and markers) I traced it with my Wacom tablet, then I gave it the basic color and after that I worked on shadows and lights. While doing these steps I usually make some changes to basic color, removing shadows or changing the transparency, etc. In this work, I noticed that the outlines traced with my Wacom were confusing the type work, so I used them just as a track for coloring. After the main work is done I start refining details, adjusting some curves, adding little elements, etc. The final step was adding the cherry characters and the ice cream lady recycled from a previous work and adjusted a bit. #### 10. Thanks for the interview Rubens! Is there any advice that you'd like to give aspiring illustrators and designer who are working hard to grow professionally? Trivial but so true: keep working hard. Try to develop your own style and work hard on promoting yourself. Thanks again for this opportunity! Also, I will be participating in the upcoming Blood Sweat Vector West Berlin Gallery show, which opened just a couple days ago with loads of other vector artists' work on display. #### Rubens Cantuni on the Web Subscribe to the Vectortuts+ RSS Feed to stay up to date with the latest vector tutorials and articles. One subscription.	2021-11-27 09:04:27	{"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": 0.176029771566391, "perplexity": 2096.489341793088}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00584.warc.gz"}
https://www.r-bloggers.com/2018/02/enter-sadman/	Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. ## Yet more Metalliplots I’m just going to tidy up some issues from my last post where I did some tidytext analysis on the legendary rockers – Metallica. It was a fun way to pass some time and keep my hand in with text mining, something I still don’t seem to get around to doing in the day job. I was pretty annoyed at not getting my faceted split of top terms by descending order to plot by actual descending order. Attempt 1: Many failed efforts later, I arrive at this: Attempt 2 -getting better, but still not right. Julia, in the unlikely event you read this, WHAT am I doing wrong? Enough of my failures (not enough time,folks) lets look at sentiment on a line by line basis, instead of just a net score by track. To do this, I used the sentimentr package, by Tyler Rinker. Stolen from the the package homepage on GitHub: “sentimentr is designed to quickly calculate text polarity sentiment at the sentence level and optionally aggregate by rows or grouping variable(s)”…“sentimentr attempts to take into account valence shifters (i.e., negators, amplifiers (intensifiers), de-amplifiers (downtoners), and adversative conjunctions) while maintaining speed.” It was certainly fast, allowing me to take my original data frame of lyrics (grouped by album, track and line number) and calculate a sentiment score. I then set upper and lower bounds for each line so I could use geom_ribbon as follows: line_sentiment <- sentimentr::sentiment(data\$text) data2 <- data %>% filter(title %notin% c("Am I Evil?", "(Anesthesia) - Pulling Teeth")) %>% mutate(linenumber = row_number()) %>% left_join(line_sentiment, by=c("linenumber"="element_id")) %>% group_by(Album,title) %>% mutate(linenumber= row_number()) %>% ungroup() %>% mutate(max = ifelse(sentiment > 0, sentiment, 0), min = ifelse(sentiment < 0, sentiment,0)) ggthemr(KEA)#custom palette filter(data2,Album =="Kill_Em_All") %>% ggplot(aes(linenumber,sentiment),fill =title,colour="grey50")+ #geom_col(show.legend = FALSE)+ geom_ribbon(aes(ymin= min, ymax = max))+ ggtitle("Kill Em All - Lyric Sentiment Polarity Line by Line")+ xlab("Song Line #") + ylab("Sentiment polarity")+ facet_wrap( track_n ~ title, ncol=5)+ theme_minimal(base_size = 8) In chronological order, and album specific colours, natch: Those of you paying attention last time might remember this plot: I tried recreating it with different seeds and limits, and that link between album one and ten still holds: So I thought I should revisit the topic modelling, this time, instead of by album, grouping them into the 90’s releases (Metallica,Load & Reload) and the first, and latest 2 (Kill Em All, Death Magnetic and Hardwired to Self Destruct). Here are the resulting plots. Alhough I can determine where most of the words are coming from, I’m still not getting much of a feel for a definite “topic” as such: The 90’s : Hardwire ‘Em All : And to finish, you write posts and hope folk like them, and this one went down quite well. Not everyone liked it (hence the title – I am “sad”, apparently) but even so, who cares, because Julia Silge made 2018 an awesome year already : Stuff like this makes staying up late battling with code worthwhile. And of course, I get to use cutting edge data science tools and techniques. That’s got to be worth a “Yeah!” R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job. Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.	2021-04-19 09:33:40	{"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": 0.2209729552268982, "perplexity": 5965.931312927146}, "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-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00154.warc.gz"}
http://mathhelpforum.com/algebra/44993-totally-stuck-print.html	# totally stuck • Jul 31st 2008, 07:04 AM DroMan totally stuck I have to find all complex number solutions and I have to leave my answer in trig form: x^5-i=0 i have no idea how to start this problem. How would I go about getting the 1st degree to plug into the trig form? • Jul 31st 2008, 07:26 AM Soroban Hello, DroMan! Quote: Solve: . $x^5-i\:=\:0$ We have: . $x^5 \:=\:i \quad\Rightarrow\quad x \:=\:i^{\frac{1}{5}}$ Convert to polar form: . $i \;=\;\cos\left(\frac{\pi}{2} + 2\pi n\right) + i\sin\left(\frac{\pi}{2} + 2\pi n\right)$ Use Demoivre's Theorem: . . $i^{\frac{1}{5}} \;=\;\bigg[\cos\left(\frac{\pi}{2} + 2\pi n\right) + i\sin\left(\frac{\pi}{2} + 2\pi n\right)\bigg]^{\frac{1}{5}}$ . . . . $= \;\cos\left(\frac{\pi}{10} + \frac{2\pi}{5}n\right) + i\sin\left(\frac{\pi}{10} + \frac{2\pi}{5}n\right)\quad\hdots\quad \text{for }n \,=\,0,1,2,3,4$ Therefore: . $x \;=\;\begin{Bmatrix}\cos\frac{\pi}{10} + i\sin\frac{\pi}{10} \\ \\[-4mm] \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} &=& i \\ \\[-4mm] \cos\frac{9\pi}{10} + i\sin\frac{9\pi}{10} \\ \\[-4mm] \cos\frac{13\pi}{10} + \sin\frac{13\pi}{10} \\ \\[-4mm] \cos\frac{17\pi}{10} + i\sin\frac{17\pi}{10}\end{Bmatrix}$ • Jul 31st 2008, 01:05 PM DroMan thnaks so much	2016-10-27 15:56:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "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": 0.9870232939720154, "perplexity": 3021.2887319635165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721347.98/warc/CC-MAIN-20161020183841-00135-ip-10-171-6-4.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/209819/a-normed-space-embedding-question	A normed space embedding question Does any normed space X can be embedded into another normed space Y, such that X is density in the Y and dim(Y)=dim(X)+1. - What does $\dim Y=\dim X+1$ mean if $X$ (and $Y$) is infinitely dimensional? – Davide Giraudo Oct 9 '12 at 11:39 Perhaps it is better write codimension(Y)=1 ? – Tomás Oct 9 '12 at 12:00 Yes, codim(Y)=1! – Strongart Oct 12 '12 at 5:47 Well, the answer is no, the usual $\mathbb R^n$ is not going to be dense in $\mathbb R^{n+1}$ (every norm on finite dimension determines the same topology). Take any proper dense subspace $Y$ of an infinite dimension normed space (I bet, such always exists, but for example $X:=L_1[0,1]$ and $Y:=C[0,1]$ with the $L_1$-norm), and extend algebraically its basis -using axiom of choice- to a basis of $X$ and leave one basis vector. That $\,\Bbb R^n\,$ is not going to be dense in $\,\Bbb R^{n+1}\,$ does not prove yet that what the OP asked cannot be attained. – DonAntonio Oct 9 '12 at 12:03	2015-03-01 19:21:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9460315108299255, "perplexity": 654.2108170497013}, "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-2015-11/segments/1424936462548.53/warc/CC-MAIN-20150226074102-00059-ip-10-28-5-156.ec2.internal.warc.gz"}
https://apollo.open-resource.org/mission:log:2012:01:13:dlink-green-ethernet-switch-review	~~SS~~ ## Hot Projects DIY Food Hacking Open Solar Power #### picoReflow DIY Reflow Soldering #### PiGI RasPi Geiger Counter Wideband Antenna Map everything! ## Mission-Tags ~~TAGCLOUD:50~~ Labeling things with the term “green” has become quite popular these days. While evaluating unmanaged SOHO Gigabit-Ethernet switches for Apollo's core net-infrastructure, a lot of “green” switches turned up. After sifting through numerous datasheets D-Link's DGS-1008D seemed to be the top choice when power consumption is an issue. The DGS-1008D employs intelligent power saving modes to actively reduce energy consumption. Unconnected ports or ports without a client in powered or standby mode are put to sleep. The switch also determines, whether the ethernet cable is longer than 20m, if it's not, the port reduces its transmit output power to save even more energy. ## Power Consumption Analysis Fortunately muCCC had the 2nd generation of these switches deployed, so it was possible to take comparative measurements with different use-scenarios and two generations of the same model in a controlled environment. The switches were measured directly on the 5V input in order to avoid misleading results due to power conversion losses. Each switch was tested in the same environment using different operation modes to determine a more real-life power consumption profile: DGS-1008D G2 DGS-1008D G3 Connected Mode Current Power Current Power 0 Clients Switch only 66-93mA 0.33-0.46W 67-111mA 0.33-0.55W 1 Client Client Stdby 100-160mA 0.5-0.8W 109-124mA 0.54-0.62W 2 Clients Client Stdby 141-194mA 0.70-0.97W 153-161mA 0.76-0.80W 2 Clients Client Idle 160-200mA 0.8-1W 170mA 0.85W 3 Clients Client SCP 250mA 1.25W 210-220mA 1.07W 4 Clients Client SCP 250mA 1.25W 210-220mA 1.07W ## Reliability & Performance The DGS-1008D test candidate has been up and running 24×7 for more than 365 days now without any glitch or the necessity to power cycle. Power saving often leads to compromises in performance but the switch performed equally well compared to a CISCO 3560, that was in use before power consumption became a major issue. ## Conclusion Grid powered devices still have the luxury to be able to consume more power than absolutely necessary, since the limiting factor is money and not power itself. Off-Grid scenarios change that point of view on costs, as the value is no longer money, but the painfully finite element of power itself. There are still a lot of devices out there with a standby-consumption of well above 10W, the DGS-1008D G3 connects 4 clients, actively transfers a lot of data and consumes about 1W. D-link obviously did a good job with this, although there is a little curiosity in StandBy: The G2 consumes a little less power in StandBy, when no clients are connected but the G3 saves considerably more power when active so overall the G3 is a very good evolutionary step in low-power consumption Gigabit-Ethernet technology. The only drawback of these switches is their un-managed nature. Hopefully, in the future even managed switches can operate with this kind of power consumption profile. Other manufacturers, who also build similar products, may always send us their products for hardcore real live evaluation as well :) ## Discussion Enter your comment. Wiki syntax is allowed: _ __ ____ ___ ___ ___ / \|/ / / _/ / _ ) / _ \| / _ \| / / _/ / / _ \| / __ \| / __ \| /_/\|_/ /___/ /____/ /_/ \|_\|/_/ \|_\|	2021-10-27 12:14:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.17276325821876526, "perplexity": 8226.812724924837}, "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/1634323588153.7/warc/CC-MAIN-20211027115745-20211027145745-00659.warc.gz"}
http://content.gpwiki.org/index.php/VB:Tutorials:Spaceship_Physics	# VB:Tutorials:Spaceship Physics "An object in motion remains in motion unless acted upon by an outside force." Thank-you Sir Isaac Newton! At first this may seem to be contrary to what you'd expect, after all, if I roll a ball along the floor, it comes to a stop eventually, doesn't it? Well yes, that is true, but there's an outside force at work: friction. The force of friction between the ground and the ball (and also the air and the ball) acts to decrease the speed of the ball, causing it to come to rest at some point. Now, if we take this discussion out into the vacuum of space (my favourite place!), we will no longer have this friction to deal with! Alright.. if there are any die-hards out there, they're probably thinking, "Ryan's stupid! There are still particles and friction in space!" Ok, well you just shut-up :) For our purposes the 3 particles per cubic meter of space (deep space) will cause negligible friction! Where was I before those die-hards so rudely interrupted me? Oh yes, frictionless space. In space, when an object is given a little push it will continue to travel in that direction infinitely (unless acted upon by ANOTHER force)! Cool, huh? How do we calculate the resultant speed and direction? A couple of physics equations are all we need: $F = ma$ "Force equals mass times acceleration." A force can be anything from gravity (which could cause our object to orbit!), to thrust from an engine (SPACESHIPS! WOOOO!). ``` Vf = Vi + at ``` "Final velocity equals initial velocity, plus acceleration times time." We will assume, for now, that the force and initial velocity have the same direction. So, do you notice any similarities between our two equations? They both contain "a" or "acceleration"... we can combine the two! Use the force! $Vf = Vi + (F/m)t$ So now, if we know the magnitude and duration of the force, the mass of the object, and the object's initial velocity, we can calculate the final velocity. Let's say we have a 10000kg spaceship capable of producing 20000N (Newtons) of thrust for 5 seconds, and the spaceship is already travelling at 5m/s: $Vf = (5) + (20000/10000) * 5 = 15m/s$ Our ship would end up travelling at 15m/s (meters per second). Note that this is essentially the same as adding two velocities.. the initial velocity, and the velocity induced by the force. However, if these two velocities do not have the same direction, we're in a bit of trouble! Bring on the trigonometry! Suppose that our ship is travelling with an initial speed of Si at an angle of Theta (with zero being straight "up"). Suppose also that our ship (mass = m) is facing an angle of Phi and is about to thrust with a force of F in that direction. The two resultant velocity vectors may not be parallel, we will have to sum their components (X and Y constituent segments) in order to determine the final velocity of the ship. Let ViSegX and ViSegY represent the component segments of the initial velocity, and FSegX and FSegY represent the component segments of the velocity due to thrust: $ViSegX = sin(Theta) * Si ViSegY = cos(Theta) * Si FSegX = sin(Phi) * (F/m) FSegY = cos(Phi) * (F/m)$ Hey baby, what's your sine? (I do love a good math joke!) Now, we can add these values together to get the components of the final velocity vector, and use this to determine the final speed and direction: $Speed = Sqr((ViSegX + FSegX)^2 + (ViSegY + FSegY)^2)$ Pythagoras rules! (Sqr = Square Root) $Direction = Atn((ViSegX + FSegX)/(ViSegY + FSegY))$ Let's hear it for ArcTangent! :) (Atn = Arctangent) Don't forget, however, that arctangent will only return values in the range from -90 to 90 degrees (-π/2 to π/2 radians). You must check the value of (ViSegY + FSegY), if it is less than zero, you must add 180 degrees (or π radians) to the arctangent result to correct. Keep in mind: all values must be converted to Radians for use in Visual Basic. To convert degrees to radians, simply multiply by (π/180).	2015-02-28 12:23:14	{"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": 0, "mathjax_asciimath": 0, "img_math": 6, "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": 0.6485295295715332, "perplexity": 708.8199008244337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936461944.75/warc/CC-MAIN-20150226074101-00190-ip-10-28-5-156.ec2.internal.warc.gz"}
https://www.beatthegmat.com/viewtopic.php?f=69&t=276283&p=739136	## The number of tests that the group should buy? ##### This topic has expert replies Legendary Member Posts: 641 Joined: 14 Feb 2012 Thanked: 11 times Followed by:8 members ### The number of tests that the group should buy? by gmattesttaker2 » Sat May 10, 2014 4:10 pm Hello, Can you please assist with this: A paleontological research group in Montana is purchasing two types of disposable fossil testing kits for its upcoming dig. Fossil Ease brand tests cost $20 each, and Dino Diagnosis brand tests cost$30 each. The group wants to spend an equal amount of money on each type of test. In the table below, identify the number of tests that the group should buy from Fossil Ease and from Dino Diagnosis so that the group spends an equal amount of money on each type of test. Make only one selection in each column. Fossil Ease Dino Diagnosis 100 150 200 250 400 OA: Fossil Ease : 150 Dino Diagnosis: 100 I tried to take the lcm of 20 and 30 i.e. 60 but I don't think this is the right approach. Sri ### GMAT/MBA Expert Elite Legendary Member Posts: 10347 Joined: 23 Jun 2013 Location: Palo Alto, CA Thanked: 2867 times Followed by:508 members GMAT Score:800 by [email protected] » Sat May 10, 2014 4:14 pm Hi Sri, You absolutely had the right idea, you just have to take it a step further. We're told that the two tests cost $20 and$30 AND that we want to spend an EQUAL AMOUNT of money on each. The LCM of 20 and 30 is 60, so imagine if you spent $60 on each test.... You'd get 3 Fossil Ease Tests and 2 Dino Diagnosis Tests. This gives us a RATIO of what the correct answer has to be.....3:2 Now, pick two numbers from the given options that end in a 3:2 ratio... Final Answer: [spoiler]FE = 150, DD = 100[/spoiler] GMAT assassins aren't born, they're made, Rich Contact Rich at [email protected] Legendary Member Posts: 641 Joined: 14 Feb 2012 Thanked: 11 times Followed by:8 members by gmattesttaker2 » Sat May 10, 2014 5:11 pm [email protected] wrote:Hi Sri, You absolutely had the right idea, you just have to take it a step further. We're told that the two tests cost$20 and $30 AND that we want to spend an EQUAL AMOUNT of money on each. The LCM of 20 and 30 is 60, so imagine if you spent$60 on each test.... You'd get 3 Fossil Ease Tests and 2 Dino Diagnosis Tests. This gives us a RATIO of what the correct answer has to be.....3:2 Now, pick two numbers from the given options that end in a 3:2 ratio... Final Answer: [spoiler]FE = 150, DD = 100[/spoiler] GMAT assassins aren't born, they're made, Rich Hi Rich, Thanks a lot for the explanation. Best Regards, Sri Newbie \| Next Rank: 10 Posts Posts: 1 Joined: 22 Jan 2015 by riakhan » Thu Jan 22, 2015 10:15 pm ok thanks gmattesttaker2 _________________________________________________ https://www.passguide.us/ www.uca.es www.hw.ac.uk Last edited by riakhan on Mon Feb 09, 2015 11:16 pm, edited 1 time in total. ### GMAT/MBA Expert Elite Legendary Member Posts: 10347 Joined: 23 Jun 2013 Location: Palo Alto, CA Thanked: 2867 times Followed by:508 members GMAT Score:800 by [email protected] » Mon Jan 26, 2015 10:56 pm Hi riakhan, Welcome to the Forums. An old IR post seems like an odd place to post your first message, but that's okay. How have your studies been going? And when are you planning to take the GMAT? GMAT assassins aren't born, they're made, Rich Contact Rich at [email protected] • Page 1 of 1	2021-02-27 12:28:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2983148396015167, "perplexity": 6034.447336220199}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358956.39/warc/CC-MAIN-20210227114444-20210227144444-00211.warc.gz"}
http://databasefaq.com/index.php/answer/13556/c-multithreading-condition-variable-condition-variables-notification-is-missed	Question: At the moment I am writing some kind of Fork/Join pattern using std::threads. Therefore I wrote a wrapper class for std::thread which uses a reference counter for all children. Whenever a child finishes its execution the reference counter is decremented and a notification is sent to all waiting threads. The waiting threads wait for the reference counter to become 0 which means all child threads finished their execution. Unfortunately, it seems that sometimes the notification is being missed. I've debugged the program using gdb which showed me that the reference counter in the deepest blocking thread was actually already 0 but it didn't recognize it. /** * * Children management is required for the fork/join model. It is realized by using an atomic reference counter. * The reference counter is initially set or changed dynamically by threadsafe operations. * It is decreased automatically whenever a child task finishes its execution. / { public: /* * Creates a new thread attachment without creating the actual thread nor starting it. / /* * Sets the counter of the child tasks. / void setChildCount (int count); /* * Increments the counter of the child tasks by one. / void incrementChildCount(); /* * Decrements the counter of the child tasks by one. * * Besides it notifies \ref m_childrenConditionVariable for all threads which means that all threads which are calling \ref joinChildren() are being awakened. / void decrementChildCount(); /* * \return Returns the counter of the child tasks. / int childCount(); /* * Waits for notifications of \ref m_childrenConditionVariable if the counter of child tasks is not already 0. * Checks on each notification for the counter to become 0. If the counter is finally 0 it stops blocking and continues the execution. / void joinChildren(); /* * Allocates the actualy std::thread instance which also starts the thread immdiately. * The thread executes the corresponding task safely when executed itself by the operating systems thread scheduler. * \note This method should only be called once. / void start(); /* * Joins the previously with \ref start() allocated and started std::thread. / void join(); /* * Detaches the previously with \ref start() allocated and started std::thread. * This releases the thread as well as any control. / void detach(); private: /* * The thread is created in \ref start(). * It must be started after all attachment properties have been set properly. / /* * This mutex protects concurrent operations on \ref m_thread. / /* * A reference counter for all existing child threads. * If this value is 0 the thread does not have any children. / std::atomic_int m_childrenCounter; /* * This mutex is used for the condition variable \ref m_childrenConditionVariable when waiting for a notification. / std::mutex m_childrenConditionVariableMutex; /* * This condition variable is used to signal this thread whenever one of his children finishes and its children counter is decreased. * Using this variable it can wait in \ref join() for something to happen. / std::condition_variable m_childrenConditionVariable; }; The method start() starts the thread: void ThreadAttachment::start() { / * Use one single attachment object only once for one single task. * Do not recycle it to prevent confusion. / / * Lock the mutex to avoid data races on writing the unique pointer of the thread which is not threadsafe itself. * When the created thread runs it can write data to itself safely. * It is okay to lock the mutex in the method start() since the creation of the thread does not block. * It immediately returns to the method start() in the current thread. / { std::lock_guard<std::mutex> lock(mutex); / * The attachment should stay valid until the task itself is destroyed. * So it can be passed safely. * * http://stackoverflow.com/a/7408135/1221159 * * Since this call does not block and the thread's function is run concurrently the mutex will be unlocked and then the thread can acquire it. / { / * Synchronize with the thread's creation. * This lock will be acquired after the method start() finished creating the thread. * It is used as simple barrier but should not be hold for any time. * Otherwise potential deadlocks might occur if multiple locks are being hold especially in decreaseParentsChildrenCounter() / { std::lock_guard<std::mutex> lock(mutex); } / * After spawning and joining in the task's logic there should be no more children left. / assert(attachment->childCount() == 0); / * Finally the children counter of the parent task has to be decreased. * This has to be done by the scheduler since it is a critical area (access of the different attachments) and therefore must be locked. / assert(scheduler); scheduler->decreaseParentsChildrenCounter(attachment); })); } } This is the method decreaseParentsChildrenCounter() of the class ThreadScheduler: void ThreadScheduler::decreaseParentsChildrenCounter(ThreadAttachment attachment) { { std::lock_guard<std::mutex> lock(this->m_mutex); assert(child != nullptr); if (parent != nullptr) { Attachment parentAttachment = this->attachment(parent); assert(parentAttachment); / * The parent's children counter must still be greater than 0 since this child is still missing. / } } } It basically calls decrementChildCount() for the parent thread. The method joinChildren() waits for all children to be finished: void ThreadAttachment::joinChildren() { / * Since the condition variable is notified each time the children counter is decremented * it will always awake the wait call. * Otherwise the predicate check will make sure that the parent thread continues work. / std::unique_lock<std::mutex> l(this->m_childrenConditionVariableMutex); this->m_childrenConditionVariable.wait(l, [this] { / * When the children counter reached 0 no more children are executing and the parent can continue its work. / return this->childCount() == 0; } ); } These are the atomic counter operations and as you can see I do send a notification whenever the value is decremented: void ThreadAttachment::setChildCount(int counter) { this->m_childrenCounter = counter; } { this->m_childrenCounter++; } { this->m_childrenCounter--; / * The counter should never be less than 0. * Otherwise it has not been initialized properly. / assert(this->childCount() >= 0); / * Notify all thread which call joinChildren() which should usually only be its parent thread. / this->m_childrenConditionVariable.notify_all(); } { } As test case I calculate a Fibonacci number recursively with the Fork/Join pattern. I thought that if the notification is missed it should check the predicate and detect the children counter to be 0. Apparently the value becomes 0 so how can it be missed? Update the variables affecting the condition (in this case the member count) only within a lock for the mutex corresponding to the condition (this->m_childrenConditionVariableMutex). See this answer for the reasoning. 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Thanks...	2017-03-26 07:12:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.17853468656539917, "perplexity": 4048.755081357661}, "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-2017-13/segments/1490218189130.57/warc/CC-MAIN-20170322212949-00390-ip-10-233-31-227.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/9071/intuitive-explanation-of-contribution-to-sum-of-two-normally-distributed-random	# Intuitive explanation of contribution to sum of two normally distributed random variables If I have two normally distributed independent random variables $X$ and $Y$ with means $\mu_X$ and $\mu_Y$ and standard deviations $\sigma_X$ and $\sigma_Y$ and I discover that $X+Y=c$, then (assuming I have not made any errors) the conditional distribution of $X$ and $Y$ given $c$ are also normally distributed with means $$\mu_{X\|c} = \mu_X + (c - \mu_X - \mu_Y)\frac{ \sigma_X^2}{\sigma_X^2+\sigma_Y^2}$$ $$\mu_{Y\|c} = \mu_Y + (c - \mu_X - \mu_Y)\frac{ \sigma_Y^2}{\sigma_X^2+\sigma_Y^2}$$ and standard deviation $$\sigma_{X\|c} = \sigma_{Y\|c} = \sqrt{ \frac{\sigma_X^2 \sigma_Y^2}{\sigma_X^2 + \sigma_Y^2}}.$$ It is no surprise that the conditional standard deviations are the same as, given $c$, if one goes up the other must come down by the same amount. It is interesting that the conditional standard deviation does not depend on $c$. What I cannot get my head round are the conditional means, where they take a share of the excess $(c - \mu_X - \mu_Y)$ proportional to the original variances, not to the original standard deviations. For example, if they have zero means, $\mu_X=\mu_Y=0$, and standard deviations $\sigma_X =3$ and $\sigma_Y=1$ then conditioned on $c=4$ we would have $E[X\|c=4]=3.6$ and $E[Y\|c=4]=0.4$, i.e. in the ratio $9:1$ even though I would have intuitively thought that the ratio $3:1$ would be more natural. Can anyone give an intuitive explanation for this? This was provoked by a Math.SE question The question readily reduces to the case $\mu_X = \mu_Y = 0$ by looking at $X-\mu_X$ and $Y-\mu_Y$. Clearly the conditional distributions are Normal. Thus, the mean, median, and mode of each are coincident. The modes will occur at the coordinates of a local maximum of the bivariate PDF of $X$ and $Y$ constrained to the curve $g(x,y) = x+y = c$. This implies the contour of the bivariate PDF at this location and the constraint curve have parallel tangents. (This is the theory of Lagrange multipliers.) Because the equation of any contour is of the form $f(x,y) = x^2/(2\sigma_X^2) + y^2/(2\sigma_Y^2) = \rho$ for some constant $\rho$ (that is, all contours are ellipses), their gradients must be parallel, whence there exists $\lambda$ such that $$\left(\frac{x}{\sigma_X^2}, \frac{y}{\sigma_Y^2}\right) = \nabla f(x,y) = \lambda \nabla g(x,y) = \lambda(1,1).$$ This analysis works for correlated $X$ and $Y$ as well and it applies to any linear constraints, not just the sum.	2022-01-28 16:34:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9215092062950134, "perplexity": 115.65225605593173}, "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/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00118.warc.gz"}
http://math.stackexchange.com/questions/315153/find-lim-n-to-infty-int-0n-frac1nn2-sin-left-fracxn2-ri	# Find $\lim_{n \to \infty} \int_{0}^n\frac{1}{n+n^2 \sin \left( \frac{x}{n^2} \right)} \mbox{d}x$ [duplicate] Find $$\lim_{n \to \infty} \int_{0}^n\frac{1}{n+n^2 \sin \left( \dfrac{x}{n^2} \right)} \mbox{d}x$$ I've defined $f_n(x) = \begin{cases} \dfrac{1}{n+n^2 \sin \left( \frac{x}{n^2} \right)}, & \text{if} \ \ x \in [0,n] \\ 0, & \text{if} \ x >n\end{cases}$ Of course $f_n \to 0$. I have to find $\displaystyle\lim_{n \to \infty} \int_{0}^\infty f_n(x) \mbox{d}x$, I'm trying to use Lebesgue theorem but i can't find function $g$ such that $\|f_n(x)\| \le g(x)$ and $\displaystyle\int _0 ^\infty g(x) \mbox{d}x < \infty$. - ## marked as duplicate by Ayman Hourieh, Amzoti, Did, Micah, Chris EagleFeb 26 '13 at 21:21 I would just rescale the integral: let $x=n u$, then the integral becomes $$\lim_{n \rightarrow \infty} \int_0^1 \frac{du}{1+n \sin{\left ( \frac{u}{n}\right)}} = \int_0^1 \frac{du}{1+u} = \log{2}$$	2014-10-24 11:32:14	{"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": 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": 0.949465811252594, "perplexity": 694.5130751719948}, "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-2014-42/segments/1414119645866.1/warc/CC-MAIN-20141024030045-00060-ip-10-16-133-185.ec2.internal.warc.gz"}
https://cms.math.ca/cjm/msc/18D05?fromjnl=cjm&jnl=CJM	location: Publications → journals Search results Search: MSC category 18D05 ( Double categories, $2$-categories, bicategories and generalizations ) Expand all Collapse all Results 1 - 1 of 1 1. CJM 2009 (vol 62 pp. 614) Pronk, Dorette; Scull, Laura Translation Groupoids and Orbifold Cohomology We show that the bicategory of (representable) orbifolds and good maps is equivalent to the bicategory of orbifold translation groupoids and generalized equivariant maps, giving a mechanism for transferring results from equivariant homotopy theory to the orbifold category. As an application, we use this result to define orbifold versions of a couple of equivariant cohomology theories: K-theory and Bredon cohomology for certain coefficient diagrams. Keywords:orbifolds, equivariant homotopy theory, translation groupoids, bicategories of fractionsCategories:57S15, 55N91, 19L47, 18D05, 18D35 top of page \| contact us \| privacy \| site map \|	2017-10-20 19:48:35	{"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": 0.4492325186729431, "perplexity": 4879.505181613805}, "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-2017-43/segments/1508187824325.29/warc/CC-MAIN-20171020192317-20171020212317-00226.warc.gz"}
https://math.stackexchange.com/questions/3233485/sequentially-continuity	# Sequentially continuity Given a map $$E:L^\infty\rightarrow C^1(\overline{\Omega}), g_n\rightarrow g$$ in $$L^\infty(\Omega)$$. $$u_n=E(g_n), n\in \mathbb{N}$$ and u=E(g) and show that every time we take a subsection we can find a subsubsection that converge to $$u \in C^1(\overline{\Omega})$$then they say the function is continuous, why? Lemma: Let $$X$$ be a topological space and let $$x_n, x \in X$$. Then $$x_n \to x$$ if and only if every subsequence $$x_{n_k}$$ of the sequence $$x_n$$ has a further subsequence $$x_{n_{k_j}}$$ such that $$x_{n_{k_j}} \to x$$ as $$j \to \infty$$. To prove the non-trivial direction in this lemma, you assume that $$x_n \not \to x$$ and use this to construct a subsequence that doesn't have any subsequence converging to $$x$$. Continuity of your map $$E$$ then follows since for maps between metric spaces sequential continuity and continuity coincide so that it is enough to check that if $$g_n \to g$$ then $$E(g_n) \to E(g)$$. This is precisely what is provided by your hypotheses and the lemma. • The lemma is really very easy to prove. You should try to do it yourself. For the non-trivial direction, if $x_n \not \to x$ then there is an open neighbourhood $u$ of $x$ such that there are infinitely many $n$ with $x_n \not \in U$. Use this to write down a subsequence that has no further subsequence that converges to $x$. – Rhys Steele May 24 at 13:10	2019-07-19 01:27:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9762687683105469, "perplexity": 53.42804229314238}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00315.warc.gz"}
http://www.qetlab.com/KpNormDual	# KpNormDual Other toolboxes required kpNormDual Computes the dual of the (k,p)-norm of a vector or matrix none kpNormKyFanNormSchattenNormTraceNorm Norms yes (convex) kpNormDual is a function that computes the dual of the (k,p)-norm of a vector or matrix. More explicitly, the (k,p)-norm of a vector $x = (x_1,x_2,\ldots,x_n)$ is $\\|x\\|_{(k,p)} := \left(\sum_{j=1}^k \big\|x_i^\downarrow\big\|^p \right)^{1/p},$ where $(x_1^\downarrow,x_2^\downarrow,\ldots,x_n^\downarrow)$ is a rearrangement of the vector $x$ with the property that $\|x_1^\downarrow\| \geq \|x_2^\downarrow\| \geq \cdots \geq \|x_n^\downarrow\|$. Similarly, the (k,p)-norm of a matrix is the (k,p)-norm of its vector of singular values. This function computes the dual of this norm, which is fairly complicated and was derived in[1]. This function works with both full and sparse vectors and matrices. ## Syntax • NRM = kpNormDual(X,K,P) ## Argument descriptions • X: A vector or matrix to have its norm computed. • K: A positive integer. • P: A real number ≥ 1, or Inf. ## Examples ### A simple 4-by-4 example The (k,p)-norm of a matrix when k = 1 is simply the operator norm. The dual of the operator norm is the trace norm, so when k = 1 this function just returns the trace norm (regardless of p): >> X = [1 1 1 1;1 2 3 4;1 4 9 16;1 8 27 64]; >> [kpNormDual(X,1,1), TraceNorm(X)] ans = 77.0015 77.0015 Similarly, if K = min(size(X)) and P = 2 then kpNorm(X,K,P) is the Frobenius norm, which is its own dual. Thus kpNormDual(X,K,2) decreases from the trace norm of X to its Frobenius norm as K increases: >> [kpNormDual(X,1,2), TraceNorm(X)] ans = 77.0015 77.0015 >> kpNormDual(X,2,2) ans = 72.6903 >> kpNormDual(X,3,2) ans = 72.6505 >> [kpNormDual(X,4,2), norm(X,'fro')] ans = 72.6498 72.6498 ## Source code Click on "expand" to the right to view the MATLAB source code for this function. 1. %% KPNORMDUAL Computes the dual of the (k,p)-norm of a vector or matrix 2. % This function has three required arguments: 3. % X: a vector or matrix 4. % K: a positive integer 5. % P: a real number >= 1, or Inf 6. % 7. % NRM = kpNormDual(X,K,P) is the dual of the (K,P)-norm of the vector or 8. % matrix X (see URL for details). 9. % 10. % URL: http://www.qetlab.com/kpNormDual 11. 12. % requires: kpNorm.m 13. % author: Nathaniel Johnston (nathaniel@njohnston.ca) 14. % package: QETLAB 15. % last updated: June 24, 2015 16. 17. function nrm = kpNormDual(X,k,p) 18. 19. sX = size(X); 20. nX = min(sX); 21. xX = max(sX); 22. 23. % There are some special cases that we can compute slightly faster than 24. % doing a full SVD, so we consider those cases separately. 25. if((nX > 1 && k >= nX) && p == 2) % dual of the Frobenius norm is the Frobenius norm 26. nrm = norm(X,'fro'); 27. elseif((nX > 1 && k >= nX) && p == 1) % dual of the trace norm is the operator norm, which is computed faster by kpNorm.m, which uses svds in this case 28. nrm = kpNorm(X,1,1); 29. elseif(p == Inf) % this case is needed to avoid errors, not for speed reasons: dual of operator norm is trace norm 30. nrm = kpNorm(X,xX,1); 31. else % in all other cases, it seems like we have to compute all singular values or just brute-force our way with CVX 32. 33. % If X is a CVX variable, try to compute the norm in a way that won't 34. % piss off MATLAB. 35. if(isa(X,'cvx') == 1) 36. % Just compute the value via CVX using the definition of what it 37. % means to be the dual norm: optimize over the unit ball in the 38. % original norm. 39. cvx_begin quiet 40. cvx_precision best; 41. variable Y(sX(2),sX(1)) complex 42. maximize real(trace(X'Y)) 43. subject to 44. kpNorm(Y,k,p) <= 1; 45. cvx_end 46. 47. nrm = real(cvx_optval); 48. else 49. if(nX == 1) 50. k = min(k,xX); 51. s = sort(X,'descend'); 52. else 53. k = min(k,nX); 54. s = svd(full(X)); 55. end 56. 57. for r=(k-1):-1:0 58. av = sum(s(r+1:end))/(k-r); 59. if(r == 0);break;end 60. if(s(r) > av);break;end 61. end 62. if(p == 1) 63. nrm = max([s(1),av]); 64. else 65. nrm = norm([s(1:r).',avones(1,k-r)],p/(p-1)); 66. end 67. end 68. end ## References 1. G.S. Mudholkar and M. Freimer. A structure theorem for the polars of unitarily invariant norms. Proc. Amer. Math. Soc., 95:331–337, 1985.	2017-09-23 19:51:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8678982257843018, "perplexity": 4795.2258808912875}, "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-2017-39/segments/1505818689775.73/warc/CC-MAIN-20170923194310-20170923214310-00132.warc.gz"}
https://ask.libreoffice.org/en/question/134253/conditional-format-column-updated-10-16-2017/	# Conditional Format Column (updated) 10-16-2017 .--------------------------------------------------------------------------------------------------------------------------. . . . . Update.....11/16/2017 Not sure what is going on or what I am doing wrong, but very frustrating. I duplicate suggestions and nothing happens.... past week I downloaded new version of LO thinking this maybe the problem but its not..... see new pic and file with Conditional Formatting . . . . . C:\fakepath\Conditional Format Columns 10-16-2017 Not Working.ods . . . . . edit retag close merge delete Anyone? .. ( 2017-10-11 06:02:49 +0200 )edit I think I got confused. Do you want each cell in column E to be red if it is greater than any column F value, that is, the maximum value of column F? So that each cell in column E is color green if it is less of any value in column G, that is, of the minimum value of column F? ( 2017-10-12 07:53:12 +0200 )edit Yes, Ty Charlie Each cell in column E to be red if it is greater than any CELL in column F value, that is, the maximum value of column F? So that each cell in column E is color green if it is less of any CELL value in column G, that is, of the minimum value of column F? ( 2017-10-13 18:46:43 +0200 )edit Because there are not values on E greater than the maximum value on F, that is way I changed on the sample some values on E to verify CF was working fine. ( 2017-10-16 22:17:35 +0200 )edit Sort by » oldest newest most voted Ciao, see my attachment please. If my answer helped you, vote it with ✔ (here on the left) Edit:I am sorry, I'm wrong in copy and paste. Replace F18:F115 with $F$18:$F$115 and G18:G115 with $G$18:$G$115. Edit ---------------------------------------------------------------------------------------- NewFile.ods End Edit ----------------------------------------------------------------------------------- 15076581288654682.ods more Thank you, I understand Cell Value is .... pertaining to another cell.... What i was looking for, to show if each Cell in column E if Greater than the value in any cell within column F (F18::F115) it will turn RED and less than any cell within Column G (G18::G115) it will turn Green. EXAMPLE: E18 will scan all of (F18::F115) and (G18::G115) if Greater than F column it will turn RED if less than column G it will turn Green and so on then next cell E19 will do the same.... next cell E20... ( 2017-10-11 14:39:23 +0200 )edit Thank you for the update file.... I tried this before posting.... when i uploaded file i see not conditional format changes to the cells in column E example what I would like to see.. Column E Row 93 159.79 should be Green since it is less than Column G Row 98 ( 2017-10-11 16:07:53 +0200 )edit I am sorry, I'm wrong in copy and paste. Replace F18:F115 with $F$18:$F$115 and G18:G115 with $G$18:$G$115. ( 2017-10-11 17:27:07 +0200 )edit thank you for the feedback, i tried this before posting. would u be able to provide .ods that shows conditional format? all attempts on my side have failed. please show screen shot of conditional format if possible ( 2017-10-11 18:51:16 +0200 )edit I think I got confused. Do you want each cell in column E to be red if it is greater than any column F value, that is, the maximum value of column F? So that each cell in column E is color green if it is less of any value in column G, that is, of the minimum value of column F? ( 2017-10-12 19:03:50 +0200 )edit Yes, Ty Charlie Each cell in column E to be red if it is greater than any CELL in column F value, that is, the maximum value of column F? So that each cell in column E is color green if it is less of any CELL value in column G, that is, of the minimum value of column F? ( 2017-10-13 18:49:14 +0200 )edit Doing the comparison against the MAX($F$6:$F$43) and MIN($G$6:$G$43), and can be done for all cells at once, instead one by one with the cell range. 15077558367905342.ods Some values was change to have some cells with the CF condition. more ## Stats Asked: 2017-10-10 19:55:37 +0200 Seen: 92 times Last updated: Oct 16 '17	2020-06-03 17:07:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2828362286090851, "perplexity": 2326.2965455355848}, "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-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00443.warc.gz"}
https://www.shaalaa.com/question-bank-solutions/show-that-relation-r-set-r-real-numbers-defined-r-a-b-b2-neither-reflexive-nor-symmetric-nor-transitive-types-relations_11714	Show that the Relation R in the Set R of Real Numbers, Defined as R = {(A, B): A ≤ B2 Neither Reflexive Nor Symmetric Nor Transitive. - Mathematics Show that the relation R in the set of real numbers, defined as R = {(ab): a ≤ b2} is neither reflexive nor symmetric nor transitive. Solution R = {(ab): a ≤ b2} It can be observed that (1/2, 1/2) in R, Since 1/2 > (1/2)^2 = 1/4 ∴R is not reflexive. Now, (1, 4) ∈ R as 1 < 42 But, 4 is not less than 12. ∴(4, 1) ∉ R ∴R is not symmetric. Now, (3, 2), (2, 1.5) ∈ R (as 3 < 22 = 4 and 2 < (1.5)2 = 2.25) But, 3 > (1.5)2 = 2.25 ∴(3, 1.5) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. Is there an error in this question or solution? APPEARS IN NCERT Class 12 Maths Chapter 1 Relations and Functions Q 2 \| Page 5	2021-03-07 09:14:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.576055109500885, "perplexity": 1950.2595702572482}, "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-2021-10/segments/1614178376206.84/warc/CC-MAIN-20210307074942-20210307104942-00197.warc.gz"}
https://kb.vshn.ch/oc4/explanations/gcp/name_lengths.html	# Names and name length on Google Cloud Platform Abstract Node names are limited to a length of 63 characters [1]. If the GCP project and region names are too long, provisioning of nodes won’t work and the installer will fail. When installing a new OpenShift cluster, the installer will create a lot of names automatically. Some parts of those names are generated by the installer, others are derived from the underlying cloud. Let us have a look at how the name of a node is built. ``````<node/hostname> ::= <machine name>.<region>-<zone>.c.<project name>.internal <machine name> ::= <machine set name>-<random> <machine set name> ::= <cluster id>-w-<zone>`````` `<cluster id>` and `<random>` are generated by the installer. `<zone>`, `<region>` and `<project>` are derived from GCP. As node names are limited to 63 characters [1], this can become an issue. The `<cluster id>` will have a length of twelve characters, `<zone>` is just one characters and `<random>` has a length of five. Summing up all the characters that are static and or are generated by the installer, we end up at 37 (see example below). This leaves us with 26 characters to be distributed between the project name and the region. The length of GCP region names vary between eight and 23. In the best case, the project can be 18 (63 - 37 - 8) characters long. In the worst case, only three (3, 63 - 37 - 23) characters are available. For Zürich (`europe-west6`), the project length must not exceed 14 (63 - 37 - 12) characters. Example from an actual cluster which exceeded the maximum. ```opensh-5g8lh-w-a-xbtr4.europe-west6-c.c.openshift-4-poc-1.internal opensh-5g8lh-w-a-xbtr4.<region>-c.c.<project>.internal opensh-5g8lh-w-a-xbtr4.-c.c..internal``` 1. What happens when the node name exceeds 63 characters? The status of the Machine object will be `Provisioned` but no Node object will show up. The CertificateSigningRequest won’t get approved (remains in `Pending`) and a new one will be created every few seconds. When SSH into the affected VM, one can observe that there is no `/etc/hostname` file and that the hostname is identified as `localhost`. The kublet log will contain something that looks like the following: `May 13 13:38:18 opensh-5g8lh-w-a-xbtr4.europe-west6-a.c.openshift-4-poc-1.intern hyperkube[5777]: E0513 13:38:18.478461 5777 kubelet_node_status.go:92] Unable to register node "opensh-5g8lh-w-a-xbtr4.europe-west6-a.c.openshift-4-poc-1.intern" with API server: Node "opensh-5g8lh-w-a-xbtr4.europe-west6-a.c.openshift-4-poc-1.intern" is invalid: metadata.labels: Invalid value: "opensh-5g8lh-w-a-xbtr4.europe-west6-a.c.openshift-4-poc-1.intern": must be no more than 63` When installing a new cluster, the installer log will look something like the following: ```[…] INFO Waiting up to 30m0s for the cluster at https://api.openshift-4-poc-1.openshift-4-poc-1.appuio-beta.ch:6443 to initialize... ERROR Cluster operator authentication Degraded is True with IngressStateEndpoints_MissingSubsets::RouteStatus_FailedHost: IngressStateEndpointsDegraded: No subsets found for the endpoints of oauth-server RouteStatusDegraded: route is not available at canonical host oauth-openshift.apps.openshift-4-poc-1.openshift-4-poc-1.appuio-beta.ch: [] INFO Cluster operator authentication Progressing is Unknown with NoData: INFO Cluster operator authentication Available is Unknown with NoData: INFO Cluster operator console Progressing is True with RouteSync_FailedHost: RouteSyncProgressing: route is not available at canonical host [] INFO Cluster operator console Available is Unknown with NoData: INFO Cluster operator image-registry Available is False with NoReplicasAvailable: The deployment doesn't have available replicas INFO Cluster operator image-registry Progressing is True with DeploymentNotCompleted: The deployment has not completed	2022-08-19 05:55:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.18463212251663208, "perplexity": 2780.857361376676}, "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-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00229.warc.gz"}
https://codegolf.stackexchange.com/questions/58891/dont-google-google/76322	We all know that if you google the word "google" it will break the internet. Your task is to create a function that accepts one string and returns its length, in the fewest possible Unicode characters. However, if the given string is google (lowercase), it will cause an error. For example, g('bing') will return 4 but g('google') will cause an error. Please provide an example of usage, and the error if possible. • @Geobits That is simply a test to see if I will google Google, which I will not. :D – rybo111 Sep 28 '15 at 16:33 • Does the function need to be case sensitive? Should it throw given 'gOOgle'? – AXMIM Sep 30 '15 at 22:07 • When I type google into google (the search bar on chrome), a message came up asking if I wanted to go to google. (Now that it is a tld, this makes sense i.e. com.google works). I clicked it and got a dns lookup error. Internet:broken! – Craig Oct 1 '15 at 4:18 • I'm voting to reopen this. I have seen no questions about what constitutes an error for this challenge and it already has 154 answers so I don't think it's fair to change the spec. This may not be an example of a good question but it's clear enough. If an answer really comes down to whether or not a certain output is an error it probably just won't get as many upvotes, anyway. – Poke Aug 1 '18 at 19:52 # Julia, 26 24 chars Two characters shorter now, thanks to Dennis. The challenge says ‘[...]in as few amount of Unicode characters as possible’, so I’m going for ≠ and ℕ in my solution. Anonymous function. Just assign it to a variable: f->f≠"google"?endof(f):ℕ In the case of "google" you get the error message: julia> (f->f≠"google"?endof(f):ℕ)("google") ERROR: UndefVarError: ℕ not defined As ordinary function (26 chars): x(f)=f≠"google"?endof(f):ℕ ℕ is an unassigned variable, so it produces an error message. If that doesn’t count or is against the rules, then just using !f (one char longer) instead would produce an error anyway. Boolean not is not defined for strings. # Fuzzy Octo Guacamole, 19 bytes (non-competing) "google".^={e}_!r_; "google' pushes the string "google" to the stack. . swaps the active stack. ^ gets input. = is the rocketship operator, pushes 0 for equality, -1 for less than, and 1 for greater than as x <=> y with x as the top of the active stack, and y as the top of the inactive one. { and } denote an if statement, like bf's goto/if thing. Skips the code inside braces if the top of the stack is truthy. e errors. That's it. _ pops the stack, this time is just to remove the equality check result and show the input. ! sets the universal register to the top of the stack, or the length of the top of the stack for strings/lists. r clears the register and pushes it back to the stack. _ pops again, and ; prints. So this checks for equality, then does a clumsy cast to int, and than prints the result. • the quotes... aren't maTCHED the song of unmatched quotes will extinguish the voices of mortal man from the sphere I can see it can you see ̲͚̖͔̙î̩́t̲͎̩̱͔́̋̀ it is beautiful the final snuffing of the lies of Man ALL IS LOŚ͖̩͇̗̪̏̈́T ALL IS LOST the pon̷y he comes he c̶̮omes he comes the ichor permeates all MY FACE MY FACE ᵒh god no NO NOO̼OO NΘ stop the an̶͑̾̾̅ͫ͏̙̤g͇̫͛͆̾ͫ̑͆l͖͉̗̩̳̟̍ͫͥͨe̠̅s ͎a̧͈͖r̽̾̈́͒͑e not rè̑ͧ̌aͨl̘̝̙̃ͤ͂̾̆ ZA̡͊͠͝LGΌ ISͮ̂҉̯͈͕̹̘̱ TO͇̹̺ͅƝ̴ȳ̳ TH̘Ë͖́̉ ͠P̯͍̭O̚N̐Y̡ H̸̡̪̯ͨ͊̽̅̾̎Ȩ̬̩̾͛ͪ̈́̀́͘ ̶̧̨̱̹̭̯ͧ̾ͬC̷̙̲̝͖ͭ̏ͥͮ͟Oͮ͏̮̪̝͍M̲̖͊̒ͪͩͬ̚̚͜Ȇ̴̟̟͙̞ͩ͌͝S̨̥̫͎̭ͯ̿̔̀ͅ – cat Mar 26 '16 at 12:10 • -1 for unmatched quotes ಠ_ಠ – cat Mar 26 '16 at 12:15 • @tac they represent the beginning and end of the string literal. That is intentional. – Rɪᴋᴇʀ Mar 26 '16 at 15:31 • I know that, I just don't know why anyone would willingly create a language that uses unmatched things. (I know many golfing languages do it and it makes my eyes bleed) – cat Mar 26 '16 at 15:32 • @tac I can make it work with matched ones also if you want. – Rɪᴋᴇʀ Mar 26 '16 at 15:39 # C++14, 42 chars For giggles, as I did not see a very good c++ solution. My TI-89 solution is superior at 34 bytes [](auto s){return s.size()/(s!="google");} # TI 89 BASIC, 34 bytes I did this in TI 89 for fun. There has got to be a more optimized method, but this is the best I could come up with for the moment. f(x) Basically abuses the fact that a string divided by itself becomes 1. 0 is necessary to remove the string from the calculation otherwise you would return ("fun" + 3) which i don't think is compliant to the rules. • Alright, wasn't sure if I could post multiple answers to a solution @lirtosiast , thanks! – STDQ Mar 26 '16 at 20:20 ## Hoon, 32 bytes \| a/* ?< (lent a) Uses a wet gate to avoid having to specify a/tape instead of a/, assert that a doesn't equal "google" or panic, return the length of the tape. > %. "abc" \| a/* ?< (lent a) 3 \|* a/* ?< (lent a) ford: build failed ~[/g/~dirdet-lasmes-digwyc-ribrux--rispyx-bitrus-bidmut-winsud/use/dojo/~dirdet-lasmes-digwyc-ribrux--rispyx-bitrus-bidmut-winsud/inn/hand /g/~dirdet-lasmes-digwyc-ribrux--rispyx-bitrus-bidmut-winsud/use/hood/~dirdet-lasmes-digwyc-ribrux--rispyx-bitrus-bidmut-winsud/out/dojo/drum/phat/~dirdet-lasmes-digwyc-ribrux--rispyx-bitrus-bidmut-winsud/dojo /d //term/1] • Um... is that supposed to link to urbit? – Destructible Lemon Nov 20 '16 at 9:55 • @DestructibleWatermelon Yup. Urbit provides the compiler/interpreter for the Hoon language. – RenderSettings Nov 21 '16 at 4:00 # Tellurium, 13 bytes This language is newer than the question, so non-competing I guess. i?google\|d]L^ This program gets input using i, and checks if the input is "google". If it is, it tries dividing "google" by zero (d) well, that doesn't work (duh) so it throws an error. If the input isn't google, it outputs the length of the selected cell's value (which is the input) using L^. ## AWK, 44 bytes func f(x){return x=="google"?f(x):length(x)} Example usages: awk 'func f(x){return x=="google"?f(x):length(x)}{print f($0)}' <<< "non-google string" awk 'func f(x){return x=="google"?f(x):length(x)}{print f($0)}' <<< "google" Print: 17 and Segmentation fault (core dumped) respectively. The second one may cause your computer to run out of memory before it segfaults if you don't have oodles of RAM. I find Segmentation fault errors to be a bit more broken than divide by zero since they don't even say where the problem is. Something that looks more interesting would be: func f(x){if(x=="google"){printf x;return f(x)}return length(x)} This would print googlegooglegooglegoogle.... wrapping around the screen until it finally produces the seg. fault, but it's not as 'golfy' NB. Yeah, I'm a bit late to the party, but nobody had an AWK answer yet. :) • +1 for "Segmentation fault (core dumped)" -- my favourite error message – cat Mar 25 '16 at 15:46 • It's my favorite message, too... if I'm not the one that has to debug it. :p – Robert Benson Mar 25 '16 at 16:04 • On my 64-bit Ubuntu 16.04 with GAWK 4.1.3, the second one doesn't segfault but just eats all my memory until one of three things happens. A) the oom-killer kills it and says Killed, B) my window manager crashes or C) my computer overheats and switches off. Thus, I am adding a note to your answer that this answer may be harmful on machines which have a big stack. – cat Jun 28 '16 at 13:34 • Aha! That's funny. It is not that your machine has so little memory space that AWK gives up and segfaults immediately, it's that you have so much memory that AWK doesn't cause the kernel to kill it but instead tries to address so much memory that its virtual machine eventually segfaults. I only have a paltry 6 GB of RAM :P – cat Jun 28 '16 at 16:32 • It's not wrong at all, I think recursing forever and possibly forcing a shutdown is a perfectly acceptable error – cat Jun 28 '16 at 16:52 # PowerShell, 36 35 Bytes param($a)$a.length/($a-cne'google') This blatantly abuses PowerShell's dynamic casting and uses xnor's trick for dividing by zero. In PowerShell, this is a terminating error and halts execution tossing a most excellently-verbose error (the error, at 242 characters, is over 6.5x the size of the function itself) Attempted to divide by zero. At line:1 char:11 + param($a);$a.length/($a-cne'google') + ~~~~~~~~~~~~~~~~~~~~~~~~~~ + CategoryInfo : NotSpecified: (:) [], RuntimeException + FullyQualifiedErrorId : RuntimeException Saved a byte thanks to ConnorLSW • you can use param($a)$a.length/($a-cne'google') to skip using the the !, saving one byte. – colsw Nov 29 '16 at 10:57 • @ConnorLSW Indeed. Thanks. – AdmBorkBork Nov 29 '16 at 15:54 # PHP, 48 bytes eval((google==$s=$argv[1])."echo strlen(\$s);"); php -r '<code>' google --> Parse error: syntax error, unexpected 'echo' (T_ECHO) [...] in eval()'d code on line 1 54 bytes for a function: function g($s){eval((google==$s).'echo strlen($s);');} # JavaScript, 25 bytes p=>p=='google'?_:p.length • “However, if the given string is google (lowercase), it will cause an error.” – Will this? – manatwork Nov 29 '16 at 15:15 • Now it matches the challenge's requirement. Sadly the generic expectation is the solutions to be either full programs or functions. In both cases the input and output has to be handle either explicitly by the code or implicitly by the interpreter. So you can not assume your code will find data in some global variable. – manatwork Nov 29 '16 at 15:38 • You can make this valid in one of two ways: 1. Make it into a full program with input and output: g=prompt();if(g!='google')alert(g.length);else throw 0 2. Make it into a function: function(g){if(g!='google')return g.length;else throw 0} – ETHproductions Nov 29 '16 at 16:44 • Also, you don't have to explicitly throw an error; calling some undefined variable, such as _, will do the trick. – ETHproductions Nov 29 '16 at 16:44 ## PHP, 56 bytes function google($i){echo strlen($i)/(__FUNCTION__!=$i);} Yes, it's a bit long, but I feel this is really in the spirit of the question. You really mustn't google google when google is the actual function's name. The constant __FUNCTION__ holds the name of the function, which in this case is google. The rest of the functiondisplays the length of the input $i divided by 1 if $i is not google, or by 0 if it is. The latter throws an error. Try it at Repl.it! # Jelly, 10 bytes ßL⁼“æ8Ụ»$? Try it online! (with "google" as input) Try it online! (with "bing" as input) ## How it works ßL⁼“æ8Ụ»$? - Main link. Argument: s (string) ? - if... ⁼ $- the input is equal to... “æ8Ụ» - "google" - then ß - call the main link (segfault) - else L - return the length of the input # Aceto, 19 bytes rdM"google"J=$L€lp Took me a while, but I finally got it. Aceto uses a hilbert curve for it's ip, which is really annoying. r grabs input d duplicates it M pops the top value and stores it in quick memory "google" pushes google on the stack, but with a lot of spaces in between. - splits it on whitespace J concatenates it = pushes a bool on the stack: whether the top two values are equal $asserts that the top value is truthy, if not it raises an error. Don't ask me why i don't have to negate it L loads the quick memory € explodes the string, putting each char as a separate val on the stack l takes the height of the stack p prints that num Try it online! # Pyth - 15 bytes ?qz"google"'0lz My favorite part of this is how I error. I use the ' function which takes a string, but pass 0, which is not a string. # Brachylog, 12 bytes "google"∧∈\|l Try it online! Since Brachylog simply fails a predicate that tries to divide by 0, the error here comes from trying to unify the output with a list containing an entirely un-instantiated variable instead. If the input can be unified with google, it errors, and otherwise this predicate outputs the length of the input (add one more byte w to the end to make it a full program that prints the length). ## Wd, 10 9 bytes This is a full program but is technically also a function. ↔XÑ║╜▄]ÿ ## Explanation Decompressed: -Fl'InakS/ After string decompression: google"nakS/ google"n # Does the input not* equal to "google"? ak # Find the length of the input S/ # Divide the length of the input by the above condition # Clojure, 47 44 bytes Old answer #(/(count %)(get{false 1 true 0}(= %"google"))) Try it online! New Answer #(/(count %)({false 1 true 0}(= %"google"))) Removed the get function call, because you can call a hashmap with a key to get the value. Try it online! • Welcome to the site, and nice first answer! Be sure to check out our Tips for golfing in Clojure page for ways you can golf your program – caird coinheringaahing Feb 20 at 20:16 • @cairdcoinheringaahing thank you^^ – Satoshi Feb 20 at 20:19 # Vyxal, 7 bytes ‛»Ǒ≠/÷L Try it Online! ‛»Ǒ≠/÷L ‛»Ǒ # "google" ≠ # ↑ != input / # split input into ↑ even pieces. If input is "google", this will be 0, consequently causing an error ÷L # Get the length of the item • You might want to clarify that it's length(^) / (^ != ^^) instead of (length(^) / ^) != ^^. – Wezl Feb 22 at 16:02 # Red, 34 bytes func[g][length? any[g ="google"g]] Try it online! (the code so dense that it crashes the program). ## CoffeeScript, 36 bytes f=(x)->throw 0if'google'==x;x.length # Emacs Lisp, 43 bytes (lambda(s)(if(equal s"google")(length s))) Throws the error (void-variable ) for any string that equals google. • @FryAmTheEggman I'm sorry, I just wrote the explanation wrong, the code is correct. – nanny Sep 28 '15 at 18:55 ## STATA, 44 bytes pr de a if"google"==0' f di length(0') end prints "unrecognized command: f" when input is "google" # Perl5, 48 bytes sub google{map{eval,-7+length}qq~&{"::$_[0]"}~} Try it: perl -e 'sub google{map{eval,-7+length}qq~&{"::$_[0]"}~} print google @ARGV' google ## 47 bytes The following is one char shorter, but more fiddly on the command line: sub google{map{eval,-6+length}qq~&{"'$_[0]"}~} # F#, 33 Characters function"google"->0/0\|s->s.Length When "google" is provided as input, it produces a DevideByZeroException. Usage: let g = function"google"->0/0\|s->s.Length g "bing" // 4 # Java 1.8, 33 bytes (s)->s.length()/(s=="google"?0:1) ## Explanation The lambda takes a String named s, finds the length, and if it isn't "google", divides it by one, otherwise dividing it by zero and causing an Exception. ## Usage Note that java.util.function.Function has to be imported. Function<String, Integer> f = (s)->s.length()/(s=="google"?0:1); //Assign function to variable //Note that java type inferencing automatically handles the String type System.out.println(f.apply("elgoog")); //Prints 6 • As far as I know, required imports need to be counted in the score, so your full code would be the import and the function. – Alex A. Sep 29 '15 at 15:53 • @AlexA. The thing with Java is that lambdas are weird- the question only asked for a function, and the lambda is a function. However, Java needs to squeeze the function into a functional interface with the same method signature (accepts string, returns int), and the Java.util.function.Function interface fits the bill. However, the full function is provided. (Also see codegolf.stackexchange.com/a/58981/41505) – Daniel M. Sep 29 '15 at 16:31 • You are going to have to change == to .equals unless I'm missing something with java 8 – jlars62 Oct 2 '15 at 16:55 • It worked for me using the provided implementation. It may be a subtle difference between String objects and literals, but it works for me – Daniel M. Oct 2 '15 at 17:08 # Lua, 47 bytes print(assert(arg[1]~="google")and arg[1]:len()) Throws "assertion failed" if it's "google" # golflua, 25 characters \g(s)?s=="google"e""$~#s$ Sample run: bash-4.3$golflua -e '\g(s)?s=="google"e""$~#s$w(g("google"))' golflua: (command line):1: stack traceback: [C]: in function 'e' (command line):1: in function 'g' (command line):1: in main chunk [C]: in ? bash-4.3$ golflua -e '\g(s)?s=="google"e""$~#s$ w(g("yahoo"))' 5 # PHP, 41 bytes Because the division by zero is overused in the already existing answers, I tried to come up with something else (sacrificing 3 bytes): <?=strlen($x=$argv[1])+log($x!="google"); It's not a function (the task explicitly asks to write a function) but it can be invoked from the command line in a functional manner: $ echo '<?=strlen($x=$argv[1])+log($x!="google");' \| php -- bing 4$ echo '<?=strlen($x=$argv[1])+log($x!="google");' \| php -- google -INF It doesn't produce an error when the argument is google but it doesn't display the length of the string google either. However it displays -INF (i.e. minus infinity) and this value can be considered an error for a function that returns a length (which, by definition, is a count, i.e. a non-negative number). ### A 38 bytes PHP solution using division by zero: <?=strlen($x=$argv[1])/($x!="google"); It can be invoked in the same way as above. When the argument is google it displays a PHP warning. • You get a +1 from me for the division-by-zero, clever! But a -1 for not making it a function as the topic states :) – Martijn Sep 29 '15 at 12:53 • -2 bytes: The quotation marks are unnecessary. – Titus Nov 29 '16 at 15:48 # Clojure - 41 40 bytes (defn g[s](if(= s"google")(g)(count s))) Attempts to call itself with zero arguments for the input "google". > (g "bing") 4 clojure.lang.ArityException: Wrong number of args (0) passed to: sandbox10419$g # Swift, 71 Bytes Short but lame: print({assert($0 != "google");return $0.characters.count}(readLine()!)) Longer (75) but not lame: print({{0/$0}($0.hash&-0x20f4f91ecf8d43)^$0.characters.count}(readLine()!)) The second one doesn't use any String literal. Works by subtracting the hash value of the input string by the "google" hashValue. Then it divides 0 by this value, resulting in a runtime error when it's 0 (0/0 = undef.) but in all other cases the result is 0 (0/x = 0). This result gets XOR'd by the character count in the String.	2021-08-01 04:29:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.22324371337890625, "perplexity": 3289.167115283016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00378.warc.gz"}
https://socratic.org/questions/how-do-you-solve-1-3x-2-5-6	# How do you solve 1/3x+2=5/6? Feb 26, 2017 See the entire solution process below: #### Explanation: First, subtract $\textcolor{red}{2}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced: $\frac{1}{3} x + 2 = \frac{5}{6}$ $\frac{1}{3} x + 2 - \textcolor{red}{2} = \frac{5}{6} - \textcolor{red}{2}$ $\frac{1}{3} x + 0 = \frac{5}{6} - \left(\frac{6}{6} \times \textcolor{red}{2}\right)$ $\frac{1}{3} x = \frac{5}{6} - \frac{12}{6}$ $\frac{1}{3} x = - \frac{7}{6}$ Now, multiply each side of the equation by $\textcolor{red}{3}$ to solve for $x$ while keeping the equation balanced: $\textcolor{red}{3} \times \frac{1}{3} x = \textcolor{red}{3} \times - \frac{7}{6}$ $\cancel{\textcolor{red}{3}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} x = \cancel{\textcolor{red}{3}} \times - \frac{7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} 2}$ $x = - \frac{7}{2}$ Feb 26, 2017 $x = - \frac{7}{2}$ or $3 \frac{1}{2}$ #### Explanation: $\frac{1}{3} x + 2 = \frac{5}{6}$ $\therefore \frac{x}{3} + \frac{2}{1} = \frac{5}{6}$ $\therefore \frac{2 x + 12 = 5}{6}$ $\therefore \frac{2 x}{6} + \frac{12}{6} = \frac{5}{6}$ multiply L.H.S. and R.H.S by 6 $\therefore 2 x + 12 = 5$ $\therefore 2 x = 5 - 12$ $\therefore 2 x = - 7$ $x = - \frac{7}{2}$ substitute $x = - \frac{7}{2}$ $\therefore \frac{1}{3} \left(- \frac{7}{2}\right) + 2 = \frac{5}{6}$ $\therefore - \frac{7}{6} + \frac{12}{6} = \frac{5}{6}$ $\therefore \frac{5}{6} = \frac{5}{6}$	2019-12-15 10:12:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "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": 0.6787847876548767, "perplexity": 879.3375710940877}, "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-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00213.warc.gz"}
https://jndl.us/online-5x5-solver.html	# Online 5x5 Solver Play online with the Professor's Cube with this online 5x5x5 Rubik's Cube simulator. Press the Scramble button to jumble the 5x5x5 Rubik's puzzle then try to solve it using the rotation buttons. You can use the same buttons which you already know from the 4x4x4 Rubik's Revenge simulator. Every capital letter marks a face rotation. If you see a small 2 after the letter that means that you turn two faces at the same time. Go ahead and test it. • Big Boggle Solver or the Boggle 5x5 Solver. Played on the 5x5 grid, this Boggle variant may be just as exciting for Boggle fans as the standard 4x4 version of Boggle. Of course, our Boggle 5x5 Solver crunches all possible answers in milliseconds. • You will need to get assistance from your school if you are having problems entering the answers into your online assignment. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. You may speak with a member of our customer support team by calling 1-800-876-1799. • Online Calculator for Determinant 5x5. The online calculator calculates the value of the determinant of a 5x5 matrix with the Laplace expansion in a row or column and the gaussian algorithm. Inverse Matrix Calculator. Here you can calculate inverse matrix with complex numbers online for free with a very detailed solution. The inverse is calculated using Gauss-Jordan elimination. Read the instructions. Matrix dimension: About the method. To calculate inverse matrix you need to do the following steps. Boggle is a word game where players race to find words hidden in a grid of letters. Enter a Boggle board and see a list of all the words that can be found! Choose a board size - 3x3, 4x4 or 5x5 - and the minimum number of letters each word must have, and get a complete listing of all of the solutions along with point totals and other statistics. Find out interesting facts and a tutorial on how to solve the 5x5x5 puzzle clicking here. Have fun! ## Methods explanation ### Laplace Expansion Theorem The Laplacian development theorem provides a method for calculating the determinant, in which the determinant is developed after a row or column. The dimension is reduced and can be reduced further step by step up to a scalar. $\mathrm{det A}=\sum _{i=1}^{n}{-1}^{i+j}\cdot {a}_{ij}\mathrm{det}{A}_{ij}\text{( Expansion on the j-th column )}$ $\mathrm{det A}=\sum _{j=1}^{n}{-1}^{i+j}\cdot {a}_{ij}\mathrm{det}{A}_{ij}\text{( Expansion on the i-th row )}$ where Aij, the sub-matrix of A, which arises when the i-th row and the j-th column are removed. #### Example of the Laplace expansion according to the first row on a 3x3 Matrix. $\mathrm{det A}=\left\|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right\|$ The first element is given by the factor a11 and the sub-determinant consisting of the elements with green background. $\left\|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right\|=>{a}_{11}\left\|\begin{array}{cc}{a}_{22}& {a}_{23}\\ {a}_{32}& {a}_{33}\end{array}\right\|$ The second element is given by the factor a12 and the sub-determinant consisting of the elements with green background. $\left\|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right\|=>{a}_{12}\left\|\begin{array}{cc}{a}_{21}& {a}_{23}\\ {a}_{31}& {a}_{33}\end{array}\right\|$ The third element is given by the factor a13 and the sub-determinant consisting of the elements with green background. ## Online 5x5 Solver Games $\left\|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right\|=>{a}_{13}\left\|\begin{array}{cc}{a}_{21}& {a}_{22}\\ {a}_{31}& {a}_{32}\end{array}\right\|$ With the three elements the determinant can be written as a sum of 2x2 determinants. ## Rubik's Cube 5x5 Solver $\mathrm{det A}=\left\|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right\|={a}_{11}\left\|\begin{array}{cc}{a}_{22}& {a}_{23}\\ {a}_{32}& {a}_{33}\end{array}\right\|-{a}_{12}\left\|\begin{array}{cc}{a}_{21}& {a}_{23}\\ {a}_{31}& {a}_{33}\end{array}\right\|+{a}_{13}\left\|\begin{array}{cc}{a}_{21}& {a}_{22}\\ {a}_{31}& {a}_{32}\end{array}\right\|$ ## 5x5 Cube Solver It is important to consider that the sign of the elements alternate in the following manner. ## Magisches Quadrat 5x5 Online Solver $\left\|\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right\|$	2021-06-24 23:58:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.4117991626262665, "perplexity": 998.6981307060136}, "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-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00080.warc.gz"}
https://web2.0calc.com/questions/help-please_56297	+0 0 82 1 +253 Suppose that 2x^2 - 5x + k = 0 is a quadratic equation with one solution for x. Express k as a common fraction. Jun 10, 2019 #1 +23041 +3 Suppose that $$2x^2 - 5x + k = 0$$ is a quadratic equation with one solution for $$x$$. Express $$k$$ as a common fraction. $$\begin{array}{\|rcll\|} \hline \mathbf{2x^2 - 5x + k} &=& \mathbf{0} \\\\ x &=& \dfrac{5\pm \sqrt{25-4\cdot 2k} }{2\cdot 2} \\ \hline 25-4\cdot 2k &=& 0 \\ 25-8k &=& 0 \\ 8k &=& 25 \\ \mathbf{k} &=& \mathbf{\dfrac{25}{8}} \\ \hline \end{array}$$ Jun 10, 2019 #1 +23041 +3 Suppose that $$2x^2 - 5x + k = 0$$ is a quadratic equation with one solution for $$x$$. Express $$k$$ as a common fraction. $$\begin{array}{\|rcll\|} \hline \mathbf{2x^2 - 5x + k} &=& \mathbf{0} \\\\ x &=& \dfrac{5\pm \sqrt{25-4\cdot 2k} }{2\cdot 2} \\ \hline 25-4\cdot 2k &=& 0 \\ 25-8k &=& 0 \\ 8k &=& 25 \\ \mathbf{k} &=& \mathbf{\dfrac{25}{8}} \\ \hline \end{array}$$ heureka Jun 10, 2019	2019-09-15 20:57:08	{"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": 0.43488049507141113, "perplexity": 414.06043520908497}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00141.warc.gz"}
http://math.stackexchange.com/questions/247422/whats-the-expected-value-of-a-lottery-ticket?answertab=oldest	# What's the expected value of a lottery ticket? Suppose there's a lottery. Each ticket sold has probability $p$ of winning, and they are all independent of each other. The size of the jackpot is $j$. If there are $n$ winners, each winner gets a payoff of $j/n$ dollars. The total number of tickets sold is $t$. What is the expected value of a lottery ticket? Also, given that I win, what is the probability that I have to share the jackpot with at least one other person? PS - I think I know the answer, but have failed to convince someone else, so I'm looking for a third-party to give an answer. - @FlybyNight $n$ is a random variable whose distribution is determined from $p$ and $t$. I believe you do have enough information. – Mark Eichenlaub Nov 29 '12 at 17:46 For the expected value, I believe you should get $$\frac{j}{t}(1-(1-p)^t)$$ unless my quick chicken-scratch has an error (which has nonnegligible probability). – cardinal Nov 29 '12 at 17:58 @cardinal I hadn't thought of it that way, but it's a good point and looks right to me, thanks. – Mark Eichenlaub Nov 29 '12 at 18:00 "and they are all independent of each other" may be the source of the disagreement, since it is not the way usual lotteries work. – André Nicolas Nov 29 '12 at 18:56 ## 1 Answer Probability that no one wins: $(1-p)^t$. Expected value for a particular ticket: $\dfrac j t (1-(1-p)^t)$. (Pot times probability that the pot is distributed.) Probability that you win alone: $p(1-p)^{t-1}$ Probability of shared win: $p-p(1-p)^{t-1}$ Probability of sharing given that you win: $\dfrac{\text{prob of shared win}}{\text{prob win}}=\dfrac{p-p(1-p)^{t-1}}p = 1-(1-p)^{t-1}$ This is just one minus the probability of everyone else losing because of independence. - Just an aside: While your parenthetical remark on the expected-value result is a good heuristic and provides useful intuition, it is not really a rigorous line of reasoning to arrive at the result. – cardinal Nov 29 '12 at 18:33 @cardinal: Oh yes, it is. (Note that every lottery ticket is created equal in this probability distribution.) – Phira Nov 29 '12 at 18:35 Yes, I agree it can be formalized via an exchangeability argument. – cardinal Nov 29 '12 at 18:41	2014-10-25 09:38:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8756744861602783, "perplexity": 293.85602910172975}, "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-2014-42/segments/1414119648008.18/warc/CC-MAIN-20141024030048-00047-ip-10-16-133-185.ec2.internal.warc.gz"}
https://codegolf.stackexchange.com/questions/28865/saying-a-number-in-the-shortest-way	# Saying a number in the shortest way A string of digits (positive integer) can be converted to plain English in various ways. For instance, 115 can be read as "hundredfifteen", "oneonefive", "elevenfive", "twoonesfive". So question's challenge is to output the shortest string representation for a given digits string. Rules: • The input string can start with a zero and the output should account for that. • Maximum length of input is 18. So you deal with numbers less than a Quintillion. • You should use only the following mathematical words for numbers in your output: zero, one, ..., nine, ten, eleven, ..., twenty, ..., ninety, hundred, million, billion, trillion. You can use their plural forms for sequences (i.e. "fivetens" for 1010101010). No synonyms (i.e. "oh" for 0) or slang (i.e. legs for 11) are allowed. You don't need to include a "one" prefix for cases like 100, just "hundred" is fine. • You don't need to put spaces between different words in your output. • The output shouldn't be ambigous. This means, from the output string one should be able to determine the input unambigously. So for 705 you can't output "seventyfive" since it may be the output of 75 also. You may use "and" if that'll remove ambiguity, otherwise you don't need to include "and" in your output. • Winner is the entry that will have the shortest code length excluding import statements and if not some user finds out a deficit in the code (i.e. a shorter representation than the program spits out for a particular input ). • Please post your output for the following test cases to make evaluation easier: 1234, 00505, 122333, 565577555555666888, 1000010, 10001000, 10101010. • No reading from an input file, so usage of a dictionary or hash map should count towards your code length. edit1: Changed the competition type to code-golf per comments and added test cases. edit2: Restricted the words that can be used in the output. Thanks to @PeterTaylor for the examples. • Could you put a hard limit on the input numbers? Or do I have to support "Septendecillion"? Furthermore, I think with a slightly more rigid spec this would be a great code-golf question. I don't really see why it has to popularity-contest. The current spec is already too tight to leave a lot of room for the kind of creativity that is rewarded by popularity contests. – Martin Ender May 30 '14 at 22:32 • @m.buettner, do you think changing the criteria to "output longest string" will enable more creativity and make it more suitable to popularity-contest? Secondly, can you elaborate on how we can make the spec more rigid to make it a good code-golf question? Having a limit makes sense for inputs like 1000000 so I'll add it. Thanks for pointing. – pembeci May 31 '14 at 1:37 • I think this would work best as a code challenge. You have a secret list of 5000 numbers, after a week you reveal the numbers and whichever program represents them with the least total characters wins. – Hovercouch May 31 '14 at 3:29 • Why isn't "18403" the shortest way to say the number 18403? – Glen O May 31 '14 at 11:15 • The spec is too imprecise as it stands, so it just sets up lots of arguments as to whether a string represents the given number or not. – Peter Taylor May 31 '14 at 11:36 # 4 This is cheeky and a cheat. I wrote this script about a year ago based off of this comic (tad of profanity). The idea goes like this. Given any number as a string, adding the number of characters will eventually condense to 4. Four itself is 4 letters long and thus the end of the line. For example: • 10 -> 'ten' • 'ten' -> 3 (letters long) • 3 -> 'three' • 'three' -> 5 (letters long) • 5 -> 'five' • 'five' -> 4 (letters long) I verified up to 100,000. Very hacky but got the job done: to20 = [ 'zero' ,'one' ,'two' ,'three' ,'four' ,'five' ,'six' ,'seven' ,'eight' ,'nine' ,'ten' ,'eleven' ,'twelve' ,'thirteen' ,'fourteen' ,'fifteen' ,'sixteen' ,'seventeen' ,'eighteen' ,'nineteen' ] by10s = ['twenty' ,'thirty' ,'forty' ,'fifty' ,'sixty' ,'seventy' ,'eighty' ,'ninety' ] by10powers = [ 'hundred' ,'thousand' ] def cosmetize(number): # the number is too damn high! if number >= 100000: return "Too high" numstr = str(number) # Do ones if number < 10: # Pshh.. who needs zero anymore? to20[0] = '' # Do tens numstr10 = numstr[-2:] if int(numstr10) < 20: numstr10 = to20[int(numstr10)] else: numstr10 = by10s[int(numstr10[0])-2] + to20[int(numstr10[1])] if number < 100: return numstr10 # Do hundreds numstr100 = numstr[-3] if numstr10 == '': numstr10 = "" numstr100 = to20[int(numstr100)] + by10powers[0] elif int(numstr[-2:]) > 19: numstr100 = to20[int(numstr100)] + by10powers[0] numstr100 += 'and' else: numstr100 = 'and' if number < 1000: return numstr100 + numstr10 # Do thousands numstr1000 = numstr[:-3] if numstr100 == 'hundred': numstr100 = "" if int(numstr1000) < 20: numstr1000 = to20[int(numstr1000)] + by10powers[1] else: numstr1000 = by10s[int(numstr1000[0])-2] + by10powers[1] return numstr1000 + numstr100 + numstr10 i = 0 mx = [] z = 0 while i < 100000: length = cosmetize(i) x = [length] while not len(length) == 4: length = cosmetize(len(length)) i += 1 # No inifinite loop? what magic is this? print "OMG" # Python - (191 - 38) = 153 bytes This code has the worst case length of 18 for input of 000000000000000000. I'm not sure if this is system dependant. If it's not, I could remove a couple characters. from string import* from math import* s=digits+lowercase+uppercase c=raw_input() b=len(s) a=int(c) print'-'(len(c)-len(c.lstrip('0')))+''.join(s[a/b*i%b]for i in range(int(log(a,b)),-1,-1)) For the last line I have to give credit to this fellow. Example outputs: > 1234 9j > 00505 --3Z > 122333 6Vn > 565577555555666888 5haShÄøŠé > 1000010 êÌõ > 10001000 48Šõ > 10101010 4ewª Disclaimer: This answer is not a serious answer and it's not trying to compete. This is taking advantage of ambiguity of the rules. • Clever one but "êÌõ" is not plain English as mentioned in the first sentence. I thought it was clear from the examples but it seems like I need to state this explicitly. – pembeci Jun 2 '14 at 7:54 • @pembeci As I said, the point of this was to show how ambiguous the rules are. – seequ Jun 2 '14 at 8:02	2019-11-20 02:40:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.40380480885505676, "perplexity": 2842.228620347264}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670389.25/warc/CC-MAIN-20191120010059-20191120034059-00430.warc.gz"}
http://tex.stackexchange.com/questions/6985/place-quotation-dictum-in-a-box?answertab=active	# Place quotation (dictum) in a box ## Background Looking to change the appearance of quotations at the start of each Chapter (or Part). The book is using KOMA Script v2. (Most searches for the quote command come back with results for typesetting quotation marks, rather than aphorisms.) ## Problem Currently, the quotation resembles: Am looking for something similar to: Good code is short, simple, and symmetrical -- the challenge is figuring out how to get there. ~Sean Parent (A double line border, a background colour, but using a serif font with the author name right-justified.) ## Source Code The quotation is defined as follows: \setchapterpreamble{\dictum[Sean Parent]{Good code...}} I have altered the style of the dictum settings as follows: \renewcommand{\dictumwidth}{.75\textwidth} \renewcommand{\dictumauthorformat}[1]{\textsc{#1}\bigskip} ## Questions • What packages allow such fine control over quotations (or quotes)? (Neither fncychap nor quotchap offer sufficient sophistication.) • How can the aphorism be placed in a \colorbox? (Using \dictumtext?) • What is a good way to add double lines to the box? ## Related - the epigraph package might be useful, but better check the way Knuth defined quotations at the end of his Chapters in the TeXBook and adapt it to your class. Is there a fine distinction between aphorisms and quotations linguistically speaking? – Yiannis Lazarides Dec 12 '10 at 20:04 @Yiannis: Easier said than done. ;-) I'm using LyX. – Dave Jarvis Dec 12 '10 at 23:49 I did not use epigraph. I wrote my own version: \usepackage{times} \definecolor{quotationcolour}{HTML}{F0F0F0} \definecolor{quotationmarkcolour}{HTML}{1F3F81} % Double-line for start and end of epigraph. \newcommand{\epiline}{\hrule \vskip -.2em \hrule} % Massively humongous opening quotation mark. \newcommand{\hugequote}{% \fontsize{42}{48}\selectfont \color{quotationmarkcolour} \textbf{} \vskip -.5em } % Beautify quotations. \newcommand{\epigraph}[2]{% \bigskip \begin{flushright} \colorbox{quotationcolour}{% \parbox{.60\textwidth}{% \epiline \vskip 1em {\hugequote} \vskip -.5em \parindent 2.2em #1\begin{flushright}\textsc{#2}\end{flushright} \epiline } } \end{flushright} \bigskip } The result: Suggestions for improvement are welcome. -	2014-07-11 22:55:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.7143113017082214, "perplexity": 8916.810682292173}, "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-2014-23/segments/1404776429991.81/warc/CC-MAIN-20140707234029-00000-ip-10-180-212-248.ec2.internal.warc.gz"}
http://www.maths-informatique-jeux.com/blog/frederic/?post/2014/01/13/Improvements-to-Mathematics-on-Wikipedia	## Introduction As mentioned during the Mozilla Summit and recent MathML meetings, progress has recently be made to the way mathematical equations are handled on Wikipedia. This work has mainly be done by the volunteer contributor Moritz Schubotz (alias Physikerwelt), Wikimedia Foundation's developer Gabriel Wicke as well as members of MathJax. Moritz has been particularly involved in that project and he even travelled from Germany to San Francisco in order to meet MediaWiki developers and spend one month to do volunteer work on this project. Although the solution is essentially ready for a couple of months, the review of the patches is progressing slowly. If you wish to speed up the integration of what is probably the most important improvements to MediaWiki Math to happen, please read how you can help below. ## Current Status The approach that has been used on Wikipedia so far is the following: • Equations are written in LaTeX or more precisely, using a specific set of LaTeX commands accepted by texvc. One issue for the MediaWiki developers is that this program is written in OCaml and no longer maintained, so they would like to switch to a more modern setup. • texvc calls the LaTeX program to convert the LaTeX source into PNG images and this is the default mode. Unfortunately, using images for representing mathematical equations on the Web leads to classical problems (for example alignment or rendering quality just to mention a few of them) that can not be addressed without changing the approach. • For a long time, registered users have been able to switch to the MathJax mode thanks to the help of nageh, a member of the MathJax community. This mode solves many of the issues with PNG images but unfortunately it adds its own problems, some of them being just unacceptable for MediaWiki developers. Again, these issues are intrinsic to the use of a Javascript polyfill and thus yet another approach is necessary for a long-term perspective. • Finally, registered users can also switch to the LaTeX source mode, that is only display the text source of equations. ## Short Term Plan Native MathML is the appropriate way to fix all the issues regarding the display of mathematical formulas in browsers. However, the language is still not perfectly implemented in Web rendering engines, so some fallback is necessary. The new approach will thus be: • The TeX equation will still be edited by hand but it will be possible to use a visual editor. • texvc will be used as a filter to validate the TeX source. This will ensure that only the texvc LaTeX syntax is accepted and will avoid other potential security issues. The LaTeX-to-PNG conversion as well as OCaml language will be kept in the short term, but the plan is to drop the former and to replace the latter with a a PHP equivalent. • A LaTeX-to-MathML conversion followed by a MathML-to-SVG conversion will be performed server-side using MathJax. • By default all the users will receive the same output (MathML+SVG+PNG) but only one will be made visible, according your browser capabilities. As a first step, native MathML will only be used in Gecko and other rendering engines will see the SVG/PNG fallback ; but the goal is to progressively drop the old PNG output and to move to native MathML. • Registered users will still be able to switch to the LaTeX source mode. • Registered users will still be able to use MathJax client-side, especially if they want to use the HTML-CSS output. However, this is will no longer be a separate mode but an option to enable. That is, the MathML/SVG/PNG/Source is displayed normally and progressively replaced with MathJax's output. Most of the features above have already been approved and integrated in the development branch or are undergoing review process. ## How can you help? The main point is that everybody can review the patches on Gerrit. If you know about Javascript and/or PHP, if you are interested in math typesetting and wish to get involved in an important Open Source project such as Wikipedia then it is definitely the right time to help the MediaWiki Math project. The article How to become a MediaWiki hacker is a very good introduction. When getting involved in a new open source project one of the most important step is to set up the development environment. There are various ways to setup a local installation of MediaWiki but using MediaWiki-Vagrant might be the simplest one: just follow the Quick Start Guide and use vagrant enable-role math to enable the Math Extension. The second step is to create a WikiTech account and to set up the appropriate SSH keys on your MediaWiki-Vagrant virtual machine. Then you can check the Open Changes, test & review them. The Gerrit code review guide may helpful, here. If you need more information, you can ask Moritz or try to reach people on the #mediawiki (freenode) or #mathml (mozilla) channels. Thanks in advance for your help!	2014-12-19 06:25:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5284702777862549, "perplexity": 1219.7291983339148}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768276.101/warc/CC-MAIN-20141217075248-00112-ip-10-231-17-201.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/629860/missing-sign-in-dirac-equation	# Missing sign in Dirac equation This is very trivial, but it's really bugging me. The ansatz for the Dirac equation in terms of $$\boldsymbol\alpha$$ and $$\beta$$ matrices is $$[i(\partial_t+\boldsymbol\alpha\cdot\boldsymbol\nabla)-\beta m]\psi=0,$$ and to get the standard form in terms of gamma matrices, one defines $$\gamma^\mu=(\beta, \beta\boldsymbol \alpha)$$ so that multiplying the equation by $$\beta$$ gives $$[i(\beta\partial_t+\beta\boldsymbol\alpha\cdot\boldsymbol\nabla)-\beta^2m]=[i\gamma^\mu\partial_\mu-m]=0$$ as long as the index on $$\gamma^\mu$$ is actually a vector index, which we can make sure of by requiring $$\psi$$ to transform in the spinor representation. But considering that the correct expansion of the Minkowski product is $$\gamma^\mu\partial_\mu=\gamma^0\partial_0-\boldsymbol\gamma\cdot\boldsymbol\nabla=\beta\partial_t-\beta\boldsymbol\alpha\cdot\boldsymbol\nabla,$$ shouldn't the expression in the first term be $$\beta\partial_t-\beta\boldsymbol\alpha\cdot\boldsymbol\nabla$$, with a minus sign rather than a plus sign? ## 1 Answer This is a classic misuse of the Einstein summation convention. The correct expression in your final equation is $$\gamma^{\mu} \partial_{\mu} = \sum_{\mu = 0}^3 \gamma^{\mu} \partial_{\mu} = \gamma^0 \partial_0 + \vec{\gamma} \cdot \vec{\nabla}$$ The reason for your confusion is because you're used to taking the Minkowski inner product between two contravariant or two covariant vectors. If $$v^{\mu} = (v^0, \vec{v})$$ and $$w^{\mu} = (w^0, \vec{w})$$, then of course $$v^{\mu} w_{\mu} = g_{\mu \nu} v^{\mu} w^{\nu} = v^0 w^0 - \vec{v} \cdot \vec{w}$$. But in your expression, $$\partial_{\mu}$$ is naturally covariant, not contravariant. • Oh, right. I should have written $\gamma^j\partial_j$ explicitly to see that there was no need to raise the index of $\partial_\mu$. Thanks! – A Quantum Field Day Apr 16 at 15:46	2021-07-27 04:28:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8872719407081604, "perplexity": 119.57608993283438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00249.warc.gz"}
https://msp.org/agt/2015/15-5/p02.xhtml	#### Volume 15, issue 5 (2015) Recent Issues The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Other MSP Journals Khovanov homology is a skew Howe $2$–representation of categorified quantum $\mathfrak{sl}_m$ ### Aaron D Lauda, Hoel Queffelec and David E V Rose Algebraic & Geometric Topology 15 (2015) 2517–2608 ##### Abstract We show that Khovanov homology (and its ${\mathfrak{s}\mathfrak{l}}_{3}$ variant) can be understood in the context of higher representation theory. Specifically, we show that the combinatorially defined foam constructions of these theories arise as a family of $2$–representations of categorified quantum ${\mathfrak{s}\mathfrak{l}}_{m}$ via categorical skew Howe duality. Utilizing Cautis–Rozansky categorified clasps we also obtain a unified construction of foam-based categorifications of Jones–Wenzl projectors and their ${\mathfrak{s}\mathfrak{l}}_{3}$ analogs purely from the higher representation theory of categorified quantum groups. In the ${\mathfrak{s}\mathfrak{l}}_{2}$ case, this work reveals the importance of a modified class of foams introduced by Christian Blanchet which in turn suggest a similar modified version of the ${\mathfrak{s}\mathfrak{l}}_{3}$ foam category introduced here. ##### Keywords Khovanov homology, categorified quantum groups, cobordism categories, foam categories, skew Howe duality, link homology ##### Mathematical Subject Classification 2010 Primary: 81R50 Secondary: 17B37, 57M25, 18G60	2019-06-16 19:29:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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": 0.5055165886878967, "perplexity": 3936.38928010684}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00017.warc.gz"}
https://docs.scikit-nano.org/generated/sknano.generators.SWNTGenerator.electronic_type.html	#### Previous topic sknano.generators.SWNTGenerator.dt #### Next topic sknano.generators.SWNTGenerator.element1 # sknano.generators.SWNTGenerator.electronic_type¶ SWNTGenerator.electronic_type SWNT electronic type. New in version 0.2.7. The electronic type is determined as follows: if $$(2n + m)\,\mathrm{mod}\,3=0$$, the nanotube is metallic. if $$(2n + m)\,\mathrm{mod}\,3=1$$, the nanotube is semiconducting, type 1. if $$(2n + m)\,\mathrm{mod}\,3=2$$, the nanotube is semiconducting, type 2. The $$x\,\mathrm{mod}\,y$$ notation is mathematical shorthand for the modulo operation, which computes the remainder of the division of $$x$$ by $$y$$. So, for example, all armchair nanotubes must be metallic since the chiral indices satisfy: $$2n + m = 2n + n = 3n$$ and therefore $$3n\,\mathrm{mod}\,3$$ i.e. the remainder of the division of $$3n/3=n$$ is always zero. Note Mathematically, $$(2n + m)\,\mathrm{mod}\,3$$ is equivalent to $$(n - m)\,\mathrm{mod}\,3$$ when distinguishing between metallic and semiconducting. However, when distinguishing between semiconducting types, one must be careful to observe the following convention: • Semiconducting, type 1 means: • $$(2n + m)\,\mathrm{mod}\,3=1$$ • $$(n - m)\,\mathrm{mod}\,3=2$$ • Semiconducting, type 2 means: • $$(2n + m)\,\mathrm{mod}\,3=2$$ • $$(n - m)\,\mathrm{mod}\,3=1$$	2020-11-29 21:56:46	{"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": 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": 0.9569058418273926, "perplexity": 1813.1097449376218}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00469.warc.gz"}
https://www.semanticscholar.org/paper/Automating-Resolution-is-NP-Hard-Atserias-M%C3%BCller/eb5f7af28ebf58c5535f97904d52dc4b0e0e198a	# Automating Resolution is NP-Hard ```@article{Atserias2019AutomatingRI, title={Automating Resolution is NP-Hard}, author={Albert Atserias and Moritz Christian M{\"u}ller}, journal={2019 IEEE 60th Annual Symposium on Foundations of Computer Science (FOCS)}, year={2019}, pages={498-509} }``` • Published 5 April 2019 • Mathematics • 2019 IEEE 60th Annual Symposium on Foundations of Computer Science (FOCS) We show that the problem of finding a Resolution refutation that is at most polynomially longer than a shortest one is NP-hard. In the parlance of proof complexity, Resolution is not automatizable unless P = NP. Indeed, we show that it is NP-hard to distinguish between formulas that have Resolution refutations of polynomial length and those that do not have subexponential length refutations. This also implies that Resolution is not automatizable in subexponential time or quasi-polynomial time… ## Tables from this paper • Zoë Bell • Mathematics Electron. Colloquium Comput. Complex. • 2020 It is shown that it is hard to find regular or even ordered Resolution proofs, extending the breakthrough result for general Resolution from Atserias and Müller to these restricted forms. • Mathematics, Computer Science Electron. Colloquium Comput. Complex. • 2020 This work extends, and gives a simplified proof of, the recent breakthrough of Atserias and Müller that established an analogous result for Resolution, and shows that algebraic proofs are hard to find. • Mathematics, Computer Science Electron. Colloquium Comput. Complex. • 2022 It is proved that the proof system OBDD( ∧, weakening) is not automatable unless P = NP, where n is the number of variables in the non-lifted formula. • Michal Garl'ik • Mathematics, Computer Science Electron. Colloquium Comput. Complex. • 2020 It is shown that if NP is not included in P (resp. QP, SUBEXP) then for every integer \$k \geq 1\$, Res (\$k\$) is not automatable in polynomial time. For any unsatisfiable CNF formula we give an exponential lower bound on the size of resolution refutations of a propositional statement that the formula has a resolution refutation. We describe three For any unsatisfiable CNF formula we give an exponential lower bound on the size of resolution refutations of a propositional statement that the formula has a resolution refutation. We describe three For any unsatisfiable CNF formula we give an exponential lower bound on the size of resolution refutations of a propositional statement that the formula has a resolution refutation. We describe three We show that tree-like resolution is not automatable in time no(log n) unless ETH is false. This implies that, under ETH, the algorithm given by Beame and Pitassi (FOCS 1996) that automates tree-like • Computer Science, Mathematics Electron. Colloquium Comput. Complex. • 2020 Two new lifting theorems are relied on: Dag-like lifting for gadgets with many output bits and tree- like lifting that simulates an r-round protocol with gadgets of query complexity O(r). • Computer Science, Mathematics ICALP • 2019 A streamlined proof is obtained of an important result by Alekhnovich and Razborov, showing that it is hard to automatize both tree-like and general Resolution. ## References SHOWING 1-10 OF 47 REFERENCES • Mathematics, Computer Science Electron. Colloquium Comput. Complex. • 2020 This work extends, and gives a simplified proof of, the recent breakthrough of Atserias and Müller that established an analogous result for Resolution, and shows that algebraic proofs are hard to find. • Michal Garl'ik • Mathematics, Computer Science Electron. Colloquium Comput. Complex. • 2020 It is shown that if NP is not included in P (resp. QP, SUBEXP) then for every integer \$k \geq 1\$, Res (\$k\$) is not automatable in polynomial time. • K. Iwama • Mathematics, Computer Science MFCS • 1997 It is shown that if there is a polynomial-time (superpolynomial-time or subexponential time, respectively) approximation algorithm that finds a nearly shortest proof of length up to S + O(n d ), where S is the length of the shortest proof and d may be any constant, then this immediately gives a positive answer to the open problem asking whether finding a shortest Resolution proof is NP-hard. For any unsatisfiable CNF formula we give an exponential lower bound on the size of resolution refutations of a propositional statement that the formula has a resolution refutation. We describe three • Computer Science, Mathematics Electron. Colloquium Comput. Complex. • 2020 Two new lifting theorems are relied on: Dag-like lifting for gadgets with many output bits and tree- like lifting that simulates an r-round protocol with gadgets of query complexity O(r). • Computer Science, Mathematics ICALP • 2019 A streamlined proof is obtained of an important result by Alekhnovich and Razborov, showing that it is hard to automatize both tree-like and general Resolution. • Computer Science, Philosophy Proceedings. Fourteenth Annual IEEE Conference on Computational Complexity (Formerly: Structure in Complexity Theory Conference) (Cat.No.99CB36317) • 1999 Abstract.In this paper, we show how to extend the argument due to Bonet, Pitassi and Raz to show that bounded-depth Frege proofs do not have feasible interpolation, assuming that factoring of Blum • Mathematics Proceedings 2001 IEEE International Conference on Cluster Computing • 2001 We show that neither Resolution nor tree-like Resolution is automatizable unless the class W[P] from the hierarchy of parameterized problems is fixed-parameter tractable by randomized algorithms with • Computer Science, Mathematics Proceedings of 37th Conference on Foundations of Computer Science • 1996 New lower bounds on the lengths of resolution proofs for the pigeonhole principle and for randomly generated formulas are given and the range of formula sizes for which non-trivial lower bounds are known is extended. If there exists an optimal proof system for propositional logic in which every tautology has a short proof, it can be formed by augmenting the usual text-book calculus by the extension rule and by an additional, polynomial time recognizable, set of tautologies as extra axioms.	2023-03-20 12:04:08	{"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": 0.8027315735816956, "perplexity": 1924.6159188777985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00532.warc.gz"}
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/How.20to.20use.20a.20type.20alias.20without.20breaking.20typeclass.20synthesis.html	## Stream: new members ### Topic: How to use a type alias without breaking typeclass synthesis #### Marko Grdinić (Oct 25 2019 at 09:58): def History : Type* := {die // die ∈ dice} × list {i // ∃ v, claims.read i = v} def ha : HistoryToActions ({die // die ∈ dice} × list {i // ∃ v, claims.read i = v}) Action \| (_, []) := ⟨ actions.begin.1, actions.begin.read ⟩ \| (_, x :: xs) := let v := actions.later.drop (x.1 + 1) in ⟨ v.1, v.read ⟩ structure Particle (α : Type) := mk :: (state : α) (probability : ℚ) (infosets : Infosets ha) (is_updateable : bool) This piece of code typechecks as written above, but when I replace {die // die ∈ dice} × list {i // ∃ v, claims.read i = v} with History in ha's type, I get an error that it cannot synthesize the typeclass instances for (infosets : Infosets ha). The particular typeclass instance is has_lt History. What should I do here? instance : has_lt History := by {unfold History, } Also, it occurred to me that I could try providing it directly, but I am not sure which tactic to use here. Even if this could work, I'd prefer it if I could instruct the compiler to inline the alias rather than doing this. Sorry about not giving a simplified example - I gave it a shot, but there are a lot of hierarchies that I do not know how to pare down without significant effort. #### Rob Lewis (Oct 25 2019 at 10:00): instance : has_lt History := by {unfold History, apply_instance} #### Rob Lewis (Oct 25 2019 at 10:00): Alternatively, @[derive has_lt] def History : Type := ... Thanks! #### Marko Grdinić (Oct 25 2019 at 12:29): maximum class-instance resolution depth has been reached (the limit can be increased by setting option 'class.instance_max_depth') (the class-instance resolution trace can be visualized by setting option 'trace.class_instances') The advice worked for my previous problem, but now I am running into this issue. I am not even sure for which typeclass instance it is looking for. How do I find that out? I tried turning on the trace, but the instance it is looking for there should be available, so I am thinking it should be something else. #### Marko Grdinić (Oct 25 2019 at 12:41): Also, is it possible to derive decidable_rel using the first form? Something like...@[derive has_lt, derive decidable_rel] def Die := {die // die ∈ dice}? The issue with decidable_rel is that it depends on has_lt. #### Marko Grdinić (Oct 25 2019 at 13:10): @[derive has_lt] def Die := {die // die ∈ dice} @[derive has_lt] def PastClaims := list {i // ∃ v, claims.read i = v} @[derive has_lt] def History := Die × PastClaims I need decidable_rel for these 3, but forget about the previous question for a bit. I thought I would be able to do it once I made up my mind, but now that I am trying I find I have absolutely no idea how Lean deals with decidability of propositions. I am not even sure how it would work for regular nats and not these slightly more complex structures. How should I do this? I could use some good examples of how decidable_rel works. #### Kevin Buzzard (Oct 25 2019 at 13:20): It's impossible to debug a max class instance error without a fully working example demonstrating the error. #### Kevin Buzzard (Oct 25 2019 at 13:21): You can turn certain traces on and look at debug output, but you have to know what you're looking for. #### Marko Grdinić (Oct 25 2019 at 13:25): @Kevin Buzzard I see. Well in this case, I am sure that the problem is because it is looking for the missing decidable_rel for History. I managed to track it down by using @ on the offending function and manually checking all the arguments. #### Kevin Buzzard (Oct 25 2019 at 13:27): Type class inference is complicated and is going to be one of the big changes in Lean 4. There are people here who know how to work with what we have, and no doubt they could sort out your specific problem if any of them had the time, but we have also been collecting lists of when typeclass inference fails or works badly, and the devs are going to see if they can come up with something better in Lean 4. It's being implemented right now, is my understanding. #### Rob Lewis (Oct 25 2019 at 13:30): History isn't a relation, is it? You can't give it a decidable_rel instance. #### Rob Lewis (Oct 25 2019 at 13:31): You might be able to derive decidable equality. @[derive [has_lt, decidable_eq]]. You'd need this to be successful on Die and PastClaims first to have any hope on History. #### Marko Grdinić (Oct 25 2019 at 13:32): @Rob Lewis History isn't a relation, is it? You can't give it a decidable_rel instance. No, I meant for the a < b where a b : History. instance History_decidable_rel [x : has_lt History] : decidable_rel x.lt := sorry Something like this. #### Marko Grdinić (Oct 25 2019 at 13:36): @[derive [has_lt, decidable_eq]] def Die := {die // die ∈ dice} @[derive [has_lt, decidable_eq]] def PastClaims := list {i // ∃ v, claims.read i = v} @[derive [has_lt, decidable_eq]] def History := Die × PastClaims This works for me. But decidable_rel is tricky in that it needs to be passed the specific has_lt instance by hand. That is how it was in the few places where I used it. So just putting @[derive [has_lt, decidable_rel]] def Die := {die // die ∈ dice} fails for me. type mismatch at application decidable_rel Die term Die has type Type : Type 1 but is expected to have type ?m_1 → ?m_1 → Prop : Sort (max ? 1) It seems this mechanism is passing it Die directly rather than the less-than function for Die. #### Marko Grdinić (Oct 25 2019 at 13:38): def response {ℍ 𝔸 : Type*} [lt : has_lt ℍ] [decidable_rel lt.lt] I have a bunch of functions like this one where I declare that a type has has_lt and then also that the relation is decidable. #### Rob Lewis (Oct 25 2019 at 13:38): Ah. Yeah, the derive handler definitely won't work for that. You'll need to define it manually and unfold enough so that apply_instance sees the decidability. Something like instance History_decidable_rel : decidable_rel ((<) : History -> History -> Prop) := ... #### Rob Lewis (Oct 25 2019 at 13:39): Where ... does some unfolding and then calls apply_instance. #### Marko Grdinić (Oct 25 2019 at 13:47): instance nat_decidable_rel : decidable_rel ((<) : nat → nat → Prop) := by { intros a b, apply_instance } As a sanity check this does work for nats. @[derive has_lt] def Die := {die // die ∈ dice} instance Die_decidable_rel : decidable_rel ((<) : Die → Die → Prop) := by { intros a b, apply_instance } Tactic State 1 goal a b : Die ⊢ decidable (a < b) dudo.lean:29:8: error tactic.mk_instance failed to generate instance for decidable (a < b) state: a b : Die ⊢ decidable (a < b) For Die not so much. It cannot find the instance. I am trying to see whether I can get it to so it only compares the first argument (which is a nat), but I cannot unfold has_lt.lt here. What should I do? #### Marko Grdinić (Oct 25 2019 at 14:14): def Die := {die // die ∈ dice} instance Die_has_lt : has_lt Die := ⟨ fun a b, a.val < b.val ⟩ instance Die_decidable_rel : decidable_rel ((<) : Die → Die → Prop) := infer_instance I get it now. I won't be sure until I go over the subtype, but it seems that the default has_lt was not good for my use case. I should be able to make this work now. #### Marko Grdinić (Oct 25 2019 at 14:17): Hmmm...I can't find anything on has_lt for subtypes in the core library's init.data.subtype folder. There is no much there. Is it somewhere else? If so, where? #### Marko Grdinić (Oct 25 2019 at 14:19): Is there any way to track down the instance that is being inferred? #### Reid Barton (Oct 25 2019 at 14:20): Hmmm...I can't find anything on has_lt for subtypes in the core library's init.data.subtype folder. There is no much there. Is it somewhere else? If so, where? Are you using mathlib? #### Reid Barton (Oct 25 2019 at 14:24): A bit convoluted, but you can write, for example, def x : has_lt nat := by apply_instance #print x which will print nat.has_lt, then #check nat.has_lt which will output something you don't care about, but now you can use jump-to-definition on nat.has_lt #### Marko Grdinić (Oct 25 2019 at 14:33): Apparently, the function is preorder.to_has_lt. class preorder (α : Type u) extends has_le α, has_lt α := (le_refl : ∀ a : α, a ≤ a) (le_trans : ∀ a b c : α, a ≤ b → b ≤ c → a ≤ c) (lt := λ a b, a ≤ b ∧ ¬ b ≤ a) (lt_iff_le_not_le : ∀ a b : α, a < b ↔ (a ≤ b ∧ ¬ b ≤ a) . order_laws_tac) When I try to track it down by Ctrl + left click, it just puts me at the class definition. I am not sure where to go from there. #### Reid Barton (Oct 25 2019 at 14:34): Ah, in that case try set_option pp.all true before the #print, and try to guess where the actual instance is, or post the output here #### Reid Barton (Oct 25 2019 at 14:37): preorder.to_has_lt extracts the has_lt instance from a preorder instance #### Reid Barton (Oct 25 2019 at 14:38): it's automatically generated by that class preorder declaration #### Marko Grdinić (Oct 25 2019 at 14:40): Ah, I see. So that is why I get sent to the typeclass definition. I guess the question now becomes of how to track down where the subtype is declared as a preorder. #### Marko Grdinić (Oct 25 2019 at 14:42): Then I would be able to find the implementation of has_lt for it. What would be the best way to do this? #### Kevin Buzzard (Oct 25 2019 at 15:06): Are you talking about why nat is a preorder? You can just use the trick Reid showed you again: def x : preorder ℕ := by apply_instance #print x -- partial_order.to_preorder ℕ The preorder on nat comes from the partial order on nat. Last updated: May 13 2021 at 21:12 UTC	2021-05-13 21:54:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2967630922794342, "perplexity": 4149.239026360267}, "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-2021-21/segments/1620243992514.37/warc/CC-MAIN-20210513204127-20210513234127-00351.warc.gz"}
https://mathoverflow.net/questions/378473/what-is-the-endgoal-of-formalising-mathematics	# What is the endgoal of formalising mathematics? Recently, I've become interested in proof assistants such as Lean, Coq, Isabelle, and the drive from many mathematicians (Kevin Buzzard, Tom Hales, Metamath, etc) to formalise all of mathematics in such a system. I am very sympathetic to some of the aims of this project, such as finding mistakes in published proofs, and definitively verifying complex and controversial proofs [Kepler conjecture (Hales), four-colour theorem (Appel/Haken), Fermat's last theorem (Wiles), ...]. However, I'd like to understand more about the ultimate goals of this project, and the vision for achieving a computer formalisation of all of mathematics. I understand a good portion of core (undergraduate) mathematics has been formalised. Yet, the bulk of research-level mathematics remains unformalised, and it seems to me that we are currently creating new mathematics at a far greater rate than we are formalising mathematics. Are we always going to be "chasing our tails" here? What needs to change so that we can catch up and achieve a complete formalisation of mathematics? Another issue is, of course, new mathematics is being created all the time. If we achieve complete formalisation, how do we sustain this? Will there be "professional formalisers" whose job it is to formalise the recent results of mathematicians? Will mathematicians themselves be expected to formalise their results as they are proved? What role will proof assistants play in 20 (or 50, or 100) years? • "What role will proof assistants play in 20 (or 50, or 100) years?" Well, in order to answer this, I guess one should possess some kind of pre-cog ability :-) – Francesco Polizzi Dec 9 '20 at 7:31 • One near-term goal motivation of formalization can be seen in Scholze's recent formalization challenge. He (and Dustin Clausen) have a result that is "logically involved" (of the form $\forall\exists\forall\exists\forall\exists$), and that could be the basis of a new approach to real functional analysis by applying the theorem in a "black box" way. The "black box" application means the details of the proof may get examined less than other foundational results. One should read the linked post for more details. – Mark Dec 9 '20 at 18:22 • Who is talking about formalizing all of mathematics, and where? That assumption of the question seems dubious, and I don't see any references for that claim. – Matt F. Dec 9 '20 at 19:30 • I would suggest, as with several questions that have surfaced concerning what Kevin has said and written in a certain rhetorical vein, that we are best off waiting for him to turn up in person, when he can either row back or double down as he sees fit :) – Yemon Choi Dec 9 '20 at 20:02 • What's the point of digitising music? Vinyl works fine. But digitising music was a game changer -- now we have Spotify. What's the point in digitising books? What's the point in LaTeX? Typewriters worked fine. What's the point of digitising mathematics? We have no idea what will happen as a consequence. People will use it to do crazy things we can't even imagine right now. Ask the machine learning people what the point of digitising mathematics is. Ask the educators. Ask Scholze. People will figure out uses. We should do it because we can. Because nobody tried to do it yet. – Kevin Buzzard Dec 10 '20 at 9:52 ## 4 Answers This answer is based only on my having watched this lecture by Kevin Buzzard, followed by my applying some skepticism and critical thinking. Buzzard has had some experiences that make him not trust some of the results that people are claiming in mathematics papers. He tells the story of refereeing a paper and fighting (and winning) a long, drawn-out battle because he didn't believe the authors had proved what they said they had. He also says there are papers coming out in which a result is said to be proved, but there are 100 pages of details that have been left out and will be supplied in a forthcoming publication. With, in my opinion, less justice, he complains that professional mathematicians "do math like Gauss and Euler," "don't know foundations," "don't know set theory," and decide what is correct mathematics based on a consensus of beliefs among the elders in the community. This seems absurd to me. I would consider a theorem by Gauss or Euler to be less reliable if it depended on ZFC. If someone told me that a theorem in analysis depended on choice, I would decide that that theorem was not really true but only true conditioned on AC. If ZFC is proved tomorrow to be inconsistent, I will still have great faith that the theorems of Gauss and Euler are correct within some more conservative foundational system. However, if you believe at least some of his complaints about real uncertainty and doubt in published mathematical results, then it does make some sense to see if you can "formalize" these results in the sense of getting a computer to verify them all the way down to some foundation. What seems to have been going on for a long time with computer theorem checkers is that because using them was so labor-intensive, people would try to code up the tiniest possible fragment of mathematics in order to achieve their goal. The trouble with this strategy seems to be the following. If you believe that a lot of math really is in doubt, then the doubtful parts are in areas where there is a big pyramid of math. At the apex you have Fermat's last theorem or something about perfectoid spaces (which I know nothing about, but he cites them as modern pieces of mathematical machinery with a lot of moving parts). The apex is supported by a whole bunch of more basic stuff, going all the way down through freshman calculus, etc. Therefore you can't do a tiny fragment of mathematics, because these aren't self-contained fragments. So I think that Buzzard's argument is: (1) there are important, significant doubts about whether results deemed true by the elders really are true; (2) theorem provers are a good way of solving this problem; and (3) in order to carry out this solution, one needs to build up a comprehensive database of the beliefs of the elders (his characterization of Hale). My own opinion is that 1 is probably true but not a threat to the enterprise of mathematics, 2 is speculative, and 3 is infeasible. Of course my opinion is worthless because I'm not even an amateur in this area. But note that even if I'm right, it doesn't mean that proof assistants and automated theorem checkers shouldn't be worked on at all -- it just means that Buzzard's grandiose dream isn't going to happen. If we achieve complete formalisation, how do we sustain this? Will there be "professional formalisers" whose job it is to formalise the recent results of mathematicians? Looking into my crystal ball, I predict that it will not be achieved, because it costs too much, and the necessary vast funding will not be allocated. It won't be allocated because not many mathematicians agree that it would be a worthwhile use of the resources. Systematic ongoing formalization will also not happen, for the same reasons. Will mathematicians themselves be expected to formalise their results as they are proved? No. By all accounts this work is tedious and extremely time-consuming. Mathematicians want to work with new ideas, not spend their time writing code to explain their work to a computer that is orders of magnitude dumber than the dumbest undergrad they've ever taught. • This brings up an interesting point. As mentioned, if the goal of formalization is to feel more secure about the cutting-edge of math by formalizing everything, that would require a huge change in mathematical practice and culture. On the other hand, if the goal is to root out bogus arguments, perhaps a different system could be designed: one which could be fed in the logical "shape" of an argument and try to detect places where the logic might fail. Notice that in the first picture, the onus is on the prover, and in the second it is on the doubter. – Sam Hopkins Dec 9 '20 at 19:03 • So, true is the theory you get from the reverse mathematics of Gauss + Euler + whatever fragments you have that don't require things like "the real numbers are not a countable union of countable sets" and other choicey consequences. – Asaf Karagila Dec 9 '20 at 19:28 • @AsafKaragila: Sure, or maybe we should call it classical-analysis-true. Seems to me that there is a whole separate Q&A implied here, which you would be more qualified than I am both to ask and to answer :-) – Ben Crowell Dec 9 '20 at 19:32 • @BenCrowell : I don't think that Buzzard's point about "elders" has anything to do with ZFC specifically. If ZFC bothers you, replace it with something much weaker, such as RCA$_0$. As I see it, Buzzard is contrasting a situation in which all the details of a proof are truly available for public inspection, with a situation where many crucial details are locked up inaccessibly in the brains of a few experts. – Timothy Chow Dec 9 '20 at 23:06 • 'professional mathematicians "do math like Gauss and Euler,'"' - Wouldn't it be great if that were true? – Jonny Evans Dec 10 '20 at 5:50 Since you mentioned 100 years out, let me suggest that the end goal is to put ourselves out of business. We just need to formalize enough mathematics so that AI programs will get the idea of what we're trying to do. Then they'll take over the job of doing mathematics research. We won't need to formalize any old math at that point because the AIs will rediscover everything worth rediscovering anyway. [EDIT: For an example of some first baby steps in this direction, see Generative language modeling for automated theorem proving by Stanislas Polu and Ilya Sutskever.] I am half joking here, but only half. I don't think that the only, or even the main, goal of formalizing is to increase the certainty of a proof from 99.9% to 99.9999999%. The main goal is to enable new approaches to doing research that don't currently exist. For that goal, complete formalization isn't necessary, but for what it's worth, I also don't believe that "chasing our tail" will be a problem. Setting aside my pie-in-the-sky vision of putting mathematicians out of business, I think that once the tools get good enough, and the mathematical populace is sufficiently trained, new proofs will be formalized as a matter of course, just as papers are now written in $$\LaTeX$$ as a matter of course. As for the older papers, formalizing them will be assigned as exercises for students who are just learning the system. • Music has been thoroughly digitized, but that has not put all musicians out of business. – Matt F. Dec 9 '20 at 22:06 • @MattF. : But in 100 years, perhaps we'll all have folded hands, that don't even play music. – Timothy Chow Dec 9 '20 at 22:56 • It's not going to take 100 years... – Andrej Bauer Dec 10 '20 at 9:01 • ... and then years later people may want to apply what is stated in print rather than "what was in the heads of the experts" – Yemon Choi Dec 10 '20 at 21:12 • @TimothyChow My understanding is that Kevin's Lean Math group has had to rewrite its category theory library several times because of this kind of thing. – Harry Gindi Jan 17 at 17:44 Other answers address several aspects of your question, but to add a point not mentioned yet — you write: it seems to me that we are currently creating new mathematics at a far greater rate than we are formalising mathematics. Are we always going to be "chasing our tails" here? What needs to change so that we can catch up and achieve a complete formalisation of mathematics? Another issue is, of course, new mathematics is being created all the time. If we achieve complete formalisation, how do we sustain this? Will there be "professional formalisers" whose job it is to formalise the recent results of mathematicians? Will mathematicians themselves be expected to formalise their results as they are proved? Certainly, at the moment, formalising maths is very laborious and time-consuming. But I think most people who work the formalisation hope that it won’t stay so bad! It’s been getting steadily better as proof assistants and their libraries have progressed — “better” both in terms of how quickly experienced users can cover ground, and in terms of the learning curve for new users. This improvement hasn’t perhaps been as fast as we’d like, or might have hoped — but it’s been steady, and continuing. If usability continues to improve, then it’s reasonable to imagine a future proof assistant with a role comparable to LaTeX today. Typesetting our work in LaTeX takes a bit of work, and has a bit of a learning curve — but most mathematicians accept the time and effort as worthwhile for the payoff of good typesetting. A very few mathematicians outsource this to professional typesetters; a larger minority leave it to their co-authors when possible. But it’s part of the expectations of the field, and there’s no problem of “tail-chasing” — no question of “can we typeset mathematics as fast as we can produce it?”, or of requiring major grant support to get work typeset. This is one possible future that I and some others in formalisation envisage, and that it seems reasonable to hope for in a timescale of a few decades. One likely difference from LaTeX is that this could well arrive in some subfields of maths sooner than others, since (so far) some topics seem much more amenable to formalisation. I've thought more about what would be required for a complete formalisation of mathematics. I believe this could only happen if mathematicians formalise their results as they go, i.e. all of them are actively using proof assistants as part of the research process. To this end, here are some properties that I think would be desirable in a proof system: 1. Syntax that is "proof-y" rather than "program-y". I played around with Lean last night, on Kevin Buzzard's great natural number game, but it still felt like I was programming, rather than writing a proof. Yes, I know they're equivalent by the Curry-Howard isomorphism, but mathematicians don't want to learn a new programming language, they want to write proofs. Why couldn't we have syntax that looks like an actual, natural-language proof, e.g. Theorem: Let U, V be vector spaces. Let f: U -> V be a linear map. Then dim(U) = rank(f) + nullity(f). Proof: Fix v in V. .... Here, Let could be a command indicating that one is defining a new object, be / be a / be an indicates the type of said object, Then indicates you are done defining things, . indicates the end of a command, etc. 2. A typing system that reflects mathematical intuition. I know very little type theory, but it seems like dependent types are necessary to the way mathematics is currently practiced. Other properties such as univalence, polymorphism, inheritance, abstract data types also seem necessary to reflect "real" mathematical practice. I know this might add complexity to a proof assistant; however, the proof assistant should be designed to meet mathematicians, rather than us having to meet the proof system. 3. I work in logic, and there are many subfields (set theory, reverse mathematics, constructive math) where axioms can vary. So to formalise these areas, we would need a way to change the axioms as necessary. E.g. $$\mathsf{ZFC}$$ would be taken as the default set of axioms, but you could declare Axioms = ZF+AD. at the start to change the axioms for your proofs. Further, you would want to be able to treat axioms as objects in their own right. 4. Many steps of reasoning will often be skipped in a mathematician's proof (as it is painstaking to write it all out in full). Therefore, I would want an "autofill" system, where one puts in the main steps of the proof (i.e. those that you would write in the natural language proof): (n + 1)^3 ... n^3 + 3n^2 + 3n + 1 and the proof assistant would search to find correct reasoning to fill the space between. 5. One of the current issues is that not enough prerequisite material has been formalised for most research mathematicians to start using proof assistants for their work. I wonder if it's possible to have a "modular" approach, where mathematicians can take the theorems they want to use as "axioms", and proceed with formalising their work. Then, someone else can come along later and prove these "axioms" from more foundational principles. This would mean that mathematicians could start using the proof system right now, rather than in 20 years when all the necessary background has been formalised (and their research has moved on). (I suppose the relevant definitions still need to be formalised beforehand). I don't know much about the different proof assistants, and whether there are any that have the above properties. • As an absolute non-expert, let me make some comments. 1. Natural language is ambiguous. Any attempt to make it unambiguous is bound to make the language more "mechanical". Sure some basic things like "let" can be simplified, but quickly it will be less feasible. 4. You are asking for (what in Lean at least is called) tactics. Many such exist, for instance there is one which will automatically prove any identity in the language of commutative rings. 5. This is already (to an extent) happening. However, the entry barrier is high, so almost everyone is first working out the basics.. – Wojowu Dec 9 '20 at 22:51 • Overall though, I do agree with the sentiment that there is some work left to be done to make proof assitants more mathematician-friendly. Kevin Buzzard is very vocal in expressing his frustration when first learning Lean and how the documentation was written "by computer scientists for computer scientists". One has to remember though that someone has to build those programs, and someone with nearly enough programming proficiency to more than likely to come from CS background than math one. – Wojowu Dec 9 '20 at 22:55 • In terms of proofs that look more like natural language, the proof assistants based on set theory tend to be more readable; see the Mizar examples in Freek Wiedijk's Notices article. Also, set-theoretic proof assistants arguably are closer to mathematicians' intuition. I discuss some of the objections to set-theoretic proof assistants in another MO answer. – Timothy Chow Dec 9 '20 at 23:35 • Historical minutiae: When the BASIC programming language first appeared in 1964, defining a new object (variable) did in fact require the keyword "LET", for precisely the math-idiom reasons seen here. But over time that was considered burdensome, and the use of LET was made optional and then dropped (instead, the first time a new variable is seen it's assumed to be declared). dartmouth.edu/basicfifty/commands.html – Daniel R. Collins Dec 10 '20 at 3:33 • I don't think being based on set theory has anything to do with what input syntax provers offer. Isabelle also has declarative proof syntax, which is probably still quite a bit away from what you are looking for, but certainly closer than most systems. (see the examples here). – Manuel Eberl Dec 10 '20 at 7:48	2021-05-06 21:53:14	{"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": 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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6297969222068787, "perplexity": 653.2750723648523}, "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-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00188.warc.gz"}
https://www.semanticscholar.org/paper/Couplings-of-the-and-'-mesons-to-the-nucleon-Nasrallah/d6c96372ff120bafab76f08203e478db414fe50d	Couplings of the ? and ? ' mesons to the nucleon @article{Nasrallah2005CouplingsOT, title={Couplings of the ? and ? ' mesons to the nucleon}, author={Nasrallah F. Nasrallah}, journal={Physics Letters B}, year={2005} } • N. Nasrallah • Published 4 December 2005 • Physics • Physics Letters B 3 Citations Quark flavor decomposition of the nucleon axial form factors • Physics Physical Review D • 2021 We present results on the isoscalar form factors including the disconnected contributions, as well as on the strange and charm quark form factors. Using previous results on the isovector form References SHOWING 1-10 OF 38 REFERENCES Two-photon decay of the pseudoscalars, the chiral symmetry breaking corrections The extrapolation of the decay amplitudes of the pseudoscalar mesons into two photons from the soft meson limit where it is obtained from the axial-anomaly to the mass shell involves the contribution Quark Structure of Pseudoscalar Mesons I review to which extent the properties of pseudoscalar mesons can be understood in terms of the underlying quark (and eventually gluon) structure. Special emphasis is put on the progress in our DECAY OF THE Pi MESON • Physics • 1956 A quantitative study of π→μ+ν decay is presented using the techniques of dispersion theory. The discussion is based on a model in which the decay occurs through pion disintegration into a Low-energy nucleon-nucleon potential from Regge-pole theory • Physics • 1978 The results from a potential model for the low-energy NN interaction based on Regge-pole theory are presented. The forces are due to the dominant parts of the New measurement of Sigma beam asymmetry for eta meson photoproduction on the proton • Physics • 1998 We present new Sigma beam asymmetry data for eta meson photoproduction on the proton, using a novel tagged, laser backscattered, linearly polarized photon beam up to 1.1 GeV. The data show large, eta - eta ' mixing angle. • Physics Physical review. D, Particles and fields • 1987 The current experimental evidence on the value of the eta-eta' mixing angle is summarized in the light of our present theoretical understanding. A value of theta/sub P/approx. =-20/sup 0/ is Effective hadron dynamics: From meson masses to the proton spin puzzle. • Physics Physical review. D, Particles and fields • 1993 The soliton sector of this improved chiral Lagrangian is used to investigate some aspects of baryon physics which are especially sensitive to symmetry breaking, and both the "matter" and "glue" contributions are small but they do tend to cancel each other.	2022-08-17 13:51:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8409123420715332, "perplexity": 3468.204853274359}, "config": {"markdown_headings": false, "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-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00480.warc.gz"}
https://brilliant.org/problems/candy-store/	# Candy Store! Algebra Level 1 I went to the candy store last weekend and I bought a total of 10 packs of candy. I bought M&Ms for $1 per pack and Twizzlers for$0.50 per pack. I spent a total of \$6. How many packs of M&Ms did I buy? ×	2017-05-28 04:57:36	{"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": 0.20015203952789307, "perplexity": 3337.234738416757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00031.warc.gz"}
https://www.winslowassociates.com/doctor-doctor-kgybf/acc945-unique-left-inverse	share. Proposition If the inverse of a matrix exists, then it is unique. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. eralization of the inverse of a matrix. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . Left-cancellative Loop (algebra) , an algebraic structure with identity element where every element has a unique left and right inverse Retraction (category theory) , a left inverse of some morphism left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. inverse. endstream endobj startxref Note that other left inverses (for example, A¡L = [3; ¡1]) satisfy properties (P1), (P2), and (P4) but not (P3). In general, you can skip the multiplication sign, so 5x is equivalent to 5x. Yes. In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. 8 0 obj One consequence of (1.2) is that AGAG=AG and GAGA=GA. Let $f \colon X \longrightarrow Y$ be a function. (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) g = finverse(f) returns the inverse of function f, such that f(g(x)) = x. h�bbdb� �� 9D�H�_ ��Dj�HE�8�,�&f��L[�z�H�W�� HU{��Z �(� �� A��O0� lZ'��{,��.�l�\��@��OL@��q�� Let e e e be the identity. (An example of a function with no inverse on either side is the zero transformation on .) This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. For any elements a, b, c, x ∈ G we have: 1. Still another characterization of A+ is given in the following theorem whose proof can be found on p. 19 in Albert, A., Regression and the Moore-Penrose Pseudoinverse, Aca-demic Press, New York, 1972. Show Instructions. 3. Matrix Multiplication Notation. Recall that $B$ is the inverse matrix if it satisfies $AB=BA=I,$ where $I$ is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. %%EOF Proof: Assume rank(A)=r. As f is a right inverse to g, it is a full inverse to g. So, f is an inverse to f is an inverse to stream However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). If the function is one-to-one, there will be a unique inverse. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If f contains more than one variable, use the next syntax to specify the independent variable. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … Theorem. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. Remark When A is invertible, we denote its inverse … Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. The Moore-Penrose pseudoinverse is deﬂned for any matrix and is unique. New comments cannot be posted and votes cannot be cast. %�� wqhh��llf�)eK�y�I��bq�(��Ã.4-�{xe��8��b�c[��ö��TBYb�ʃ4��&�1��o[{cK�sAt��3�'vp=�$��$�i.��j8@�g�UQ��>��g�lI&�OuL����wCu�0 �]l� One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, …) = (b 2, b 3, …). Two-sided inverse is unique if it exists in monoid 2. In a monoid, if an element has a right inverse… /Filter /FlateDecode << /S /GoTo /D [9 0 R /Fit ] >> Theorem A.63 A generalized inverse always exists although it is not unique in general. (We say B is an inverse of A.) It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Theorem 2.16 First Gyrogroup Properties. In general, you can skip the multiplication sign, so 5x is equivalent to 5x. Thus both AG and GA are projection matrices. See the lecture notesfor the relevant definitions. A.12 Generalized Inverse Deﬁnition A.62 Let A be an m × n-matrix. inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). There are three optional outputs in addition to the unique elements: 53 0 obj <> endobj Then 1 (AB) ij = A i B j, 2 (AB) i = A i B, 3 (AB) j = AB j, 4 (ABC) ij = A i BC j. u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. If the function is one-to-one, there will be a unique inverse. Let (G, ⊕) be a gyrogroup. Note that other left 5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 By using this website, you agree to our Cookie Policy. /Length 1425 Hello! Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. The equation Ax = b always has at least one solution; the nullspace of A has dimension n − m, so there will be The left inverse tells you how to exactly retrace your steps, if you managed to get to a destination – “Some places might be unreachable, but I can always put you on the return flight” The right inverse tells you where you might have come from, for any possible destination – “All places are reachable, but I can't put you on the I know that left inverses are unique if the function is surjective but I don't know if left inverses are always unique for non-surjective functions too. h�b�y�� cca�� ِ� q��#�!�A�ѬQ�a��[�50�F��3&9'��0 qp�(R�&�a�s4�p�[��f^'w�P&޶ 7��,��[T�+�J��9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w In gen-eral, a square matrix P that satisﬂes P2 = P is called a projection matrix. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. best. '+o�f P0��'�,�\� y��bf\�; wx.��";MY�}��إ� Subtraction was defined in terms of addition and division was defined in terms ofmultiplication. Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). (4x1�@�y�,(��.�BY��⧆7G�߱Zb�?��,��T��9o��H0�(1q��D� �;:��vK{Y�wY�/��5��c�iZl�B\\��L�bE��8;�!�#�)�L�{�M��dUт6��%�V^��ZW��f�4R�p�p�b��x��.L��1sh��Y�U��! In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. When working in the real numbers, the equation ax=b could be solved for x by dividing bothsides of the equation by a to get x=b/a, as long as a wasn't zero. We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. An associative on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same . Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. �� Sort by. Ask Question Asked 4 years, 10 months ago. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). Hence it is bijective. If A is invertible, then its inverse is unique. This thread is archived. Let $f \colon X \longrightarrow Y$ be a function. Viewed 1k times 3. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. 36 0 obj << This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. x��XKo#7��W�hE�[ע��E��:v�4q��/)�c��>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG��8>��'��x�ټrc��>?��[��?�'��(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ��N)&Q ��M��Yy��?A��4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi��e��A��(;p��V�Jڛ,�t~�d��̘H9��/��_a��v�68gq"��D�\|a5��P\|Jv��l1j��x��&޺N��V"��"��}! numpy.unique¶ numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] ¶ Find the unique elements of an array. This preview shows page 275 - 279 out of 401 pages.. By Proposition 5.15.5, g has a unique right inverse, which is equal to its unique inverse. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. See Also. endobj Actually, trying to prove uniqueness of left inverses leads to dramatic failure! Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). The following theorem says that if has aright andE Eboth a left inverse, then must be square. A.12 Generalized Inverse Deﬁnition A.62 Let A be an m × n-matrix. This is no accident ! From this example we see that even when they exist, one-sided inverses need not be unique. Theorem A.63 A generalized inverse always exists although it is not unique in general. Stack Exchange Network. Then they satisfy $AB=BA=I \tag{}$ and Active 2 years, 7 months ago. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Remark Not all square matrices are invertible. 87 0 obj <>/Filter/FlateDecode/ID[<60DDF7F936364B419866FBDF5084AEDB><33A0036193072C4B9116D6C95BA3C158>]/Index[53 73]/Info 52 0 R/Length 149/Prev 149168/Root 54 0 R/Size 126/Type/XRef/W[1 3 1]>>stream 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Show Instructions. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. 125 0 obj <>stream If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). example. Let (G, ⊕) be a gyrogroup. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. It's an interesting exercise that if$a$is a left unit that is not a right uni Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Recall also that this gives a unique inverse. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. 6 comments. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. Yes. Generalized inverse Michael Friendly 2020-10-29. 11.1. g = finverse(f,var) ... finverse does not issue a warning when the inverse is not unique. Let G G G be a group. Proof: Assume rank(A)=r. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. U-semigroups �n��r��6��d}��wF>�G�/��k� K�T�SE�� &ʬ�Rbl�j��\|�Tx��)��Rdy�Y ? If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. %PDF-1.4 Theorem 2.16 First Gyrogroup Properties. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective 100% Upvoted. If is a left inverse and a right inverse of , for all ∈, () = ((()) = (). >> save hide report. If BA = I then B is a left inverse of A and A is a right inverse of B. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. If E has a right inverse, it is not necessarily unique. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Proof. 0 In mathematics, and in particular, algebra, a generalized inverse of an element x is an element y that has some properties of an inverse element but not necessarily all of them. %PDF-1.6 %�� A i denotes the i-th row of A and A j denotes the j-th column of A. Some easy corollaries: 1. If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. Note the subtle difference! ��E�O]{z^��h%�w�-�B,E�\J��\|�Y\2z)��ME��5��@5��q��\|7P��@��&��5�9�q#��h�>Rҹ�/�Z1�&�cu6��B��e�^BXx��r��=�E�_� ��Tm��z��8g�~t.i}��߮:>;�PG�paH�T. For any elements a, b, c, x ∈ G we have: 1. h��[[�۶�+\|l\wp��ߝ�N\��&�䁒�]��%"e��{>��HJZi�k�m� �wnt.I�%. Returns the sorted unique elements of an array. Let A;B;C be matrices of orders m n;n p, and p q respectively. B$ and $c$ of the matrix $a$ exists in monoid 2 years 10. We denote its inverse is because matrix multiplication is not unique in,... Unique left inverse and the unique left inverse inverse, then must be unique denotes the i-th row of.... Its transpose has a right inverse ( which is the zero transformation on. months.... Then does it necessarily have a unique right inverse ( which is the transformation. $c$ of the matrix $a$ unique left inverse be a unique right inverse, it is if. )... finverse does not issue a warning when the inverse of a and a is invertible we... 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Variable, use the next syntax to specify the independent variable use the next syntax to specify independent! Zero transformation on. the zero transformation on. other conditions that damage brain! Not unique in general the independent variable i-th row of a matrix,. 4 years, 10 months ago has a nonzero nullspace unique if it exists must!	2021-02-25 19:04:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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": 0.8751460909843445, "perplexity": 720.5046601983116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351454.16/warc/CC-MAIN-20210225182552-20210225212552-00177.warc.gz"}
https://saket-choudhary.me/blog/2022/03/10/fast-sparsespearman/	# Fast & memory efficient spearman correlation on sparse matrices #### Or why you should avoid loops The previous post formulated a solution for calculating spearman correlation on sparse matrices. The trick was simple - we exploited the sparsity structure in our matrix to prevent it from being converted to a dense form at any step. For spearman correlation, following this line of thought is tricky, because the ranks are not sparse, at least not by default. We then used the property of covariance, where adding a constant quantity to all entries of a vector (or matrix) does not change its variance (or covariance) structure to come up with a solution that works great in theory and solves our purpose of keeping the memory foot print low. A notebook with an implementation here. However, the time benchmarks look awful - though we ended up saving memory, SparseSpearmanCor() was atleast 2 times slower than the naive approach of densifying the matrix and calculating correlation using cor(as.matrix(X.sparse), method="spearman"). This in practise, defeats the motivation - we are saving memory at the cost of speed. ## Solution The costliest step in my original implementation of SparseSpearmanCor() was a simple lookup operation: which(j == column) , where I fetch the non-zero entries in a column for calculating the rank, and this happens for all the columns (j stores the index of columns where there are non-zero entries). I tried other ways of making this faster, such as by using fastmatch. But the actual speedup came from a simple thought - if we care about the non zero entries, I should just deal with them separately. So instead of doing repeated lookups, I just separate the non-zero entries out, do the rank sparsification operations on them and put them back into the sparse matrix. I call this implementation SparseSpearmanCor2() and you can find the implementation in the notebook, but here are some comparisons with the dense approach and the previous implementation SparseSpearmanCor(). The result is a function that calculates values 10x faster than any approach on large matrices (10000 x 5000): SparseSpearmanCor2() and time benchmarks are available in this noteobok.	2022-05-23 21:12:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41724440455436707, "perplexity": 783.8948777349592}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00438.warc.gz"}
http://citizendia.org/Exponential_utility	In economics exponential utility refers to a specific form of the utility function, used in many contexts because of its convenience when uncertainty is present. Economics is the social science that studies the production distribution, and consumption of goods and services. In Economics, utility is a measure of the relative satisfaction from or desirability of Consumption of various Goods and services. Uncertainty is a term used in subtly different ways in a number of fields including Philosophy, Statistics, Economics, Finance, Insurance Formally, exponential utility is given by: u(c) = − e ac, where c is consumption and a is a constant. Exponential utility implies constant absolute risk aversion, with coefficient of absolute risk aversion equal to $\frac{-u''(c)}{u'(c)}=a.$ Though isoelastic utility, exhibiting constant relative risk aversion, is considered more plausible (as are other utility functions exhibiting decreasing absolute risk aversion), exponential utility is particularly convenient for many calculations. Risk aversion is a concept in Economics, Finance, and Psychology related to the behaviour of consumers and investors under uncertainty Risk aversion is a concept in Economics, Finance, and Psychology related to the behaviour of consumers and investors under uncertainty Specifically, under exponential utility, expected utility is given by: E(u(c)) = E( − e a(c + ε)), where E is the expectation operator. In Economics, Game theory, and Decision theory the expected utility theorem or expected utility hypothesis predicts that the "betting preferences" With normally distributed noise, ie, $\epsilon \sim N(\mu, \sigma^2),\!$ E(u(c)) can be calculated easily using the fact that $E(e^{-a \epsilon})=e^{-a \mu + \frac{a^2}{2}\sigma^2}.$ The normal distribution, also called the Gaussian distribution, is an important family of Continuous probability distributions applicable in many fields © 2009 citizendia.org; parts available under the terms of GNU Free Documentation License, from http://en.wikipedia.org network: \| \|	2013-05-25 23:52:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "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": 0.9096559286117554, "perplexity": 1181.523663958493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706472050/warc/CC-MAIN-20130516121432-00058-ip-10-60-113-184.ec2.internal.warc.gz"}
https://mathematik.univie.ac.at/eventsnews/nachrichtenvolldarstellung/news/subshifts-of-finite-type-on-amenable-baumslag-solitar-groups/?no_cache=1&tx_news_pi1%5Bcontroller%5D=News&tx_news_pi1%5Baction%5D=detail&cHash=049298abc88d56c8342d51bcdb44943c	# Subshifts of finite type on amenable Baumslag-Solitar groups 08.06.2021 15:00 - 17:00 Nathalie Aubrun (Paris-Saclay) Introductory talk: Wang’s tilings and the Domino problem Abstract: Wang tiles are unit squares with colored edges. They can be assembled into a tiling of the plane iff two adjacent Wang tiles wear the same color on their common edge.This tiling model was defined in the 60’s by Hao Wang, in order to study some fragment of monadic second order logics. The Domino problem asks whether there exists or not an algorithm that takes as input a finite number of Wang tiles and outputs YES if there exists a valid tiling of the plane, and NO otherwise. In this talk I will explain the links between the decidability of the Domino problem and the existence of aperiodic Wang tile sets. We will also briefly review known results about the Domino problem generalized to other structures than the infinite grid (here Cayley graphs of groups). Research talk: Subshifts of finite type on amenable Baumslag-Solitar groups Abstract: Amenable Baumslag-Solitar groups are two generators one relator groups with presentation $\langle a,t \| at=ta^n \rangle$ with $n$ a positive integer. In the first part of this talk I will present in details these groups and their Cayley graph, and how to embed them in $\mathbb{R}^2$. I will then review recent results about subshifts of finite type (SFTs) on amenable Baumslag-Solitar groups: indecidability of the emptiness problem for SFTs and existence of aperiodic SFTs. Based on joint works with Jarkko Kari and with Michael Schraudner. Organiser: G. Arzhantseva, A. Evetts Location: Join online at link below, using Chrome. Password: the order of the smallest non-abelian simple group.	2021-07-27 08:51:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5289517641067505, "perplexity": 640.4258421308841}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00408.warc.gz"}
https://www.gamedev.net/forums/topic/569742-quaternion-to-rotation-matrix-problem/	# Quaternion to rotation matrix problem... This topic is 3093 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Hi, I contact you because in my software I use quaternion to do so rotation and in a specific context I need to convert my quaternion to a rotation matrix. In fact, my software use a formula for this and it works fine for the work I have do until now. Last week I have added some code of someone else, he is also doing so rotation with quaternion but with another formula. What I have discover is that my formula and his formula are rotating in the inverse direction !!! And I don't know why because both should have the same rotation result. This stuff is driving me nuts... so if someone understand what is happening ? Here is the pseudo-code of my formula : Matrix3 rotation = Matrix3.Identity; float length = (x * x) + (y * y) + (z * z) + (w * w); float s = (length > 0f) ? 2f / length : 0f; float xx = x * x; float yy = y * y; float zz = z * z; float xw = x * w; float yw = y * w; float zw = z * w; float xz = x * z; float xy = x * y; float yz = y * z; rotation.M00 = 1f - (s * yy + s * zz); rotation.M10 = s * xy - s * zw; rotation.M20 = s * xz + s * yw; rotation.M01 = s * xy + s * zw; rotation.M11 = 1f - (s * xx + s * zz); rotation.M21 = s * yz - s * xw; rotation.M02 = s * xz - s * yw; rotation.M12 = s * yz + s * xw; rotation.M22 = 1f - (s * xx + s * yy); The pseudo code of his formula Matrix4 mat = Matrix4.Identity; float ww = w * w; float xx = x * x; float yy = y * y; float zz = z * z; // Diagonal elements mat.M00 = ww + xx - yy - zz; mat.M11 = ww - xx + yy - zz; mat.M22 = ww - xx - yy + zz; // 0,1 and 1,0 elements float q03 =w * z; float q12 =x * y; mat.M01 = 2.0f * (q12 - q03); mat.M10 = 2.0f * (q03 + q12); // 0,2 and 2,0 elements float q02 = w * y; float q13 = x * z; mat.M02 = 2.0f * (q02 + q13); mat.M20 = 2.0f * (q13 - q02); // 1,2 and 2,1 elements float q01 = w * x; float q23 = y * z; mat.M12 = 2.0f * (q23 - q01); mat.M21 = 2.0f * (q01 + q23); ##### Share on other sites Here's what I've coded: void to_rotation_matrix( Quaternion const & src, Matrix3 & dest ) { Real wx = src.get_w() * src.get_x(); Real wy = src.get_w() * src.get_y(); Real wz = src.get_w() * src.get_z(); Real xx = src.get_x() * src.get_x(); Real xy = src.get_x() * src.get_y(); Real xz = src.get_x() * src.get_z(); Real yy = src.get_y() * src.get_y(); Real yz = src.get_y() * src.get_z(); Real zz = src.get_z() * src.get_z(); dest.set_00( (Real)1.0 - (Real)2.0 * ( yy + zz ) ); dest.set_01( (Real)2.0 * ( xy - wz ) ); dest.set_02( (Real)2.0 * ( xz + wy ) ); dest.set_10( (Real)2.0 * ( xy + wz ) ); dest.set_11( (Real)1.0 - (Real)2.0 * ( xx + zz ) ); dest.set_12( (Real)2.0 * ( yz - wx ) ); dest.set_20( (Real)2.0 * ( xz - wy ) ); dest.set_21( (Real)2.0 * ( yz + wx ) ); dest.set_22( (Real)1.0 - (Real)2.0 * ( xx + yy ) ); } [Edited by - johnstanp on April 29, 2010 7:07:09 AM] ##### Share on other sites If the rotations appear to be in opposite directions, it is most likely because one version is intended for use with row vectors, and the other is intended for use with column vectors. Another possibility is that it's a problem with matrix layout conventions (i.e. 'majorness'), which can also have the end effect of matrices appearing to be transposes of one another. Assuming it's the former (which I'm guessing it is, from what you've posted), then you can simply use the version that produces the correct results in the context in which it's being used. For what it's worth, it looks to me like your version is set up for row vectors and his is set up for column vectors, so you should use whichever version matches the vector notation convention that you're using. 1. 1 2. 2 3. 3 Rutin 22 4. 4 5. 5 • 12 • 19 • 14 • 9 • 9 • ### Forum Statistics • Total Topics 632929 • Total Posts 3009286 • ### Who's Online (See full list) There are no registered users currently online ×	2018-10-17 16:31:27	{"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": 0.20264245569705963, "perplexity": 4111.702752489426}, "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-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00295.warc.gz"}
http://riotorto.users.sourceforge.net/Maxima/gnuplot/omar/index.html	Back to draw-Gnuplot # Omar Khayyam's method This plot explains Omar Khayyam's graphical method to solve the cubic equation $$x^3+a^2 x = b$$, for $$a, b \gt 0$$: a: 3$b: 26$ r: b/(2a^2)$draw2d(/ General settings / xrange = [-3,3], yrange = [-2,4], axis_top = false, axis_right = false, axis_left = false, axis_bottom = false, transparent = true, user_preamble = "unset tics", title = "Omar Khayyam solution to the cubic x^3 + a^2 x = b", dimensions = [450,450], terminal = svg, file_name = "omar", / draw parabola y=x^2/a / explicit(x^2/a,x,-3,3), head_length = 0.1, vector([-1.626,2],[-0.5,-0.5]), label(["y=x^2/a",-1.5,2.1]), / draw circle: radius r & center (r,0) / ellipse(r,0,r,r,0,360), vector([r,0],[0,r]), label(["r=b/(2 a^2)",r-0.5,r2/3]), /* draw axis / points_joined = true, color = black, point_size = 0, points([[0,3.5],[0,-1.5]]), points([[-2.5,0],[2.5,0]]), / Mark points P e Q/ color = black, point_type = 7, point_size = 1, points([[2,4/3]]), label(["P",2.25,1.42]), points([[2,0]]), label(["Q",2.25,0.23]), / Draw parallel to parabola axis / point_size = 0, color = blue, line_type = dots, points([[2,2],[2,-0.5]]), / Get the solution / line_width = 4, color = red, line_type = solid, points([[0,0],[2,0]]), label(["The length of this red segment is the solution",1,-0.5]))$ In words: 1. Draw parabola $$y=\frac{x^2}{a}$$. 2. Draw circumference $$(x-\frac{b}{2 a^2})^2 + y^2 = \frac{b^2}{4a^4}$$. 3. Draw a segment from P, the intersecting point, paralel to the axis of the parabola, and get point Q. 4. Measure the distance from Q to the vertex of the parabola. This is the solution of the equation. We can make use of Maxima to verify this result. Point P is the instersection of two curves, the parabola $$y=\frac{x^2}{a}$$ and the circumference $$(x-\frac{b}{2 a^2})^2 + y^2 = \frac{b^2}{4a^4}$$: kill(all)$sol: solve([y=x^2/a, (x-b/(2a^2))^2+y^2=b^2/(4a^4)], [x,y])$ The result is too long and we don't reproduce it here. Since Khayyam claims that the length of the red segment is the solution, taking into account that OQ is equal to the abscissa of P, we only show the first coordinates of the solutions saved in variable sol. absc: map(first,sol); $\left[ x=\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}}\right)\, \left({{\sqrt{27\,b^2+4\,a^6}}\over{2\,3^{{{3}\over{2}}}}}+{{b }\over{2}}\right)^{{{1}\over{3}}}-{{\left({{\sqrt{3}\,i}\over{2}}+{{ -1}\over{2}}\right)\,a^2}\over{3\,\left({{\sqrt{27\,b^2+4\,a^6} }\over{2\,3^{{{3}\over{2}}}}}+{{b}\over{2}}\right)^{{{1}\over{3}}}}} , \\ x=\left({{\sqrt{3}\,i}\over{2}}+{{-1}\over{2}}\right)\,\left({{ \sqrt{27\,b^2+4\,a^6}}\over{2\,3^{{{3}\over{2}}}}}+{{b}\over{2}} \right)^{{{1}\over{3}}}-{{\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2 }}\right)\,a^2}\over{3\,\left({{\sqrt{27\,b^2+4\,a^6}}\over{2\,3^{{{ 3}\over{2}}}}}+{{b}\over{2}}\right)^{{{1}\over{3}}}}} , \\ x=\left({{ \sqrt{27\,b^2+4\,a^6}}\over{2\,3^{{{3}\over{2}}}}}+{{b}\over{2}} \right)^{{{1}\over{3}}}-{{a^2}\over{3\,\left({{\sqrt{27\,b^2+4\,a^6} }\over{2\,3^{{{3}\over{2}}}}}+{{b}\over{2}}\right)^{{{1}\over{3}}}}} , \\ x=0 \right]$ We are looking for real numbers, since at the time Omar Khayyam lived, complex number were not known. Therefore we don't pay attention to the first and second abscissas, which contain the imaginary unit i. The fourth abscissas is also discarded. So only the third can be the first coordinate of the intersecting point P, a real positive number, namely: realsol: third(absc); $x=\left({{\sqrt{27\,b^2+4\,a^6}}\over{2\,3^{{{3}\over{2}}}}}+{{b }\over{2}}\right)^{{{1}\over{3}}}-{{a^2}\over{3\,\left({{\sqrt{27\,b ^2+4\,a^6}}\over{2\,3^{{{3}\over{2}}}}}+{{b}\over{2}}\right)^{{{1 }\over{3}}}}}$ Is this a valid solution of equation $$x^3+a^2 x = b$$? Let's ask Maxima substituting this expressions in the cubic equation and simplifying: radcan(subst(realsol, x^3+a^2x=b)); $b=b$ Which is an identity, and Khayyam was right.	2017-09-26 09:00:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8977220058441162, "perplexity": 745.2486874903368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818695375.98/warc/CC-MAIN-20170926085159-20170926105159-00270.warc.gz"}
http://www.plustwophysics.com/transistor-as-a-switch/	# Transistor as a Switch (This is one of the Expected Question for AISSCE Physics 2011) Explain the action of transistor as a switch. To answer this question (with proper understanding) you should have an idea of the three regions in the VI characteristics of a transistor. viz; Cutoff region, Active region and the saturation region. When the input bias of a CE transistor (Vbe) is very low, the transistor does not conduct. So the output voltage (Vce) will be almost equal to Vcc, the output bias voltage. By increasing the input voltage, the transistor can be brought to saturation region when the transistor will be conducting well. Now the pd across the transistor (Vce) will be very low. This is the working principle of transistor as a switch.	2017-06-24 03:32:41	{"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": 0.8406561017036438, "perplexity": 1960.7710084561527}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320215.92/warc/CC-MAIN-20170624031945-20170624051945-00381.warc.gz"}
http://mathhelpforum.com/discrete-math/224617-combinatorics-print.html	Combinatorics • November 25th 2013, 10:42 AM davidciprut Combinatorics Hi, I uploaded two statements (I guess), and I didn't understand them , can someone help me understand why do they do k times Q(n-1,k-1) or k times Q(n-1,k)??? I don't understand the k times part.. And the lecturer wrote to the properties of Q, Q(n,1)=1.... shouldn't that be equal to n? Q is just a letter he picked for bijection functions...Does he mean by Q(n,k), n!/(n-k)!??? I hope my questions are clear enogh • November 25th 2013, 11:15 AM emakarov Re: Combinatorics Quote: Originally Posted by davidciprut Q is just a letter he picked for bijection functions... Do you mean he denotes bijective functions by Q? • November 25th 2013, 11:42 AM davidciprut Re: Combinatorics Not the bijective functions, the operation of permutation on bijective functions he denoted as Q.. Q(n,k)=n!/(n-k)! I think... because for one to one functions he used A(n,k) but still its the same operation I think... • November 25th 2013, 12:11 PM emakarov Re: Combinatorics The whole problem seems strange. I am not familiar with the function Q(n, k). A bijection between A and B exists iff n = k. If f is a bijection, then $f^{-1}(f(a))=a$, and $\|f^{-1}(f(a))\|>1$ is impossible. Not to say that when the vertical bar in the set-builder notation is followed by another bar from an absolute value, the first bar should be changed into, say, a colon. • November 25th 2013, 12:20 PM Plato Re: Combinatorics Quote: Originally Posted by davidciprut Not the bijective functions, the operation of permutation on bijective functions he denoted as Q.. Q(n,k)=n!/(n-k)! I think... because for one to one functions he used A(n,k) but still its the same operation I think... I fear that you are at the mercy of your instructor, only he/she knows what that means. I will suggest if you can explain basically what is being asked. Is it about the number of injections, surjections and/or bijections between sets? Or some variation of that?	2016-05-27 20:31:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9001996517181396, "perplexity": 1457.7333838980771}, "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-2016-22/segments/1464049277091.27/warc/CC-MAIN-20160524002117-00029-ip-10-185-217-139.ec2.internal.warc.gz"}
https://quantumcomputing.stackexchange.com/questions/20675/measure-langle-hatx-rangle-and-langle-haty-rangle-from-counts	# Measure $\langle \hat{X}\rangle$ and $\langle \hat{Y}\rangle$ from counts I'm confused about how I can measure $$\langle \hat{X}\rangle$$ and $$\langle \hat{Y}\rangle$$ using counts. Here's my code for X: x-basis: # Measurement in x-basis. quanc_x = QuantumCircuit(1) quanc_x.u(1,2,3,0) # prepare some random state quanc_x.h(0) quanc_x.measure_all() quanc_x.draw(output='mpl') # number of repetitions N = 10000 backend = Aer.get_backend( 'qasm_simulator' ) job = execute( quanc_x, backend, shots=N ) result = job.result() measurement_result = result.get_counts( quanc_x ) print( measurement_result ) plot_histogram( measurement_result ) cos_phi_est = ( measurement_result['0'] - measurement_result['1'] ) / N #<--Question print( "cos(phi) estimated: ", cos_phi_est ) My question of this code is marked above. I'm not pretty sure if that looks correct. For Pauli X, we have $$\langle \hat{X}\rangle=\langle\psi\|0\rangle\langle1\|\psi\rangle+\langle\psi\|1\rangle\langle0\|\psi\rangle$$ Can I simplify that further? Should that correspond to my code with the question mark? How can I apply that to $$\langle \hat{Y}\rangle$$? Thanks for the help! That looks right to me. Since, $$HZH = X$$ then we have that $$\langle \psi \| X \| \psi \rangle = \langle \psi \| HZH \| \psi \rangle = \langle \psi H \| Z \| H\psi \rangle$$. In your code, you generate $$\|\psi \rangle$$ with a $$U_3(\theta, \phi, \lambda)$$ gate applied to $$\|0\rangle$$. Then you applied the Hadamard gate ($$H$$) before measuring which is what needed to measure in the $$X$$ basis as discussed above. For $$\langle Y \rangle$$ you should note that $$(SH)Z(HS^\dagger) = Y$$ $$\langle \psi \|Y\| \psi \rangle = \langle \psi \| (SH)Z (HS^\dagger) \| \psi \rangle = \langle \psi SH \| Z \| H S^\dagger \psi \rangle$$ Thus, here you want to apply $$S^\dagger$$ follow by the Hadamard gate $$H$$ before measurement. • Thanks for the answer! If I want to find the expectation value from counts, is my code also correct for $X$? (( measurement_result['0'] - measurement_result['1'] ) / N)Should that be the same as $Y$? – ZR- Jul 30 at 23:25 • Yes. The counts remain the same as you wrote it. Jul 30 at 23:28 • Thanks, can I understand the measured expectation value as the difference of some probabilities? Why it is not result[1]- result[0]? – ZR- Jul 30 at 23:31 • You already rotate to the computational basis once you apply the rotation $H$ or $H S^\dagger$ to your state $\|\psi \rangle$. In the $Z$ basis, $\|0\rangle$ has eigenvalue of $+1$ and $\|1\rangle$ has eigenvalue of $-1$ since $Z\|0\rangle = 1\|0\rangle$ and $Z\|1\rangle = -1 \|1\rangle$. Jul 30 at 23:42	2021-09-18 02:00:08	{"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7810496091842651, "perplexity": 1218.0541425387385}, "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-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00204.warc.gz"}
https://smorbieu.gitlab.io/accuracy-from-classification-to-clustering-evaluation/	# Accuracy: from classification to clustering evaluation Posted on Tue 04 June 2019 in machine learning • 4 min read This post is part 7 of the "clustering" series: Accuracy is often used to measure the quality of a classification. It is also used for clustering. However, the scikit-learn accuracy_score function only provides a lower bound of accuracy for clustering. This blog post explains how accuracy should be computed for clustering. Let's first recap what accuracy is for a classification task. ## Accuracy for classification For $$n$$ samples in a dataset, let $$y_i$$ be the class label for the $$i$$-th sample and $$\hat{y}_i$$ the predicted value. Accuracy between the class labels $$y$$ and the predicted values $$\hat{y}$$ is defined by: \begin{equation} accuracy(y, \hat{y}) = \frac{1}{n} \sum_{i=0}^{n-1} 1(\hat{y}_i = y_i) \end{equation} where $$x \rightarrow 1(x)$$ is the indicator function: $$1(\hat{y}_i = y_i) = 1$$ if $$\hat{y}_i = y_i$$ and $$0$$ else. It is computed as the sum of the diagonal elements of the confusion matrix, divided by the number of samples to get a value between 0 and 1. For clustering, there is however no association provided by the clustering algorithm between the class labels and the predicted cluster labels. We therefore need to change the formula a little. ## Accuracy for clustering For clustering, we have to find the best match between the class labels and the cluster labels, so accuracy is defined by: \begin{equation} accuracy(y, \hat{y}) = \max_{perm \in P} \frac{1}{n} \sum_{i=0}^{n-1} 1(perm(\hat{y}_i) = y_i) \end{equation} where $$P$$ is the set of all permutations in $$[1; K]$$ where $$K$$ is the number of clusters. It is important to note that there are $$K !$$ permutations in $$P$$ which is quite high and the computation of accuracy is therefore prohibitive if we apply blindly this formula. The Hungarian algorithm can help us compute it in $$O(K^3)$$ which is significantly faster. Let's see in practice how to compute accuracy on clustering results in a polynomial time. ## Hands-on In this part, we are going to first get some textual data to process, then cluster the documents and compute the accaracy between the clustering results and the known classes. ### Get some data Let's get some documents from the 20 Newsroups data set. We retrieve only five classes from it: from sklearn.datasets import fetch_20newsgroups categories = [ 'rec.motorcycles', 'rec.sport.baseball', 'comp.graphics', 'sci.space', 'talk.politics.mideast' ] ng5 = fetch_20newsgroups(categories=categories, shuffle=True) labels = ng5.target documents = ng5.data Latent Semantic Indexing is a well known method to transform documents into low dimensional vectors (low compared to the size of the vocabulary, i.e. the number of unique words in the dataset): from sklearn.pipeline import Pipeline from sklearn.feature_extraction.text import CountVectorizer from sklearn.feature_extraction.text import TfidfTransformer from sklearn.decomposition import TruncatedSVD from sklearn.preprocessing import Normalizer pipeline = Pipeline([ ('vect', CountVectorizer()), ('tfidf', TfidfTransformer()), ('svd', TruncatedSVD()), ('normalizer', Normalizer()) ]) pipeline.set_params(svd__n_components=300) A = pipeline.fit_transform(documents) Each line of the matrix A is a document represented by a vector. We can therefore apply a clustering algorithm on this matrix. ### Clustering Spherical k-means is a good algorithm to cluster textual data. One implementation is given by the Coclust Python library: from coclust.clustering import SphericalKmeans skm = SphericalKmeans(n_clusters=5) skm.fit(A) predicted_labels = skm.labels_ We are now ready to compute the accuracy between labels and predicted_labels. As described before, we can do this by first computing the confusion matrix. ### Confusion matrix Scikit-learn library provides a function called confusion_matrix to create a Numpy array containing the values of the confusion matrix: from sklearn.metrics import confusion_matrix cm = confusion_matrix(labels, predicted_labels) Let's visualize it with Seaborn visualization library: import seaborn as sns; sns.set() ax = sns.heatmap(cm, annot=True, fmt="d", cmap="Blues") To compute the accuracy, one need to sum the values on the diagonal, but for clustering the rows (or columns) of the confusion matrix should be permuted to obtain the maximum value for the sum. This is not the case here. Let's see how to find the best permutation. ### Permutation maximizing the sum of the diagonal elements Scikit-learn provides a function linear_assignment to apply the Hungarian algorithm. One can use it to create our reordered confusion matrix cm2: from sklearn.utils.linear_assignment_ import linear_assignment import numpy as np def _make_cost_m(cm): s = np.max(cm) return (- cm + s) indexes = linear_assignment(_make_cost_m(cm)) js = [e[1] for e in sorted(indexes, key=lambda x: x[0])] cm2 = cm[:, js] Since linear_assigment minimizes a cost which should be positive, we defined the function _make_cost_m to transform the confusion matrix into a cost matrix. The reordered confusion matrix cm2 can be visualized: We are almost done. Let's finally compute the accuracy. ### Compute accuracy The accuracy is the sum of the diagonal elements divided by the number of samples: np.trace(cm2) / np.sum(cm2) Instead of implementing all this stuff ourselves, we could just use accuracy function provided by Coclust: from coclust.evaluation.external import accuracy accuracy(labels, predicted_labels) It should be noted that scikit-learn accuracy_score function does not give the accuracy score for the clustering since it does not permute the rows of the confusion matrix: from sklearn.metrics.classification import accuracy_score accuracy_score(labels, predicted_labels) gives the same value as: np.trace(cm) / np.sum(cm) ## Take away Here is a summary of the key elements of this post: • accuracy_score provided by scikit-learn is meant to deal with classification results, not clustering. • Computing accuracy for clustering can be done by reordering the rows (or columns) of the confusion matrix so that the sum of the diagonal values is maximal. • The linear assignment problem can be solved in $$O(n^3)$$ instead of $$O(n!)$$. • Coclust library provides an implementation of the accuracy for clustering results.	2023-03-22 03:25:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7043976187705994, "perplexity": 1287.5809126840923}, "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/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00776.warc.gz"}
https://repository.uantwerpen.be/link/irua/105214	Title Measurement of the underlying event in the Drell-Yan process in proton-proton collisions at $\sqrt{s}=7TeV$ Author Chatrchyan, S. Khachatryan, V. Sirunyan, A. M. Bansal, S. Cornelis, T. de Wolf, E.A. Janssen, X. Luyckx, S. Mucibello, L. Roland, B. Rougny, R. Selvaggi, M. van Haevermaet, H. Van Mechelen, P. Van Remortel, N. Van Spilbeeck, A. et al. Faculty/Department Faculty of Sciences. Physics Publication type article Publication 2012 Berlin , 2012 Subject Physics Source (journal) European physical journal : C : particles and fields. - Berlin Volume/pages 72(2012) :9 , p. 1-24 ISSN 1434-6044 Article Reference 2080 ISI 000309707300006 Carrier E-only publicatie Target language English (eng) Full text (Publishers DOI) Affiliation University of Antwerp Abstract A measurement of the underlying event (UE) activity in proton-proton collisions at a center-of-mass energy of 7 TeV is performed using Drell-Yan events in a data sample corresponding to an integrated luminosity of 2.2 fb(-1), collected by the CMS experiment at the LHC. The activity measured in the muonic final state (q (q) over bar -> mu(+) mu(-)) is corrected to the particle level and compared with the predictions of various Monte Carlo generators and hadronization models. The dependence of the UE activity on the dimuon invariant mass is well described by PYTHIA and HERWIG++ tunes derived from the leading jet/track approach, illustrating the universality of the UE activity. The UE activity is observed to be independent of the dimuon invariant mass in the region above 40 GeV/c(2), while a slow increase is observed with increasing transverse momentum of the dimuon system. The dependence of the UE activity on the transverse momentum of the dimuon system is accurately described by MADGRAPH, which simulates multiple hard emissions. Full text (open access) https://repository.uantwerpen.be/docman/irua/aacb0c/3614.pdf E-info http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000309707300006&DestLinkType=RelatedRecords&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000309707300006&DestLinkType=FullRecord&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000309707300006&DestLinkType=CitingArticles&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 Handle	2017-04-26 12:13:11	{"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": 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": 0.5857293009757996, "perplexity": 5584.248231534083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00326-ip-10-145-167-34.ec2.internal.warc.gz"}
https://live.poshenloh.com/zh/past-contests/amc10/2022A/solutions	### 2022 AMC 10A Solutions Problems used with permission of the Mathematical Association of America. Scroll down to view solutions, print PDF solutions, view answer key, or: Try Exam 1. What is the value of $3+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3}}}?$ $$\dfrac{31}{10}$$ $$\dfrac{49}{15}$$ $$\dfrac{33}{10}$$ $$\dfrac{109}{33}$$ $$\dfrac{15}{4}$$ ###### Solution(s): We can simplify this expression as follows: \begin{align} 3 +& \dfrac{1}{3 + \dfrac{1}{3 + \dfrac{1}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{1}{\dfrac{10}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{3}{10}} \\ &= 3 + \dfrac{1}{\dfrac{33}{10}} \\ &= 3 + \dfrac{10}{33} \\ &= \dfrac{109}{33}. \end{align} Thus, D is the correct answer. 2. Mike cycled $$15$$ laps in $$57$$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $$27$$ minutes? $$5$$ $$7$$ $$9$$ $$11$$ $$13$$ ###### Solution(s): We can set up a proportion to solve this problem: $\dfrac{15}{57} = \dfrac{x}{27}.$ Cross multiplying, we get $x = \dfrac{15}{57} \cdot 27 = \dfrac{135}{19} \approx 7.$ Thus, B is the correct answer. 3. The sum of three numbers is $$96.$$ The first number is $$6$$ times the third number, and the third number is $$40$$ less than the second number. What is the absolute value of the difference between the first and second numbers? $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ ###### Solution(s): Let $$x, y,$$ and $$z$$ be the three numbers. The conditions from the problem give us the following relations: \begin{gather} x + y + z = 96 \tag{(1)} \\ x = 6z \tag{(2)} \\ z = y - 40 \tag{(3)}. \end{gather} Rearranging $$(3),$$ we get $$y = z + 40.$$ Plugging this new equation and $$(2)$$ into $$(1),$$ we get $6z + z + 40 + z = 96$ $8z + 40 = 96$ $8z = 56 \Rightarrow z = 7.$ From this, we get that $x = 6 \cdot z = 6 \cdot 7 = 42$ and $y = z + 40 = 7 + 40 = 47.$ Therefore, $$y - x = 47 - 42 = 5.$$ Thus, E is the correct answer. 4. In some countries, automobile fuel efficiency is measured in liters per $$100$$ kilometers while other countries use miles per gallon. Suppose that $$1$$ kilometer equals $$m$$ miles, and $$1$$ gallon equals $$l$$ liters. Which of the following gives the fuel efficiency in liters per $$100$$ kilometers for a car that gets $$x$$ miles per gallon? $$\dfrac{x}{100lm}$$ $$\dfrac{xlm}{100}$$ $$\dfrac{lm}{100x}$$ $$\dfrac{100}{xlm}$$ $$\dfrac{100lm}{x}$$ ###### Solution(s): We can do the following conversions to get the desired answer. \begin{align} \dfrac{x \text{ mi}}{1 \text{ gal}} \cdot \dfrac{1 \text{ km}}{m \text{ mi}} &= \dfrac{x \text{ km}}{m \text{ gal}} \\ \dfrac{x \text{ km}}{m \text{ gal}} \cdot \dfrac{1 \text{ gal}}{l \text{ L}} &= \dfrac{x \text{ km}}{ml \text{ L}} \\ \dfrac{x \text{ km}}{ml \text{ L}} \cdot \dfrac{100 \text{ km}}{100 \text{ km}} &= \dfrac{100x \text{ km}}{100ml \text{ L}} \end{align} Taking the reciprocal, we find that the equivalent fuel efficiency would be $\dfrac{100ml \text{ L}}{100x \text{ km}} = \dfrac{\dfrac{100ml}{x} \text{ L}}{100 \text{ km}}.$ Thus, E is the correct answer. 5. Square $$ABCD$$ has side length $$1.$$ Points $$P,$$ $$Q,$$ $$R,$$ and $$S$$ each lie on a side of $$ABCD$$ such that $$APQCRS$$ is an equilateral convex hexagon with side length $$s.$$ What is $$s?$$ $$\dfrac{\sqrt{2}}{3}$$ $$\dfrac{1}{2}$$ $$2 - \sqrt{2}$$ $$1 - \dfrac{\sqrt{2}}{4}$$ $$\dfrac{2}{3}$$ ###### Solution(s): Consider the diagram: Since $$AP = QC = s,$$ we know that $$PB = PQ.$$ This shows that $$\triangle PBQ$$ is a right triangle. Using the Pythagorean theorem, we get that $$PB = \dfrac{s}{\sqrt{2}}.$$ We also know that $1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$ This equation simplifies to $1 = (1 + \dfrac{1}{\sqrt{2}})s$ Which implies that $s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$ We can rationalize this fraction to get $\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$ Thus, C is the correct answer. 6. Which expression is equal to $\left\|a-2-\sqrt{(a-1)^2}\right\|$ for $$a < 0?$$ $$3 - 2a$$ $$1 - a$$ $$1$$ $$a + 1$$ $$3$$ ###### Solution(s): By the definition of square root, we get that $\sqrt{(a - 1)^2} = \|a - 1\|.$ Since $$a < 0,$$ we get that $$a - 1 < 0,$$ which means that $$\|a - 1\| = 1 - a.$$ The whole expression therefore simplifies to $\|a - 2 - (1 - a)\| = \|2a - 3\|.$ Since $$a < 0,$$ we know that $$2a - 3 < 0.$$ This means that $$\|2a - 3\| = 3 - 2a.$$ Thus, A is the correct answer. 7. The least common multiple of a positive integer $$n$$ and $$18$$ is $$180,$$ and the greatest common divisor of $$n$$ and $$45$$ is $$15.$$ What is the sum of the digits of $$n?$$ $$3$$ $$6$$ $$8$$ $$9$$ $$12$$ ###### Solution(s): We can see that $$180 = 2^2 \cdot 3^2 \cdot 5$$ and $$18 = 2 \cdot 3^2.$$ This means that $$n$$ must include a factor of $$2^2$$ and $$5.$$ We also know that $$45 = 3^2 \cdot 5$$ and $$15 = 3 \cdot 5,$$ which means that $$n$$ has a factor of $$3^2$$ and $$5.$$ Therefore, we can set $n = 2^2 \cdot 3 \cdot 5 = 60.$ Thus, B is the correct answer. 8. A data set consists of $$6$$ (not distinct) positive integers: $$1,$$ $$7,$$ $$5,$$ $$2,$$ $$5,$$ and $$X.$$ The average (arithmetic mean) of the $$6$$ numbers equals a value in the data set. What is the sum of all positive values of $$X?$$ $$10$$ $$26$$ $$32$$ $$36$$ $$40$$ ###### Solution(s): The average of the $$6$$ numbers is $\dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}.$ This value can equal any of the terms in the set, so we can case on what it equals. $\dfrac{20 + X}{6} = 1 \iff X = -14$ $\dfrac{20 + X}{6} = 7 \iff X = 22$ $\dfrac{20 + X}{6} = 5 \iff X = 10$ $\dfrac{20 + X}{6} = 2 \iff X = -8$ $\dfrac{20 + X}{6} = X \iff X = 4$ Adding up all the positive values for $$X,$$ we get $$36.$$ Thus, D is the correct answer. 9. A rectangle is partitioned into $$5$$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible? $$120$$ $$270$$ $$360$$ $$540$$ $$720$$ ###### Solution(s): There are $$5$$ choices for the color of the bottom left rectangle. This forces there to be $$4$$ choices for the top left rectangle. The middle bottom rectangle touches both of the previous ones, so there are $$3$$ color options for this rectangle. The rectangle in the top right is also limited to $$3$$ colors since it touches the two previous rectangles. Finally, the rectangle in the bottom right also has $$3$$ color options. Multiplying these together, we get $$540$$ total colorings. Thus, D is the correct answer. 10. Daniel finds a rectangular index card and measures its diagonal to be $$8$$ centimeters. Daniel then cuts out equal squares of side $$1$$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $$4 \sqrt{2}$$ centimeters, as shown below. What is the area of the original index card? $$14$$ $$10 \sqrt{2}$$ $$16$$ $$12 \sqrt{2}$$ $$18$$ ###### Solution(s): We can label $$a$$ and $$b$$ as the width and height as in the diagram. Then we get that $a^2 + b^2 = 64$ and $(a - 2)^2 + (b - 2)^2 = 32.$ The latter expression simplifies to $a^2 + b^2 - 4a - 4b + 4 + 4 = 32,$ which is the same as $72 - 4(a + b) = 32.$ From this we get $a + b = 10.$ Squaring this, we get $a^2 + b^2 + 2ab = 100,$ which gets us that $2ab = 36,$ which means that the area $$(ab)$$ is $$18.$$ Thus, E is the correct answer. 11. Ted mistakenly wrote $2^m\cdot\sqrt{\dfrac{1}{4096}}$ as $2\cdot\sqrt[m]{\dfrac{1}{4096}}.$ What is the sum of all real numbers $$m$$ for which these two expressions have the same value? $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ ###### Solution(s): We can rewrite $$4096$$ as $$2^{12},$$ so $$\dfrac{1}{4096} = 2^{-12}.$$ Then if we equate the given expressions, we get $2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}.$ Equating the exponents, we get $m - 6 = 1 + \dfrac{-12}{m}.$ Multiplying by $$m,$$ we get $m^2 - 6m = m - 12$ and so $m^2 - 7m + 12 = 0$$(m-4)(m-3)$$m=4,~m=3$ Therefore, we can see that the sum of the solutions is $$7.$$ Thus, C is the correct answer. 12. On Halloween $$31$$ children walked into the principal's office asking for candy. They can be classified into three types: Some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order. "Are you a truth-teller?" The principal gave a piece of candy to each of the $$22$$ children who answered yes. "Are you an alternater?" The principal gave a piece of candy to each of the $$15$$ children who answered yes. "Are you a liar?" The principal gave a piece of candy to each of the $$9$$ children who answered yes. How many pieces of candy in all did the principal give to the children who always tell the truth? $$7$$ $$12$$ $$21$$ $$27$$ $$31$$ ###### Solution(s): For the first question, the truth-tellers will respond yes, the liars will respond yes, and the alternaters who decided to lie first will say yes. The alternaters who decide to tell the truth first will say no. Denote this as $22 = t + l + a_l.$ For the second questions, the liars will respond yes, and the alternaters who decided to lie first will say yes (they are forced to tell the truth for this question). The truth-tellers will respond no, and the alternaters who told the truth first would lie this round, responding no. Denote this as $15 = l + a_l.$ From this, we get that $$t = 7.$$ The principal only gives candy to children who always tell the truth in the first round, therefore only giving them $$7$$ candies total. Thus, A is the correct answer. 13. Let $$\triangle ABC$$ be a scalene triangle. Point $$P$$ lies on $$\overline{BC}$$ so that $$\overline{AP}$$ bisects $$\angle BAC.$$ The line through $$B$$ perpendicular to $$\overline{AP}$$ intersects the line through $$A$$ parallel to $$\overline{BC}$$ at point $$D.$$ Suppose $$BP = 2$$ and $$PC = 3.$$ What is $$AD?$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ ###### Solution(s): Consider the following diagram: By the Angle Bisector Theorem, we can label $$AB$$ as $$2x$$ and $$AY$$ as $$3x.$$ We also get that $$\triangle ABY$$ is isosceles since $$AP \perp BY.$$ Therefore, $$AY = AB = 2x.$$ Since $$AD$$ and $$BC$$ are parallel, we know that $\triangle BYC \sim \triangle DYA.$ $YC = AC - AY = 3x - 2x = x,$ so $$AY = 2YC.$$ Using similar triangles, we get that $$AD = 2BC = 10.$$ Thus, C is the correct answer. 14. How many ways are there to split the integers $$1$$ through $$14$$ into $$7$$ pairs such that in each pair, the greater number is at least $$2$$ times the lesser number? $$108$$ $$120$$ $$126$$ $$132$$ $$144$$ ###### Solution(s): We can see that the numbers $$1-8$$ must be in different pairs. $$7$$ must also be paired with $$14$$ since no other number is at least twice $$7.$$ Now let's look at what the other numbers can pair with. $$8$$ and $$9$$ can pair with any number $$1-4.$$ $$10$$ and $$11$$ can pair with any number $$1-5,$$ and $$12$$ and $$13$$ can pair with any number $$1-6.$$ $$8$$ can pair with $$4$$ numbers, but then $$9$$ only has $$3$$ options since $$8$$ took one. $$10$$ then has $$3$$ options, since $$2$$ choices are taken, but it has one more to choose from $$(5)$$. $$11$$ then has $$2$$ options, $$12$$ has $$2$$ options, and $$13$$ only has $$1.$$ Multiplying these together yields $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$ Thus, E is the correct answer. 15. Quadrilateral $$ABCD$$ with side lengths $$AB = 7, BC = 24, CD = 20,$$ $$DA = 15$$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $$\dfrac{a \pi - b}{c},$$ where $$a, b,$$ and $$c$$ are positive integers such that $$a$$ and $$c$$ have no common prime factor. What is $$a + b + c?$$ $$260$$ $$855$$ $$1235$$ $$1565$$ $$1997$$ ###### Solution(s): Notice that $$7^2 + 24^2$$ and $$15^2 + 20^2$$ are both the same. This forces $$AC = 25$$ since otherwise $$\angle B$$ and $$\angle D$$ would both be acute or obtuse, violating the fact that their sum is $$180^{\circ}.$$ Also since $$\angle B$$ is right, we know that $$AC$$ is the diameter of the circle. The area of the circle is then $$\dfrac{625}{4} \pi.$$ To find the area of the quadrilateral, we can find the area of each of the triangles, which is $\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$ $84 + 150 = 234.$ To find the area outside the quadrilateral, we subtract to get $\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$ Therefore, $a + b + c = 625 + 936 + 4$ $= 1565.$ Thus, D is the correct answer. 16. The roots of the polynomial $$10x^3 - 39x^2 + 29x - 6$$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $$2$$ units. What is the volume of the new box? $$\dfrac{24}{5}$$ $$\dfrac{42}{5}$$ $$\dfrac{81}{5}$$ $$30$$ $$48$$ ###### Solution(s): Let $$h, l,$$ and $$w$$ be the dimensions of the old box. Then the volume of the new box is $(h + 2)(l + 2)(w + 2).$ Expanding, we get $hlw + 2(hl + hw + lw)$$+ 4(l + h + w) + 8.$ We can use Vieta's formulas to find the terms in this expression. We get that $hlw = -\dfrac{D}{A} = \dfrac{3}{5},$ $hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$ and $l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$ Plugging these values into the expression, we get $\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$ Thus, D is the correct answer. 17. How many three-digit positive integers $$\underline{a} \ \underline{b} \ \underline{c}$$ are there whose nonzero digits $$a,b,$$ and $$c$$ satisfy $0.\overline{\underline{a}~\underline{b}~\underline{c}} = \dfrac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?$ (The bar indicates repetition, thus $$0.\overline{\underline{a}~\underline{b}~\underline{c}}$$ in the infinite repeating decimal $$0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$$) $$9$$ $$10$$ $$11$$ $$13$$ $$14$$ ###### Solution(s): Let's find a closed form expression for each of the repeating decimals. We can write $$0.\overline{\underline{a}~\underline{b}~\underline{c}}$$ as $0.\underline{a} \ \underline{b} \ \underline{c} + 0.000 \ \underline{a} \ \underline{b} \ \underline{c} + \cdots.$ From this, we can see that this an infinite geometric sequence with first term $$0.\underline{a} \ \underline{b} \ \underline{c}$$ and ratio $$\dfrac{1}{1000}.$$ Using the formula for the sum of a infinite geometric sequence, we get that this equals $\dfrac{0.\underline{a} \ \underline{b} \ \underline{c}}{1 - \dfrac{1}{1000}} = \dfrac{\underline{a} \ \underline{b} \ \underline{c}}{999}.$ Similarly, note that we can write $$0.\overline{a}$$ as $0.a + 0.0a + 0.00a + \cdots.$ As above, this equals $\dfrac{0.a}{1 - \dfrac{1}{10}} = \dfrac{a}{9}.$ Therefore, $0.\overline{b} = \dfrac{b}{9} \text{ and } 0.\overline{c} = \dfrac{c}{9}.$ Substituting all these values into the condition, we get $\dfrac{\underline{a} \ \underline{b} \ \underline{c}}{999} = \dfrac{1}{3} \cdot \dfrac{a + b + c}{9}.$ Multiplying through by $$999$$ yields $\underline{a} \ \underline{b} \ \underline{c} = 37(a + b +c).$ Note that we can express $$\underline{a} \ \underline{b} \ \underline{c}$$ as $$100a + 10b + c.$$ Substituting this in, we get $100a + 10b + c = 37(a + b + c),$ which simplifies to $63a = 27b + 36c \Rightarrow 7a = 3b + 4c.$ All the solutions where $$a = b = c$$ work. The expression $$3b + 4c$$ remains constant if we increase $$b$$ by $$4$$ and decrease $$c$$ by $$4.$$ We could also decrease $$b$$ by $$4$$ and increase $$c$$ by $$3.$$ Applying this principles to the first $$9$$ triples yields $(4, 8, 1), (5, 1, 8), (5, 9, 2), (6, 2, 9)$ as $$4$$ more solutions. Therefore, there are a total of $$13$$ solutions. Thus, D is the correct solution. 18. Let $$T_k$$ be the transformation of the coordinate plane that first rotates the plane $$k$$ degrees counterclockwise around the origin and then reflects the plane across the $$y$$-axis. What is the least positive integer $$n$$ such that performing the sequence of transformations $$T_1, T_2, T_3, \cdots, T_n$$ returns the point $$(1,0)$$ back to itself? $$359$$ $$360$$ $$719$$ $$720$$ $$721$$ ###### Solution(s): Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with. Let $$(r, \theta)$$ be a polar coordinate. Rotating this by $$k$$ degrees counterclockwise maps the point to $$(r, \theta + k^{\circ})$$ and then reflecting it maps it to $$(r, 180 - \theta - k^{\circ}.)$$ Therefore, we have that $T_k(r, \theta) = (r, 180 - \theta - k^{\circ}).$ From this, we can see that $T_{k + 1}(T_k(r, \theta)) =$$T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) =$$(r, \theta - 1^{\circ}).$ Now, let's analyze what happens to the point $$(1, 0^{\circ}).$$ After $$T_1,$$ we get $$(1, 179^{\circ}).$$ After $$T_2,$$ we get $$(1, -1^{\circ}).$$ After $$T_3,$$ we get $$(1, 178^{\circ}).$$ After $$T_4,$$ we get $$(1, -2^{\circ}).$$ $\vdots$ After $$T_{2n - 1},$$ we get $$(1, 180^{\circ} - k^{\circ}).$$ After $$T_{2n},$$ we get $$(1, -k^{\circ}).$$ From this, we can see that the first time the angle is back to $$0^{\circ}$$ is when $$n = 180$$ and $$k = 359.$$ Thus, A is the correct answer. 19. Define $$L_n$$ as the least common multiple of all the integers from $$1$$ to $$n$$ inclusive. There is a unique integer $$h$$ such that $\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} \cdots + \dfrac{1}{17} = \dfrac{h}{L_{17}}$ What is the remainder when $$h$$ is divided by $$17?$$ $$1$$ $$3$$ $$5$$ $$7$$ $$9$$ ###### Solution(s): We can combine all the addends on the left side into one fraction by making all their denominators $$17!.$$ $\dfrac{1}{1} + \dfrac{1}{2} + \cdots + \dfrac{1}{17} =$$\dfrac{\dfrac{17!}{1} + \dfrac{17!}{2} + \cdots + \dfrac{17!}{17}}{17!}$ Therefore, $h = L_{17} \dfrac{\dfrac{17!}{1} + \dfrac{17!}{2} + \cdots + \dfrac{17!}{17}}{17!}.$ We want to find $$h$$ mod $$17,$$ so we can analyze the expression for $$h$$ mod $$17.$$ Each of $\dfrac{\dfrac{17!}{1}}{17}, \dfrac{\dfrac{17!}{2}}{17!}, \cdots, \dfrac{\dfrac{17!}{16}}{17!}$ do not have a factor of $$17$$ in the denominator. This means that when we multiply each of them by $$L_{17},$$ the resulting product will have a factor of $$17$$ ($$17$$ divides $$L_{17}$$). That means all of these terms evaluate to $$0$$ mod $$17.$$ Now we just need to evaluate $$\dfrac{L_{17}}{17}$$ mod $$17.$$ To calculate $$L_{17},$$ note that it must contain the highest power of every prime less than or equal to $$17.$$ Therefore, $L_{17} = 16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17.$ Now we simplify: \begin{align} & 16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \; (\text{mod } 17)\\ &\equiv -1 \cdot 9 \cdot 35 \cdot 11 \cdot 13 \; (\text{mod } 17) \\ &\equiv 9 \cdot 11 \cdot -13 \; (\text{mod } 17) \\ &\equiv 99 \cdot 4 \; (\text{mod } 17) \\ &\equiv -3 \cdot 4 \; (\text{mod } 17) \\ &\equiv 5 \; (\text{mod } 17) \end{align} Thus, C is the correct answer. 20. A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $$57,$$ $$60,$$ and $$91.$$ What is the fourth term of this sequence? $$190$$ $$194$$ $$198$$ $$202$$ $$206$$ ###### Solution(s): Let the arithmetic sequence be $a, a + d, a + 2d, a + 3d$ and the geometric sequence be $b, br, br^2, br^3.$ Then $a + b = 57 \tag{(1)},$ $a + d + br = 60 \tag{(2)},$ and $a + 2d + br^2 = 91 \tag{(3)}.$ Subtracting $$(1)$$ from $$(2)$$ and $$(2)$$ from $$(3),$$ we get $d + b(r - 1) = 3$ and $d + br(r - 1) = 31.$ Subtracting these, we get $b(r - 1)^2 = 28.$ Since every variable is an integer, we get that $$b = 28$$ or $$b = 7.$$ If $$b = 28,$$ then $$r = 2,$$ $$a = 29,$$ and $$d = 25.$$ This forces the arithmetic sequence to be $29, 4, -21, -46,$ which is a contradiction. Therefore, $$b = 7.$$ Then $$r = 3,$$ $$a = 50,$$ and $$d = -11.$$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ The desired answer is $$17 + 189 = 206.$$ Thus, E is the correct answer. 21. A bowl is formed by attaching four regular hexagons of side $$1$$ to a square of side $$1.$$ The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? $$6$$ $$7$$ $$5 + 2 \sqrt{2}$$ $$8$$ $$9$$ ###### Solution(s): We can extend line segments $$l, m,$$ and $$n$$ as follows. They are concurrent since $$l$$ and $$m$$ intersect and $$l$$ and $$n$$ intersect (they are on the same plane). Since $$l$$ only intersects the plane of $$m$$ and $$n$$ once, it must intersect them both at that one point. The dashed red lines create equilateral triangles on the lateral faces of the bowl, which all have side length $$1.$$ In the top plane, we know that $$m \perp n,$$ so the dashed red lines create an isosceles right triangle with leg length $$1.$$ The octagon looks like the diagram below. The area of the octagon is the area of the square minus each of the four corner triangles. This is equal to $3^2 - 4(\dfrac{1}{2} \cdot 1^2) = 7.$ Thus, B is the correct answer. 22. Suppose that $$13$$ cards numbered $$1, 2, 3, \cdots, 13$$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $$1, 2, 3$$ are picked up on the first pass, $$4$$ and $$5$$ on the second pass, $$6$$ on the third pass, $$7, 8, 9, 10$$ on the fourth pass, and $$11, 12, 13$$ on the fifth pass. For how many of the $$13!$$ possible orderings of the cards will the $$13$$ cards be picked up in exactly two passes? $$4082$$ $$4095$$ $$4096$$ $$8178$$ $$8191$$ ###### Solution(s): Let $$n$$ be the number of cards picked up on the first pass, where $$1 \leq n \leq 12.$$ If we choose the spaces that the $$n$$ cards occupy, the positions of the remaining cards are determined since they must be placed in order. There are $$\binom{13}{n}$$ ways to choose where the $$n$$ cards go, but if the $$n$$ cards are placed at the very beginning, then all the cards will be picked up on the first pass. Therefore, for a given $$n$$ there are $$\binom{13}{n} - 1$$ ways to arrange the cards. We need to now find the sum over all possible $$n,$$ which equals $\sum_{i = 1}^{12} \binom{13}{n} - 1 =$$\left(\sum_{i = 0}^{13} \binom{13}{n}\right) - 14 = 2^{13} - 14$$= 8192 - 14 = 8178.$ Thus, D is the correct answer. 23. Isosceles trapezoid $$ABCD$$ has parallel sides $$\overline{AD}$$ and $$\overline{BC},$$ with $$BC < AD$$ and $$AB = CD.$$ There is a point $$P$$ in the plane such that $$PA=1, PB=2, PC=3,$$ and $$PD=4.$$ What is $$\tfrac{BC}{AD}?$$ $$\dfrac{1}{4}$$ $$\dfrac{1}{3}$$ $$\dfrac{1}{2}$$ $$\dfrac{2}{3}$$ $$\dfrac{3}{4}$$ ###### Solution(s): Let $$P'$$ be the reflection of $$P$$ across the perpendicular bisector of $$\overline{BC}.$$ This forms two new isosceles trapezoids: $$CBPP'$$ and $$DAPP'.$$ Therefore, we get \begin{gather} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3. \end{gather} Using Ptolemy's theorem, we know that the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: \begin{gather} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9. \end{gather} This gets us $$PP' \cdot AD = 15$$ and $$PP' \cdot BC = 5.$$ Dividing these two equations yields $$\dfrac{BC}{AD} = \dfrac{1}{3}.$$ Thus, B is the correct answer. 24. How many strings of length $$5$$ formed from the digits $$0,$$ $$1,$$ $$2,$$ $$3,$$ $$4,$$ are there such that for each $$j \in \{1,2,3,4\},$$ at least $$j$$ of the digits are less than $$j?$$ (For example, $$02214$$ satisfies this condition because it contains at least $$1$$ digit less than $$1,$$ at least $$2$$ digits less than $$2,$$ at least $$3$$ digits less than $$3,$$ and at least $$4$$ digits less than $$4.$$ The string $$23404$$ does not satisfy the condition because it does not contain at least $$2$$ digits less than $$2.$$) $$500$$ $$625$$ $$1089$$ $$1199$$ $$1296$$ ###### Solution(s): Note that there must be at least one $$0$$ to satisfy the condition. We can proceed by casework on the number of distinct digits in the string. $$1$$ digit The only possible digit is just $$0.$$ This can be done in $$1$$ way. $$2$$ digits This gives us a total of $20 + 30 + 20 + 5 = 75$ strings. $$3$$ digits This gives us a total of $120 + 150 + 90 + 60 + 60$$+ 20 = 500$ strings. $$4$$ digits This gives us a total of $10 \cdot \dfrac{5!}{2! \cdot 1! \cdot 1! \cdot 1!} = 600$ strings. $$5$$ digits This gives us $$5! = 120$$ strings. All together, we have $1 + 75 + 500 + 600$ $+ 120 = 1296$ strings. Thus, E is the correct answer. 25. Let $$R,$$ $$S,$$ and $$T$$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $$x$$-axis. The left edge of $$R$$ and the right edge of $$S$$ are on the $$y$$-axis, and $$R$$ contains $$\dfrac{9}{4}$$ as many lattice points as does $$S.$$ The top two vertices of $$T$$ are in $$R \cup S,$$ and $$T$$ contains $$\dfrac{1}{4}$$ of the lattice points contained in $$R \cup S.$$ See the figure (not drawn to scale). The fraction of lattice points in $$S$$ that are in $$S \cap T$$ is $$27$$ times the fraction of lattice points in $$R$$ that are in $$R \cap T.$$ What is the minimum possible value of the edge length of $$R$$ plus the edge length of $$S$$ plus the edge length of $$T?$$ $$336$$ $$337$$ $$338$$ $$339$$ $$340$$ ###### Solution(s): Let $$r$$ be the number of lattice points on the side length of $$R.$$ Similarly define $$s$$ for $$S$$ and $$t$$ for $$T.$$ Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length. The first conditions gives us that $r^2 = \dfrac{9}{4} \cdot s^2$$r = \dfrac{3}{2} \cdot s \tag{(1)}$ The number of lattice points in $$R \cup T$$ is the sum of the lattice points in each of the regions, but there is overlap along the $$y$$-axis where $$S$$ touches it. The second condition, therefore, yields $t^2 = \dfrac{1}{4}(r^2 + s^2 - s)$ $t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s)$ $t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4}$ $16t^2 = s(13s - 4).$ From $$(1),$$ we get that $$s$$ is a multiple of $$2.$$ We can substitute $$s$$ with $$2j$$ to get $16t^2 = 2j(26j - 4)$ $4t^2 = j(13j - 2).$ For the product to be divisible by $$4,$$ $$j$$ must be divisible by $$2.$$ We can again substitute $$j$$ with $$2k$$ to get $4t^2 = 2k(26k - 2)$ $t^2 = k(13k - 1) \tag{(2)}$ Let $$x$$ be the number of lattice points along the bottom of the rectangle formed by $$S \cap T$$ and $$y$$ be the number of lattice points along the bottom of the rectangle formed by $$R \cap T.$$ Using these variable, we get that the number of lattice points in $$S \cap T$$ is $$xt$$ and in $$R \cap T$$ is $$yt.$$ The third condition gives us that $\dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2}$ $\dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} x = 12y.$ We also know that $$t = x + y - 1$$ (accounting for overlap), and this yields $t = 13y - 1 \tag*{(3)}$ $$(3)$$ gives us that $t \equiv -1 \bmod{13} t^2 \equiv 1 \bmod{13}.$ However, by $$(2),$$ we get that $t^2 \equiv k \cdot -1 \bmod{13}$ $k \equiv -1 \bmod{13}.$ By $$(2),$$ we also get that $$k$$ is a perfect square since it is relatively prime to $$13k - 1,$$ and they must multiply to a perfect square. Using these restrictions on $$k,$$ we can try to find the smallest $$k$$ that works. We get that $$k = 25$$ satisfies both conditions. From this value of $$k,$$ we get that $$j = 2 \cdot 25 = 50,$$ $$s = 2 \cdot 50 = 100,$$ and $$r = \dfrac{3}{2} \cdot 100 = 150.$$ We can also find that $t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324$ $t = 5 \cdot 18 = 90.$ Therefore, $r + s + t = 340.$ The question, however, asked for the sum of the side lengths. The side lengths of the squares are $$1$$ less than the number of lattice points on the side, so we have to subtract $$3.$$ Therefore, the desired answer is $$340 - 3 = 337.$$ Thus, B is the correct answer.	2023-01-29 16:12:17	{"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": 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": 5, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9937453866004944, "perplexity": 288.951921113647}, "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/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00453.warc.gz"}
https://www.jobilize.com/online/course/4-2-the-binary-filter-tree-filter-banks-and-wavelets-by-openstax?qcr=www.quizover.com	# 4.2 The binary filter tree Page 1 / 1 This module introduces the binary filter tree. Recall that for image compression (see The 2-band Filter Bank ), the purpose of the 2-band filter bank in the Haar transform is tocompress most of the signal energy into the low-frequency band. We may achieve greater compression if the low brand is further split into two. This may be repeated a number of times to givethe binary filter tree, shown with 4 levels in . In 1-D, this is analogous to the way the 2-D Haar transform was extended to the multi-level Haar transform . For an $N$ -sample input vector $x$ , the sizes and bandwidths of the signals of the 4-level filter tree are: Signal No. of samples Approximate pass band $x$ $N$ $0\to \frac{1}{2}{f}_{s}$ ${y}_{1}$ $\frac{N}{2}$ $\frac{1}{4}\to \frac{1}{2}{f}_{s}$ ${y}_{01}$ $\frac{N}{4}$ $\frac{1}{8}\to \frac{1}{4}{f}_{s}$ ${y}_{001}$ $\frac{N}{8}$ $\frac{1}{16}\to \frac{1}{8}{f}_{s}$ ${y}_{0001}$ $\frac{N}{16}$ $\frac{1}{32}\to \frac{1}{16}{f}_{s}$ ${y}_{0000}$ $\frac{N}{16}$ $0\to \frac{1}{32}{f}_{s}$ Because of the downsampling (decimation) by 2 at each level, the total number of output samples = $N$ , regardless of the number of levels in the tree. The ${H}_{0}$ filter is normally designed to be a lowpass filter with a passband from 0 to approximately $\frac{1}{4}$ of the input sampling frequency for that stage; and ${H}_{1}$ is a highpass (bandpass) filter with a pass bandapproximately from $\frac{1}{4}$ to $\frac{1}{2}$ of the input sampling frequency. When formed into a 4-level tree, the filter outputs have the approximate pass bands given in . The final output ${y}_{0000}$ is a lowpass signal, while the other outputs are all bandpass signals, each covering a band of approximately oneoctave. An inverse tree, mirroring , may be constructed using filters ${G}_{0}$ and ${G}_{1}$ instead of ${H}_{0}$ and ${H}_{1}$ , as shown for just one level in part (b) of this figure . If the PR conditions of this previous equation and this previous equation are satisfied, then the output of each level will be identical to the input of the equivalent level in , and the final output will be a perfect reconstruction of the input signal. ## Multi-rate filtering theorem To calculate the impulse and frequency responses for a multistage network with downsampling at each stage, as in , we must first derive an important theorem for multirate filters. The downsample-filter-upsample operation of (a) is equivalent to either the filter-downsample-upsample operation of (b) or the downsample-upsample-filter operation of (c), if the filter is changed from $H(z)$ to $H(z^{2})$ . From (a): Take z-transforms: Reverse the order of summation and let $m=n-2i$ : therefore, where $Y(z)=H(z^{2})X(z)$ This describes the operations of (b). Hence the first result is proved. The result from line 3 in $\stackrel{^}{Y}(z)=\frac{1}{2}(X(z)+X(-z))H(z^{2})=\stackrel{^}{X}(z)H(z^{2})$ shows that the filter $H(z^{2})$ may be placed after the down/up-sampler as in (c), which proves the second result. ## General results for m:1 subsampling It can be shown that: • $H(z)$ becomes $H(z^{M})$ if shifted ahead of an M:1 downsampler or following an M:1 upsampler. • M:1 down/up-sampling of a signal $X(z)$ produces: $\stackrel{^}{X}(z)=\frac{1}{M}\sum_{m=0}^{M-1} X(ze^{\frac{i\times 2\pi m}{M}})$ ## Transformation of the filter tree Using the result of , can be redrawn as in with all downsamplers moved to the outputs. (Note requires much more computation than .) We can now calculate the transfer function to each output (before the downsamplers) as: ${H}_{01}(z)={H}_{0}(z){H}_{1}(z^{2})$ ${H}_{001}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{1}(z^{4})$ ${H}_{0001}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{0}(z^{4}){H}_{1}(z^{8})$ ${H}_{0000}(z)={H}_{0}(z){H}_{0}(z^{2}){H}_{0}(z^{4}){H}_{0}(z^{8})$ In general the transfer functions to the two outputs at level $k$ of the tree are given by: ${H}_{k,1}=\prod_{i=0}^{k-2} {H}_{0}(z^{2^{i}}){H}_{1}(z^{2^{(k-1)}})$ ${H}_{k,0}=\prod_{i=0}^{k-1} {H}_{0}(z^{2^{i}})$ For the Haar filters of this equation and this equation from our discussion of the 2-band filter bank, the transfer functionsto the outputs of the 4-level tree become: ${H}_{01}(z)=\frac{1}{2}(z^{-3}+z^{-2}-z^{-1}+1)$ ${H}_{001}(z)=\frac{1}{2\sqrt{2}}(z^{-7}+z^{-6}+z^{-5}+z^{-4}-z^{-3}+z^{-2}+z^{-1}+1)$ ${H}_{0001}(z)=\frac{1}{4}(z^{-15}+z^{-14}+z^{-13}+z^{-12}+z^{-11}+z^{-10}+z^{-9}+z^{-8}-z^{-7}+z^{-6}+z^{-5}+z^{-4}+z^{-3}+z^{-2}+z^{-1}+1)$ ${H}_{0000}(z)=\frac{1}{4}(z^{-15}+z^{-14}+z^{-13}+z^{-12}+z^{-11}+z^{-10}+z^{-9}+z^{-8}+z^{-7}+z^{-6}+z^{-5}+z^{-4}+z^{-3}+z^{-2}+z^{-1}+1)$ are nano particles real yeah Joseph Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master? no can't Lohitha where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine has a lot of application modern world Kamaluddeen yes narayan what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!	2021-05-08 19:57:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 52, "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": 0.6311051249504089, "perplexity": 2040.122501998405}, "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-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00239.warc.gz"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=17&t=46803	## Homework Problem A11 H-Atom ($E_{n}=-\frac{hR}{n^{2}}$) Michael Du 1E Posts: 117 Joined: Sun Sep 22, 2019 12:16 am ### Homework Problem A11 For this problem, the question is "In the spectrum of atomic hydrogen, several lines are generally classified together as belonging to a series (For example, Balmer series or Lyman series). What is common to the lines within a series that makes grouping them together logical?". Can anyone please explain to me why the lower energy level is n =1 for the Lyman series and n = 2 for the Balmer series (according to the solutions manual)? thank you! Deepika Reddy 1A Posts: 125 Joined: Thu Jul 11, 2019 12:15 am ### Re: Homework Problem A11 I believe the Lyman series is the ultraviolet one, which is the part of the electromagnetic spectrum with the highest energy/frequency. Since n=1 has the farthest gap from the other levels, an electron requires a lot more energy to jump up or has to emit a lot more energy to drop down, so it corresponds to the Lyman series. AKatukota Posts: 100 Joined: Thu Jul 25, 2019 12:18 am ### Re: Homework Problem A11 For the Lyman series, the lower energy level is n=1. For the Balmer series it is n=2. Sydney Myers 4I Posts: 100 Joined: Fri Aug 09, 2019 12:17 am ### Re: Homework Problem A11 The gaps between energy levels get larger the further from the nucleus you go, so the jumps (or emissions) with the largest energy will have to do with the largest gap because it would require the most energy to jump. Since the largest gap will always involve n=1, it is said to be associated with the Lyman series. The Lyman series is the set of emissions with the highest energy and therefore smallest wavelength (they're classified as uv radiation). As for the Balmer series, it's the grouping of emissions with the second shortest wavelength (and second highest energy) and it's associated with the second largest jump an electron can make. The second largest gap is from n=3 to n=2, so this series involves the energy level of n=2.	2020-11-25 23:24:15	{"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": 0.6921942234039307, "perplexity": 1103.8716033898374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141184870.26/warc/CC-MAIN-20201125213038-20201126003038-00656.warc.gz"}
http://meetings.aps.org/Meeting/MAR07/Event/59667	### Session L27: Glassy Systems 2:30 PM–5:06 PM, Tuesday, March 6, 2007 Chair: Dimitrios Papaconstantopoulos, George Mason University Abstract ID: BAPS.2007.MAR.L27.11 ### Abstract: L27.00011 : X-ray and Neutron Scattering Studies of Methyl Iodide Confined in GelTech$_{\copyright}$ Glass 4:30 PM–4:42 PM Preview Abstract MathJax On \| Off Abstract #### Authors: Yvonne Glanville (Penn State University) Paul Sokol (Indiana University Cyclotron Facility) Steven Ehrlich (National Synchotron Light Source) X-ray diffraction and neutron scattering studies of methyl iodide confined in 200 {\AA} GelTech$_{\copyright }$ glass have revealed a never before observed intermediate solid phase of methyl iodide. Bulk methyl iodide has one phase transition below room temperature, at 207 K where it transitions from a liquid to an orthorhombic solid. Neutron scattering studies of the diffusion of methyl iodide confined in the 200 {\AA} pores show a transition from a liquid to a solid at 203 K. X-ray diffraction measurements support this finding and identify the transition as one from a liquid to the never before observed amorphous solid. The amorphous solid remains down to 168 K upon cooling at which point a second transition appears from the amorphous solid to an orthorhombic solid, which upon indexing is identical to the bulk. To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2007.MAR.L27.11	2013-05-21 19:15:18	{"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": 0.6828433871269226, "perplexity": 10392.821960468558}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700477029/warc/CC-MAIN-20130516103437-00090-ip-10-60-113-184.ec2.internal.warc.gz"}
https://ptfonseca.github.io/inspector/	## Overview The inspector package provides utility functions that implement and automate common sets of validation tasks, namely: • inspect_prob() checks if an object is a numeric vector of valid probability values. • inspect_log_base() checks if an object is a valid logarithmic base. • inspect_true_or_false() checks if an object is a non-missing logical value. • inspect_bfactor() checks if an object is a numeric vector of valid Bayes factors values. • inspect_bfactor_log() checks if an object is a numeric vector of valid logarithmic Bayes factors values. • inspect_bfactor_scale() validates Bayes factor interpretation scales (from the pcal package). • inspect_categories() validates factor levels. • inspect_character() validates character vectors. • inspect_character_match() validates character values with predefined allowed values. • inspect_data_dichotomous() validates dichotomous data • inspect_data_categorical() and inspect_data_cat_as_dichotom() validate categorical data. • inspect_par_bernoulli() validates parameters for the Bernoulli and Binomial distributions. • inspect_par_multinomial() validates parameters for the Multinomial distribution. • inspect_par_beta() validates parameters for the Beta distribution. • inspect_par_dirichlet() validates parameters for the Dirichlet distribution. • inspect_par_haldane() validates parameters for the Haldane distribution. These functions are particularly useful to validate inputs, intermediate objects and output values in user-defined functions, resulting in tidier and less verbose functions. ## Installation The development version of inspector can be installed from Github with the devtools package: # install.packages("devtools") devtools::install_github("ptfonseca/inspector") ## Usage Imagine we want to write a function that simulates n flips of the same coin. Assuming that bias is the probability of the “heads” outcome: set.seed(123) flip_coins <- function(n, bias) { sample(x = c("heads", "tails"), size = n, replace = TRUE) } flip_coins(n = 5, bias = 0.5) #> [1] "heads" "heads" "heads" "tails" "heads" Since bias is a probability, it is natural that we require flip_coins() to only accept values of bias between 0 and 1. Furthermore, we may want to ensure that bias is not null, not missing, and is a numeric vector of length 1. This results an a quite verbose function body: set.seed(123) flip_coins <- function(n, bias) { if (is.null(bias)) { stop(paste("Invalid argument: bias is NULL.")) } if (any(isFALSE(is.atomic(bias)), isFALSE(is.vector(bias)))) { stop(paste("Invalid argument: bias must be an atomic vector.")) } if (isFALSE(length(bias) == 1)) { stop(paste("Invalid argument: bias must be of length 1.")) } if (is.na(bias)) { stop(paste("Invalid argument: bias is NA or NaN.")) } if (isFALSE(is.numeric(bias))) { stop(paste("Invalid argument: bias must be numeric.")) } if (any(bias >= 1, bias <= 0)) { stop(paste("Invalid argument: bias must be in the (0, 1) interval.")) } sample(x = c("heads", "tails"), size = n, replace = TRUE) } flip_coins(n = 5, bias = 0.5) #> [1] "heads" "heads" "heads" "tails" "heads" The inspector package was built to automate this kind of validation task. In the flip_coins() example, to perform an equivalent validation of inputs we can use inspect_par_bernoulli, since bias is the parameter of a Bernoulli distribution: set.seed(123) flip_coins <- function(n, bias) { inspect_par_bernoulli(bias) sample(x = c("heads", "tails"), size = n, replace = TRUE) } flip_coins(n = 5, bias = 0.5) #> [1] "heads" "heads" "heads" "tails" "heads" This results in a tidier function body since the validation of bias is abstracted away from the body of the function. Now imagine we want to implement equation 4 from Berger and Delampady (1987), a formula that calculates posterior probabilities using prior probabilities and Bayes factors as input. In this case we need to validate a vector of Bayes factors, lets call it bf, and a vector of prior probabilities, lets call it prior_prob. Since bf is expected to contain valid Bayes factor values, we need to ensure that only non-empty numeric vectors, containing only non-negative values, are accepted. Since prior_prob is a vector of probabilities, we need to check if it is a non-empty numeric vector containing only values between 0 and 1. Since we are now validating two inputs, the function body would be even more verbose than in the flip_coins() example: bfactor_to_prob <- function(bf, prior_prob = .5) { if (is.null(bf)) { stop(paste("Invalid argument: bf is NULL.")) } if (any(isFALSE(is.atomic(bf)), isFALSE(is.vector(bf)))) { stop(paste("Invalid argument: bf must be an atomic vector.")) } if (length(bf) == 0) { stop(paste("Invalid argument: bf is empty.")) } if (all(is.na(bf))) { stop(paste("Invalid argument: all elements of bf are NA or NaN.")) } if (isFALSE(is.numeric(bf))) { stop(paste("Invalid argument: the type of bf must be numeric.")) } if (any(bf[!is.na(bf)] < 0)) { stop(paste("Invalid argument: all elements of bf must be non-negative.")) } if (is.null(prior_prob)) { stop(paste("Invalid argument:", output_name, "is NULL.")) } if (any(isFALSE(is.atomic(prior_prob)), isFALSE(is.vector(prior_prob)))) { stop(paste("Invalid argument:", output_name, "must be an atomic vector.")) } if (length(prior_prob) == 0) { stop(paste("Invalid argument:", output_name, "is empty.")) } if (all(is.na(prior_prob))) { stop(paste("Invalid argument: all elements of", output_name, "are NA or NaN.")) } if (isFALSE(is.numeric(prior_prob))) { stop(paste("Invalid argument: the type of", output_name, "must be numeric.")) } if (any(prior_prob[!is.na(prior_prob)] < 0, prior_prob[!is.na(prior_prob)] > 1)) { stop(paste("Invalid argument: all elements of", output_name, "must be in the [0, 1] interval.")) } (1 + (1 - prior_prob) / prior_prob * (1 / bf)) ^(-1) } bfactor_to_prob(c(2.1, 0.5, 11)) #> [1] 0.6774194 0.3333333 0.9166667 Now lets use inspector instead. To perform an equivalent validation of inputs we can use inspect_bfactor() to validate bf and inspect_prob() to validate prior_prob: bfactor_to_prob <- function(bf, prior_prob = .5) { inspect_bfactor(bf) inspect_prob(prior_prob) (1 + (1 - prior_prob) / prior_prob * (1 / bf)) ^ (-1) } bfactor_to_prob(c(2.1, 0.5, 11)) #> [1] 0.6774194 0.3333333 0.9166667 ## Getting Help If you find a bug, please file an issue with a minimal reproducible example on GitHub. Feature requests are also welcome. You can find me at . ## References Berger, James O., and Mohan Delampady. 1987. “Testing Precise Hypotheses.” Statistical Science 2 (3): 317–35.	2021-10-19 02:15:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.40087369084358215, "perplexity": 11320.95065482545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585231.62/warc/CC-MAIN-20211019012407-20211019042407-00179.warc.gz"}
http://ctan.mirror.garr.it/cran/web/packages/trialr/vignettes/LevyCaseStudy.html	CRM Case Study 1 - Levy et al (2006) 2020-04-06 The Continual Reassessment Method (CRM) for dose-finding is introduced in its own article. This vignette reproduces the analysis in a real dose-finding trial that used the CRM by Levy et al. (2006). The chosen example is particularly attractive because the authors present sequential analyses of their dose decisions in their Table 1, allowing us the opportunity to reproduce their analysis. The authors investigate five doses of semisynthetic homoharringtonine (ssHHT), seeking a dose associated with a dose-limiting toxicity (DLT) probability of approximately 33%. The specific doses and the investigators’ initial beliefs on the probabilities of DLT are: library(dplyr) tibble( Dose-level = 1:5, Dose (mg/m2 per day) = c(0.5, 1, 3, 5, 6), Prior Pr(DLT) = c(0.05, 0.1, 0.15, 0.33, 0.5) ) %>% knitr::kable() Dose-level Dose (mg/m2 per day) Prior Pr(DLT) 1 0.5 0.05 2 1.0 0.10 3 3.0 0.15 4 5.0 0.33 5 6.0 0.50 We are not told the exact details of the model form or the parameter prior(s) used. However, we are told that the BPCT software was used to calculate the recommended doses. Zohar et al. (2003) describe the software as using a one-parameter logistic model with fixed-value intercept taking possible values 1, 2, 3 or 4, and exponential or uniform priors on the gradient term with expected value 0.5, 1 or 2. For the purposes of this article, we assume that the investigators used a one-parameter logistic CRM with the intercept term fixed at $$a_0 = 1$$ and an Exponential(1) = Gamma(1, 1) prior on $$\beta$$. In the appendix, we describe an attempt to identify the exact parameterisation used by comparing a broad array of model fits. Prior beliefs library(trialr) target <- 0.33 skeleton <- c(0.05, 0.1, 0.15, 0.33, 0.5) fit0 <- crm_prior_beliefs(skeleton, target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1) fit0 #> No patients have been treated. #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 0 0 0.243 0.147 0.3703 #> 2 2 0.10 0 0 0.282 0.225 0.0658 #> 3 3 0.15 0 0 0.313 0.286 0.1070 #> 4 4 0.33 0 0 0.412 0.451 0.1830 #> 5 5 0.50 0 0 0.520 0.575 0.2740 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 3. #> The dose most likely to be the MTD is 1. #> Model entropy: 1.45 Notice that both the prior mean (ProbTox) and median (MedianProbTox) probabilities of toxicity are quite far from the skeleton. In an ordinary regression model, fixing the intercept will affect the gradient. The same is true here: library(tidyr) library(purrr) library(ggplot2) get_prior_fit <- function(a0) { crm_prior_beliefs(skeleton, target, model = 'logistic_gamma', a0 = a0, beta_shape = 1, beta_inverse_scale = 1) } tibble(a0 = c(-1, 0, 1, 2, 3, 4)) %>% mutate(Mod = map(a0, get_prior_fit)) %>% mutate( Dose = Mod %>% map("dose_indices"), ProbTox = Mod %>% map("prob_tox"), ) %>% select(-Mod) %>% unnest(cols = c(Dose, ProbTox)) %>% mutate(a0 = factor(a0)) %>% ggplot(aes(x = Dose, y = ProbTox, group = a0, col = a0)) + geom_line() + ylim(0, 1) + labs(title = 'Prior Prob(DLT) location is affected by the fixed intercept, a0') Please note that the behaviour of unnest changed in v1.0 of the tidyr package. For older versions, the command should be: tibble(a0 = c(-1, 0, 1, 2, 3, 4)) %>% mutate(Mod = map(a0, get_prior_fit)) %>% mutate( Dose = Mod %>% map("dose_indices"), ProbTox = Mod %>% map("prob_tox"), ) %>% select(-Mod) %>% unnest(cols = c(Dose, ProbTox)) %>% mutate(a0 = factor(a0)) %>% ggplot(aes(x = Dose, y = ProbTox, group = a0, col = a0)) + geom_line() + ylim(0, 1) + labs(title = 'Prior Prob(DLT) location is affected by the fixed intercept, a0') Cohort 1 The trial starts at the lowest dose, 0.5 mg/m2 per day. Three patients are treated and none experiences DLT. fit1 <- stan_crm(outcome_str = '1NNN', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit1 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.083 0.0214 0.0865 #> 2 2 0.10 0 0 0.117 0.0517 0.0398 #> 3 3 0.15 0 0 0.149 0.0876 0.1120 #> 4 4 0.33 0 0 0.266 0.2520 0.2940 #> 5 5 0.50 0 0 0.414 0.4446 0.4677 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 5. #> Model entropy: 1.30 We see that the cohort has a great bearing on the predicted rates of toxicity. The trialists report estimated DLT probabilities of 0.001, 0.003, 0.006, 0.035, 0.11. The values calculated by trialr diverge somewhat from those values. It is difficult to know why this is given such a small sample size. Dose-level 5 is recommended for the next cohort. The trialists understandably resist the desire to skip three doses in going from the lowest to highest dose, electing instead to skip the second dose and treat the next cohort at dose-level 3. Cohort 2 This cohort of three was treated at 3 mg/m2 per day. One DLT was seen. fit2 <- stan_crm(outcome_str = '1NNN 3NNT', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit2 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> 4 4 3 0 1 #> 5 5 3 0 1 #> 6 6 3 1 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.147 0.101 0.1588 #> 2 2 0.10 0 0 0.203 0.171 0.0963 #> 3 3 0.15 3 1 0.250 0.230 0.2190 #> 4 4 0.33 0 0 0.397 0.407 0.3842 #> 5 5 0.50 0 0 0.533 0.548 0.1417 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 4. #> Model entropy: 1.49 The trialists report Prob(DLT) = (0.07, 0.14, 0.19, 0.39, 0.55). The median estimates from trialr are now quite close to these values. Despite the observation of a DLT and the observed DLT-rate at dos-level 3 matching the target of 33%, the model advocates escalation to dose-level 4. That is what the investigators did. Cohort 3 This cohort of three was treated at dose-level 4, which corresponds to 5 mg/m2 per day. Once again, one DLT was seen. fit3 <- stan_crm(outcome_str = '1NNN 3NNT 4NNT', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit3 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> 4 4 3 0 1 #> 5 5 3 0 1 #> 6 6 3 1 1 #> 7 7 4 0 1 #> 8 8 4 0 1 #> 9 9 4 1 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.129 0.0904 0.1027 #> 2 2 0.10 0 0 0.186 0.1569 0.0875 #> 3 3 0.15 3 1 0.235 0.2153 0.2370 #> 4 4 0.33 3 1 0.389 0.3935 0.4557 #> 5 5 0.50 0 0 0.531 0.5402 0.1170 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 4. #> Model entropy: 1.40 The trialists report Prob(DLT) = (0.07, 0.13, 0.19, 0.38, 0.54) after this cohort, barely shifting from the previous estimates. This time the model advocates to remain at dose-level 4 and that is exactly what the trialists did. Cohort 4 This cohort of three was also treated at dose-level 4. No DLTs were seen in this cohort. fit4 <- stan_crm(outcome_str = '1NNN 3NNT 4NNT 4NNN', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit4 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> 4 4 3 0 1 #> 5 5 3 0 1 #> 6 6 3 1 1 #> 7 7 4 0 1 #> 8 8 4 0 1 #> 9 9 4 1 1 #> 10 10 4 0 1 #> 11 11 4 0 1 #> 12 12 4 0 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.0723 0.0428 0.0328 #> 2 2 0.10 0 0 0.1184 0.0886 0.0338 #> 3 3 0.15 3 1 0.1616 0.1361 0.1452 #> 4 4 0.33 6 1 0.3191 0.3145 0.5320 #> 5 5 0.50 0 0 0.4842 0.4896 0.2562 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 4. #> Model entropy: 1.19 The trialists report Prob(DLT) = (0.03, 0.07, 0.11, 0.27, 0.45) after this cohort. ProbMTD shows the implied probability that each dose is the maximum tolerable dose, that is, the dose with Prob(DLT) closest to the toxicity target, 33%. The observation of no DLTs in this cohort means it is now very unlikely that dose-levels 1 and 2 are the true MTD. The amount of entropy in the experiment has fallen to reflect this. Once again, the model advocates to remaining at dose-level 4. Cohort 5 This cohort was also treated at dose-level 4. One-out-of-three DLTs were seen. fit5 <- stan_crm(outcome_str = '1NNN 3NNT 4NNT 4NNN 4NTN', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit5 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> 4 4 3 0 1 #> 5 5 3 0 1 #> 6 6 3 1 1 #> 7 7 4 0 1 #> 8 8 4 0 1 #> 9 9 4 1 1 #> 10 10 4 0 1 #> 11 11 4 0 1 #> 12 12 4 0 1 #> 13 13 4 0 1 #> 14 14 4 1 1 #> 15 15 4 0 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.0726 0.0501 0.0255 #> 2 2 0.10 0 0 0.1212 0.1001 0.0260 #> 3 3 0.15 3 1 0.1667 0.1502 0.1507 #> 4 4 0.33 9 2 0.3288 0.3302 0.6085 #> 5 5 0.50 0 0 0.4927 0.5001 0.1893 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 4. #> Model entropy: 1.09 The trialists report Prob(DLT) = (0.04, 0.08, 0.12, 0.29, 0.46). Cohort 6 This cohort was also treated at dose-level 4. Two-out-of-three DLTs were seen. fit6 <- stan_crm(outcome_str = '1NNN 3NNT 4NNT 4NNN 4NTN 4TNT', skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99)) fit6 #> Patient Dose Toxicity Weight #> 1 1 1 0 1 #> 2 2 1 0 1 #> 3 3 1 0 1 #> 4 4 3 0 1 #> 5 5 3 0 1 #> 6 6 3 1 1 #> 7 7 4 0 1 #> 8 8 4 0 1 #> 9 9 4 1 1 #> 10 10 4 0 1 #> 11 11 4 0 1 #> 12 12 4 0 1 #> 13 13 4 0 1 #> 14 14 4 1 1 #> 15 15 4 0 1 #> 16 16 4 1 1 #> 17 17 4 0 1 #> 18 18 4 1 1 #> #> Dose Skeleton N Tox ProbTox MedianProbTox ProbMTD #> 1 1 0.05 3 0 0.104 0.0751 0.0490 #> 2 2 0.10 0 0 0.161 0.1364 0.0653 #> 3 3 0.15 3 1 0.211 0.1925 0.2313 #> 4 4 0.33 12 4 0.374 0.3728 0.5647 #> 5 5 0.50 0 0 0.524 0.5275 0.0897 #> #> The model targets a toxicity level of 0.33. #> The dose with estimated toxicity probability closest to target is 4. #> The dose most likely to be the MTD is 4. #> Model entropy: 1.20 The trialists report Prob(DLT) = (0.06, 0.12, 0.17, 0.36, 0.53). Prob(MTD) actually decreased and the scenario entropy increased after the evaluation of these three patients, because the observation of two-in-three DLTs was slightly surprising to the model. Nevertheless, the trialists stopped here and concluded that 5 mg/m2 per day was probably the MTD. They reported a 95% credibility interval for the DLT rate at this dose to be (15.8, 58.6%). We can verify this: apply(as.data.frame(fit6, pars = 'prob_tox'), 2, quantile, probs = c(0.025, 0.975)) #> prob_tox[1] prob_tox[2] prob_tox[3] prob_tox[4] prob_tox[5] #> 2.5% 0.007936399 0.02356285 0.04562227 0.1788696 0.3828175 #> 97.5% 0.339954375 0.41377685 0.46175719 0.5694487 0.6406686 The 95% CI at dose-level 4 is very close to that reported. They say “The estimated DLT probability associated with the dose level of 5mg/m2 would have been expected to change by less than 5% even if three further patients were included.” It is possible to show this with DTPs. The prevailing probability of toxicity given the patients evaluated thus far is: prob_tox_mtd <- fit6$prob_tox[fit6$recommended_dose] prob_tox_mtd #> [1] 0.3743222 We calculate the future pathways for a single additional cohort of 3 patients, conditional on the outcomes observed hitherto: paths <- crm_dtps(skeleton = skeleton, target = target, model = 'logistic_gamma', cohort_sizes = c(3), previous_outcomes = '1NNN 3NNT 4NNT 4NNN 4NTN 4TNT', a0 = 1, beta_shape = 1, beta_inverse_scale = 1, seed = 123, control = list(adapt_delta = 0.99), refresh = 0) library(tibble) df <- as_tibble(paths) The putative future inferences on the probability of toxicity at dose-level 4 conditional on each of the four possible cohort outcomes are: library(dplyr) library(purrr) library(tidyr) df %>% mutate(prob_tox = map(fit, 'prob_tox')) %>% select(-fit, -parent_fit) %>% unnest(cols = c(dose_index, prob_tox)) %>% filter(dose_index == 4) #> New names: #> * -> ...1 #> * -> ...2 #> * -> ...3 #> * -> ...4 #> * -> ...5 #> # A tibble: 5 x 7 #> .node .parent .depth outcomes next_dose dose_index prob_tox #> <dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl> #> 1 1 NA 0 "" 4 4 0.374 #> 2 2 1 1 "NNN" 4 4 0.334 #> 3 3 1 1 "NNT" 4 4 0.370 #> 4 4 1 1 "NTT" 4 4 0.410 #> 5 5 1 1 "TTT" 3 4 0.443 We can put this data into a format conducive to reproducing the trialists’ claim. df %>% filter(.depth > 0) %>% mutate(prob_tox = map(fit, 'prob_tox')) %>% select(-fit, -parent_fit) %>% unnest(cols = c(dose_index, prob_tox)) %>% filter(dose_index == 4) %>% select(outcomes, prob_tox) %>% bind_cols( lik = dbinom(x = 0:3, size = 3, prob = prob_tox_mtd)) -> future_scenario future_scenario #> # A tibble: 4 x 3 #> outcomes prob_tox lik #> <chr> <dbl> <dbl> #> 1 NNN 0.334 0.245 #> 2 NNT 0.370 0.440 #> 3 NTT 0.410 0.263 #> 4 TTT 0.443 0.0524 Above we see the likelihood of the four possible outcomes, inferred using binomial probabilities and the prevailing probability of toxicity at dose-level 4. The outcome TTT would increase the probability by over 5% but that scenario is unlikely, given what we know. The expected change is indeed less than 5%: future_scenario %>% mutate(prob_tox_change = abs(prob_tox - prob_tox_mtd)) %>% summarise(expected_change = sum(lik * prob_tox_change)) #> # A tibble: 1 x 1 #> expected_change #> <dbl> #> 1 0.0244 This concludes the main case study on Levy et al. (2006). The section below details an honest but ultimately unsuccessful attempt to infer the precise parameterisation used by the trialists. Trying to identify the exact model Levy et al. used At the end of the trial, the investigators reported the estimated probabilities of DLT (0.06, 0.12, 0.17, 0.36, 0.53). We will fit models using an exponential prior and each combination of a_0 = 1, 2, 3, 4 and beta_inverse_scale = 0.5, 1, 2 to the complete set of all outcomes observed. We seek the fit that yields inference closest to that of the investigators. The investigators concluded: levy_reported <- tibble( Dose = 1:5, ProbTox = c(0.06, 0.12, 0.17, 0.36, 0.53), ) We also define a helper function to fit the models: fit_levy_crm <- function(outcomes, a0, beta_inverse_scale) { stan_crm(outcome_str = outcomes, skeleton = skeleton, target = target, model = 'logistic_gamma', a0 = a0, beta_shape = 1, beta_inverse_scale = beta_inverse_scale, seed = 123, refresh = 0) } This code block calculates the parameter combinations, fits the model to each, and extracts the posterior mean probability of DLT: expand.grid(a0 = 1:4, beta_inverse_scale = c(0.5, 1, 2)) %>% mutate(Series = rownames(.)) %>% mutate(Mod = map2(a0, beta_inverse_scale, fit_levy_crm, outcomes = '1NNN 3NNT 4NNT 4NNN 4NTN 4TNT')) %>% mutate( Dose = Mod %>% map("dose_indices"), ProbTox = Mod %>% map("prob_tox"), ) %>% select(-Mod) %>% unnest(cols = c(Dose, ProbTox)) %>% mutate(a0 = factor(a0), beta_inverse_scale = factor(beta_inverse_scale)) -> all_fits We then plot our inferences with the investigators’ inferences superimposed in bright green. Plots are grouped by the value for beta_inverse_scale in columns, and values for a0 are reflected by colour: all_fits %>% ggplot(aes(x = Dose, y = ProbTox)) + geom_line(aes(group = Series, col = a0)) + geom_line(data = levy_reported, col = 'green', size = 1.2) + facet_wrap(~ beta_inverse_scale) + ylim(0, 0.7) + labs(title = "None of the exponential models quite matches the investigators' inferences") We see that there is broad agreement at the higher doses but none of the series quite matches the investigators’ inferences. There are many possible explanations for the difference. The investigators might not have used an exponential prior. They might have use parameters we have not tested. They might have reported some statistic other than the posterior mean. trialr or their code or indeed both might be wrong. Either way, there is enough agreement to agree on the probable identity of the MTD. Other CRM vignettes There are many vignettes illuminating the CRM in trialr: trialr trialr is available at https://github.com/brockk/trialr and https://CRAN.R-project.org/package=trialr References Levy, V, S Zohar, C Bardin, A Vekhoff, D Chaoui, B Rio, O Legrand, et al. 2006. “A Phase I Dose-Finding and Pharmacokinetic Study of Subcutaneous Semisynthetic Homoharringtonine (ssHHT) in Patients with Advanced Acute Myeloid Leukaemia.” British Journal of Cancer 95 (3): 253–59. https://doi.org/10.1038/sj.bjc.6603265. Zohar, Sarah, Aurelien Latouche, Mathieu Taconnet, and Sylvie Chevret. 2003. “Software to Compute and Conduct Sequential Bayesian Phase I or II Dose-Ranging Clinical Trials with Stopping Rules.” Computer Methods and Programs in Biomedicine 72 (2): 117–25. https://doi.org/10.1016/S0169-2607(02)00120-7.	2020-04-07 16:38:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.4031454622745514, "perplexity": 7088.305001998631}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00257.warc.gz"}
https://math.stackexchange.com/questions/2164452/let-n-be-a-cyclic-normal-subgroup-of-a-group-g-and-h-is-any-subgroup-of-n	# Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$. Prove $H$ is a normal subgroup of $G$ [duplicate] Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$. Prove $H$ is a normal subgroup of $G$ Guessinng that exists $a\in G$ where $\langle a\rangle=N$ of some order. $H \subset N$, $H$ is a subgroup of $N$. $gH =Hg \equiv g h_1 =h_2 g \equiv g a^{k_1} =a^{k_2}g$. Not sure if on the right track just spit balling ## marked as duplicate by Dietrich Burde, Michael Burr, Juniven, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 28 '17 at 1:46 • In a cyclic group of order $n$, for each $k$ dividing $n$, there is a unique subgroup of that order. • Let $g\in G$. $gHg^{-1}$ is a cyclic group of the same order as $H$ (and $H$ is a cyclic group as it is a subgroup of a cyclic group). • $gHg^{-1}$ is a subgroup of $N$ by the normality of $K$ ($gHg^{-1}$ is a subset of $N$ by normality of $N$ and $gHg^{-1}$ is a group, so its a subgroup of $K$). Can you use these three facts to prove that $gHg^{-1}=H$?	2019-10-17 22:36:43	{"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": 0.3706095516681671, "perplexity": 487.48699573004217}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00344.warc.gz"}
https://mathoverflow.net/questions/401018/the-4th-lemoine-circle	# The 4th Lemoine circle The first and second Lemoine circles are well-known to geometers. According to this article the third Lemoine circle has been first discovered by Jean-Pierre Ehrmann in 2002. It is worth noting that the centres of these circles belong to a line that is going through the Lemoine point and the circumcenter of the triangle. I would like to present the construction of another circle, which I believe might be called the 4th Lemoine circle. The idea behind its construction is slightly similar to that proposed by Ehrmann: Symmedians AA', BB', CC' intersect each other at the point M (the Lemoine point of the triangle ABC). The circumcircle of the triangle A'B'C' was drawn (also known as the 'symmedial circle'). A'', B'', C'' are the first intersection points of the symmedians with the symmedial circle. Finally, the circumcircles of the triangles A''B''M, B''C''M, A''C''M always intersect the sides of the original triangle ABC at six points, that are conclycic. Theorem illustration As far as I can tell the point O (the center of our six point circle) is not included into the C.Kimberling's Encyclopedia, so it must be unknown or uncatalogued. O belongs to the line that contains the centres of the first three Lemoine circles, so by my reckoning, the dotted red circle that is shown in the picture above perfectly qualifies for being called the 4th Lemoine circle. Could you please give a synthetic proof of this theorem ? Geogebra dynamic sketch. • Interesting! Is this 4th Lemoine circle a Tucker circle? Aug 4 at 20:37 • So, I've just checked on Euklid Dynageo that the circle is not a Tucker circle. This makes it even more surprising that its center seemingly lies on the Tucker line (i.e., on the line joining the circumcenter with the symmedian point)! Aug 5 at 7:30 • @darijgrinberg, I'm not sure what specifically causes your surprise, so this may or may not affect it, but my analysis of Kimberling's ETC finds that 493 of the first 10000 triangle centres are on the Brocard axis, which makes it second only to the Euler line (960 of the first 10000). Aug 22 at 6:56 • @darij grinberg mathoverflow.net/questions/402310/lemoine-lozada-circles Aug 22 at 19:27 A notational preamble: Writing $$K$$ for the circumcenter, and $$L_1$$, $$L_2$$ (OP's $$M$$), $$L_3$$ for the centers of the First, Second, and (Ehrmann's) Third Lemoine Circles, all of which are collinear, we have this happy terminological coincidence: $$\frac{KL_i}{KL_1}=i \tag{1}$$ (This invites dubbing the circumcenter $$L_0$$, and the circumcircle itself the "Zero-th Lemoine Circle"; but I digress.) To the problem at hand ... Some ugly coordinate bashing confirms that OP's center $$O$$ lies on the line-of-$$L$$s, a rather remarkable property that provides the corresponding circle some "Numbered Lemoine Circle" credibility. Further bashing shows that the counterpart of $$(1)$$ is $$\frac{KO}{KL_1}=\frac{3 (a^2 + b^2 + c^2)^3 - (-a^2 + b^2 + c^2) (a^2 - b^2 + c^2) (a^2 + b^2 - c^2)}{8\,(a^2 + b^2) (a^2 + c^2) (b^2 + c^2)} \tag2$$ The dependence of this value upon the shape of the triangle distinguishes it from the $$L_i$$, perhaps so much so that OP's circle establishes a new category, "Numbered Non-Tucker Lemoine Circles". I'll take this opportunity describe another point on the line-of-$$L$$s with a shape-independent ratio $$(1)$$: For each vertex $$V=A, B, C$$, construct the circle through $$V$$ and $$L_2$$ whose center lies on $$\overleftrightarrow{VK}$$. The six "other" points where the three circles meet the triangle's side-lines are concyclic, and their circumcenter —known to Kimberling as $$X(585)$$wants to be denoted $$L_{3/2}$$, because $$\frac{KL_{3/2}}{KL_1} = \frac32 \tag3$$ This may-or-may-not earn the circle (which, incidentally, is a Tucker circle) the title of "$$\frac32$$-th Lemoine Circle". By the Power of a Point theorem, $$B C_B \cdot B C_A = B B'' \cdot BM = B A_B \cdot B A_C$$ and thus again by the Power of a Point theorem, the points $$C_B, C_A, A_B, A_C$$ all lie on one circle, and symmetrically the points $$A_B, A_C, B_A, B_C$$ lie on one circle, and the points $$B_A, B_C, C_B, C_A$$ lie on one circle. If these are not all the same circle, their radical axes intersect at one point, that is the lines $$AB, AC, BC$$ intersect in one point which is a contradiction. • This is very nice, but does not show anything about the center of the circle. Aug 4 at 20:39 • Thanks. I should have rather asked, whether it is a Tucker circle and what are its parameters or about the collinearity of the points O, R and M... Aug 4 at 20:42 • Oh sorry, I misread the question. I'll think about that Aug 4 at 21:17	2021-11-29 13:51:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6668278574943542, "perplexity": 356.91186825106945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00382.warc.gz"}
http://shpsoftware.com/error-function/inverse-error-function-tables.php	Loading... Home > Error Function > Inverse Error Function Tables # Inverse Error Function Tables ## Contents MR0167642. Comp. 23 (107): 631–637. Wolfram Problem Generator» Unlimited random practice problems and answers with built-in Step-by-step solutions. At the real axis, erf(z) approaches unity at z→+∞ and −1 at z→−∞. navigate here PARI/GP: provides erfc for real and complex arguments, via tanh-sinh quadrature plus special cases. Online Integral Calculator» Solve integrals with Wolfram\|Alpha. Contact the MathWorld Team © 1999-2016 Wolfram Research, Inc. \| Terms of Use THINGS TO TRY: inverse erf 135/216 - 12/25 det {{a, b, c}, {d, e, f}, {g, h, j}} For complex, the Faddeeva package provides a C++ complex implementation. ## Inverse Error Function Excel Sloane, N.J.A. y0≦y≦1 6dgt10dgt14dgt18dgt22dgt26dgt30dgt34dgt38dgt42dgt46dgt50dgt Privacy Policy Terms of use FAQ Contact us © 2016 CASIO COMPUTER CO., LTD. For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special.[21] A complex-argument erf is also in the arbitrary-precision arithmetic • Another approximation is given by erf ⁡ ( x ) ≈ sgn ⁡ ( x ) 1 − exp ⁡ ( − x 2 4 π + a x 2 1 • Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. • W. • Step-by-step Solutions» Walk through homework problems step-by-step from beginning to end. • ISBN978-1-4020-6948-2. ^ Winitzki, Sergei (6 February 2008). "A handy approximation for the error function and its inverse" (PDF). • The error function is related to the cumulative distribution Φ {\displaystyle \Phi } , the integral of the standard normal distribution, by[2] Φ ( x ) = 1 2 + 1 • This series diverges for every finite x, and its meaning as asymptotic expansion is that, for any N ∈ N {\displaystyle N\in \mathbb Γ 2 } one has erfc ⁡ ( • Please try the request again. Daniel Soper. Compute the inverse error function for x = -1, x = 0, and x = 1. Sorry for the inconvenience but we’re performing some maintenance at the moment. Inverse Erfc Press, William H.; Teukolsky, Saul A.; Vetterling, William T.; Flannery, Brian P. (2007), "Section 6.2. More Aboutcollapse allInverse Error FunctionThe inverse error function is defined as erf-1(x), such that erf(erf-1(x))=erf-1(erf(x))=x. Erf(2) The integrand ƒ=exp(−z2) and ƒ=erf(z) are shown in the complex z-plane in figures 2 and 3. New York: Dover, 1972. Wolfram Education Portal» Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Cambridge, England: Cambridge University Press, 1998. Inverse Error Function Python Another form of erfc ⁡ ( x ) {\displaystyle \operatorname ⁡ 2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ⁡ ( x \| x ≥ 0 The inverse error function is also known as the Gauss inverse error function.Please enter the necessary parameter values, and then click 'Calculate'. For any complex number z: erf ⁡ ( z ¯ ) = erf ⁡ ( z ) ¯ {\displaystyle \operatorname − 0 ({\overline ⁡ 9})={\overline {\operatorname ⁡ 8 (z)}}} where z ## Erf(2) MathWorks does not warrant, and disclaims all liability for, the accuracy, suitability, or fitness for purpose of the translation. http://www.ams.org/mcom/1976-30-136/S0025-5718-1976-0421040-7/S0025-5718-1976-0421040-7.pdf Go: Provides math.Erf() and math.Erfc() for float64 arguments. Inverse Error Function Excel Wolfram Language» Knowledge-based programming for everyone. Inverse Erf Function Some authors discuss the more general functions:[citation needed] E n ( x ) = n ! π ∫ 0 x e − t n d t = n ! π ∑ Schöpf and P. http://shpsoftware.com/error-function/inverse-error-function-ti-89.php By using this site, you agree to the Terms of Use and Privacy Policy. Mathematica: erf is implemented as Erf and Erfc in Mathematica for real and complex arguments, which are also available in Wolfram Alpha. Your cache administrator is webmaster. Inverse Error Function Matlab Continued fraction expansion A continued fraction expansion of the complementary error function is:[11] erfc ⁡ ( z ) = z π e − z 2 1 z 2 + a 1 Home/ Special Function/ Error function Inverse error function Calculator Calculates the inverse error function erf -1(y) and inverse complementary error function erfc -1(y). Compute the inverse error function for complex numbers. http://shpsoftware.com/error-function/inverse-error-function-c.php Incomplete Gamma Function and Error Function", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN978-0-521-88068-8 Temme, Nico M. (2010), "Error Functions, Dawson's and Fresnel Integrals", Translate erfinvInverse error functioncollapse all in page Syntaxerfinv(X) exampleDescriptionexampleerfinv(X) computes the inverse error function of X. Inverse Complementary Error Function If L is sufficiently far from the mean, i.e. μ − L ≥ σ ln ⁡ k {\displaystyle \mu -L\geq \sigma {\sqrt {\ln {k}}}} , then: Pr [ X ≤ L Matlab provides both erf and erfc for real arguments, also via W. ## Haskell: An erf package[18] exists that provides a typeclass for the error function and implementations for the native (real) floating point types. Wolfram\|Alpha» Explore anything with the first computational knowledge engine. Math. Excel: Microsoft Excel provides the erf, and the erfc functions, nonetheless both inverse functions are not in the current library.[17] Fortran: The Fortran 2008 standard provides the ERF, ERFC and ERFC_SCALED Inverse Error Function C++ is the double factorial: the product of all odd numbers up to (2n–1). Cody's algorithm.[20] Maxima provides both erf and erfc for real and complex arguments. The Q-function can be expressed in terms of the error function as Q ( x ) = 1 2 − 1 2 erf ⁡ ( x 2 ) = 1 2 Given random variable X ∼ Norm ⁡ [ μ , σ ] {\displaystyle X\sim \operatorname {Norm} [\mu ,\sigma ]} and constant L < μ {\displaystyle L<\mu } : Pr [ X weblink Please try the request again. ISBN 978-0-486-61272-0. See [2]. ^ http://hackage.haskell.org/package/erf ^ Commons Math: The Apache Commons Mathematics Library ^ a b c Cody, William J. (1969). "Rational Chebyshev Approximations for the Error Function" (PDF). Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view TweetOnline Tools and Calculators > Math > Error Function Calculator Error Function Calculator Number: About This Tool The online In order of increasing accuracy, they are: erf ⁡ ( x ) ≈ 1 − 1 ( 1 + a 1 x + a 2 x 2 + a 3 x The inverse complementary error function is defined as erfc − 1 ⁡ ( 1 − z ) = erf − 1 ⁡ ( z ) . {\displaystyle \operatorname ζ 8 ^{-1}(1-z)=\operatorname Washington D.C., USA; New York, USA: United States Department of Commerce, National Bureau of Standards; Dover Publications. Google search: Google's search also acts as a calculator and will evaluate "erf(...)" and "erfc(...)" for real arguments. Asymptotic expansion A useful asymptotic expansion of the complementary error function (and therefore also of the error function) for large real x is erfc ⁡ ( x ) = e − When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usually discussed in scaled form as the Faddeeva function: w ( z ) = Applications When the results of a series of measurements are described by a normal distribution with standard deviation σ {\displaystyle \textstyle \sigma } and expected value 0, then erf ( a Similarly, the En for even n look similar (but not identical) to each other after a simple division by n!. Generated Wed, 19 Oct 2016 06:12:19 GMT by s_wx1011 (squid/3.5.20) Softw., 19 (1): 22–32, doi:10.1145/151271.151273 ^ Zaghloul, M. SEE ALSO: Confidence Interval, Erf, Inverse Erfc, Probable Error RELATED WOLFRAM SITES: http://functions.wolfram.com/GammaBetaErf/InverseErf/, http://functions.wolfram.com/GammaBetaErf/InverseErf2/ REFERENCES: Bergeron, F.; Labelle, G.; and Leroux, P. To use these approximations for negative x, use the fact that erf(x) is an odd function, so erf(x)=−erf(−x). Derivative and integral The derivative of the error function follows immediately from its definition: d d z erf ⁡ ( z ) = 2 π e − z 2 . {\displaystyle © Copyright 2017 shpsoftware.com. All rights reserved.	2018-05-26 04:23:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7125378847122192, "perplexity": 4061.2878777043816}, "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-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00283.warc.gz"}
https://math.stackexchange.com/questions/2879562/epsilon-delta-study-material	# Epsilon delta study material [duplicate] I want to practice epsilon-delta method for problems in analysis. Does anyone have any suggestions for books with solutions, or solution books available, for undergraduate analysis? Edit: I didn't write book because I cant buy books. Was referring to free material. ## marked as duplicate by Jyrki Lahtonen, Namaste, user99914, Sil, TaroccoesbroccoAug 12 '18 at 8:27 The explanation of $\epsilon$-$\delta$ continuity in this book is exactly what you want!(Read Section $2.1$ with exercises. In this section, the author explains how to find a $\delta$ for a given $\epsilon$ and give lot of problems for practice )	2019-10-17 18:25:44	{"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": 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": 0.20788294076919556, "perplexity": 1013.2575911188123}, "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/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00002.warc.gz"}
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=emj&paperid=214&option_lang=eng	RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB General information Latest issue Archive Impact factor Search papers Search references RSS Latest issue Current issues Archive issues What is RSS Eurasian Math. J.: Year: Volume: Issue: Page: Find Eurasian Math. J., 2016, Volume 7, Number 1, Pages 28–49 (Mi emj214) Inequalities between the norms of a function and its derivatives A. S. Kochurov Department of Mechanics and Mathematics, Moscow State University, Leninskie gory, Moscow 119991, Russia Abstract: The paper is devoted to the problem of finding the maximum of the norm $\|\|x\|\|_q$ with the constraints $\|\|x\|\|_p=\eta$, $\|\|\dot{x}\|\|_r=\sigma$, $x(0)=a$, $a, \sigma, \eta>0$, for functions $x\in L_p(\mathbb{R}_-)$ with derivatives $\dot{x}\in L_r(\mathbb{R_-})$, $0 < p \leqslant q < \infty$, $r > 1$. The arguments employed are based on the standard machinery of the calculus of variations. Keywords and phrases: inequalities for derivatives, necessary conditions for an extremum, Weierstrass formula, Euler equation. Funding Agency Grant Number Russian Foundation for Basic Research 14-01-00744_à This research was carried out with the financial support of the Russian Foundation for Basic Research (grant no. 14-01-00744). Full text: PDF file (440 kB) References: PDF file HTML file Bibliographic databases: MSC: 26D10 Language: Citation: A. S. Kochurov, “Inequalities between the norms of a function and its derivatives”, Eurasian Math. J., 7:1 (2016), 28–49 Citation in format AMSBIB \Bibitem{Koc16} \by A.~S.~Kochurov \paper Inequalities between the norms of a function and its derivatives \jour Eurasian Math. J. \yr 2016 \vol 7 \issue 1 \pages 28--49 \mathnet{http://mi.mathnet.ru/emj214} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000380176800002}	2019-10-22 11:02:54	{"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": 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": 0.3637382984161377, "perplexity": 4753.319471922913}, "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/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00029.warc.gz"}
http://www.oovee.co.uk/forum/topic/12549-recent-game-crashes/	Oovee will not accept mods/code etc ripped from other games or made from models with illegal/unsuitable copyright or not in accordance with licenses and/or forum rules. Any Mods will be removed and the user may be banned. Only statements relating to Spintires that are posted here by Oovee staff are official statements from Oovee limited and Spin Tires # Recent game crashes 435 replies to this topic ### #1 David Jones David Jones Member • Experienced Member • 84 posts • LocationMerthyr Tydfil , Wales Posted 23 February 2016 - 03:14 PM Steam seamed to have done an update to te game today and ever since it crashes out after about a minute or so. Looking on Facebook it seams i'm not the only one with this. Can anyone on here shine a bit of light on this please. ### #2 stetson stetson Veteran Member • Community Contributor • 1,698 posts • LocationU.S.A, Fla Posted 23 February 2016 - 03:19 PM i never noticed the update but this morning I crashed out quite a few times(around the same point everytime) with huntermod I don't think he ever crashed Edited by stetson, 23 February 2016 - 03:19 PM. ### #3 newbie_master newbie_master Member • Experienced Member • 70 posts • LocationAuvergne, France Posted 23 February 2016 - 03:20 PM Hi! After several crashes while trying bregel's map, I noticed that game version has switched back to 17.11.15. I tried a complete cleanup and install, but without success. has there been a regression or am i missing something? "Stinky?! Mudhole?! My home this is!!" ### #4 MarijnS95 MarijnS95 Raytrace-em-all • Senior Moderator • 3,527 posts • LocationThe Netherlands Posted 23 February 2016 - 03:35 PM Multiple people reported this a week ago as well, however there has been no change to the game whatshowever. For the people who don't believe me: check steamdb, last change to the Spintires depot was 21 days ago. Edit: Merged topics. Apparently my game displays the same now. I'll investigate. • David Jones, Sledge Hammer (Inspector) and Axonn like this Please use the search function or Google before posting a question! It might've been asked already, and you'll get your answer way quicker than waiting for a response. Keep in mind that I'm a voluntary moderator, not an official Oovee representative. Things I say may or may not represent the view of Oovee themselves. Create and send log 1. Start Spintires while holding Shift. (Start it by clicking play in Steam or (double)clicking the icon on your desktop/start-menu/taskbar) 2. Try to reproduce the bug. Make sure you get the behavior that you are reporting before proceeding to the next step. 3. Open up the forum thread, right to the button there is another button called . Click that. 4. On the bottom there is a button called . A file selection window should appear. Paste type in %appdata%\Spintires\Log.txt and press enter. Now the log file should upload. If it gives an error, manually check if the file Log.txt exists. If not, try again 5. Finish your post and send it. Clean up your Spintires installation 1. Go to the installation directory of Spintires, usually in C:\Program Files (x86)\Steam\SteamApps\common\Spintires . This is the default location. If you told Steam to install at another location, you probably know where it is. If you can't find it, right-click the game in Steam, click on the tab then click on . 2. Delete all files that end with the .mdmp extension. If there are other files besides _CommonRedist , Media , User , Media.zip , Shaders.zip , SpinTires.exe , steam_api.dll , steam_appid.txt and TexturesCache.zip , delete these too. 3. Go into the Media folder and delete everything except the videos folder. 4. Press and simultaneously, type in (or copy) %appdata% and press . Now delete the Spintires folder. Warning: this deletes your settings and saves! If you want to keep this, backup this folder to another location. 5. Now, verify the Steam Appcache for Spintires. To do so click . Don't be surprised, this button will fire up Steam and do the validation! Delete Spintires from your computer Are you sure? Sure, delete Spintires. Create DxDiag 1. Press and simultaneously, type in (or copy) dxdiag and press . Click yes if a warning about signatures pops up. 2. Click the save button on the bottom right. Choose a location you can easily find! Wait for the saving process to complete. 3. Upload the file you created to the forum. Click next to the button to upload the file. 4. Finish your post (write something descriptive) and it. ### #5 newbie_master newbie_master Member • Experienced Member • 70 posts • LocationAuvergne, France Posted 23 February 2016 - 03:37 PM Multiple people reported this a week ago as well, however there has been no change to the game whatshowever. For the people who don't believe me: check steamdb, last change to the Spintires depot was 21 days ago. Version switched back to 17.11.15 here Edited by newbie_master, 23 February 2016 - 03:38 PM. "Stinky?! Mudhole?! My home this is!!" ### #6 hoanns hoanns Member • Experienced Member • 115 posts Posted 23 February 2016 - 03:56 PM no beta activated was 25.12.15c before ### #7 MarijnS95 MarijnS95 Raytrace-em-all • Senior Moderator • 3,527 posts • LocationThe Netherlands Posted 23 February 2016 - 04:01 PM no beta activated was 25.12.15c before Merged your topic here. We're investigating the case. • David Jones likes this Please use the search function or Google before posting a question! It might've been asked already, and you'll get your answer way quicker than waiting for a response. Keep in mind that I'm a voluntary moderator, not an official Oovee representative. Things I say may or may not represent the view of Oovee themselves. Create and send log 1. Start Spintires while holding Shift. (Start it by clicking play in Steam or (double)clicking the icon on your desktop/start-menu/taskbar) 2. Try to reproduce the bug. Make sure you get the behavior that you are reporting before proceeding to the next step. 3. Open up the forum thread, right to the button there is another button called . Click that. 4. On the bottom there is a button called . A file selection window should appear. Paste type in %appdata%\Spintires\Log.txt and press enter. Now the log file should upload. If it gives an error, manually check if the file Log.txt exists. If not, try again 5. Finish your post and send it. Clean up your Spintires installation 1. Go to the installation directory of Spintires, usually in C:\Program Files (x86)\Steam\SteamApps\common\Spintires . This is the default location. If you told Steam to install at another location, you probably know where it is. If you can't find it, right-click the game in Steam, click on the tab then click on . 2. Delete all files that end with the .mdmp extension. If there are other files besides _CommonRedist , Media , User , Media.zip , Shaders.zip , SpinTires.exe , steam_api.dll , steam_appid.txt and TexturesCache.zip , delete these too. 3. Go into the Media folder and delete everything except the videos folder. 4. Press and simultaneously, type in (or copy) %appdata% and press . Now delete the Spintires folder. Warning: this deletes your settings and saves! If you want to keep this, backup this folder to another location. 5. Now, verify the Steam Appcache for Spintires. To do so click . Don't be surprised, this button will fire up Steam and do the validation! Delete Spintires from your computer Are you sure? Sure, delete Spintires. Create DxDiag 1. Press and simultaneously, type in (or copy) dxdiag and press . Click yes if a warning about signatures pops up. 2. Click the save button on the bottom right. Choose a location you can easily find! Wait for the saving process to complete. 3. Upload the file you created to the forum. Click next to the button to upload the file. 4. Finish your post (write something descriptive) and it. ### #8 King Unique King Unique Armed with Music • Experienced Member • 141 posts • LocationLithuania Posted 23 February 2016 - 04:03 PM 2 russian players causing crash too, since game version updated. Newbie • Experienced Member • 8 posts Posted 23 February 2016 - 04:09 PM Same for me! ### #10 David Jones David Jones Member • Experienced Member • 84 posts • LocationMerthyr Tydfil , Wales Posted 23 February 2016 - 04:14 PM Just to confirm mines reverted back to 17.11.15 , also resetting all the game settings. Thanks for looking into it . • drizz786 likes this ### #11 junkers.lubo junkers.lubo Newbie • Experienced Member • 1 posts Posted 23 February 2016 - 04:27 PM Same for me! ### #12 Schlammspringer Schlammspringer • Community Contributor • 576 posts • LocationBavaria Posted 23 February 2016 - 04:35 PM Upcoming update? And this is an open betatest..... ### #13 stetson stetson Veteran Member • Community Contributor • 1,698 posts • LocationU.S.A, Fla Posted 23 February 2016 - 04:36 PM care for the mdmp files from my crash outs this morning? (note I have no idea what they are.i just know they show up after crashes so I assume they have some info in them about the crash) Edited by stetson, 23 February 2016 - 04:39 PM. ### #14 YONs YONs Newbie • Experienced Member • 1 posts Posted 23 February 2016 - 05:03 PM I get updates spintires version 17.11.15 and always crashes when I play, why ...? an updated version which seemed strange to me ### #15 vistastang vistastang Newbie • Experienced Member • 3 posts Posted 23 February 2016 - 05:20 PM I completely uninstalled spintires and then reinstalled it. It has switched back to 17.11.15 but no luck, it continues to crash. This is really a let down as I just purchased the game yesterday. I played it for hours yesterday and I was looking forward to playing it again today. ### #16 Oliver Duborg Oliver Duborg Newbie • Experienced Member • 1 posts Posted 23 February 2016 - 05:57 PM Steam seamed to have done an update to te game today and ever since it crashes out after about a minute or so. Looking on Facebook it seams i'm not the only one with this. Can anyone on here shine a bit of light on this please. Yeah it crashes to me now it sucks.... when i open lobby i drive like 2-3 Mins then it crashes • hamtme2 and David Jones like this ### #17 Schlammspringer Schlammspringer • Community Contributor • 576 posts • LocationBavaria Posted 23 February 2016 - 07:09 PM Delete workshop folder and it will work again! ;-) • newbie_master and LucaT like this ### #18 Schlammspringer Schlammspringer • Community Contributor • 576 posts • LocationBavaria Posted 23 February 2016 - 07:13 PM I think, that's only a faulty startscreen! How can program change without any download??????? ### #19 RockSurfer RockSurfer Newbie • Experienced Member • 6 posts Posted 23 February 2016 - 07:21 PM Yes, constant crashing, it does not run for more than a couple of minutes :-( ### #20 newbie_master newbie_master Member • Experienced Member • 70 posts • LocationAuvergne, France Posted 23 February 2016 - 08:15 PM Delete workshop folder and it will work again! ;-) Seems to work! "Stinky?! Mudhole?! My home this is!!" ### Also tagged with one or more of these keywords: undergoing admin observation #### 0 user(s) are reading this topic 0 members, 0 guests, 0 anonymous users	2016-10-22 17:47:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2540956139564514, "perplexity": 8158.657386404664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719033.33/warc/CC-MAIN-20161020183839-00201-ip-10-171-6-4.ec2.internal.warc.gz"}
https://crypto.stackexchange.com/questions/12661/what-could-look-for-the-constants-that-you-might-find-in-rc6-mean/12670	# What could "look for the constants that you might find in RC6" mean? In this presentation at the 30c3 (long but extremely interesting and well-documented), Jacob Appelbaum gave this verbal advice (circa 43'30"): look for the constants that you might find in RC6 I'm trying to make sense out of that; my problem is, there are very few constants that I could find in the description of RC6. Any idea? • I only see constants $P$ and $Q$ in the key schedule and in the paper they are descibed as nothing up my sleeve, i.e.,derived from $e$ and the the Golden ratio, but I didn't check that. Dec 30 '13 at 21:54 • @DrLecter: I missed these constants on first reading! That could well be the answer, or at least part of it. – fgrieu Dec 30 '13 at 22:28 • I just checked the hex values for $P$ and $Q$ and they are correct as given in the paper as $P=\tt B7E15163$ and $Q=\tt 9E3779B9$ when taking rounding into account :/ Dec 31 '13 at 15:29 • I haven't watched the video, but could he be talking about the fact that RC6 was used by some NSA malware (likely just to evade heuristic detection, since a more popular cipher's constants might set off red flags)? Feb 16 at 2:19 The Key schedule appendix of the RC6 paper (Rivest et al.) defines two arbitrary constants in hexadecimal notation. They also appear in RC5. • $P_{32} = \tt{B7E15163}$ • $Q_{32} = \tt{9E3779B9}$ Be aware that the binary coding of these constants will vary based on the endian-ness of the computer architecture. There seems to be nothing 'special' about these constants in relation to the construction of the RC6 algorithm. They can be replaced with any random value, producing an incompatible variant of RC6 with presumably identical security properties, and the authors make note of this. (The 'official' constants were chosen based on open criteria, described in the paper, in order to dispel any notion that they induce a back door.) Thus a serious search for RC6 would include RC6-alikes and look for algorithm structure rather than constants. There are no other arbitrary numeric constants defined in RC6, unless you would go so far as to consider the number of rounds etc. as such. By the way, Appelbaum's talk wasn't "well documented" enough to provide any justification for this RC6 hunt. Typically of Appelbaum, he takes the position, "Trust me, I've seen the secret documents." Sounds like some senators I know. • Regarding your last paragraph: These are the Snowden documents, and he is not the only person that processed them. Glenn Greenwald and some authors of Der Spiegel among others agree with Jacob (e.g spiegel.de/international/germany/…). However common sense does the job here, if you just open your eyes and see the manhunt and harassment all those people have suffered. If they were just lying, why all the trouble? Nov 15 '18 at 8:50 • Also most of these documents are available online, e.g: eff.org/files/2014/01/06/20131230-appelbaum-nsa_ant_catalog.pdf However it seems that most people in tech community tend to believe anything that supports their ideology and dismiss anything else, regardless the clues... Nov 15 '18 at 8:58 • @chefarov The ANT Catalog is not a Snowden document, its source has never been revealed or vetted or even openly discussed, and the Der Spiegel article does not mention RC6. Nov 15 '18 at 15:37 • 1) I wasn't talking about RC6, but for your broad generalization (last sentence) about a person that you don't know, and who has been harassed/attacked by US in such a way (I would say top10 in EU/US) that almost proves his sayings. 2) We don't know the source of ANT (although Snowden is still a possible candidate), but I was referring to presentation docs (accepted by eff & dozen researchers). In any case Snowden's docs describe similar things. Nov 15 '18 at 16:51	2021-11-27 08:54:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5718097686767578, "perplexity": 1663.2144101101906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00292.warc.gz"}
https://www.wasyresearch.com/low-frequency-radio-navigation-system-for-use-in-natural-disaster-situations/	Low frequency radio navigation system for use in natural disaster situations It is critical to have a navigation system for search and rescue of victims and important assets in a disaster situation. It is critical to have a navigation system for search and rescue of victims and important assets in a disaster situation. This post is based on a paper “Low RF-based Navigation for Emergencies in Difficult Environments”. All details can be found in the paper [1]. Both terrestrial and global navigation satellite system (GNSS) infrastructures have limitations for use as a navigation tool in a disaster situation. In this post, we will present and discuss what the limitations of these two systems are and discuss a solution for navigation in such disaster environment. Motivation: Navigation in disaster environments In a disaster or hostile environments, such as after earthquake, flood, volcano or storm phenomena, common wireless communication system infrastructures are prone to destruction. For example, cellular network and other terrestrial communication system as well as supporting infrastructures, such as electricity supply, can be destroyed due to a natural disaster. Due to the destruction, these communication systems cannot function properly for communication and also for navigation. One simple communication system that can be used in disaster environment is walkie-talkie. This device is battery operated and is portable. However, this device cannot provide a navigation functionality. Navigation functionality is needed in disaster environment. For example, to find victims or critical assets in a wrecked building, a rescue team need to carry a navigation system to guide them finding the victims or assets. Figure 1 below shows some examples of disaster environment or emergency conditions, such as wrecked buildings due to tornado and earthquake as well as destroyed buildings or houses due to flood. GNSS system can provide global navigation that can be accessed anywhere and GNSS infrastructure is free from disasters occurring on earth. However, the signal power of received GNSS signals are very weak due to the signals come from satellites, typically about 23000 km above the earth (the height of medium-earth orbit satellite). Also, since the transmission frequency used by GNSS system is very high at GHz level (for example GPS L1 and GALILEO E1 signals are transmitted at 1575.42 MHz), the propagation power of GNSS signals to penetrate obstacles, such as walls of a building, is low. Based on the disadvantages of terrestrial and GNSS radio frequency (RF) transmission, a radio navigation system using low frequency transmission is developed. The main idea of the system is that it uses low radio frequency transmission (sub-GHz) to have a high propagation power and transmits ranging signal for navigation. Figure 2 below shows the final goal of the low frequency system to provide radio navigations in disaster environment. In figure 2, the main unit of the system is a transmission (Tx) unit. The Tx unit should be compact and portable and battery-operated. Because, the Tx unit will be mounted on a moving vehicle, especially drones. The Tx unit receives its position from open-sky GNSS signals and re-transmits its position and ranging signals to a receiver (Rx) unit located in obstacle environments such as wrecked buildings. That is, the Tx signal should be able to perform outdoor-to-indoor transmission. Based on the Tx position and range, the Rx unit can estimate its position. Similar to GNSS positioning, the low frequency radio navigation system requires at least four Tx units (four drones) to be able to robustly estimate the position of the Rx unit. In addition, the Tx unit of the system implements a GPS L1 like direct sequence spread-spectrum ranging signals with low number of message bits to reduce computational demands and energy use (prolong the battery usage). RF signal propagation in cluttered area or through obstacles Radio frequency (RF) signals will be attenuated when the signals travel via an obstacle. Theoretically, the lower the transmission frequency, the higher the RF signal propagation capability. Figure 3 below shows the schematic view of signal propagation mechanisms in cluttered environment with obstacles. In figure 3, propagation losses can be caused by the walls, windows and doors of buildings, trees and other obstacles. In general, the higher the Tx unit height above the ground, the less obstacles it will have and the less signal propagation loss it will experience. The main propagation loss due to obstacles experienced by transmitted signals from a Tx unit are classified into several conditions, that are: • Propagation loss due to propagation in-between buildings or houses or trees • Propagation loss due to propagation above the rooftops of buildings or houses • Propagation loss through low-attenuating window and grazing incident angle • Propagation loss due to guided indoor propagation and walls/doors/multi-floor openings The total propagation-loss $PL_{out-to-in}$ of an outdoor-to-indoor transmission is formulated as: Where: • $PL_{out}$ is the total propagation loss for outdoor-to-outdoor transmission and is a function of Tx height $h_{TX}$, Rx height $h_{RX}$, the linear distance from the Tx to an incident point of a building $d_{out}$ and transmission frequency $f$. • $L_{BEL}$ is the propagation loss for outdoor-to-indoor transmission and is a function of building materials $B_{m}$, transmission frequency $f$ and incident angle $\theta$. • $PL_{in}$ is the propagation loss for indoor-to-indoor transmission and is a function of the linear distance from the incident point on the building to the Rx unit $d_{in}$ and number of openings $n_{walls}$, such as doors, corridors and walls. It is important to note that different building materials will significantly affect the propagation loss of RF signals when by passing the wall of buildings. In addition, the propagation loss is also affected by the RF frequency. Low frequency radio navigation system (LowRF) development The development of the low frequency radio navigation system is based on commercial high-grade software defined radio (SDR) from Ettus. For the Tx unit USRP E312 is used. Meanwhile, for the Rx unit, USRP X310 is used. However, the low frequency system can be built by using any other sufficient SDR instruments with low cost, such as Hack-RF SDR. This HACK-RF is well-known and readily available for home use and for hobbyist with community supported opensource software. The E312 used as the Tx unit is a battery-operated SDR system having a compact size. This E312 Tx unit can be mounted onto a medium size drone with payload of around 0.8 Kg. Figure 4 below shows the schematic view of the low radio frequency navigation system. In figure 4, the Tx unit is mounted on a drone so that the Tx unit can get open-sky GNSS signals and construct its coordinate to send to the Rx unit. The Tx unit is battery operated and has an integrated computer to process and generate ranging signals with messages containing the Tx position. The X310 Rx unit is a high-grade SDR receiver to receive the signal form the Tx unit. Special for Rx unit, even a low-grade SDR can be used since the signal processing will be performed in a computer that can be in small or compact size. Any SDR with at least 2 MHz sampling rate capability will be sufficient to be used as the Rx unit of the system, for example, RTL-SDR. The Rx unit is meant to be inside buildings since its main goal is to locate victims or assets in building or obstacle environments. The transmitted message sent by the Tx unit contains 200 bits. This message contains pseudorandom number (PRN) code as well as the position and velocity of the Tx unit mounted on a drone. The software development in the E312 Tx unit use USRP UHD 4.0 libraries with RFNOC 4 framework. READ MORE: TUTORIAL: RFNoC 4 environment development setup for USRP devices. The main transmission frequency used in the implementation of the system is 500 Mhz. The antenna used for the system is a low-cost of-the-self antenna with operating bandwidth of 400 MHz-500 MHz. Figure 5 below shows the implementation of the Tx unit with USRP E312 and the Rx unit with USRP X310. In figure 5, initial test of the system is performed to test the signal operability and the signal processing at the receiver side. The test is performed in a laboratory. From the test, the receiver can get the code delay to be converted for ranging and the position of the Rx unit. In figure 5 for the initial laboratory test, both Rx and Tx units are static. However, Doppler frequency shift is observed at the Rx side. This Doppler shift is due to the TCXO clock drift of the receiver that is usually 1-2 ppm of the frequency, that is around 500 Hz – 75- Hz Doppler shift. Detailed system designed and implementation can be found in [1]. Low RF navigation system field testing After testing the functionality of the Tx and Rx units in the laboratory. Two real tests, also operating at 500 MHz frequency, are performed. The two tests are as follow. The Tx and Rx unit are both in the same building but in different rooms The Tx unit and Rx unit is separated for about 10 m. Since the Tx and Rx units are placed in different rooms, transmitted signals will experience obstacle and multipath. Figure 6 below shows the received signals at the Rx unit and the signal evolution in different processing stage: acquisition and tracking. From figure 6, the raw I/Q signals are distorted due to channel impairments (multipath and obstacles). However, good acquisition can be obtained as the signal peak can be distinctively observed and the correlation peak shows a near perfect triangle shape. From figure 6, the signal can be tracked by observing the IQ constellation plot is clustered in two groups around ‘+’ and ‘-‘ values at the I-axis and around ‘0’ values at the Q-axis. Also, we can observe that the tracked I signal have a rectangular shape representing data bits and the tracked Q signal mostly contains noise centred at zero value. From the tracked I signal, message bit demodulation can be successfully performed. The Tx is mounted on a drone (outdoor) and the Rx is inside a building In this test, the Tx unit is outdoor and is mounted on a drone. The drone is around 20 m height above the ground. Meanwhile, the Rx unit is placed inside a building. Since the distance between Tx and Rx is about 2 km and the drone height is quite low, this transmission is more challenging then at the first case presented above. More obstacles will affect the transmission, especially from trees and other infrastructures (other surrounding buildings and cars). Figure 7 below shows the signal acquisition and tracking processes of received signals by the Tx unit inside the building. From figure 7, we can see clearly that more noises on the acquisition search space are obtained. The noises are due to more obstacles causing high propagation losses of the transmitted signals. The correlation peak in figure 7 has a not-near perfect triangle shape (compared to the one in figure 6). As expected, the noise effects become significant in this type of transmission test. From the tracking process (figure 7), the data or message bit of the signals can be recovered. The tracked I channel shows rectangular shapes representing the data or message bit. Meanwhile, the tracked Q channel only shows noise centred at zero values. Hence, message bit demodulation processes can be successfully performed from this tracked I signal, From these results, the developed system shows promising results to be use for radio navigation system in disaster environments. Conclusion A low frequency radio navigation (LowRF) system specifically designed for disaster environment has been presented and discussed in this post. In disaster environment, a navigation functionality is needed for search and rescue operations to find victims and critical assets, for example, inside a wrecked or destroyed buildings. Common wireless communication system infrastructures are prone to destruction and become unavailable in disaster environments. Other available communication systems, such as walkie talkie, cannot provide navigation services. The low frequency radio navigation system can provide navigation functionalities in disaster environment, for example, navigation into wrecked buildings. One of the main features of the low frequency system is that it has a high signal penetration power due to its low frequency (sub-GHz) transmission. In addition, the system transmits ranging signal containing position data of the transmitter. References [1] Syam, W. P., Scott, D., Conesa, A. Pérez, Rodríguez, I., López, M., Juan, E., Ioannidis, R. T., "Low RF-Based Navigation for Emergencies in Difficult Environments," Proceedings of the 35th International Technical Meeting of the Satellite Division of The Institute of Navigation (ION GNSS+ 2022), Denver, Colorado, September 2022, pp. 1729-1745. https://doi.org/10.33012/2022.18499 You may find some interesting items by shopping here.	2022-11-27 05:00:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3271019458770752, "perplexity": 1929.1068120826835}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00277.warc.gz"}
https://wikimili.com/en/Dynamical_pictures	Dynamical pictures Last updated In quantum mechanics, dynamical pictures (or representations) are the multiple equivalent ways to mathematically formulate the dynamics of a quantum system. Contents The two most important ones are the Heisenberg picture and the Schrödinger picture . These differ only by a basis change with respect to time-dependency, analogous to the Lagrangian and Eulerian specification of the flow field: in short, time dependence is attached to quantum states in the Schrödinger picture and to operators in the Heisenberg picture. There is also an intermediate formulation known as the interaction picture (or Dirac picture) which is useful for doing computations when a complicated Hamiltonian has a natural decomposition into a simple "free" Hamiltonian and a perturbation. Equations that apply in one picture do not necessarily hold in the others, because time-dependent unitary transformations relate operators in one picture to the analogous operators in the others. Not all textbooks and articles make explicit which picture each operator comes from, which can lead to confusion. Schrödinger picture Background In elementary quantum mechanics, the state of a quantum-mechanical system is represented by a complex-valued wavefunction ψ(x, t). More abstractly, the state may be represented as a state vector, or ket, \|ψ⟩. This ket is an element of a Hilbert space , a vector space containing all possible states of the system. A quantum-mechanical operator is a function which takes a ket \|ψ⟩ and returns some other ket \|ψ⟩. The differences between the Schrödinger and Heiseinberg pictures of quantum mechanics revolve around how to deal with systems that evolve in time: the time-dependent nature of the system must be carried by some combination of the state vectors and the operators. For example, a quantum harmonic oscillator may be in a state \|ψ⟩ for which the expectation value of the momentum, ${\displaystyle \langle \psi \|{\hat {p}}\|\psi \rangle }$, oscillates sinusoidally in time. One can then ask whether this sinusoidal oscillation should be reflected in the state vector \|ψ⟩, the momentum operator ${\displaystyle {\hat {p}}}$, or both. All three of these choices are valid; the first gives the Schrödinger picture, the second the Heisenberg picture, and the third the interaction picture. The Schrödinger picture is useful when dealing with a time-independent Hamiltonian H, that is, ${\displaystyle \partial _{t}H=0}$. The time evolution operator Definition The time-evolution operator U(t, t0) is defined as the operator which acts on the ket at time t0 to produce the ket at some other time t: ${\displaystyle \|\psi (t)\rangle =U(t,t_{0})\|\psi (t_{0})\rangle .}$ ${\displaystyle \langle \psi (t)\|=\langle \psi (t_{0})\|U^{\dagger }(t,t_{0}).}$ Properties Unitarity The time evolution operator must be unitary. This is because we demand that the norm of the state ket must not change with time. That is, ${\displaystyle \langle \psi (t)\|\psi (t)\rangle =\langle \psi (t_{0})\|U^{\dagger }(t,t_{0})U(t,t_{0})\|\psi (t_{0})\rangle =\langle \psi (t_{0})\|\psi (t_{0})\rangle .}$ Therefore, ${\displaystyle U^{\dagger }(t,t_{0})U(t,t_{0})=I.}$ Identity When t = t0, U is the identity operator, since ${\displaystyle \|\psi (t_{0})\rangle =U(t_{0},t_{0})\|\psi (t_{0})\rangle .}$ Closure Time evolution from t0 to t may be viewed as a two-step time evolution, first from t0 to an intermediate time t1, and then from t1 to the final time t. Therefore, ${\displaystyle U(t,t_{0})=U(t,t_{1})U(t_{1},t_{0}).}$ Differential equation for time evolution operator We drop the t0 index in the time evolution operator with the convention that t0 = 0 and write it as U(t). The Schrödinger equation is ${\displaystyle i\hbar {\frac {\partial }{\partial t}}\|\psi (t)\rangle =H\|\psi (t)\rangle ,}$ where H is the Hamiltonian. Now using the time-evolution operator U to write ${\displaystyle \|\psi (t)\rangle =U(t)\|\psi (0)\rangle }$, we have ${\displaystyle i\hbar {\partial \over \partial t}U(t)\|\psi (0)\rangle =HU(t)\|\psi (0)\rangle .}$ Since ${\displaystyle \|\psi (0)\rangle }$ is a constant ket (the state ket at t = 0), and since the above equation is true for any constant ket in the Hilbert space, the time evolution operator must obey the equation ${\displaystyle i\hbar {\partial \over \partial t}U(t)=HU(t).}$ If the Hamiltonian is independent of time, the solution to the above equation is [1] ${\displaystyle U(t)=e^{-iHt/\hbar }.}$ Since H is an operator, this exponential expression is to be evaluated via its Taylor series: ${\displaystyle e^{-iHt/\hbar }=1-{\frac {iHt}{\hbar }}-{\frac {1}{2}}\left({\frac {Ht}{\hbar }}\right)^{2}+\cdots .}$ Therefore, ${\displaystyle \|\psi (t)\rangle =e^{-iHt/\hbar }\|\psi (0)\rangle .}$ Note that ${\displaystyle \|\psi (0)\rangle }$ is an arbitrary ket. However, if the initial ket is an eigenstate of the Hamiltonian, with eigenvalue E, we get: ${\displaystyle \|\psi (t)\rangle =e^{-iEt/\hbar }\|\psi (0)\rangle .}$ Thus we see that the eigenstates of the Hamiltonian are stationary states: they only pick up an overall phase factor as they evolve with time. If the Hamiltonian is dependent on time, but the Hamiltonians at different times commute, then the time evolution operator can be written as ${\displaystyle U(t)=\exp \left({-{\frac {i}{\hbar }}\int _{0}^{t}H(t')\,dt'}\right),}$ If the Hamiltonian is dependent on time, but the Hamiltonians at different times do not commute, then the time evolution operator can be written as ${\displaystyle U(t)=\mathrm {T} \exp \left({-{\frac {i}{\hbar }}\int _{0}^{t}H(t')\,dt'}\right),}$ where T is time-ordering operator, which is sometimes known as the Dyson series, after F.J.Dyson. The alternative to the Schrödinger picture is to switch to a rotating reference frame, which is itself being rotated by the propagator. Since the undulatory rotation is now being assumed by the reference frame itself, an undisturbed state function appears to be truly static. This is the Heisenberg picture (below). Heisenberg picture The Heisenberg picture is a formulation (made by Werner Heisenberg while on Heligoland in the 1920s) of quantum mechanics in which the operators (observables and others) incorporate a dependency on time, but the state vectors are time-independent. Definition In the Heisenberg picture of quantum mechanics the state vector, ${\displaystyle \|\psi \rangle }$, does not change with time, and an observable A satisfies ${\displaystyle {\frac {d}{dt}}A(t)={\frac {i}{\hbar }}[H,A(t)]+{\frac {\partial A(t)}{\partial t}},}$ where H is the Hamiltonian and [•,•] denotes the commutator of two operators (in this case H and A). Taking expectation values yields the Ehrenfest theorem featured in the correspondence principle. By the Stone–von Neumann theorem, the Heisenberg picture and the Schrödinger picture are unitarily equivalent. In some sense, the Heisenberg picture is more natural and convenient than the equivalent Schrödinger picture, especially for relativistic theories. Lorentz invariance is manifest in the Heisenberg picture. This approach also has a more direct similarity to classical physics: by replacing the commutator above by the Poisson bracket, the Heisenberg equation becomes an equation in Hamiltonian mechanics. Derivation of Heisenberg's equation The expectation value of an observable A, which is a Hermitian linear operator for a given state ${\displaystyle \|\psi (t)\rangle }$, is given by ${\displaystyle \langle A\rangle _{t}=\langle \psi (t)\|A\|\psi (t)\rangle .}$ In the Schrödinger picture, the state ${\displaystyle \|\psi \rangle }$ at time t is related to the state ${\displaystyle \|\psi \rangle }$ at time 0 by a unitary time-evolution operator, ${\displaystyle U(t)}$: ${\displaystyle \|\psi (t)\rangle =U(t)\|\psi (0)\rangle .}$ If the Hamiltonian does not vary with time, then the time-evolution operator can be written as ${\displaystyle U(t)=e^{-iHt/\hbar },}$ where H is the Hamiltonian and ħ is the reduced Planck constant. Therefore, ${\displaystyle \langle A\rangle _{t}=\langle \psi (0)\|e^{iHt/\hbar }Ae^{-iHt/\hbar }\|\psi (0)\rangle .}$ Define, then, ${\displaystyle A(t):=e^{iHt/\hbar }Ae^{-iHt/\hbar }.}$ It follows that ${\displaystyle {d \over dt}A(t)={i \over \hbar }He^{iHt/\hbar }Ae^{-iHt/\hbar }+e^{iHt/\hbar }\left({\frac {\partial A}{\partial t}}\right)e^{-iHt/\hbar }+{i \over \hbar }e^{iHt/\hbar }A\cdot (-H)e^{-iHt/\hbar }}$ ${\displaystyle ={i \over \hbar }e^{iHt/\hbar }\left(HA-AH\right)e^{-iHt/\hbar }+e^{iHt/\hbar }\left({\frac {\partial A}{\partial t}}\right)e^{-iHt/\hbar }}$ ${\displaystyle ={i \over \hbar }\left(HA(t)-A(t)H\right)+e^{iHt/\hbar }\left({\frac {\partial A}{\partial t}}\right)e^{-iHt/\hbar }.}$ Differentiation was according to the product rule, while ∂A/∂t is the time derivative of the initial A, not the A(t) operator defined. The last equation holds since exp(iHt/ ħ) commutes with H. Thus ${\displaystyle {d \over dt}A(t)={i \over \hbar }[H,A(t)]+e^{iHt/\hbar }\left({\frac {\partial A}{\partial t}}\right)e^{-iHt/\hbar },}$ whence the above Heisenberg equation of motion emerges, since the convective functional dependence on x(0) and p(0) converts to the same dependence on x(t), p(t), so that the last term converts to ∂A(t)/∂t . [X, Y] is the commutator of two operators and is defined as [X, Y] := XY YX. The equation is solved by the A(t) defined above, as evident by use of the standard operator identity, ${\displaystyle {e^{B}Ae^{-B}}=A+[B,A]+{\frac {1}{2!}}[B,[B,A]]+{\frac {1}{3!}}[B,[B,[B,A]]]+\cdots .}$ which implies ${\displaystyle A(t)=A+{\frac {it}{\hbar }}[H,A]-{\frac {t^{2}}{2!\hbar ^{2}}}[H,[H,A]]-{\frac {it^{3}}{3!\hbar ^{3}}}[H,[H,[H,A]]]+\dots }$ This relation also holds for classical mechanics, the classical limit of the above, given the correspondence between Poisson brackets and commutators, ${\displaystyle [A,H]\leftrightarrow i\hbar \{A,H\}}$ In classical mechanics, for an A with no explicit time dependence, ${\displaystyle \{A,H\}={d \over dt}A~,}$ so, again, the expression for A(t) is the Taylor expansion around t = 0. Commutator relations Commutator relations may look different from in the Schrödinger picture, because of the time dependence of operators. For example, consider the operators x(t1), x(t2), p(t1) and p(t2). The time evolution of those operators depends on the Hamiltonian of the system. Considering the one-dimensional harmonic oscillator, ${\displaystyle H={\frac {p^{2}}{2m}}+{\frac {m\omega ^{2}x^{2}}{2}}}$ , the evolution of the position and momentum operators is given by: ${\displaystyle {d \over dt}x(t)={i \over \hbar }[H,x(t)]={\frac {p}{m}}}$ , ${\displaystyle {d \over dt}p(t)={i \over \hbar }[H,p(t)]=-m\omega ^{2}x}$ . Differentiating both equations once more and solving for them with proper initial conditions, ${\displaystyle {\dot {p}}(0)=-m\omega ^{2}x_{0},}$ ${\displaystyle {\dot {x}}(0)={\frac {p_{0}}{m}},}$ ${\displaystyle x(t)=x_{0}\cos(\omega t)+{\frac {p_{0}}{\omega m}}\sin(\omega t)}$ , ${\displaystyle p(t)=p_{0}\cos(\omega t)-m\omega \!x_{0}\sin(\omega t)}$ . Direct computation yields the more general commutator relations, ${\displaystyle [x(t_{1}),x(t_{2})]={\frac {i\hbar }{m\omega }}\sin(\omega t_{2}-\omega t_{1})}$ , ${\displaystyle [p(t_{1}),p(t_{2})]=i\hbar m\omega \sin(\omega t_{2}-\omega t_{1})}$ , ${\displaystyle [x(t_{1}),p(t_{2})]=i\hbar \cos(\omega t_{2}-\omega t_{1})}$ . For ${\displaystyle t_{1}=t_{2}}$, one simply recovers the standard canonical commutation relations valid in all pictures. Interaction Picture The interaction Picture is most useful when the evolution of the observables can be solved exactly, confining any complications to the evolution of the states. For this reason, the Hamiltonian for the observables is called "free Hamiltonian" and the Hamiltonian for the states is called "interaction Hamiltonian". Definition Operators and state vectors in the interaction picture are related by a change of basis (unitary transformation) to those same operators and state vectors in the Schrödinger picture. To switch into the interaction picture, we divide the Schrödinger picture Hamiltonian into two parts, ${\displaystyle H_{S}=H_{0,S}+H_{1,S}~.}$ Any possible choice of parts will yield a valid interaction picture; but in order for the interaction picture to be useful in simplifying the analysis of a problem, the parts will typically be chosen so that ${\displaystyle H_{0,S}}$ is well understood and exactly solvable, while ${\displaystyle H_{1,S}}$ contains some harder-to-analyze perturbation to this system. If the Hamiltonian has explicit time-dependence (for example, if the quantum system interacts with an applied external electric field that varies in time), it will usually be advantageous to include the explicitly time-dependent terms with ${\displaystyle H_{1,S}}$, leaving ${\displaystyle H_{0,S}}$ time-independent. We proceed assuming that this is the case. If there is a context in which it makes sense to have ${\displaystyle H_{0,S}}$ be time-dependent, then one can proceed by replacing ${\displaystyle e^{\pm iH_{0,S}t/\hbar }}$ by the corresponding time-evolution operator in the definitions below. State vectors A state vector in the interaction picture is defined as [2] ${\displaystyle \|\psi _{I}(t)\rangle =e^{iH_{0,S}t/\hbar }\|\psi _{S}(t)\rangle ~,}$ where ${\displaystyle \|\psi _{S}(t)\rangle }$ is the same state vector as in the Schrödinger picture. Operators An operator in the interaction picture is defined as ${\displaystyle A_{I}(t)=e^{iH_{0,S}t/\hbar }A_{S}(t)e^{-iH_{0,S}t/\hbar }.}$ Note that ${\displaystyle A_{S}(t)}$ will typically not depend on t, and can be rewritten as just ${\displaystyle A_{S}}$. It only depends on t if the operator has "explicit time dependence", for example due to its dependence on an applied, external, time-varying electric field. Hamiltonian operator For the operator ${\displaystyle H_{0}}$ itself, the interaction picture and Schrödinger picture coincide, ${\displaystyle H_{0,I}(t)=e^{iH_{0,S}t/\hbar }H_{0,S}e^{-iH_{0,S}t/\hbar }=H_{0,S}.}$ This is easily seen through the fact that operators commute with differentiable functions of themselves. This particular operator then can be called H0 without ambiguity. For the perturbation Hamiltonian H1,I, however, ${\displaystyle H_{1,I}(t)=e^{iH_{0,S}t/\hbar }H_{1,S}e^{-iH_{0,S}t/\hbar },}$ where the interaction picture perturbation Hamiltonian becomes a time-dependent Hamiltonian—unless [H1,s, H0,s] = 0 . It is possible to obtain the interaction picture for a time-dependent Hamiltonian H0,s(t) as well, but the exponentials need to be replaced by the unitary propagator for the evolution generated by H0,s(t), or more explicitly with a time-ordered exponential integral. Density matrix The density matrix can be shown to transform to the interaction picture in the same way as any other operator. In particular, let ${\displaystyle \rho _{I}}$ and ${\displaystyle \rho _{S}}$ be the density matrix in the interaction picture and the Schrödinger picture, respectively. If there is probability ${\displaystyle p_{n}}$ to be in the physical state ${\displaystyle \|\psi _{n}\rangle }$, then ${\displaystyle \rho _{I}(t)=\sum _{n}p_{n}(t)\|\psi _{n,I}(t)\rangle \langle \psi _{n,I}(t)\|=\sum _{n}p_{n}(t)e^{iH_{0,S}t/\hbar }\|\psi _{n,S}(t)\rangle \langle \psi _{n,S}(t)\|e^{-iH_{0,S}t/\hbar }=e^{iH_{0,S}t/\hbar }\rho _{S}(t)e^{-iH_{0,S}t/\hbar }.}$ Time-evolution equations States Transforming the Schrödinger equation into the interaction picture gives: ${\displaystyle i\hbar {\frac {d}{dt}}\|\psi _{I}(t)\rangle =H_{1,I}(t)\|\psi _{I}(t)\rangle .}$ This equation is referred to as the SchwingerTomonaga equation. Operators If the operator ${\displaystyle A_{S}}$ is time independent (i.e., does not have "explicit time dependence"; see above), then the corresponding time evolution for ${\displaystyle A_{I}(t)}$ is given by: ${\displaystyle i\hbar {\frac {d}{dt}}A_{I}(t)=\left[A_{I}(t),H_{0}\right].\;}$ In the interaction picture the operators evolve in time like the operators in the Heisenberg picture with the Hamiltonian ${\displaystyle H'=H_{0}}$. Density matrix Transforming the Schwinger–Tomonaga equation into the language of the density matrix (or equivalently, transforming the von Neumann equation into the interaction picture) gives: ${\displaystyle i\hbar {\frac {d}{dt}}\rho _{I}(t)=\left[H_{1,I}(t),\rho _{I}(t)\right].}$ Existence The interaction picture does not always exist. In interacting quantum field theories, Haag's theorem states that the interaction picture does not exist. This is because the Hamiltonian cannot be split into a free and an interacting part within a superselection sector. Moreover, even if in the Schrödinger picture the Hamiltonian does not depend on time, e.g. H = H0 + V, in the interaction picture it does, at least, if V does not commute with H0, since ${\displaystyle H_{\rm {int}}(t)\equiv e^{{(i/\hbar })tH_{0}}\,V\,e^{{(-i/\hbar })tH_{0}}}$. Comparison of pictures The Heisenberg picture is closest to classical Hamiltonian mechanics (for example, the commutators appearing in the above equations directly correspond to classical Poisson brackets). The Schrödinger picture, the preferred formulation in introductory texts, is easy to visualize in terms of Hilbert space rotations of state vectors, although it lacks natural generalization to Lorentz invariant systems. The Dirac picture is most useful in nonstationary and covariant perturbation theory, so it is suited to quantum field theory and many-body physics. Summary comparison of evolutions Evolution Picture () of: Heisenberg Interaction Schrödinger Ket state constant ${\displaystyle \|\psi _{\rm {I}}(t)\rangle =e^{iH_{0,\mathrm {S} }~t/\hbar }\|\psi _{\rm {S}}(t)\rangle }$ ${\displaystyle \|\psi _{\rm {S}}(t)\rangle =e^{-iH_{\rm {S}}~t/\hbar }\|\psi _{\rm {S}}(0)\rangle }$ Observable ${\displaystyle A_{\rm {H}}(t)=e^{iH_{\rm {S}}~t/\hbar }A_{\rm {S}}e^{-iH_{\rm {S}}~t/\hbar }}$ ${\displaystyle A_{\rm {I}}(t)=e^{iH_{0,\mathrm {S} }~t/\hbar }A_{\rm {S}}e^{-iH_{0,\mathrm {S} }~t/\hbar }}$ constant Density matrix constant ${\displaystyle \rho _{\rm {I}}(t)=e^{iH_{0,\mathrm {S} }~t/\hbar }\rho _{\rm {S}}(t)e^{-iH_{0,\mathrm {S} }~t/\hbar }}$ ${\displaystyle \rho _{\rm {S}}(t)=e^{-iH_{\rm {S}}~t/\hbar }\rho _{\rm {S}}(0)e^{iH_{\rm {S}}~t/\hbar }}$ Equivalence It is evident that the expected values of all observables are the same in the Schrödinger, Heisenberg, and Interaction pictures, ${\displaystyle \langle \psi \mid A(t)\mid \psi \rangle =\langle \psi (t)\mid A\mid \psi (t)\rangle =\langle \psi _{I}(t)\mid A_{I}(t)\mid \psi _{I}(t)\rangle ~,}$ as they must. Notes 1. Here we use the fact that at t = 0, U(t) must reduce to the identity operator. 2. The Interaction Picture, online lecture notes from New York University (Mark Tuckerman) Related Research Articles In quantum mechanics, the Hamiltonian of a system is an operator corresponding to the total energy of that system, including both kinetic energy and potential energy. Its spectrum, the system's energy spectrum or its set of energy eigenvalues, is the set of possible outcomes obtainable from a measurement of the system's total energy. Due to its close relation to the energy spectrum and time-evolution of a system, it is of fundamental importance in most formulations of quantum theory. The mathematical formulations of quantum mechanics are those mathematical formalisms that permit a rigorous description of quantum mechanics. This mathematical formalism uses mainly a part of functional analysis, especially Hilbert space which is a kind of linear space. Such are distinguished from mathematical formalisms for physics theories developed prior to the early 1900s by the use of abstract mathematical structures, such as infinite-dimensional Hilbert spaces, and operators on these spaces. In brief, values of physical observables such as energy and momentum were no longer considered as values of functions on phase space, but as eigenvalues; more precisely as spectral values of linear operators in Hilbert space. The quantum harmonic oscillator is the quantum-mechanical analog of the classical harmonic oscillator. Because an arbitrary smooth potential can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point, it is one of the most important model systems in quantum mechanics. Furthermore, it is one of the few quantum-mechanical systems for which an exact, analytical solution is known. The Schrödinger equation is a linear partial differential equation that governs the wave function of a quantum-mechanical system. It is a key result in quantum mechanics, and its discovery was a significant landmark in the development of the subject. The equation is named after Erwin Schrödinger, who postulated the equation in 1925, and published it in 1926, forming the basis for the work that resulted in his Nobel Prize in Physics in 1933. A wave function in quantum physics is a mathematical description of the quantum state of an isolated quantum system. The wave function is a complex-valued probability amplitude, and the probabilities for the possible results of measurements made on the system can be derived from it. The most common symbols for a wave function are the Greek letters ψ and Ψ. In physics, the Heisenberg picture is a formulation of quantum mechanics in which the operators incorporate a dependency on time, but the state vectors are time-independent, an arbitrary fixed basis rigidly underlying the theory. In physics, the Schrödinger picture is a formulation of quantum mechanics in which the state vectors evolve in time, but the operators are constant with respect to time. This differs from the Heisenberg picture which keeps the states constant while the observables evolve in time, and from the interaction picture in which both the states and the observables evolve in time. The Schrödinger and Heisenberg pictures are related as active and passive transformations and commutation relations between operators are preserved in the passage between the two pictures. Creation and annihilation operators are mathematical operators that have widespread applications in quantum mechanics, notably in the study of quantum harmonic oscillators and many-particle systems. An annihilation operator lowers the number of particles in a given state by one. A creation operator increases the number of particles in a given state by one, and it is the adjoint of the annihilation operator. In many subfields of physics and chemistry, the use of these operators instead of wavefunctions is known as second quantization. The adiabatic theorem is a concept in quantum mechanics. Its original form, due to Max Born and Vladimir Fock (1928), was stated as follows: In quantum mechanics, the interaction picture is an intermediate representation between the Schrödinger picture and the Heisenberg picture. Whereas in the other two pictures either the state vector or the operators carry time dependence, in the interaction picture both carry part of the time dependence of observables. The interaction picture is useful in dealing with changes to the wave functions and observables due to interactions. Most field-theoretical calculations use the interaction representation because they construct the solution to the many-body Schrödinger equation as the solution to the free-particle problem plus some unknown interaction parts. In quantum mechanics, a two-state system is a quantum system that can exist in any quantum superposition of two independent quantum states. The Hilbert space describing such a system is two-dimensional. Therefore, a complete basis spanning the space will consist of two independent states. Any two-state system can also be seen as a qubit. In quantum mechanics, the Hellmann–Feynman theorem relates the derivative of the total energy with respect to a parameter, to the expectation value of the derivative of the Hamiltonian with respect to that same parameter. According to the theorem, once the spatial distribution of the electrons has been determined by solving the Schrödinger equation, all the forces in the system can be calculated using classical electrostatics. In quantum mechanics, the momentum operator is the operator associated with the linear momentum. The momentum operator is, in the position representation, an example of a differential operator. For the case of one particle in one spatial dimension, the definition is: The Ehrenfest theorem, named after Paul Ehrenfest, an Austrian theoretical physicist at Leiden University, relates the time derivative of the expectation values of the position and momentum operators x and p to the expectation value of the force on a massive particle moving in a scalar potential , The Jaynes–Cummings model is a theoretical model in quantum optics. It describes the system of a two-level atom interacting with a quantized mode of an optical cavity, with or without the presence of light. It was originally developed to study the interaction of atoms with the quantized electromagnetic field in order to investigate the phenomena of spontaneous emission and absorption of photons in a cavity. In quantum physics, unitarity is the condition that the time evolution of a quantum state according to the Schrödinger equation is mathematically represented by a unitary operator. This is typically taken as an axiom or basic postulate of quantum mechanics, while generalizations of or departures from unitarity are part of speculations about theories that may go beyond quantum mechanics. A unitarity bound is any inequality that follows from the unitarity of the evolution operator, i.e. from the statement that time evolution preserves inner products in Hilbert space. In quantum mechanics, given a particular Hamiltonian and an operator with corresponding eigenvalues and eigenvectors given by , then the numbers are said to be good quantum numbers if every eigenvector remains an eigenvector of with the same eigenvalue as time evolves. This is a glossary for the terminology often encountered in undergraduate quantum mechanics courses. In quantum mechanics, particularly Perturbation theory, a transition of state is a change from an initial quantum state to a final one. In quantum mechanics, the Schrödinger equation describes how a system changes with time. It does this by relating changes in the state of system to the energy in the system. Therefore, once the Hamiltonian is known, the time dynamics are in principle known. All that remains is to plug the Hamiltonian into the Schrödinger equation and solve for the system state as a function of time. References • Cohen-Tannoudji, Claude; Bernard Diu; Frank Laloe (1977). Quantum Mechanics (Volume One). Paris: Wiley. pp. 312–314. ISBN 0-471-16433-X. • Albert Messiah, 1966. Quantum Mechanics (Vol. I), English translation from French by G. M. Temmer. North Holland, John Wiley & Sons. • Merzbacher E., Quantum Mechanics (3rd ed., John Wiley 1998) p. 430-1 ISBN 0-471-88702-1 • L.D. Landau, E.M. Lifshitz (1977). Quantum Mechanics: Non-Relativistic Theory. Vol. 3 (3rd ed.). Pergamon Press. ISBN 978-0-08-020940-1. • R. Shankar (1994); Principles of Quantum Mechanics, Plenum Press, ISBN 978-0306447907 . • J. J. Sakurai (1993); Modern Quantum Mechanics (Revised Edition), ISBN 978-0201539295 .	2021-10-22 07:01:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 86, "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": 0.8743675351142883, "perplexity": 304.67531356447085}, "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-2021-43/segments/1634323585460.87/warc/CC-MAIN-20211022052742-20211022082742-00013.warc.gz"}
http://mymathforum.com/algebra/39221-two-players-b-play-game-dice.html	My Math Forum Two players A and B play a game of dice . Algebra Pre-Algebra and Basic Algebra Math Forum October 30th, 2013, 01:23 PM #1 Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1 Two players A and B play a game of dice . Two players A and B play a game of dice . They roll a pair of dice alternately . The player who rolls 7 first wins . If A starts then find the probability of B winning the game ? please help me how I can solve this question October 30th, 2013, 01:47 PM #2 Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Two players A and B play a game of dice . Well, the probability of rolling a 7 is 1/6, so the probability of not rolling a 7 is 5/6. If A goes first, he must not roll a 7 in order for B to win. So this has a 5/6 chance of occurrence. Then, if B rolls a 7, B wins. This has a chance of 1/6. So the probability of B winning after one round is $\frac{5}{6} \cdot \frac{1}{6}= \frac{5}{36}$. But this is the probability of B winning in any round, given that neither player has won yet. The probability that neither player wins in a given round is $\frac{5}{6} \cdot \frac{5}{6}= \frac{25}{36}$. Hence, the probability that B wins in the nth round is $(\frac{25}{36})^{n-1} \cdot \frac{5}{36}$. The probability that B wins is then $\frac{5}{36}(1 + \frac{25}{36} + (\frac{25}{36})^2 + (\frac{25}{36})^3 + ... )= \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{5}{36} \cdot \frac{36}{11} = \frac{5}{11}$. October 30th, 2013, 01:59 PM #3 Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Two players A and B play a game of dice . How you got A ) that the probability of rolling a 7 is 1/6 ? B ) the probability that neither player wins in a given round is 25/36 thanks October 31st, 2013, 11:37 AM #4 Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Two players A and B play a game of dice . A) There are 36 possible rolls for two dice, since there are 6 possibilities for the first die and 6 possibilities for the second die. Of these, there are 6 combinations that yield a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). This gives a $\frac{6}{36}= \frac{1}{6}$ chance of rolling a 7. B) Since there is a 1/6 chance of either player rolling a 7, there is a 5/6 chance that A will not roll a 7, and a 5/6 chance that B will not roll a 7. Since these two events are independent, the probability of both of them occurring in one round is the product of both chances, which is $\frac{5}{6} \cdot \frac{5}{6}= \frac{25}{36}$. October 31st, 2013, 12:44 PM #5 Math Team Joined: Oct 2011 Posts: 9,457 Thanks: 640 Re: Two players A and B play a game of dice . Quote: Originally Posted by r-soy How you got A ) that the probability of rolling a 7 is 1/6 ? Example: 1st roll : 3 2nd roll: only a 4 will make 7, so 1/6 Works similarly for any 1st roll result: 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 October 31st, 2013, 03:49 PM #6 Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Two players A and B play a game of dice . thanks all now is clear Tags dice, game, play, players , , , , , , , , , , , , , , # two dices thrown if a gets 5 he win probability of b as winner if a starts firstg Click on a term to search for related topics. Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Math Events 0 February 20th, 2014 12:39 PM PrayForStan Algebra 3 November 30th, 2012 07:10 AM DamballahWeddo New Users 3 August 4th, 2010 03:31 PM rv74 Algebra 0 October 30th, 2008 05:16 AM PrayForStan Number Theory 0 January 1st, 1970 12:00 AM Contact - Home - Forums - Cryptocurrency Forum - Top	2017-06-28 14:11:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 6, "/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": 0.5453783869743347, "perplexity": 560.0943792935857}, "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-2017-26/segments/1498128323682.21/warc/CC-MAIN-20170628134734-20170628154734-00176.warc.gz"}
https://www.sarthaks.com/2723987/find-the-value-of-rm-int-5-6-frac-dx-x-2-16	# Find the value of $\rm \int_{5}^{6} \frac{dx}{x^{2}-16}$ 13 views in Calculus closed Find the value of $\rm \int_{5}^{6} \frac{dx}{x^{2}-16}$ 1. $\rm \frac{1}{4}log\frac{9}{5}$ 2. $\rm \frac{1}{8}log\frac{9}{7}$ 3. $\rm \frac{1}{8}log\frac{ 9}{5}$ 4. $\rm \frac{1}{8}log\frac{7}{5}$ 5. None of these by (54.3k points) selected Correct Answer - Option 3 : $\rm \frac{1}{8}log\frac{ 9}{5}$ Concept: • $\rm \int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}log\frac{\left \|x-a\right \|}{\left \|x+a\right \|}+C$ • $\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)$, where F(x) is the anti-derivative of f(x). Calculation: Given: $\rm \int_{5}^{6} \frac{dx}{x^{2}-16}$ Using the formula, $\rm \int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}log\frac{\left \|x-a\right \|}{\left \|x+a\right \|}+C$ $\rm \Rightarrow \int_{5}^{6} \frac{dx}{x^{2}-16}=\int_{5}^{6} \frac{dx}{x^{2}-4^{2}}=\left [ \frac{1}{8}log\frac{\left \|x-4\right \|}{\left \|x+4\right \|} \right ]_{5}^{6}$ $\rm\Rightarrow \int_{5}^{6} \frac{dx}{x^{2}-16}=\frac{1}{8}(log\frac{\left \|6-4\right \|}{\left \|6+4\right \|}-log\frac{\left \|5-4\right \|}{\left \|5+4\right \|})=\frac{1}{8}(log\frac{\left \|2\right \|}{\left \|10\right \|}-log\frac{\left \|1\right \|}{\left \|9\right \|})$ It is known that log a - log b = log (a / b) $\rm\Rightarrow \int_{5}^{6} \frac{dx}{x^{2}-16}=\frac{1}{8}(log\frac{\left \|2 \times9\right \|}{\left \|10\right \|})=\frac{1}{8}log\frac{9}{5}$ Hence, the correct answer is option 3.	2022-09-26 21:58:45	{"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": 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": 0.7961408495903015, "perplexity": 2098.8699790157275}, "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-2022-40/segments/1664030334942.88/warc/CC-MAIN-20220926211042-20220927001042-00035.warc.gz"}
http://mathoverflow.net/questions/90320/formally-smooth-definition-in-sga-1/90350	# formally smooth definition in SGA 1 SGA 1 introduces formally smooth in a very non-canonical way. The way I usually saw it introduced was through the universal lifting property, i.e., for all $A$-algebra $C$ and all $J\subset C$ nilpotent, every homomorphism $B\to C/J$ lifts to a homomorphism $B\to C$. Grothendieck defers this definition to section 2, however, and instead spends extensive time treating the definition of formally smooth given by: Let $u: A\to B$ be a local homomorphism of local rings, and suppose the residue field of $B$ is finite over the residue field of $A$. Then $u$ is formally smooth (or, Grothendieck states, $B$ is formally smooth over $A$) if there exists a locally finite $\hat{A}$-algebra $A'$ which is free over $\hat{A}$ such that the (and I hope I translated the French correctly here) localizations of the semi-local ring $\hat{B}\otimes_{\hat{A}}A'$ are $A'$-isomorphic to the formal series over $A'$. I guess this definition is deferred to EGA for more intuition in the footnote, but I was wondering why this helps with our understanding of formally smooth, and how this relates to previous concepts Grothendieck introduced which would help with our understanding (e.g., does it generalize, in some sense, quasi-finite?) -	2016-05-31 00:25:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9946531057357788, "perplexity": 326.4402424694373}, "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-2016-22/segments/1464051151584.71/warc/CC-MAIN-20160524005231-00186-ip-10-185-217-139.ec2.internal.warc.gz"}
https://www.springerprofessional.de/actualities-and-development-of-heavy-duty-cnc-machine-tool-therm/13328168?fulltextView=true	main-content ## Weitere Artikel dieser Ausgabe durch Wischen aufrufen 25.07.2017 \| Review \| Ausgabe 5/2017 Open Access # Actualities and Development of Heavy-Duty CNC Machine Tool Thermal Error Monitoring Technology Zeitschrift: Chinese Journal of Mechanical Engineering > Ausgabe 5/2017 Autoren: Zu-De Zhou, Lin Gui, Yue-Gang Tan, Ming-Yao Liu, Yi Liu, Rui-Ya Li Wichtige Hinweise Supported by National Natural Science Foundation of China (Grant No. 51475343), and International Science and Technology Cooperation Program of China (Grant No. 2015DFA70340). ## 1 Introduction The history of the study of the machine tool thermal error is close to a century long. There is still no solution to the thermal error problem with modern high precision CNC machine tools. Most research about the machine tool thermal error has focused on establishing the relationship between the temperature field and thermal error of machine tools, but no solutions have presented themselves well in industry application. Since there have been no new technological breakthroughs in the experimental studies on the thermal error, traditional electrical testing and laser measurement technology are commonly used. The research objects in thermal error testing are usually small and medium-sized CNC machine tools. There is relatively less research on heavy-duty CNC machine tools. Heavy-duty CNC machine tools are pivotal pieces of equipment in many advanced manufacturing industries, such as aerospace, energy, petrochemicals, rail transport, shipbuilding, and ocean engineering. They are widely used in the machining of large parts and high-end equipment, such as steam turbines, large nuclear pumps, marine propellers, and large aircraft wings [ 1]. Improving the machining precision of heavy-duty CNC machine tools is of great significance to comprehensively improving the efficiency of steam turbine units, extending the life of the nuclear power shaft system, reducing the noise of submarine propulsion, reducing the resistance of flight, and so on. The machining error of heavy-duty CNC machine tools can be classified into five parts: (1) Geometric errors produced by machine parts’ manufacturing and assembly; (2) Thermal-induced deformation errors caused by internal and external heat sources; (3) Force-induced deformation errors caused by the cutting force, clamping force, machine tool’s own gravity, etc.; (4) Control errors caused by issues such as the response lag, positioning detection error of the servo system, CNC interpolation algorithm, etc.; (5) Tool wear and the high frequency flutter of the machine tool. The proportion of the thermal deformation error is often the largest for high precision CNC machine tools. In precision manufacturing, the thermal deformation error accounts for about 40% – 70% of the total machining errors [ 2]. In 1933, the influence of heat on precision part processing was noticed for the first time [ 3]. A number of qualitative analysis and contrast tests were carried out between the 1930s and the 1960s. Until the 1970s, researchers used the finite element method (FEM) for machine tool thermal deformation calculations and the optimization of the design of machine tools. The CNC thermal error compensation technology appeared in the late 1970s. After the 1990s, thermal error compensation technology rapidly developed, and many research institutions conducted in-depth studies on the thermal error compensation technology of CNC machine tools based on temperature measurements [ 414]. As shown in Figure 1, the ideology of real-time compensation of thermal error for CNC machine tool consists of two steps. First, extensive experiments are carried out on the CNC machine tools, that collect the CNC data, body temperature of the machine tool, ambient temperature, and the thermal error of the cutting tool tip, in order to establish the thermal error prediction models that are always the multiple linear regression (MLR) model, artificial neural network(ANN) model, and genetic algorithm(GA) model, etc.(shown in Figure 1(a)). Then, the established thermal error prediction model is applied on the CNC machine tool to make error compensation at the tool center point (TCP) through the real-time CNC data and temperature data (shown in Figure 1(b)). Over the past few decades, the International Organization for Standardization (ISO) promulgated a series of standards: ISO 230-3 (thermal deformation of the machine tool) [ 15], ISO 10791-10 (thermal deformation of the machining center) [ 16], and ISO 13041-8 (thermal distortion of the turning center) [ 17]. These standards provide systemic analysis methods for machine tool thermal behavior. Compared to small and medium-sized CNC machine tools, heavy-duty CNC machine tools have unique structural and thermal characteristics, including the following items: (1) Larger and heavier moving parts, like the spindle box, moving beam, and moving workbench; (2) Larger and more complex support structures, such as the machine tool base, column, and beam; (3) More decentralized internal heat sources in 3-D (3-dimensional) space; (4) Greater susceptibility to environmental temperature shifts. As the temperature varies over time and the moving parts are heavy, the thermal and mechanical errors exist a strong coupling effect, making the thermal deformation mechanism more complicated and the optimization of the structural design more difficult. As heavy-duty CNC machine tools are more susceptible to environmental temperature shifts (due to the large volume, small changes of environmental temperature can cause noteworthy accumulations of thermal expansion of the machine tool structure in 3-D space), the robustness of the thermal error prediction model of heavy-duty CNC machine tools is more difficult to control. Monitoring technologies related to the thermal error study of heavy-duty CNC machine tools are important foundation of the research on the machine tool thermal error mechanism and the establishment of a thermal error prediction model. These monitoring technologies include the temperature field monitoring technologies and the thermal deformation monitoring technologies. Further, the thermal deformation monitoring consists of the position error monitoring of the cutting tool tip and the thermal deformation field monitoring of the large structural parts of the machine tool. Because of the unique structural and thermal characteristics of the heavy-duty CNC machine tools mentioned above, there are lots of differences between the heavy-duty machine tool and other machine tools for thermal error monitoring, which can be concluded in three aspects: (1) In terms of temperature field monitoring, as heavy-duty CNC machine tools have a large volume and dispersive heat sources, more temperature measuring points are needed in order to establish an accurate temperature field distribution. Additionally, the installation positions of temperature sensors are more difficult to determine, and the optimization of the temperature measuring points is more complex; (2) In terms of thermal deformation monitoring, there are lots of similarity in position error monitoring of the cutting tool tip between the heavy-duty machine tool and other machine tools. However, for thermal deformation field monitoring of the large structural parts, the Heavy-duty CNC machine tools face greater challenges. The existing machine tool deformation detection techniques are mostly based on the displacement detection instruments, which only detect one point or a few points’ displacement of the machine tool structure. These methods estimate the deformation by the interpolation method. As the structural parts of the heavy-duty CNC machine tool are larger, more conventional displacement sensors or displacement measurement instruments with wide measurement range in the space are needed to reconstruct the whole thermal deformation of the structures. Additionally, as the moving parts of the heavy-duty CNC machine tool are rather heavier than the small and medium-sized CNC machine tools, when the machine tool works, the sedimentation deformation and vibration of the reinforced concrete foundation is more serious and intractable, which reduces the displacement measurement accuracy directly; (3) The processing environment of heavy-duty CNC machine tool is generally worse than the small and medium-sized CNC machine tools. Traditional electric sensors can be easily influenced by the work environment. Humidity, dust, oil pollution, and electromagnetic interference all reduce the sensors’ performance stability and reliability. The long-term thermal error monitoring of the heavy-duty CNC machine tools requires better environmental adaptability and higher reliability to the related sensors. In order to solve the thermal issues of heavy-duty CNC machine tools, we need to analyze the causes of thermal error of machine tool and then carry out in-depth study on the thermal deformation mechanism based on the existing theory and thermal deformation detection technology. In addition, we need to conclude the existing monitoring technologies and provides new technical support for thermal error research on heavy-duty CNC machine tools. Currently, there are many review literatures on the thermal error of CNC machine tools [ 2, 1826], but these papers mainly focus on the thermal issues in small and medium-sized CNC machine tools and seldom introduce thermal error monitoring technologies. This paper focuses on the study of thermal error of the heavy-duty CNC machine tool and emphasizes on its thermal error monitoring technology. First, the causes of thermal error of the heavy-duty CNC machine tool are discussed in Section 2, where the heat generation in the spindle and feed system and the environmental temperature influence are introduced. Then, the temperature monitoring technology and thermal deformation monitoring technology are reviewed in detail in Sections 3 and 4, respectively. Finally, in Section 5, the application of the new optical measurement technology, the “fiber Bragg grating distributed sensing technology” for heavy-duty CNC machine tools is discussed. This technology is an intelligent sensing and monitoring system for heavy-duty CNC machine tools and opens up new areas of research on the heavy-duty CNC machine tool thermal error. ## 2 Causes of Thermal Error of Heavy-Duty CNC Machine Tool ### 2.1 Classification of the Heat Sources The fundamental causes of thermal error of the heavy-duty CNC machine tool are related to the internal and external heat sources. (1) Internal heat sources Heat generated from friction in the spindle, ball screws, gearbox, guides, and other machine parts; Heat generated from the cutting process; Heat generated from energy loss in the motors, electric circuits, and hydraulic system; Cooling influences provided by the various cooling systems. (2) External heat sources Environmental temperature variation; Thermal radiation from the sun and other light sources. For the internal heat sources, the heat generated from the spindle and ball screws has a significant influence on the heavy-duty CNC machine tools and appears frequently in the literatures. The heating mechanism, thermal distribution, and thermal-induced deformation are often researched by theoretical and experimental methods. For the external heat sources, the dynamic change regularity of the environmental temperature and its individual influence and combined effects with internal heat sources on thermal error of heavy-duty CNC machine tools are studied. ### 2.2 Heat Generated in the Spindle #### 2.2.1 Thermal Model of the Supporting Bearing The thermogenesis of the rolling bearings, namely the rolling bearing power loss N f is generally calculated by taking the rolling bearing as a whole. It is the scalar product of the rolling bearing friction torque M f and angular velocity of the inner ring of the bearing. $$N_{\text{f}} = M_{\text{f}} \cdot \pi \cdot n_{i} /30{\kern 1pt} {\kern 1pt},$$ (1) where n i is the rotating speed of the spindle in Eq. ( 1). Palmgren [ 27, 28] developed the experiential formula for the rolling bearing friction torque M f based on experimental tests: $$M_{\text{f}} = M_{\text{l}} + M_{\text{v}} ,$$ (2) $$M_{\text{l}} = f_{1} P_{1} D_{\text{m}} ,$$ (3) M_{\text{v}} = \left\{ \begin{aligned} &10^{3} (vn_{i} )^{{\frac{2}{3}D_{\text{m}}^{3} }} ,\quad vn_{i} \ge 2 \times 10^{ - 3} , \hfill \\ &16f_{0} D_{\text{m}}^{3} {\kern 1pt} {\kern 1pt} ,\quad \quad vn_{i} < 2 \times 10^{ - 3} , \hfill \\ \end{aligned} \right. (4) where M l and M v are the load friction torque and viscous friction torque respectively in Eq. ( 2), D m is the pitch diameter of the bearing, and v is the kinematic viscosity of the lubricating oil. f 0, f 1 and P 1 are related to the bearing type in Eqs. ( 3) and ( 4). Atridage [ 29] modified Palmgren’s equation to take the effect of the lubricating oil flow into consideration. Stein and Tu [ 30] modified Palmgren’s equation to consider the effect of the induced thermal preload. The above models calculate the thermogenesis value of the rolling bearing as a whole, and they do not involve the the surface friction power loss calculation of the inner concrete components of the bearing. Rumbarger, et al. [ 31], used the established fluid traction torque model to calculate the friction power loss of the bearing roller, cage, and inner and outer ring raceway respectively. But their model ignored the heating mechanism differences between the local heat sources. Chen, et al. [ 32], calculated the total thermogenesis of the bearing from the local heat sources with different heating mechanisms. Moorthy and Raja [ 33] calculated the thermogenesis value of the local heat sources, they also took into consideration the change in the diametral clearance after the assembly and during operation that was attributed to the thermal expansion of the bearing parts, which influenced the gyroscopic and spinning moments contributing to the heat generation. Hannon [ 34] detailed the existing thermal models for the rolling-element bearing. #### 2.2.2 Thermal Distribution in the Spindle When studying the temperature field distribution of the spindle, the cause-and-effect model and power flow model should first be analyzed, to determine the heat sources and heat transfer network. Then the heat transfer parameters should be determined, including heat transfer coefficients of the materials, thermal contact resistances between the contact surface, and heat transfer film coefficients. The heat transfer coefficients are easier obtained relatively. The thermal contact resistances between the contact surfaces are concerned with the surface roughness and contact force, and are often obtained by experimental methods [ 35]. Heat convection within the housing is the most difficult to describe, so a rough approximation is often used for the heat transfer film coefficient [ 28]. $$h_{\text{v}} = 0.0332kPr^{1/3} \left( {\frac{{u_{\text{s}} }}{{v_{0} x}}} \right)^{1/2} ,$$ (5) where u s equals the bearing cage surface velocity, x equals the bearing pith diameter, v 0 represents the kinematic viscosity and Pr is the Prandtl number of the oil in Eq. ( 5). It is important to note that for different heat convection objects, the transfer film coefficients have different expression formulas. Bossmanns and Tu [ 36, 37] illustrated the detailed causes and effects of the spindle variables (shown in Figure 2), presented the power flow model (shown in Figure 3), and developed a finite difference thermal model to characterize the power distribution of a high speed motorized spindle. The heat transfer coefficients, thermal contact resistances, and heat convection coefficients were all calculated in their analytical method in detail. More research about the thermal resistance network of the bearing assembly can be found in Refs. [ 38, 39]. As the real structure of a spindle box is complicated, the finite difference method (FDM) and FEM are often preferred to obtain accurate results. Jedrzejewski, et al. [ 40], set up a thermal analysis model of a high precision CNC machining center spindle box using a combination of the FEM and the FDM. Refs. [ 41, 42] created an axially symmetric model for a single shaft system with one pair of bearings using the FEM to estimate the temperature distribution of the whole spindle system. ### 2.3 Heat Generated in the Feed Screw Nuts The thermal deformation of the feed screw nuts effects the linear position error of heavy-duty machine tools. The axial thermal errors become bigger as the runtime of the feed system increases. However, after running for a period of time, the feed system approaches the thermal balance and reaches the approximation steady state, and the variations of thermal errors ease. The variation of radial thermal expansion of the feed screw nuts is so minor that it may be ignored [ 43]. In the ball screw system, the heat generation sources are the nuts and 2 bearings, and the heat loss sources are liquid cooling and surface convection (shown in Figure 4) [ 44, 45]. The thermal balance equation can be expressed by $$Q_{{{\text{b}}1}} + Q_{{{\text{b}}2}} + Q_{\text{n}} - Q_{\text{sc}} - Q_{\text{c}} = \rho cV\frac{\partial T}{\partial t},$$ (6) where Q b1 and Q b2 are the conduction heat from the 2 support bearings, Q n is the conduction heat from the nut, Q sc is the convection heat from rotation of the ball screw shaft, and Q c is the convection heat lost from the cooling liquid. The material density is noted as ρ, c is the specific heat, V is the volume, T is the temperature, and t is the time. The conduction heat from the 2 support bearings, Q b1 and Q b2 can be caculated as shown in section 2.2. The conduction heat from the nut Q sc can be defined as below: $$Q_{\text{sc}} = 0.12\pi f_{{ 0 {\text{n}}}} v_{{ 0 {\text{n}}}} n_{\text{n}} M_{\text{n}} ,$$ (7) where f 0n is a factor related to the nut type and method of lubrication, v 0n is the kinematic viscosity of the lubricant, n n is the screw rotation velocity, and M n is the total frictional torque of the nut (preload and dynamic load) [ 44]. Mayr, et al. [ 46], established the equivalent thermal network model of the ball screw with an analytical method. Xu, et al. [ 44, 47], discovered that, in the case of a large stroke, the heat produced by the moving nut was dispersed on a larger scale than in other cases, so the screw cooling method has better deformation performance than the nut cooling method. Conversely, in the case of a small stroke, the thermal deformation performance of the nut cooling method is better than that of the screw cooling method. Some researchers [ 4, 48, 49] developed the FEM model for the screw, in which the strength of the heat source measured by the temperature sensors was applied to the FEM model to calculate the thermal errors of the feed drive system. Jin, et al. [ 5052], presented an analytical method to calculate the heat generation rate of a ball bearing in the ball screw/nut system with respect to the rotational speed and load applied to the feed system. ### 2.4 Environmental Temperature Effects Environmental temperature fluctuation changes the temperature of the heavy-duty CNC machine tool globally, and affects its machining accuracy greater compared with the small and medium-sized machine tools [ 1]. Environmental temperature fluctuations have daily periodicity and seasonal periodicity simultaneously. Tan, et al. [ 53], decomposed the environmental temperature fluctuations into Fourier series form (shown in Figure 5). x represents time in minutes. The basic angular frequency is ω 0 =2π/ T 0, where T 0=1440 min. A 0 is the average value of the daily cycle temperature that the current temperature belongs to, and it can be obtained from the temperature history through time series analysis. A n represents the amplitude of the temperature fluctuation for each order, and the orders are multiples of the basic frequency ω 0. ϕ n is the initial phase of each order. Tan, et al’s experiment verifies that the environmental temperature has a significant impact on the thermal error of the heavy-duty CNC machine tool, and there exists hysteresis time between the environmental temperature and the corresponding thermal deformation that changes with climate and seasonal weather. Zhang, et al. [ 54], established the thermal error transfer function of each object of the machine tool based on the heat transfer mechanism. Then, based on the assembly dimension chain principle, the thermal error transfer function of the whole machine tool was obtained. As the thermal error transfer function can be deduced using Laplace transform, the thermal error characteristic of the machine tool can be studied with both time domain and frequency domain methods. Taking the environmental temperature fluctuations as input, based on the thermal error transfer function, the environmental temperature induced thermal error can be obtained. ### 2.5 Thermal Analysis of the Global Machine Tool The heat generated in the spindle and feed screw nuts is discussed in Sections 2.1 and 2.2. The global thermal deformation of heavy-duty machine tools is influenced by varieties of heat sources as noted at the beginning of Section 2. The FEM was utilized for the thermal analysis of the global machine tool. Mian, et al. [ 55], presented a novel offline technique using finite element analysis (FEA) to simulate the effects of the major internal heat sources, such as bearings, motors and belt drives, and the effects of the ambient temperature variation during the machine’s operation. For this FEA model, the thermal boundary conditions were tested using 71 temperature sensors. To ensure the accuracy of the results, experiments were conducted to obtain the thermal contact conductance values. Mian, et al. [ 56], further studied the influence of the ambiance temperature variation on the deformation of the machine tool using FEM. The validation work was carried out over a period of more than a year to establish the robustness to the seasonal changes and daily changes in order to improve the accuracy of the thermal simulation of machine tools. Zhang, et al. [ 57], proposed a whole-machine temperature field and thermal deformation modeling with a simulation method for vertical machining centers. Mayr, et al. [ 5860], combined the advantages of FDM and FEA to simulate the thermo-mechanical behavior of machine tools (shown in Figure 6). The transient 3-D temperature distribution at discrete points of time during the simulated period was calculated using the FDM. Those were then used as temperature field for the FEM to calculate the thermally induced deformations. ## 3 Temperature Field Monitoring Technology for Heavy-Duty CNC Machine Tools The formation process of thermal errors in heavy-duty CNC machine tools occurs in the following steps: heat sources→temperature field→thermal deformation field→thermal error. It is obvious that the relationship between the thermal deformation field and thermal error is more relevant than the relationship between the temperature field and thermal error. However, it is quite difficult to measure the micro-thermal-deformation of the whole machine structure directly and the surface temperature of the machine tool is easier to obtain inversely. Existing thermal error prediction models are mostly based on the temperature measurement from the surface of the machine tool, establishing the relationship between the thermal drift of the cutting tool tip and the temperature at critical measuring point. Therefore, the temperature monitoring of the machine tool is a key technology in the thermal error research of CNC machine tools. It can be divided into the contact-type temperature measurements and non-contact temperature measurements, according to the installation form of the temperature sensor. ### 3.1 Contact-Type Temperature Measurement of Heavy-Duty CNC Machine Tools The contact-type surface temperature measurement sensors used in the temperature monitoring of CNC machine tools are mainly thermocouples and platinum resistance temperature detector (RTD). Their installation can be divided into the paste-type, pad-type, and screw-type. Thermocouples and platinum resistance temperature detectors are mostly used for discrete surface temperature measurement. Heavy-duty CNC machine tools have a large volume and decentralized internal heat sources. Delbressine, et al. [ 61], realized that it was difficult to determine the locations and qualities of the temperature sensors, so numerous temperature sensors should be arranged on the surface of the machine tools. Mian, et al. [ 55], used 65 temperature sensors to measure the detailed temperature gradient caused by the internal heat sources, and applied them to FEM. Zhang, et al. [ 57], used 32 platinum resistance temperature sensors to establish a machine tool temperature field. In 2014, Tan, et al. [ 53], installed 33 temperature sensors on the heavy-duty gantry type machine tool XK2650 (shown in Figure 7), and established a thermal error prediction model considered the influence of the environmental temperature. This model can predict 85% of the thermal error and has a good robustness. In addition, Refs. [ 6268] used the thermal resistance to measure the machine tool surface temperature, and Refs. [ 69, 70] used the thermocouples. Werschmoeller and Li [ 71] embedded 10 mini flaky thermocouples into the cutting tool to monitor the cutting tool temperature field. Liu, et al. [ 72], embedded thermocouples into the workpiece to investigate the workpiece temperature variations that resulted from helical milling. These electrical temperature-sensing technologies mainly utilize the linear relationship between the values of the potential, resistance, or other electrical parameters of the sensing materials and temperature to detect the temperature. They have a simple structure, fast response capability, high sensitivity, and good stability. So they play an important role in the thermal error experimental research of heavy-duty CNC machine tools. However, there are some common flaws in the electrical temperature measurement sensors. These include the following: (1) Many parts of the heavy-duty CNC machine tools are in environments exposed to oil, metal cutting chip dust, and coolant. The wires and sensitive components of electrical temperature sensors are all made from metal materials, which are susceptible to corrosion and damage, and have a short working life in a relatively harsh environment. (2) Weak ability to resist electromagnetic interference Heavy-duty CNC machine tools have many inductance elements, like the motors and electric control cabinet, which form strong time-varying electromagnetic fields. The testing signals of electrical temperature sensors are easily interfered with electromagnetic fields during transmission, reducing the signal-to-noise ratio (SNR), accuracy and reliability of the test data. (3) Wide variety of signal transmission wires The principle of electrical temperature sensors is that of an electrically closed circuit. A single electrical temperature sensor has two conductor wires, and a plurality of electrical sensors cannot be connected in a series connection. If there are N electrical sensors, there are 2 N wires. So it is difficult to create the layout of large amounts of wires in heavy-duty CNC machine tools. The testing results for the above electrical temperature sensors show the discrete-point temperature of the heavy-duty CNC machine tool’s surface. The whole temperature field can be reconstructed by using the FDM. Due to the use of few discrete temperature points, it is difficult to establish an accurate integral temperature field of a heavy-duty CNC machine tool, particularly to calculate its internal temperature. Currently, prediction models such as multiple regression or neural networks, are all established based on discrete temperature points, so there is little research on the integral temperature field reconstruction of heavy-duty CNC machine tools. However, it is of great significance for the study of the thermal error mechanism to obtain the 3-D temperature field of CNC machine tools. ### 3.2 Non-Contact Type Temperature Measurement of Heavy-Duty CNC Machine Tools Currently, infrared thermal imaging technology is a non-contact type temperature measurement method that is often applied to thermal error study of heavy-duty CNC machine tools, and it is part of the radiation temperature measurement method. A thermal infrared imager gathers infrared radiant energy and delivered it to an infrared detector through the optical system, in order to process the infrared thermal image. Using the thermal infrared imager test results, one can select the key temperature points to establish a thermal error model. Qiu, et al. [ 73], measured the spindle box temperature field through FLIR thermal imager, and selected 18 temperature points symmetrically to establish the model of the spindle thermal components using the multiple linear regression method. Infrared thermal imaging is suitable for the study of the thermal characteristics of key parts of the heavy-duty CNC machine tools as it visualizes the global temperature field of the surface with a high temperature resolution. Wu, et al. [ 74], researched the thermal behaviors of the support bearing (shown in Figure 8) and screw nut of the ball screws (shown in Figure 9) by using infrared thermographs. Uhlmann and Hu [ 75], captured the temperature field when the spindle running was at 15000 r/min for 150 min, and compared their data with emulational temperature fields (shown in Figure 10). Xu, et al. [ 47], examined the heat generation and conduction of the ball screws and investigated how the different cooling methods affects the temperature distribution using infrared thermal imaging technology. Zhang, et al. [ 76], studied temperature variable optimization for precision machine tool thermal error compensation using infrared thermometer. The infrared thermal imager can visualize the temperature distribution of CNC machine tools, and plays an important role in thermal error study of CNC machine tools. However, the infrared thermal imager is a two-dimensional plane imaging infrared system. One infrared thermal imager cannot measure the overall global temperature field of heavy-duty CNC machine tools. Even with the use of multiple expensive infrared cameras for measuring the global temperature field of heavy-duty CNC machine tools, it is still difficult to track the temperature field of the moving parts when heavy-duty CNC machine tools are involved in actual processing. The shortcomings mentioned in Section 3.1 and Section 3.2 limit the electrical temperature sensors and infrared sensing technology for monitoring the real-time temperature over the long-term in heavy-duty CNC machine tools. There must be some breakthroughs in the temperature field measurement of heavy-duty CNC machine tools in order to develop a highly intelligent temperature measurement and thermal error compensation system that is suitable for heavy-duty CNC machine tools for commercialization. ## 4 Thermal Deformation Monitoring Technology for Heavy-Duty CNC Machine Tools ### 4.1 Thermal Error Monitoring of the Cutting Tool Tip For the thermal error detecting of the cutting tool tip, three categories of sensors are mainly used, that are non-contact displacement detection sensors, high precision double ball gauge, and laser interferometer. The non-contact displacement detection sensors utilized in machine tools include eddy current transducers, capacitive transducers and laser displacement sensors. Though their sensing principles are different, their installation and error detection method are consistent with each other. The high precision double ball gauge and laser interferometer are mainly used to detect the dynamic geometric error of the machine tool, and they can also be competent at thermal error detecting. #### 4.1.1 Five-Point Detection Method The five-point detection method (shown in Figure 11) is only applicable to monitoring the thermal error of a machine tool when the spindle box is not moving. It detects the thermal deformation caused by the ambient temperature or by the rotation of the spindle in ISO230-3. This method measures the three position errors δ px , δ py , and δ pz in the X, Y, and Z direction and two angle errors ε px and ε py rotating around the X and Y axes of the tool cutting tip. Their values can be calculated by Eq. ( 8): \left\{ \begin{aligned} \delta_{{{\text{p}}x}} = \delta_{x1} + L \times \varepsilon_{{{\text{p}}x}} , \hfill \\ \delta_{{{\text{p}}y}} = \delta_{y1} + L \times \varepsilon_{{{\text{p}}y}} , \hfill \\ \delta_{\text{pz}} = \delta_{z} {\kern 1pt} , \hfill \\ \varepsilon_{{{\text{p}}x}} = (\delta_{y1} - \delta_{y2} )/d, \hfill \\ \varepsilon_{{{\text{p}}y}} = (\delta_{x1} - \delta_{x2} )/d, \hfill \\ \end{aligned} \right. (8) where δ x1, δ x2, δ y1, δ y2, and δ z are the displacements detected by the displacement sensors S X1, S X2, S Y1, S Y2, and S Z . d is the sensor distance between sensor S X1 and sensor S X2, and L represents the effective length of the test mandrel that is often made by steel, or invar alloy to oatain higher testing precision. As the test mandrel is a cylinder, a shift in one direction will cause a test error in the other direction. Using δ x1 and δ y1 for an example (shown in Figure 12), when the cutting tool tip moves from point O to point O’ in the XOY plane, the real position errors are δ x1 and δ y1. However, the position errors detected by the displacement sensors are δ x1 and δ y1, correspondingly. The relationship between them can be expressed by Eq. ( 9): \left\{ \begin{aligned} \delta_{x1}^{2} + (R - \delta^{\prime}_{y1} + \delta_{y1} )^{2} = R^{2} , \hfill \\ \delta_{y1}^{2} + (R - \delta^{\prime}_{x1} + \delta_{x1} )^{2} = R^{2} , \hfill \\ \end{aligned} \right. (9) where R represents the radius of the test mandrel. The real position errors δ x1 and δ y1 can be expressed by the displacement sensors’s detected data. It has to be noted that δ px , δ py , δ pz , ε px , and ε py interact with each other, and Eq. ( 9) does not consider ε px and ε py . #### 4.1.2 High Precision Double Ball Gauge Method A double ball gauge consists of two precision metal spheres and a telescoping bar installed on a grating ruler that can detect the displacement (shown in Figure 13). The double ball gauge method is recommended in the ASME B5.54 [ 77] to detect comprehensive error in machine tools. The advantage of this method is that it can detect the tool tip trajectory error caused by the geometric error and thermal deformation. However, the heat expansion and bending deformation of the telescoping bar or a small displacement of the stand affect the test accuracy of this method. #### 4.1.3 Laser Measurement Method The laser interferometer instrument utilizes the Doppler effect caused by the frequency shift to detect the machine tool’s linear position error (shown in Figure 14) and angle error (shown in Figure 15) moving along the guide [ 78]. A dual frequency laser interferometer is a heterodyne interferometer based on single frequency laser interferometers. It has a large gain and high SNR and it is especially suitable for measuring the thermal error of heavy-duty CNC machine tools. The laser measurement methods are widely used for geometric accuracy calibration and the heat-induced position error, angle error and straightness error of heavy-duty CNC machine tools. Ruiz, et al. [ 79], designed a set of optical measuring systems based on the laser interference principle to track and locate the tool tip of the machine tool. ### 4.2 Thermal Deformation Monitoring of Large Structural Parts of the Machine Tool Currently, the displacement detection apparatus, which detects the deformation of large structural parts of heavy-duty CNC machine tools is based on the laser displacement sensors, eddy current sensors, and capacitive sensors. For instance, Gomez-Acedo, et al. [ 80], utilized the inductive sensors array to measure the thermal deformation of a large gantry-type machine tools(shown in Figure 16). Additionally, the laser interferometer with different accessories can measure a range of values, including the precision position, straightness, verticality, yaw angle, parallelism, flatness, and turntable accuracy, and it plays an important role in the detection of the thermal deformation of a heavy-duty CNC machine tool. However, since some of these instruments mentioned above are very large, or in demanding environments, or have small measurement range, it is difficult to engage in the long-term monitoring of heavy-duty CNC machine tools. The direct measurement method requires the installation of displacement sensors on a fixed base as a benchmark, but for CNC machine tools, especially heavy-duty CNC machine tools, it is difficult to find a large constant benchmark (any large base will incur thermal deformation or force-induced deformation, both of which affect the measurement accuracy). It is difficult for the direct displacement measurement method to completely reconstruct the real-time deformation of heavy-duty CNC machine tools. Therefore, researchers are trying to find a more reliable and practical measuring principal and method to monitor the deformation of heavy-duty CNC machine tool structures using laser interferometer [ 81]. ## 5 Application of FBG Sensors in Heavy-Duty CNC Machine Tools ### 5.1 Principle and Characteristics of Fiber Bragg Grating Sensors A fiber Bragg grating sensor is a type of optical sensitive sensor that has been utilized and studied for nearly forty years. A fiber Bragg grating sensor has a number of unparalleled characteristics. It is small and explosion-proof, has electrical insulation, and is immune to electromagnetic interference. It offers high precision, and high reliability. Multiple FBG sensors can be arranged in one single fiber. Therefore, it has been widely used in many engineering fields and mechanical system [ 82]. The sensing principle of a FBG is fundamentally based on a periodic perturbation of the refractive index along the fiber axis formed by exposing the fiber core to the illumination of an intense ultraviolet interference pattern. When a broad-band light propagates along the optical fiber to a grating, a single wavelength is reflected back while therest of the signal is transmitted with a small attenuation (shown in Figure 17). The reflected wavelength is the Bragg wavelength and it can be expressed by the following equation: $$\lambda_{\text{B}} = 2n_{\text{eff}} \Lambda,$$ (10) where λ B is the Bragg wavelength, n eff is the effective refractive index of the fiber core, and Λ is the grating period. A FBG shows great sensitivity to various external perturbations, especially strain and temperature. Any change of stain or temperature will cause the change of n eff or Λ, and lead to the shift of λ B. Hence, by monitoring the Bragg wavelength shift, the value of the strain or temperature is determined. The wavelength variation response to the axial strain change Δ ε and temperature change Δ T is given by: $$\frac{{\Delta \lambda_{\text{B}} }}{{\lambda_{\text{B}} }} = (1 - p_{\text{e}} )\Delta \varepsilon + (\alpha_{\text{f}} + \zeta )\Delta T{,}$$ (11) where p e, α f, and ζ are, respectively, the effective photoelastic coefficient, thermal expansion coefficient, and thermal-optic coefficient of the fused silica fiber. In the literatures, there is little application of fiber grating sensors in the manufacturing industry and almost nothing concerning the machine tool temperature detection and thermal error monitoring. The detection technology based on fiber Bragg grating sensing is especially suitable for the thermal error monitoring of heavy-duty CNC machine tools. It offers a number of advantages over traditional detection technologies, as shown follows: (1) A fiber Bragg grating sensor has a small volume, light weight, and high measurement precision. It is especially unparalleled when a series of FBG sensors that detect a variety of physical parameters distribute in a single fiber. It is suitable for heavy-duty CNC machine tool’s large volume, multiple heat sources, and complex structure. (2) A fiber Bragg grating sensor is highly resistant to corrosion, and high temperature. It is especially suitable for the processing under conditions of high temperature, high humidity, excessive vibration, dust, and other harsh environment. It meets the requirements of the long-term stability and reliability for machine tool detection. (3) A fiber Bragg grating sensor has electrical insulation, and is immune to electromagnetic interference (EMI), making it suitable for harsh processing conditions of the heavy-duty CNC machine tool. It can achieve accurate measurement of the thermal error of machine tools. ### 5.2 Temperature Field Monitoring of Heavy-Duty CNC Machine Tool Based on Fiber Bragg Grating Sensors #### 5.2.1 Fiber Bragg Grating Temperature Sensors for the Surface Temperature Measurement of Machine Tools The fiber Bragg grating temperature measurement technology has become more mature, but the research in this field is mainly concentrated on the extremely high or low temperature measurement and the temperature sensitive enhancing technology. Currently, fiber Bragg grating temperature sensors can be divided into 5 parts by the form of packaging: tube-type fiber Bragg grating temperature sensor [ 83, 84], substrate-type fiber Bragg grating temperature sensor [ 85, 86], polymer packaged fiber Bragg grating temperature sensor [ 87], metal-coated fiber Bragg grating temperature sensor [ 8893], and sensitization-type fiber Bragg grating temperature sensor [ 94]. In order to easily install the temperature sensor and not destroy the internal structure of the machine tools, the temperature of the surface of the machine tools is usually tested and used as the basic data for the thermal error compensation. Measuring the surface temperature accurately in the high gradient temperature field is an challenging technical problem. In the present study, the traditional electrometric method for measuring the temperature of the surface of the machine tool rarely considers the precision problem of measurement. Fiber Bragg grating has been widely used in the field of temperature measurement, but there is little research on the measurement error of the surface temperature measurement. The machine tool surface temperature measurement error can be divided into 3 parts: (1) When the temperature sensor’s surface makes contact with the machine tool, the heat flow will be more concentrated at the testing point. It results in temperature measurement error Δ T 1. (2) The thermal contact resistance between a temperature sensor’s surface and machine tool surface results in a temperature drop Δ T 2. (3) There is a certain distance between the temperature sensor’s sensing point and the surface of the machine tool, which creates the temperature measurement error Δ T 3. Optical fibers are mainly made of quartz and organic resin material. Their thermal conductivity is less than the metal material wire of the thermocouple and thermal resistance. The first temperature measurement error Δ T 1 of the fiber Bragg grating is significantly smaller than that of the latter two, and the main error factors are the thermal contact resistance and the distance of the temperature sensing point. A high gradient temperature field model of the heating surface is established by the FEM [ 95]. In this model, when the hot surface temperature and the air temperature are 90.2 °C and 22 °C, the temperature falling gradient near the hot surface is −46.4 °C/mm (shown in Figure 18). Due to the existence of the coating layer on the surface of the fiber Bragg grating sensor, there is about a 0.15 mm gap between the machine tool surface and the fiber Bragg grating temperature sensing point. In the temperature gradient of −46.4 °C/mm, the small space is sufficient to produce a large temperature test error. By using thermal conductive paste, the uniformity of the surface temperature can be improved, and the error of the surface temperature measurement by FBG can be significantly reduced compared to that from a commercial thermal resistance surface temperature sensor. Ref. [ 96] studied influence of the installation types on surface temperature measurement by a FBG sensor. The surface temperature measurement error of the FBG sensor with single-ended fixation, double-ended fixation and fully-adhered fixation are theoretical analyzed and experimental studied. The single-ended fixation results in a positive linear error with increasing surface temperature, while the double-ended fixation and fully-adhered fixation both result in non-linear error with increasing surface temperature that are affected by thermal expansion strain of the tested surface’s material. Due to its linear error and strain-resistant characteristics, the single-ended fixation will play an important role in the FBG surface temperature sensor encapsulation design field . #### 5.2.2 Temperature Measurement of the Machine Tool Spindle Bearing Based on the Fiber Bragg Grating Sensors The spindle is the core component with complex assembly mechanical structure in heavy-duty CNC machine tool. The spindle consists of the rotating shaft, the front and rear bearings, and the spindle base. For the motorized spindle, it also includes the rotor and stator. As the structure of the spindle is very compact and narrow, to fix the temperature sensor inside the spindle is difficult. The thermogenesis of spindle’s front bearings is a research hotspot that has great influence to the thermal error of the heavy-duty CNC machine tool. Liu, et al. [ 97], installed two FBG temperature sensors (position 1 and 3) and four thermal resistors (position 1, 2, 3, and 4) on the bearing support surface of the spindle (shown in Figure 19). With the shaft rotating freely, the temperature rise amplitudes in position 1, 2, 3, and 4 are consistent with each other. The measured temperature by the FBG temperature sensors and the thermal resistors are the same. As the volume of the FBG temperature sensor is rather smaller than the commercial thermal resistors and thermocouples, it has natural advantages to measure the internal temperature of the spindle. Dong, et al. [ 98], embedded six FBGs connected in one fiber into the spindle housing. These FBGs were installed on the outer ring surface of the front bearing equidistantly in the circumferential direction (shown in Figure 20). With the shaft rotating freely or under radius force, the corresponding uniform or non-uniform temperature field of the outer ring was measured. Additionally, based on the testing, the influence of bearing preload on the temperature rise of the bearing was studied. #### 5.2.3 Thermal Error Measurement of a Heavy-Duty CNC Machine Tool Based on the Fiber Bragg Grating Sensors Huang, et al. [ 99], measured the surface temperature field of a heavy-duty machine tool using the fiber Bragg grating temperature sensor (shown in Figure 21). Three fibers engraved with 27 fiber Bragg grating sensors were arranged on the bed, column, motor, spindle box, and gear box (shown in Figure 22). The temperature was monitored for 24 h. The laser displacement sensors were utilized to measure the offset of the tool cutting tip in the directions of X, Y, and Z. In Figure 23, CH1-10, CH2-1 CH2-7, and CH3-6, show the air temperature changes in the different parts of the environment near to the machine tool. The rest show the temperature in different parts of the structure surface of the machine tool. The parts all had the same change trend, but the temperature of the surrounding environment had an effect on the different parts of the machine tool. There was a large temperature gradient on the surface of the structure. Figure 24 shows the relationship of the thermal drift of the tool tip in three directions and the variation of the environmental temperature and the surface temperature of machine tool. The thermal drift in three directions shifted with environmental temperature and machine tool surface temperature. The thermal drift in the Y direction was the largest. When the ambient temperature shift reached about –6 °C, the error in the direction of Y reached about 15 μm. The fiber Bragg grating has the characteristics of multi-point temperature measurement. It can realize the layout of the temperature measurement points in the large surface area of the heavy-duty CNC machine tool, which can realize the reconstruction of the temperature field of the machine tool more accurately. ### 5.3 Heavy-Duty CNC Machine Tool Thermal Deformation Monitoring Based on Fiber Bragg Grating Sensors There have been a number of achievements made in the application of the fiber Bragg grating strain sensor to large structural deformation measurements. By applying the classical beam theory, Kim and Cho [ 100] rearranged the formula to estimate the continuous deflection profile by using strains measured directly from several points equipped with the fiber Bragg sensor. Their method can be used to measure the deflection curve of bridges, which represents the global behavior of civil structures [ 101]. Kang, et al. [ 102], investigated the dynamic structural displacements estimation using the displacement–strain relationship and measured the strain data using fiber Bragg grating. It is confirmed that the structural displacements can be estimated using strain data without displacement measurement. Kang, et al. [ 103], presented an integrated monitoring scheme for the maglev guideway deflection using wavelength-division-multiplexing (WDM) based fiber Bragg grating sensors, which can effectively avoid EMI in the maglev guideway. Yi, et al. [ 104], proposed a spatial shape reconstruction method using an orthogonal fiber Bragg grating sensor array. Fiber Bragg grating sensing technology opens up a new area of study for the real-time thermal deformation monitoring of heavy-duty CNC machine tool structures. The earliest work was done by Bosetti, et al. [ 105107], who put forward a kind of reticular displacement measurement system (RDMS) based on a reticular array of fiber Bragg strain sensors to realize the real-time monitoring of deformations in the structural components of the machine tools (shown in Figure 25). For a planar and isostatic reticular structure (using the numbering conventions shown in Figure 26), the position of the ith node n i = ( x i , y i) can be expressed as a function of the coordinates of the nodes n i−1 and n i−2 and of the length of the 2 connecting beams L 2i−3 and L 2i−4 : \left\{ \begin{aligned} (x_{i} - x_{i - 1} )^{2} + (y_{i} - y_{i - 1} )^{2} = L_{2i - 3}^{2} \hfill \\ (x_{i} - x_{i - 2} )^{2} + (y_{i} - y_{i - 2} )^{2} = L_{2i - 4}^{2}. \hfill \\ \end{aligned} \right. (12) Figure 27 shows that the bending deformation of the RDMS prototype reconstructed by the measurement system respectively, which shows good consistency. In order to allow for the development of more general and 3-dimensional structures, a new algorithm was proposed [ 105]. The problem of calculating the nodal positions on the basis of their distances as measured by the FBG sensors can be reformulated as the a minimization problem. Liu, et al. [ 108], detected the thermal deformation of the column of a heavy-duty CNC machine tool with the integral method based on the FBG sensor array (shown in Figure 28). The strain data was gauged by multiple FBG sensors glued on the specified locations of the machine tool, and then transformed into the deformation. The displacement of the machine tool spindle was also gauged for evaluation. The calculation results show consistency with the testing results obtained from the laser displacement sensor (shown in Figure 29). Refs. [ 109, 110], studied the deformation measurement of heavy-duty CNC machine tool base using fiber Bragg grating array and designed a FBG-based force transducer for the anchor supporting force measurement of the heavy-duty machine tool base. These research can be extended to the analysis of the thermal deformation mechanism and thermal deformation measurement of heavy-duty CNC machine tool. ## 6 Conclusions and Outlook The thermal error compensation technology of CNC machine tools has been developed over decades, but its successful application to commercial machine tools is limited. To some extent, it is still in the laboratory stage. Heavy-duty CNC machine tools play an important role in the national economic development and national defense modernization. However, due to the more complex thermal deformation mechanism and difficulty in the monitoring technology caused by a huge volume, overcoming its thermal error problems is extremely difficult. The fiber Bragg grating sensing technology opens up a new areas of research for thermal error monitoring of heavy-duty CNC machine tools. We need to take advantage of the fiber Bragg grating sensing technology in global temperature fields and thermal deformation field measurements for heavy-duty CNC machine tool to study the thermal error mechanism of heavy-duty CNC machine tool. These can provide technological support for thermal structure optimization design of heavy-duty CNC machine tools. We also need to improve the thermal error prediction model, especially in regards to the robustness problem. Intelligent manufacturing is an important trend in manufacturing technology, and the Industry 4.0 promises to create smart factory [ 111, 112]. Intelligent sensing technology is one of the indispensable foundations for the realization of intelligent manufacturing. The fusion of optical fiber sensing technology and high-end manufacturing technology is an important research direction that will play an important role in the Industry 4.0. ## Unsere Produktempfehlungen ### Premium-Abo der Gesellschaft für Informatik Sie erhalten uneingeschränkten Vollzugriff auf alle acht Fachgebiete von Springer Professional und damit auf über 45.000 Fachbücher und ca. 300 Fachzeitschriften. Literatur Über diesen Artikel Zur Ausgabe	2021-01-19 11:04:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6657072305679321, "perplexity": 2055.136986057391}, "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-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00650.warc.gz"}
https://zbmath.org/?q=an:1245.60013	## Wasserstein geometry of Gaussian measures.(English)Zbl 1245.60013 In this paper the geometric structure of Gaussian measures is considered. The space $$\mathcal{N}^d$$ of Gaussian measures on $$\mathbb{R}^d$$ is a space of finite dimension, and this allows to write down the explicit Riemannian metric, which, in turn, induces the $$L^2$$-Wasserstein distance function. After introducing the $$L^2$$-Wasserstein geometry, results concerning the $$L^2$$-Wasserstein geometry on $$\mathcal{N}^d$$ are determined, and the $$L^2$$-Wasserstein metric is analyzed. Then the completion $$\overline{\mathcal{N}_0^d}$$ of the space $$\mathcal{N}_0^d$$ (consisting of Gaussian measures with mean $$0$$) is studied as an Alexandrov space. Stratification and tangent cones are investigated. In particular, it turns out that the singular set is stratified according to the dimension of the support of the Gaussian measures, providing an explicit nontrivial example of Alexandrov space with extremal sets. ### MSC: 60D05 Geometric probability and stochastic geometry 53C23 Global geometric and topological methods (à la Gromov); differential geometric analysis on metric spaces Full Text: ### References: [1] S. Amari: Differential-Geometrical Methods in Statistics, Lecture Notes in Statistics 28 , Springer, New York, 1985. · Zbl 0559.62001 [2] L. Ambrosio, N. Gigli and G. Savaré: Gradient Flows in Metric Spaces and in the Space of Probability Measures, second edition, Lectures in Mathematics ETH Zürich, Birkhäuser, Basel, 2008. · Zbl 1145.35001 [3] P. Billingsley: Probability and Measure, third edition, Wiley Series in Probability and Mathematical Statistics, Wiley, New York, 1995. · Zbl 0822.60002 [4] Y. Brenier: Polar factorization and monotone rearrangement of vector-valued functions , Comm. Pure Appl. Math. 44 (1991), 375-417. · Zbl 0738.46011 [5] D. Burago, Y. Burago and S. Ivanov: A Course in Metric Geometry, Graduate Studies in Mathematics 33 , Amer. Math. Soc., Providence, RI, 2001. · Zbl 0981.51016 [6] Y. Burago, M. Gromov and G. Perelman: A.D. Aleksandrov spaces with curvatures bounded below , Uspekhi Mat. Nauk 47 (1992), 3-51, 222. [7] D.C. Dowson and B.V. Landau: The Fréchet distance between multivariate normal distributions , J. Multivariate Anal. 12 (1982), 450-455. · Zbl 0501.62038 [8] S. Gallot, D. Hulin and J. Lafontaine: Riemannian Geometry, third edition, Universitext, Springer, Berlin, 2004. [9] W. Gangbo and R.J. McCann: The geometry of optimal transportation , Acta Math. 177 (1996), 113-161. · Zbl 0887.49017 [10] \begingroup C.R. Givens and R.M. Shortt: A class of Wasserstein metrics for probability distributions , Michigan Math. J. 31 (1984), 231-240. \endgroup · Zbl 0582.60002 [11] M. Knott and C.S. Smith: On the optimal mapping of distributions , J. Optim. Theory Appl. 43 (1984), 39-49. · Zbl 0519.60010 [12] J. Lott: Some geometric calculations on Wasserstein space , Comm. Math. Phys. 277 (2008), 423-437. · Zbl 1144.58007 [13] J. Lott: Optimal transport and Perelman’s reduced volume , Calc. Var. Partial Differential Equations 36 (2009), 49-84. · Zbl 1171.53318 [14] J. Lott and C. Villani: Weak curvature conditions and functional inequalities , J. Funct. Anal. 245 (2007), 311-333. · Zbl 1119.53028 [15] J. Lott and C. Villani: Ricci curvature for metric-measure spaces via optimal transport , Ann. of Math. (2) 169 (2009), 903-991. · Zbl 1178.53038 [16] R.J. McCann: A convexity principle for interacting gases , Adv. Math. 128 (1997), 153-179. · Zbl 0901.49012 [17] R.J. McCann: Polar factorization of maps on Riemannian manifolds , Geom. Funct. Anal. 11 (2001), 589-608. · Zbl 1011.58009 [18] R.J. McCann and P.M. Topping: Ricci flow, entropy and optimal transportation , Amer. J. Math. 132 (2010), 711-730. · Zbl 1203.53065 [19] I. Olkin and F. Pukelsheim: The distance between two random vectors with given dispersion matrices , Linear Algebra Appl. 48 (1982), 257-263. · Zbl 0527.60015 [20] B. O’Neill: The fundamental equations of a submersion , Michigan Math. J. 13 (1966), 459-469. · Zbl 0145.18602 [21] F. Otto: The geometry of dissipative evolution equations: the porous medium equation , Comm. Partial Differential Equations 26 (2001), 101-174. · Zbl 0984.35089 [22] G. Perelman: Elements of Morse theory on Aleksandrov spaces , Algebra i Analiz 5 (1993), 232-241. · Zbl 0815.53072 [23] G. Perelman: The entropy formula for the ricci flow and its geometric applications , · Zbl 1130.53001 [24] G. Perelman and A.M. Petrunin: Extremal subsets in Aleksandrov spaces and the generalized Liberman theorem , Algebra i Analiz 5 (1993), 242-256. · Zbl 0802.53019 [25] A. Petrunin: Semiconcave functions in Alexandrov’s geometry ; in Surveys in Differential Geometry 11, Surv. Differ. Geom. 11 , Int. Press, Somerville, MA, 137-201, 2007. · Zbl 1166.53001 [26] K.-T. Sturm: On the geometry of metric measure spaces , I, Acta Math. 196 (2006), 65-131. · Zbl 1105.53035 [27] K.-T. Sturm: On the geometry of metric measure spaces , II, Acta Math. 196 (2006), 133-177. · Zbl 1106.53032 [28] A. Takatsu and T. Yokota, Cone structure of $$\mathit{L}^{2}$$-Wasserstein spaces , · Zbl 1253.28001 [29] P. Topping: $$\mathrsfs{L}$$-optimal transportation for Ricci flow , J. Reine Angew. Math. 636 (2009), 93-122. · Zbl 1187.53072 [30] C. Villani: Topics in Optimal Transportation, Graduate Studies in Mathematics 58 , Amer. Math. Soc., Providence, RI, 2003. · Zbl 1106.90001 [31] C. Villani: Optimal Transport, Grundlehren der Mathematischen Wissenschaften 338 , Springer, Berlin, 2009. This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2022-10-02 18:24:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7905939817428589, "perplexity": 1086.9479373203083}, "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-2022-40/segments/1664030337339.70/warc/CC-MAIN-20221002181356-20221002211356-00103.warc.gz"}
http://forwardscattering.org/post/33	Take a function $$f$$ that maps from some bit string of length $$N$$ to another bit string also of length $$N$$. Suppose that this function is a 'hash function' with some properties we want: 1) Given an output value, it takes on average $$2^{N-1}$$ evaluations of different inputs passed to $$f$$ before we get the given output value. Note that if we try all $$2^N$$ different inputs, we will definitely find the input that gives the output we are searching for, if it exists. So we want our hash function to require half than number of evaluations on average before we find it, e.g. $${2^N}/2 = 2^{N-1}$$. 2) $$f$$ runs in polynomial time on the length of the input $$N$$. All practical hash functions have this property. So let's consider a search problem — call it $$S$$. Instances of $$S$$ consist of an output string $$y$$ (of $$N$$ bits length), and the problem is to find the input $$x$$ such that $$f(x) = y$$. The first property we are assuming of $$f$$ implies that any algorithm that solves this problem runs in time $$2^{N-1}$$, which is exponential in $$N$$, so not polynomial in $$N$$. So $$S$$ is not an element of $$P$$, the set of problems that are solvable in polynomial time. Now, let's suppose that we have some proposed input value $$x$$. We can verify that $$f(x) = y$$ in polynomial time, since we have assumed that $$f$$ runs in polynomial time. If $$f(x) \neq y$$, the proposed solution will be rejected in polynomial time. This means that we have an algorithm (the verification algorithm) that verifies a proposed solution in polynomial time. But we know that the complexity class $$NP$$ (non-deterministic polynomial) is the class of problems that can be verified in polynomial time. Therefore $$S$$ is in $$NP$$. Since we already proved (given our assumptions about $$f$$) that $$S$$ is not in $$P$$, we therefore have $$P \neq NP$$. So we have the following situation: The existence of a hash function with the properties we want implies $$P \neq NP$$. Since no-one knows if $$P = NP$$, we can deduce that no-one knows if hash functions with the properties we want exist. ### Fixed length outputs Note that this proof only applies for hash functions from strings of N bits to strings of N bits. Some cryptographic hash functions in the real world are defined like this, such as the Skein hash function. However, most cryptographic functions produce a fixed length output, such as SHA-256 that has a 256 bit output length. Regardless, I think the same general principle applies. As far as I know, no hash function with a fixed length output has been proved to have the analogous properties as listed above. In other words, no hash function with fixed length output N has been proved to take exponential time on N to find an input that maps to the given output. ### One-way functions This property of hash functions we are thinking about is also called the 'one-way' function property. A hash function with this property is one-way in the following sense: we can compute the output $$f(x)$$, from the input $$x$$, easily and efficiently. However going in the 'other direction', e.g. computing the inverse, $$f^{-1}(y)$$, is not computable efficiently. Remarkably we don't know if such one-way functions exist. ### Implications The implication of the above proof and comments is that no-one knows if cryptographic hash functions like the SHA hash functions really have the secure properties that people hope they do. Many aspects of modern software depend on such hash functions, such as message authentication codes, code signing of executables, and even systems such as bitcoin, which uses the SHA-256 hash function in its proof-of-work scheme. That these hash functions are not proved to be cryptographically secure is somewhat well known, but it's not that uncommon to hear people assuming that the security or properties of such systems have actually been proved. This is quite similar to the situation with factoring of large numbers which is used in public key cryptography, which is suspected, or commonly treated as if it is not solvable in polynomial time, even though it has not been proved.	2021-12-04 19:54:17	{"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": 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": 0.7146929502487183, "perplexity": 232.12893267101435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00174.warc.gz"}
https://math.stackexchange.com/questions/1937409/existence-of-a-metric-which-is-complete-and-equivalent-to-standard-metric	# Existence of a metric which is complete and equivalent to standard metric Does there exist a metric $d: (-1,1)\times (-1,1) \to \mathbb R$ which is complete and is equivalent to the standard metric (Euclidean metric) on $(-1,1)$? I think there does not exist such a metric and my current approach has been to assume both conditions on the metric and try to find a contradiction in terms of a Cauchy sequence not converging to a point in $(-1,1)$ but this hasn't been very succesful so far. I think I am supposed to use the fact that $(-1,1)$ and are $\mathbb R$ are topologically equivalent but I dont know exactly how. Yes, there is; this is immediate from the fact that $(-1,1)$ is homeomorphic to $\Bbb R$, and the standard metric on $\Bbb R$ is complete. To get a concrete example of such a metric, let $$f:(-1,1)\to\Bbb R:x\mapsto\tan\frac{\pi x}2\;,$$ and for $x,y\in(-1,1)$ define $$d(x,y)=\|f(x)-f(y)\|\;.$$ • How can I infer from this that the metric $d$ and the standard metric are equivalent in the sense that they determine the same family of open sets on $(-1,1)$? – Joogs Sep 22 '16 at 18:57 • @Joogs: Use the fact that $f$ and $f^{-1}$ take open intervals to open intervals. – Brian M. Scott Sep 22 '16 at 19:07 • Okay, $(-1,1)$ is homeomorphic to $\mathbb R$ so there exist a bijection $f$ between them, both $f$ and $f^{-1}$ are continuous so they map open intervals to open intervals. But then I cant seem to connect the dots any further. – Joogs Sep 22 '16 at 19:22 • @Joogs: Let $\tau_d$ be the topology generated by $d$, and let $\tau$ be the usual topology on $(-1,1)$. Each open $d$-ball is an open interval, so $\tau_d\subseteq\tau$. Now show that each open interval in $(-1,1)$ is a union of open $d$-balls, so that $\tau\subseteq\tau_d$. – Brian M. Scott Sep 22 '16 at 19:33 • How does it follow from the completeness of $\mathbb R$ that $(-1,1)$ is also complete with respect to $d$ ? – Joogs Sep 22 '16 at 20:08	2019-09-15 07:47:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8868553638458252, "perplexity": 84.70192746026211}, "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/1568514570830.42/warc/CC-MAIN-20190915072355-20190915094355-00349.warc.gz"}
https://www.studyadda.com/solved-papers/jee-main-paper-held-on-10-4-2019-morning_q65/817/374669	• # question_answer If the circles ${{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0$and $2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y1=0,(K\in R),$ intersect at the points P and Q, then the line $4x+5yK=0$ passes through P and Q for : [JEE Main 10-4-2019 Morning] A) exactly two values of KB) exactly one value of KC) no value of K.D) infinitely many values of K Equation of common chord $4kx+\frac{1}{2}y+k+\frac{1}{2}=0$ ....(1) and given line is $4x+5yk=0$ .....(2) On comparing (1) & (2), we get $k=\frac{1}{10}=\frac{k+\frac{1}{2}}{-k}$$\Rightarrow$No real value of k exist	2020-09-26 11:57:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6275057792663574, "perplexity": 1681.3163916237936}, "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-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00578.warc.gz"}
https://crypto.stackexchange.com/digest/preview	## Top new questions this week: ### "Memory-Hard" vs. "Memory-Bound" Functions? One of the approaches in order to prevent Sybil or DoS attacks is CPU-bound PoW. However, because of the influence of Moore’s law, the memory-based approaches are suggested. As actually there are two ... proof-of-work memory-hard ### Are these two methods of differential cryptanalysis different? I'm trying to understand how differential analysis works. I'm reading this and this tutorials, but it looks like they do different things. In the first paper author tells us to bruteforce last round ... cryptanalysis differential-analysis ### Repeated modular square roots to recover original base I'm given a large prime $p$ and $c \equiv m^e \pmod p$, and $e = 2^{64}$. Typical RSA rules don't apply here, since $\phi(p) = p - 1$ is even, and $e$ is a power of two, so they share a common factor, ... rsa modular-arithmetic prime-numbers ### Questions regarding the one-wayness and collision-resistance of a hash function based on RSA-like problem Problem statement: "Bob is a paranoid cryptographer who does not trust dedicated hash functions such as SHA1 and SHA-2. Bob decided to build his own hash function based on some ideas from number ... hash collision-resistance one-way-function ### Curve25519 choice of the prime In the paper for curve25519 is described how the prime was chosen. But I don't understand why the biggest 255-Bit prime was chosen instead of a 256 Bit prime. Can someone explain this to me? elliptic-curves ### Is it possible to create a ciphersuite for TLS 1.3 using only the Keccak-sponge/duplex? Now that Keccak has been chosen as SHA-3 we see that there have been defined quite a few related modes, including authenticated cipher modes, the KMAC message authentication mode, XOF's and whatnot. ... tls sha-3 keccak sponge ### Mapping the hash of message to a point of elliptic curve for signature Let the subgroup $G$ of elliptic curve constructed with point $P$ with prime order $q$ by $G=\langle P\rangle$. The $h(x)$ is a hash function. We want to map the hash of arbitrary message $m$ to a ... elliptic-curves signature hash-signature bls-signature ## Greatest hits from previous weeks: ### How can I use asymmetric encryption, such as RSA, to encrypt an arbitrary length of plaintext? RSA is not designed to be used on long blocks of plaintext like a block cipher, but I need to use it to send a large (encrypted) message. How can I do this? encryption rsa public-key ### How do ciphers change plaintext into numeric digits for computing? For example, in RSA, we use this for encryption: $ciphertext = (m^e \mod n)$ and for decryption. If our message is "hello world", then what number do we have to ... rsa I found this example online: In the elliptic curve group defined by $$y^2 = x^3 + 9x + 17 \quad \text{over } \mathbb{F}_{23},$$ what is the discrete logarithm $k$ of $Q = (4,5)$ to the base $... elliptic-curves asked by Keith Lau Si Keit 14 votes ### Calculating RSA private exponent when given public exponent and the modulus factors using extended euclid When given$p = 5, q = 11, N = 55$and$e = 17$, I'm trying to compute the RSA private key$d$. I can calculate$\varphi(N) = 40$, but my lecturer then says to use the extended Euclidean algorithm to ... rsa number-theory modular-arithmetic asked by DougalMaguire 35 votes answered by Thomas 40 votes ### What is the "shared secret" used for in IPSec VPN? Can somebody explain what the "shared secret" and "password" do when opening/creating a VPN tunnel? In this specific case I setup a VPN to my Fritz!Box and I had to provide a shared secret (which was ... key-exchange transport-security ipsec asked by Krumelur 8 votes ### Impacts of not using RSA exponent of 65537 This RFC says the RSA Exponent should be 65537. Why is that number recommended and what are the theoretical and practical impacts & risks of making that number higher or lower? What are the ... public-key rsa asked by goodguys_activate 65 votes answered by fgrieu 60 votes ### 7zip : Why does encrypting the same file with AES-256 not give the same output? Using 7-zip 19.00, on Windows 10 1909, build 18363.592, I encrypted a text file with the contents "hello there" using AES-256 and the password "123". I did this two times, the exact same procedure, ... encryption aes cbc asked by super 31 votes answered by kelalaka 44 votes ## Can you answer these questions? ### Is there an interactive proof protocol for proving that a preimage in a map$g: Z_{q}^{n} → G_{T}$is mapped to a given public value, say y? There already exist interactive proof protocols with logarithmic communication for proving that a secret multi-exponent$x ∈ Z^{n}_{q}$for a public multi-exponentiation$P = \textbf{g}^{\textbf{x}} ∈ ... protocol-design ### Elliptic curve subgroup with $p$ elements in field of characteristic $p$ Are there any elliptic curves defined over a finite field $\mathrm{GF}(p^k)$ with a subgroup of order $p$ where the discrete log (and preferably DDH) problem is hard? Elliptic curve with prime ... elliptic-curves paillier elliptic-curve-generation ### Importance of supersingularity of elliptic curves I'm struggling to understand the high-level idea of "Verifiable Delay Functions from Supersingular Isogenies and Pairings" (https://eprint.iacr.org/2019/166.pdf) by De Feo et al. I will ... elliptic-curves pairings asked by Max Beikirch 1 vote	2020-09-21 17:06:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3870534896850586, "perplexity": 1727.8083370379757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400201826.20/warc/CC-MAIN-20200921143722-20200921173722-00139.warc.gz"}
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-3-p-s-problem-solving-page-308/10e	## Algebra and Trigonometry 10th Edition The formula: $m=\frac{f(2+h)-4}{h}$ At $h=-1$: $\frac{(2+(-1))^2-4}{-1}=\frac{1-4}{-1}=\frac{-3}{-1}=3$ At $h=1$: $\frac{(2+(1))^2-4}{1}=\frac{9-4}{1}=\frac{5}{1}=5$ At $h=0.1$: $\frac{(2+(0.1))^2-4}{0.1}=\frac{4.41-4}{0.1}=\frac{0.41}{0.1}=4.1$ These values match the ones found in the earlier parts.	2022-09-30 15:03:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8174867630004883, "perplexity": 597.0354637198661}, "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-2022-40/segments/1664030335491.4/warc/CC-MAIN-20220930145518-20220930175518-00337.warc.gz"}
https://tex.stackexchange.com/questions/451150/why-do-routines-of-plain-tex-standard-insert-get-broken-when-using-mark-in-the	# Why do routines of plain TeX standard insert get broken when using \mark in the midst of running text? I use Edmac in plain XeTeX for typesetting scholarly editions, and for the first time have come across a problem that may be intractable. The text in question has frequently changing section numbers, which can occur either at the start of paragraphs or within the body of paragraphs. These are given as bold numerals, which ideally should be picked up by a \mark (or rather \marks) command so that they can be given in the running headlines, along with other changing elements. The problem is that within text that is line-numbered by Edmac all of plain TeX's standard insert routines are broken. A \mark in the midst of running text causes the line to burst apart, so that the only thing that's right is the headline - everything else is messed up. Any suggestions as to how this might be overcome? I vaguely thought of \write, but I don't think it can be used to do the necessary - viz. in the left headline place the section number that is in operation in the first line of that page and in the right headline place the section number that is in operation in the last line of the page. I think I'm going to have to do these manually once the pagination has settled down. Or is there another solution? • did not you receive the email I send to you when you wrote to me with this question? – Maïeul Sep 16 '18 at 20:34 This answer is not fully completed, especially because you didn't provide any MWE. The problem is not caused but the modification of the output routine, but by the fact that edmac (and its derivateds ledmac, eledmac and reledmac) use \vsplit to split the content of each \pstart...\pstart in individual line in order to add line number and to manage critical notes and so on, add do also some addition. I don't know where exactly. The solution is indeed to use \write. More precisly 1. Instead of directly call \mark{content}, write it inside the numbered auxiliary file, wraped in your own command. In your numebred auxiliary file, you should get something like this \@l \@insidethislinehook{\mark{content}} \@l You can use \unexpanded to help you with the problem of expansion. \@l is written by edtex at each line, and, at secund run, it increase line counter. 2. When reading the .1 file, \@insidethislinehook would store its argument in a macro which name includes line number (\absline@num, to define such macro, use \csname) 3. \do@line must be redefined to call the command defined on step 2 in one of the hbox. I don't know well edmac, as I started to work with ledmac, but I think it should be at the begining of this hbox \hbox to \hsize{\affixline@num{. That is the method I used for reledmac, cf https://github.com/maieul/ledmac/commit/14aaca05ff72c9adf1ab35923df36fdb2173b6bb of course, this code as content which does not interest you, as it is specific to reledmac. But see l. 9063 (step 1); l. 9075 (step 2) and 8919 (step 3).	2019-09-21 09:10:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.768369197845459, "perplexity": 1306.7556594226621}, "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-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00021.warc.gz"}
http://stochastix.wordpress.com/2011/08/21/wythoff-sequences/	## Wythoff sequences On November 20, 2007, the Putnam problem of the day was: The set of all positive integers is the union of two disjoint subsets $\{f(1), f(2), f(3), \dots\}$, $\{g(1), g(2), g(3), \dots\}$, where $f(1) < f(2) < f(3) < \dots$, and $g(1) < g(2) < g(3) < \dots$, and $g(n) = f(f(n))+1$ for $n=1, 2, 3,\dots$. Determine $f(240)$. Back in November 2007 I could not solve the problem. Now, almost four years later, I will give it another try. __________ My “solution”: Let $\mathbb{N} := \{1, 2, 3, \dots\}$ be the set of natural numbers (positive integers). We define $F := \{f(n)\}_{n \in \mathbb{N}}$ and $G := \{g(n)\}_{n \in \mathbb{N}}$. Note that we must have $F \cup G = \mathbb{N}$ and $F \cap G = \emptyset$. Let $f(1) = 1$. Then, we obtain $g(1) = f(f(1)) + 1 = f(1) + 1 = 2$. Next, let $f(2) = 3$ and $f(3) = 4$. Then, $g(2) = f(f(2)) + 1 = f(3) + 1 = 5$. Let $f(4) = 6$. Then, we obtain $g(3) = f(f(3)) + 1 = f(4) + 1 = 7$. Next, let $f(5) = 8$ and $f(6) = 9$. Then, $g(4) = f(f(4)) + 1 = f(6) + 1 = 10$. Next, let $f(7) = 11$ and $f(8) = 12$. Then, $g(5) = f(f(5)) + 1 = f(8) + 1 = 13$. Let $f(9) = 14$. Then, we obtain $g(6) = f(f(6)) + 1 = f(9) + 1 = 15$. Next, let $f(10) = 16$ and $f(11) = 17$. Then, $g(7) = f(f(7)) + 1 = f(11) + 1 = 18$. Let $f(12) = 19$. Then, we obtain $g(8) = f(f(8)) + 1 = f(12) + 1 = 20$. Next, let $f(13) = 21$ and $f(14) = 22$. Then, $g(9) = f(f(9)) + 1 = f(14) + 1 = 23$. Let us stop iterating here. So far, we have $\left(f(n)\right)_{n \in \mathbb{N}} = \left(1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, \dots\right)$ $\left(g(n)\right)_{n \in \mathbb{N}} = \left(2, 5, 7, 10, 13, 15, 18, 20, 23, \dots\right)$. Consulting the On-Line Encyclopedia of Integer Sequences (OEIS), we find out that $\left(f(n)\right)_{n \in \mathbb{N}}$ and $\left(g(n)\right)_{n \in \mathbb{N}}$ are the lower Wythoff (A000201) and upper Wythoff (A001950) sequences, respectively. These two are Beatty sequences generated by the golden ratio $\varphi := \frac{1 + \sqrt{5}}{2}$, as follows $f (n) = \lfloor n \varphi \rfloor$, $g (n) = \lfloor n \varphi^2 \rfloor$ where $\lfloor \cdot \rfloor$ is the floor function. Recall that $\varphi^2 = \varphi + 1$, which allows us to write $g (n) = \lfloor n (\varphi + 1) \rfloor$. The following Python script produces the first twenty terms of the lower and upper Wythoff sequences: from math import floor from math import sqrt # golden ratio phi = (1 + sqrt(5)) / 2.0 # define function f def f (n): return int(floor(phi * n)) # define function g def g (n): return int(floor((phi+1) * n)) print "Lower and upper Wythoff sequences:\n" for n in range(1,21): print "f(%d) = %d \t g(%d) = %d" % (n, f(n), n, g(n)) This script produces the output: Lower and upper Wythoff sequences: f(1) = 1 g(1) = 2 f(2) = 3 g(2) = 5 f(3) = 4 g(3) = 7 f(4) = 6 g(4) = 10 f(5) = 8 g(5) = 13 f(6) = 9 g(6) = 15 f(7) = 11 g(7) = 18 f(8) = 12 g(8) = 20 f(9) = 14 g(9) = 23 f(10) = 16 g(10) = 26 f(11) = 17 g(11) = 28 f(12) = 19 g(12) = 31 f(13) = 21 g(13) = 34 f(14) = 22 g(14) = 36 f(15) = 24 g(15) = 39 f(16) = 25 g(16) = 41 f(17) = 27 g(17) = 44 f(18) = 29 g(18) = 47 f(19) = 30 g(19) = 49 f(20) = 32 g(20) = 52 Finally, we have $f (240) = \lfloor 240 \varphi \rfloor = 388$. __________ I am not happy with this “solution”. I obviously cheated when I consulted the OEIS. Supposing one is not acquainted with Beatty sequences, is there any hope of solving the problem? If one happens to be acquainted with Beatty sequences and postulates that $f (n) = \lfloor n r \rfloor$, $g (n) = \lfloor n s \rfloor$ where $\frac{1}{r} + \frac{1}{s} = 1$, how does one arrive at the conclusion that $r = \varphi$ and $s = \varphi^2$? In other words, what would a proper solution look like?	2014-03-09 13:20:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 51, "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": 0.7579837441444397, "perplexity": 1830.3530225299028}, "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/1393999678747/warc/CC-MAIN-20140305060758-00061-ip-10-183-142-35.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/probability-question.489771/	# Probability Question ## Homework Statement This question relates to a famous court case in the USA where for the first time results of statistical analysis were presented as evidence of discrimination. In 1972 Mr Partida, a Mexican-American, was convicted of burglary by a jury in a particular county in Texas. Mr Partida’s lawyers, appealed the conviction on the grounds that the jury must have been selected in a discriminatory (and thus non-random fashion) because it had a disproportionately small number of Mexican-Americans on it. They based their argument on the following two pieces of information. First, a recent census had shown Mexican- Americans made up 79.1% of adults in the population in that county and that this fact was widely accepted. Second, in a sample of court cases in the county involving a total of 870 jury members, only 39% were Mexican-Americans. Assume that we will conclude that jury selection in the county was discriminatory if there is less than a 1% chance that a sample proportion of 39% (or less) would result from random selection from a population where the proportion was 79.1%. If you were a Justice of the US Supreme Court, what would your finding be? Would you find that there probably was discrimination in jury selection in the county? ## Homework Equations lets put Pr[A] = 0.791 Pr = 0.39 number of Mexican American Jury is 870*0.39=339 ## The Attempt at a Solution i was thinking use Pr[A\|B] to work it out, but since event A and B are dependent, so i can't keep going also i try using 870C339 (0.791)^339 (0.209)^(870-339) but this one is at zero.. so can anyone give me some hint? i am totally lost on "less than a 1% chance that a sample proportion of 39% (or less) would result from random selection from a population where the proportion was 79.1%" HallsofIvy Homework Helper With large numbers like those, I would be inclined to use the Normal approximation to the binomal distribution. If Mexican-Americans make up 79.1% of the population, then p= 0.791 and q= 1-0.791= 0.209. Fair selection would be binomial distribution with mean p= 0.791 and standard deviation $\sqrt{pq}= \sqrt{(0.791)(0.209)}= \sqrt{0.165319}= 0.4066$. The "standard variable" would be $z= \frac{x- \mu}{\sigma}= \frac{0.39- 0.791}{0.4066}$. Look that up in a table of the normal distribution to see if the probability of getting that, or less, is less than 1%. A good table is at http://www.math.unb.ca/~knight/utility/NormTble.htm [Broken] Last edited by a moderator:	2021-12-01 18:29:11	{"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": 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": 0.7307032942771912, "perplexity": 2159.541097079211}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00531.warc.gz"}
https://questions.examside.com/past-years/year-wise/jee/jee-main/jee-main-2021-online-18th-march-morning-shift	NEW New Website Launch Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc... ## JEE Main 2021 (Online) 18th March Morning Shift Exam Held on Thu Mar 18 2021 03:30:00 GMT+0000 (Coordinated Universal Time) Click View All Questions to see questions one by one or you can choose a single question from below. ## Chemistry <picture><source media="(max-width: 500px)" srcset="https://imagex.cdn.examgoal.... The ionic radius of Na<sup>+</sup> ions is 1.02 $$\mathop A\limits^o$$. The ion... Reaction of Grignard reagent, C<sub>2</sub>H<sub>5</sub>MgBr with C<sub>8</sub>H... Reagent, 1-napthylamine and sulphanilic acid in acetic acid is used for the dete... A non-reducing sugar "A" hydrolyses to give two reducing mono saccharides. Sugar... 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She is not able to see clearly a distan... A loop of flexible wire of irregular shape carrying current is placed in an exte... The time period of a satellite in a circular orbit of radius R is T. The period ... A particle performs simple harmonic motion with a period of 2 second. The time t... The circuit shown in the figure consists of a charged capacitor of capacity 3 $$... The voltage across the 10$$\Omega$$resistor in the given circuit is x volt.<br>... Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when the... A person is swimming with a speed of 10 m/s at an angle of 120$$^\circ$$with th... A parallel plate capacitor has plate area 100 m<sup>2</sup> and plate separation... A ball of mass 10 kg moving with a velocity 10$$\sqrt 3 $$m/s along the x-axis,... As shown in the figure, a particle of mass 10 kg is placed at a point A. When th... An npn transistor operates as a common emitter amplifier with a power gain of 10... 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A radioactive sample disintegrates via two independent decay processes having ha... ### Joint Entrance Examination JEE Main JEE Advanced WB JEE ### Graduate Aptitude Test in Engineering GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN NEET Class 12	2022-06-25 01:10:29	{"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": 0.8787915706634521, "perplexity": 3360.4672939440848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00068.warc.gz"}