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http://mathschallenge.net/view/a_radical_proof	## A Radical Proof #### Problem The radical of $n$, $\text{rad}(n)$, is the product of distinct prime factors of $n$. For example, $504 = 2^3 \times 3^2 \times 7$, so $\text{rad}(504) = 2 \times 3 \times 7 = 42$. Given any triplet of relatively prime positive integers $(a, b, c)$ for which $a + b = c$ and with $a \lt b \lt c$, it is conjectured, but not yet proved, that the largest element of the triplet, $c \lt \text{rad}(abc)^2$. Assuming that this conjecture is true, prove that $x^n + y^n = z^n$ has no integer solutions for $n \ge 6$. Problem ID: 284 (23 Jul 2006) Difficulty: 3 Star Show Problem & Solution	2016-08-27 13:20:25	{"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": 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": 0.5374997854232788, "perplexity": 341.49273944566556}, "config": {"markdown_headings": true, "markdown_code": false, "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-2016-36/segments/1471982911825.82/warc/CC-MAIN-20160823200831-00258-ip-10-153-172-175.ec2.internal.warc.gz"}
https://questions.examside.com/past-years/jee/question/the-equation-of-the-line-through-the-point-0-1-2-and-per-jee-main-mathematics-trigonometric-functions-and-equations-c9svxokg1vxdrttf	1 JEE Main 2021 (Online) 25th February Morning Shift +4 -1 The equation of the line through the point (0, 1, 2) and perpendicular to the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}$$ is : A $${x \over 3} = {{y - 1} \over { - 4}} = {{z - 2} \over 3}$$ B $${x \over 3} = {{y - 1} \over 4} = {{z - 2} \over { - 3}}$$ C $${x \over { - 3}} = {{y - 1} \over 4} = {{z - 2} \over 3}$$ D $${x \over 3} = {{y - 1} \over 4} = {{z - 2} \over 3}$$ 2 JEE Main 2021 (Online) 25th February Morning Shift +4 -1 Let $$\alpha$$ be the angle between the lines whose direction cosines satisfy the equations l + m $$-$$ n = 0 and l2 + m2 $$-$$ n2 = 0. Then the value of sin4$$\alpha$$ + cos4$$\alpha$$ is : A $${{3 \over 8}}$$ B $${{3 \over 4}}$$ C $${{1 \over 2}}$$ D $${{5 \over 8}}$$ 3 JEE Main 2021 (Online) 24th February Evening Shift +4 -1 Let a, b$$\in$$R. If the mirror image of the point P(a, 6, 9) with respect to the line $${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then \| a + b \|, is equal to : A 88 B 90 C 86 D 84 4 JEE Main 2021 (Online) 24th February Evening Shift +4 -1 The vector equation of the plane passing through the intersection of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$$, and the point (1, 0, 2) is : A $$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = {7 \over 3}$$ B $$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = 7$$ C $$\overrightarrow r .\left( {3\widehat i + 7\widehat j + 3\widehat k} \right) = 7$$ D $$\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}$$ JEE Main Subjects Physics Mechanics Electricity Optics Modern Physics Chemistry Physical Chemistry Inorganic Chemistry Organic Chemistry Mathematics Algebra Trigonometry Coordinate Geometry Calculus EXAM MAP Joint Entrance Examination GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN Medical NEET	2023-03-31 09:26:10	{"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": 0.7515305876731873, "perplexity": 3797.2100639408045}, "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-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00750.warc.gz"}
https://physics.stackexchange.com/questions/147490/ambiguity-in-ordering-of-isospin-states-for-clebsch-gordan-coefficients	# Ambiguity in ordering of isospin states for Clebsch-Gordan coefficients In studying isospin for nuclear physics, I am confused a bit by an ambiguity I found. If a process that goes from $K^- + p \rightarrow \Sigma^0+ \pi^0$, I can write the isospin for the left hand side as $\|K^- + p \rangle = \| \frac{1}{2} \frac{-1}{2} \rangle \| \frac{1}{2} \frac{1}{2} \rangle$ but decomposing this with Clebsch-Gordan coefficients gives us $\| \frac{1}{2} \frac{-1}{2} \rangle \| \frac{1}{2} \frac{1}{2} \rangle=\sqrt{\frac{1}{2}}\| 1 0 \rangle-\sqrt{\frac{1}{2}}\| 0 0 \rangle$. However, if I change the ordering, which seems completely arbitrary to me, we get $\|p + K^- \rangle = \| \frac{1}{2} \frac{1}{2} \rangle\| \frac{1}{2} \frac{-1}{2} \rangle$ but decomposing this with Clebsch-Gordan coefficients gives us $\| \frac{1}{2} \frac{1}{2} \rangle\| \frac{1}{2} \frac{-1}{2} \rangle=\sqrt{\frac{1}{2}}\| 1 0 \rangle+\sqrt{\frac{1}{2}}\| 0 0 \rangle$. Maybe this ends up not mattering, but if for instance we want to calculate a scattering amplitude, where $\|\Sigma^0+ \pi^0 \rangle = \sqrt{\frac{2}{3}}\| 2 0 \rangle-\sqrt{\frac{1}{3}}\| 0 0 \rangle$, $M(K^- + p \rightarrow \Sigma^0+ \pi^0)=-\sqrt{\frac{1}{6}}\langle 0 0 \| M \| 0 0 \rangle$, but $M(p + K^- \rightarrow \Sigma^0+ \pi^0)=\sqrt{\frac{1}{6}}\langle 0 0 \| M \| 0 0 \rangle$. Obviously this still leaves the cross section invariant, but I can imagine processes where we add something else to get something like $M= A-B$ and $M=A+B$ when I flip the ordering of two initial state particles, so then my issue is that since $\sigma \sim \|M\|^2$, it's not necessarily true that $\sigma=\|A-B\|^2$ and $\sigma= \|A+B\|^2$ are equal, whereas intituitely they should be. Is there a general rule for ordering of particles? I feel that the choice in ordering of particles is arbitrary, but can give different results. This is an old subject so I must be missing something fundamentally. The notation $\lvert \phi \rangle \lvert \psi \rangle$ is shorthand for $\lvert \phi \rangle \otimes \lvert \psi \rangle$. What you are doing is flipping a tensor product around. Though, in general, $A \otimes B = B \otimes A$, it is not the case that $a \otimes b = b \otimes a$ for $a \in A, b \in B$ (since the left and right hand side don't even live in the same space. You have to label the spaces and stick with it. In this case, there is the isospin space $\mathcal{H}_K$ for the $K$ and the isospin space $\mathcal{H}_p$ for the $p$. Though they are isomorphic, they are not the same. In one case, you are Clebsch-Gordan decomposing $\mathcal{H}_K \otimes \mathcal{H}_p$, in the other, it is $\mathcal{H}_p \otimes \mathcal{H}_K$. If you stay consistent with this labeling, and write all states in the same way, i.e. you have to choose whether $\mathcal{H}_K \otimes \mathcal{H}_p$ or $\mathcal{H}_p \otimes \mathcal{H}_K$ is the isospin space for the initial states (and similarily for the final ones), the abstract isomorphy of the tensor products guarantees that you will make no error with relative signs as you fear. If you do not keep track of that, it is indeed possible to incur sign errors, since you basically use two different spaces of states for the same thing.	2021-11-28 21:25:29	{"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": 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": 0.8936492800712585, "perplexity": 202.4723860852246}, "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/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00247.warc.gz"}
https://www.biostars.org/p/497590/	Mapping after WGS show a parabel form 0 2 Entering edit mode 7 weeks ago Assa Yeroslaviz ★ 1.5k I would like to hear your opinion to the results of my Whole-Genome-Sequencing (WGS) I did with samples from a S. cerevisiae strain. After mapping the samples with bwa, marking duplicate (Picard) and doing variant calling (bcftools) and CNV discovery (cnv.kit and CNVpytor) I have plotted the results and got a strange behavior of my results. The mapping looks very consistent and the mapping results are >99% with good quality. But somehow I haven't seen this kind of mapping before. It looks like a parabel, but for all the sequences (s. attached image). CAn anyone explain this behavior? I am not sure how to interpret this results thanks WGS mapping SNP CNV • 159 views 0 Entering edit mode Hi Assa, I recently started using CNVpytor myself and got different results. My read depth is a lot more even, but I only have used human samples this far. CNV-calling is new for me but I read something about GC-bias. Maybe the GC content is lower or higher at the start and end of chromosomes in S. cerevisiae? (just guessing) The problem I am encountering is that my data is not looking normalized and that might also be the problem in your case? I feel read depth data should be normalized to 1 so that you could properly indicate deletions or duplications. But I am not sure how to do normalization for this data... (maybe someone else could help with that?) Please let me know your follow-up steps because that could help me aswell. Have a nice day!	2021-05-07 10:20:48	{"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": 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": 0.34572532773017883, "perplexity": 1763.7196849091454}, "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/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00413.warc.gz"}
https://modularformsboy.livejournal.com/	### Ah the (Digital) Humanity! Today I'm sitting in O'Hare; tomorrow I get to meet some of the investigators in the burgeoning academic field of the Digital Humanities. These folks are all pondering the question "What sort of neat things can you do if you apply computing to understanding culture and history?" The nascent field has already borne some interesting fruit: Tonight, however, I just hope to find someone in Urbana who might like to go to Papa Del's with me when I get in around 8pm... ### Dolan v United States In a tragic day for jurisprudence, the Supreme Court of the United States Monday approved "fill in the blank" sentencing, allowing a court to decide details of a sentence (in this case the amount awarded for restitution) an indeterminate time after the defendant is convicted. Dissenting, Chief Justice Roberts wrote: [If] a trial court can “leave open, say, the amount of a fine,” ante, at 12, why not, say, the number of years? Thus, after a defendant like Dolan has served his entire sentence—and who knows how long after?—a court might still order additional imprisonment, additional restitution, an additional fine, or an additional condition of supervised release. BREYER, J., delivered the opinion of the Court, in which THOMAS, GINSBURG, ALITO, and SOTOMAYOR, JJ., joined. ROBERTS, C. J., filed a dissenting opinion, in which STEVENS, SCALIA, and KENNEDY, JJ., joined. ### "Lifting Map" I feel like a college freshman again. There's a one-to-one correspondence between the set of circles in the plane and intersections of planes with the parabaloid $z = x^2 + y^2$. So you can turn the InCircle question into a half space test by passing your test point through the "Lifting Map" $\left(x, y\right) -> \left(x, y, x^2+y^2\right)$ ### Comedia dell'Arte in Paris Arlecchino, Pantalone, Dottore and a Zanni A month ago, Lynn and I took a vacation to Paris where we stopped in at the Musée Carnavalet. Mostly devoted to the History of Paris, one room contained the above statuettes of characters from Comedia dell'Arte. Thank you I Genisii (, , Brandough, , , ) for making me smile. ### Consistency, Survivability, Speed: Pick Two Legal Disclaimer: The views and opinions expressed in this article are mine only and do not necessarily reflect the views of Google. I spent my first two years at Google working on Google Checkout, and I walked away with two great lessons. One was the lesson I learned from Pat Helland on building scalable systems. The other was admitting a fundamental trade off in system design. Like many things, it seems obvious in retrospect, but you would be surprised at the number of people who keep trying for everything [cf, OceanStore]. When building a computing system, you only get to pick two: (1) Consistency (2) Survivability (3) Speed. What do I mean by these? Consistency: Someone in Tokyo and someone in Chicago make a query on a piece of data and both observe the same state. Survivability: The system continues to function even in the face of a catastrophic outage such as the loss of power in New York state in 2003. Speed: How long does it take to mutate a single piece of data. ## Scenario 1: Consistency and Survivability If you want consistency and survivability, then you are forced to make each write in your system to backends on different failure domains. Failure domains may be quite large, as shown by the New York State 2003 power outage of 2003. This implies geographically distributed backends, and you can't beat the speed of light. A ping from Mountain View to Atlanta takes 60 ms on Google's network; and the speed of light puts the lower bound at 22 ms. At Google, systems which choose consistency and survivability include Google Docs and our global Chubby cell. This trade off has the advantage of requiring no operator interaction when one of our clusters goes offline. However, it comes at a cost: writes which could have taken 2 ms take 200 ms. An application which makes 100 writes for each user action might work well enough in the former case but won't in the latter. Therefore, Google Docs and users of Global Chubby must be much more frugal with their writes. Brewer's Law: In a network partition, there is always a losing side. If two systems are independently modifying a piece of data, it may be difficult (or impossible) to resolve the combined effects of those modifications. What does it mean for an order to be canceled and shipped (so-called "forked state")1 ? To maintain consistency during a network partition, at most one side will win ownership of each piece of data. The other side will at best have a read-only version that may be stale. ## Scenario 2: Consistency and Speed If you're willing to forgo survivability, you can get consistency and speed in a variety of ways. The simplest way is to have one computer and a single copy of the data (a global "master"). To reduce your mean time between failure, you can put a quorum system like RAID or GFS in place so that when individual components fail, your system keeps humming along. However, if your entire site goes offline due to a fiber cut or other catastrophic failure, you have an outage. The common approximation to survivability chosen by people who choose consistency and speed is to have a replica which lags a certain distance behind the master and is located elsewhere. The replica has a consistent view of the world: it's just a bit out of date. Such replicas are often used to serve read-only versions of the data, and there are approaches to force-promote a replica to be master if the master is lost, but care must be taken regarding the possibly-stale state of existing objects. ## Scenario 3: Speed and Survivability How does Google provision web search for the world? We don't synchronize the index. Google has S search clusters each of which can serve Q search queries per second. Each search cluster receives a continuous stream of background updates that bring it to a close approximation of the "current" state of the Internet. However, there's no need for each cluster to be strictly in sync with the others. Maybe you have the blog post that was made ten minutes ago in your search results and I don't. In return for such small inconsistencies, Google provides web search which is always on and always fast. Postscript Even though I've framed the survivability versus speed trade off here as one involving geographically distributed backends, the trade off arises at lower levels as well. GFS trades some survivability for speed: write completion does not wait for disk commit, only memory commit to three chunk servers. If all three chunk servers lose power before their disk caches are flushed, the writes are lost. We've found this trade off acceptable given the speed up: disk seeks take about 10 ms, whereas memory references take only 10 ns. 1 Maybe Amazon knows what it means to have an order be independently shipped and canceled. Amazon's Dynamo Paper indicates that many of their objects are written to have mergable histories. -David Eger, November 2008 ### Media List 2007 I've stumbled upon a variety of good media this year which ended up falling into three broad categories: horror, curiosities and fantasy. The best of which I list below. ### Horror Both real and imagined this genre has the ability to make us want to turn our eyes, hoping not to see the degradation and inhumanity that lays before us. ### Curiosities Curiosities give us something interesting to think about. They are full of details from which we can learn, or which we may simply enjoy. ### Fantasy The realm of fantasy is the realm of our dreams: worlds we can explore, people we wish we knew (or were!), or stories that are just plain fun. Key: Fiction, Non-Fiction ### New Year's in Atlanta It's been five months and I'm settling in to life in the Bay Area. The San Francisco Bay Area is about as expensive as New York, but you get half again as much space for your money. Lots has happened this year: a cross country move, three weddings, a funeral, a new piano, getting promoted, moving projects, seeing asia for the first time, and listening to some good books. I'll be in the Atlanta around the New Year (Dec 28 @ 7PM - Jan 6 @ 7AM), staying with Hammad and hopefully getting a chance to see you. Let me know if you're around. ### Owl visit drove up from her new home as resident linguist in the desert to visit me this weekend, hooray! Her new boy is a lot of fun; and jalen continues to be the most fun girl on the dance floor. Skinny czech girls approve. ### Dark and Imaginative Fiction I've been thinking of a lot lately, ever since I picked up a copy of Borderlands 3. Dark fiction is something that brings back memories of Terrell Street -- the place where I was first introduced to the Borderlands series. My friend Pynk kept all manner of macabre games and books, so it's no surprise that he discovered some of the more imaginative books of the genre. Not all of the stories in Monteleone's series are great, but he does a good job of finding fresh ideas. The Borderlands series is a limited edition run, however. So it's a rare pleasure to find a copy of one of the five books. Slightly easier to come by is the good stuff over at pseudopod.org. Pynk edits the podcast under very much the same rules as Monteleone, and there are enough up and coming writers to keep a good dark story coming out every week or so. ### US Presidential Candidate Ron Paul My sister told me the other day about a candidate for president: Ron Paul. He's one of the 133 congressmen who voted against the authorization of a war in Iraq, which is a nice quality in any candidate for president. Of the other popular candidates, Biden, Clinton, Edwards, Lieberman, and McCain all voted for the war. Feingold voted against it. Barack Obama didn't get elected to congress until 2004, two years after he might have voted. Senate Bill for Iraq War Authorization House Bill for Iraq War Authorization House Roll Call Ron Paul is a reasonable candidate. His stances play out as follows according to his web site: Agree • The Iraq war was a mistake, a war sold on a folio of lies.	2020-10-30 03:47:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.20764100551605225, "perplexity": 2480.1463780744566}, "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-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00088.warc.gz"}
http://computationalnonlinear.asmedigitalcollection.asme.org/article.aspx?articleid=1396421	0 Research Papers # A Penalty Formulation for Dynamics Analysis of Redundant Mechanical Systems [+] Author and Article Information Bilal Ruzzeh Department of Mechanical Engineering, and Centre for Intelligent Machines, McGill University, 817 Sherbrooke Street West, Montréal, QC, H3A 2K6, Canadabilal.ruzzeh@mail.mcgill.ca József Kövecses Department of Mechanical Engineering, and Centre for Intelligent Machines, McGill University, 817 Sherbrooke Street West, Montréal, QC, H3A 2K6, Canadajozsef.kovecses@mcgill.ca The condition of invertibility of the augmented matrix in Eq. 6 is generally known to be the independence of the constraints; i.e., the constraint Jacobian has to be of full row rank. By contact group, it is meant to be a set of mobile objects with a chain of contacting mobile objects between any two members of the group. A good initial guess can be the minimizer of the objective function $g(x)$ in unconstrained optimization. For further illustration, detailed derivations, and proofs, the reader can consult specialized books on mathematical optimization such as Ref. 21. It will be shown in the next subsection that the actuating constraints are always holonomic. The calculation of the different quantities in the actuating constraint equations at the acceleration level will also be illustrated. It is noteworthy that algorithms, which depend on model-based parameters to calculate an appropriate penalty factors, are not available in literature yet, but it might be possible to develop them in the future. The star $(∗)$ superscript signifies the state of not yet converged or nonfinal solutions. The insertion point is on the part with a smaller contact surface between the muscle and the bone, while the attachment point (or origin) is on the part with a larger contact or attachment area. The first four constraint equations are associated with the redundant link 3. The other remaining four constraint equations associated with the second redundant link 4 exhibit the same behavior. The mathematical algorithm may not succeed in achieving the required precision or tolerance in the presence of a singularity due to the limitation of the approximate estimation of the Lagrange multipliers (21). J. Comput. Nonlinear Dynam 6(2), 021008 (Oct 28, 2010) (12 pages) doi:10.1115/1.4002510 History: Received December 09, 2009; Revised June 29, 2010; Published October 28, 2010; Online October 28, 2010 ## Abstract Redundancy in the constraining of mechanical systems achieves more stability and larger load capacity for the system, while in actuation it provides better robustness against singularities and higher maneuverability. Few techniques have been developed with the aim to handle redundancy and singularities in dynamics analysis, and further research is still needed in this area. In this paper, we illustrate the concept of actuating and passive constraints. Then, we expand on the existing penalty techniques by incorporating the concept of actuating and passive constraints to present a penalty formulation that is capable of efficiently handling singularities and redundancy in constraining and actuation and can carry out either forward or inverse dynamics analysis of mechanical systems. As such, the proposed approach is referred to as the actuating-passive constraints penalty approach. <> ## Figures Figure 1 Example of a translational actuator Figure 2 Five-bar mechanism (redundantly constrained) Figure 14 The driving moments at joints L and B and the corresponding vertical components of accelerations of the centers of mass GL and GU of the model Figure 3 Plots of θ1 (rad), x3 (m), and x4 (m) Figure 4 Violations of acceleration level constraints Figure 5 Violations of configuration level constraints Figure 6 Individual constraint forces for μ=107I, where I is the m×m identity matrix with suitable mass units and m is the number of kinematic constraints Figure 7 Individual constraint forces for μ=1010I, where I is the m×m identity matrix with suitable mass units and m is the number of kinematic constraints Figure 8 Sagital view of the simplified human lower limb with two actuators Figure 9 The driving forces represented as individual driving constraint reactions Figure 10 The generalized driving forces represented as joint torques Figure 11 Plot of configuration coordinates α,β Figure 12 The deviation of β̈ from the reference solution Figure 13 The individual actuating efforts λa,1 and λa,2 are associated with the translational actuators a1 and a2; λa,3 is associated with a motor actuator at joint L ## Discussions Some tools below are only available to our subscribers or users with an online account. ### Related Content Customize your page view by dragging and repositioning the boxes below. Related Journal Articles Related Proceedings Articles Related eBook Content Topic Collections	2018-06-21 23:37:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.3155632019042969, "perplexity": 1928.8997113122275}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00395.warc.gz"}
https://socratic.org/questions/aneesha-travels-at-a-rate-of-50-miles-per-hour-morris-is-traveling-3-feet-per-se#631481	# Aneesha travels at a rate of 50 miles per hour. Morris is traveling 3 feet per second less than Aneesha. Which is the best estimate of the speed Morris is traveling? Jun 17, 2018 A reasonable estimate of Morris's speed would be $58$ miles/hour #### Explanation: The phrasing of the question implies that there should have been a list of values to choose from but no such list was provided. $3 \left(\text{feet")/("second}\right)$ $\textcolor{w h i t e}{\text{XXX")=3("feet")/("second") xx 60("seconds")/("minute")xx60("minutes")/("hour}}$ $3 \times \left(60 \times 60\right) = 3 \times 3600$ which is a little more that $10 , 000$ So $3 \left(\text{feet")/("second}\right)$ is a little more that $10 , 000 \left(\text{feet")/("hour}\right)$ There are $5 , 280$ feet in a mile. If we round this down to $5 , 000$ feet per mile. then we have an estimated speed difference of color(white)("XXX")(10,000 "feet"/"hour")/(5,000 "feet"/"mile")=2 "miles"/"hour" So Morris is travelling at approximately $2 \text{miles"/"hour}$ less than Aneesha's given $60 \text{miles"/"hour}$ or at approximately $58 \text{miles"/"hour}$ As a point of interest, using a spreadsheet to calculate the difference in traveling at 60 miles per hour versus 58 miles per hour gave a value of $2.93 \overline{3}$ feet per second; pretty close to the requested $3$ feet per second.	2021-10-27 13:59:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "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": 0.7849251627922058, "perplexity": 527.0741990952885}, "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-20211027115745-20211027145745-00238.warc.gz"}
http://wiki.math.toronto.edu/DispersiveWiki/index.php/NLS_with_potential	# NLS with potential (Thanks to Remi Carles for much help with this section. - Ed.) One can ask what happens to the NLS when a potential is added, thus $i\partial_t u + Du + \|u\|^{p-1}u= Vu\,$ where $V\,$ is real and time-independent. The behavior depends on whether $V\,$ is positive or negative, and how $V\,$ grows as $\|x\| \rightarrow \infty\,$. In the following results we suppose that $V\,$ grows like some sort of power of $x\,$ (this can be made precise with estimates on the derivatives of $V\,$, etc.) A particularly important case is that of the quadratic potential $V = \pm \|x\|^2;\,$ this can be used to model a confining magnetic trap for Bose-Einstein condensation. Most of the mathematical research has gone into the isotropic quadatic potentials, but anisotropic ones (given by quadratic forms other than $\|x\|^2\,$) are also of physical interest. • If $V\,$is linear, i.e. $V(x) = Ex\,,$ then the potential can in fact be eliminated by a change of variables [CarNky-p] • If $V\,$is smooth, positive, and has bounded derivatives up to order 2 (i.e. is quadratic or subquadratic), then the theory is much the same as when there is no potential - one has decay estimates, Strichartz estimates, and the usual local and global well posedness theory (see Fuj1979, Fuj1980, Oh1989) • When $V\,$is exactly a positive quadratic potential $V= w^2 \|x\|^2,\,$ then one has blowup for the focusing nonlinearity for negative energy in the $L^2\,$ supercritical or critical, $H^1\,$ subcritical case Zhj2000a, Car2002b, Zhj2005. • In the $L^2\,$ critical case one can in fact eliminate this potential by a change of variables Car2002c. One consequence of this change of variables is that one can convert the usual solitary wave solution for NLS into a solution that blows up in finite time (cf. how the pseudoconformal transform is used to achieve a similar effect without the potential). • When $V\,$is exactly a negative quadratic potential, one can prevent blowup even in the focusing case if the potential is sufficiently strong [Car-p]. Indeed, one has a scattering theory in this case [Car-p] • If $V\,$grows faster than quadratic, then there are significant problems due to the failure of smoothness of the fundamental solution; if $V\,$is also negative, then even the linear theory fails (for instance, the Hamiltonian need not be essentially self-adjoint on test functions). However for positive superquadratic potentials partial results are still possible YaZgg2001. Much work has also been done on the semiclassical limit of these equations; see for instance BroJer2000, Ker2002, [CarMil-p], Car2003. For work on standing waves for NLS with quadratic potential, see Zhj2000a, Zhj2000b,Fuk2001, Fuk2003, FukOt2003, FukOt2003b. Furthermore, the work about sharp criteria of global existence for NLS with quadratic potential, see Zhj2005, ChgZhj2006a, ChgZhj2006b, ShjZhj2006, ChgZhj2007. One component of the theory of NLS with potential is that of Strichartz estimates with potential, which in turn may be derived from dispersive estimates with potential, although it is possible to obtain Strichartz estimates without a dispersive inequality. Both types of estimates can only be expected to hold after first projecting to the absolutely continuous part of the spectrum (although this is not necessary if the potential is positive). The situation for dispersive estimates (which imply Strichartz), and related estimates such as local $L^2\,$ decay, and of $L^p\,$ boundedness of wave operators (which implies both the dispersive inequality and Strichartz) is as follows. Here we consider potentials which could oscillate; relying mostly on magnitude bounds on $V\,$rather than on symbol-type bounds. • When $d=1\,$ one has dispersive estimates whenever $V\,$is $L^1\,$ [GbScg-p]. • For potentials such that $^{3/2+}V\,$is in $L^1\,$, this is essentially in Wed2000; the stronger $L^p\,$ boundedness of wave operators in this case was established in Wed1999, ArYa2000. • When $d=2\,,$ relatively little is known. • $L^p\,$ boundedness of wave operators for potentials decaying like $^{-6-}\,,$ assuming 0 is not a resonance nor eigenvalue, is in Ya1999, JeYa2002. The method does not quite extend to $p=1,\infty\,$ and thus does not directly imply the dispersive estimate although it does give Strichartz estimates for $1 < p < \infty\,.$ • Local $L^2\,$ decay and resolvent estimates for low frequencies for polynomially decaying potentials are obtained in JeNc2001 • When $d=3\,$ one has dispersive estimates whenever $V\,$decays like $^{-3-}\,$ and 0 is neither a eigenvalue nor resonance [GbScg-p] • For potentials which decay like $^{-7-}\,$ and whose Fourier transform is in $L^1\,,$ a version of this estimate is in JouSfSo1991 • A related local $L^2\,$ decay estimate was obtained for exponentially decaying potentials in Ra1978; this was refined to polynomially decaying potentials (e.g. $^{-3-}\,$) (with additional resolvent estimates at low frequencies) in JeKa1980. • $L^p\,$ boundedness of wave operators was established in Ya1995 for potentials decaying like $^{-5-}\,$ and for which 0 is neither an eigenvalue nor a resonance. • If 0 is a resonance one cannot expect to obtain the optimal decay estimate; at best one can hope for $t^{-1/2}\,$ (see JeKa1980). • Dispersive estimates have also been proven for potentials whose Rollnik and global Kato norms are both smaller than the critical value of $4\pi\,$ [RoScg-p]. Indeed their arguments partly extend to certain time-dependent potentials (e.g. quasiperiodic potentials), once one also imposes a smallness condition on the $L^{3/2}\,$ norm of $V\,$ • If the potential is in $L^2\,$ and has finite global Kato norm, then one has dispersive estimates for high frequencies at least [RoScg-p]. • Strichartz estimates have been obtained for potentials decaying like $^{-2-}\,$ if 0 is neither a zero nor a resonance [RoScg-p] 1. This has been extended to potentials decaying exactly like $\|x\|^2\,$ and $d \ge 3\,$ assuming some radial regularity and if the potential is not too negative [BuPlStaTv-p], [BuPlStaTv-p2]; in particular one has Strichartz estimates for potentials $V= a/\|x\|^2, d \ge 3,\,$ and $a > -(n-2)^2/4\,$ (this latter condition is necessary to avoid bound states). • For $d > 3\,,$ most of the $d=3\,$ results should extend. For instance, the following is known. • For potentials which decay like $^{-d-4-}\,$ and whose Fourier transform is in $L^1\,,$ dispersive estimates are in JouSfSo1991 • Local $L^2\,$ decay and resolvent estimates for low frequencies for polynomially decaying potentials are obtained in Je1980, Je1984. For finite rank perturbations of the Laplacian, where each rank one perturbation is generated by a bump function and the bump functions are sufficiently far apart in physical space, decay and Schrodinger estimates were obtained in NieSf2003.The bounds obtained grow polynomially in the number of rank one perturbations. Local smoothing estimates seem to be more robust than dispersive estimates, holding in a wider range of situations.For instance, in $R^d\,,$ any potential in $L^p\,$ for $p \ge d/2\,,$ as well as inverse square potentials $1/\|x\|^2\,,$ and linear combinations of these, have local smoothing RuVe1994.Note one does not need to project away the bound states for such estimates (as the bound states tend to already be rather smooth).However, for $p < d/2\,,$ one can have breakdown of local smoothing (and also dispersive and Strichartz estimates) [Duy-p] For time-dependent potentials, very little is known.If the potential is small and quasiperiodic in time (or more generally, has a highly concentrated Fourier transform in time) then dispersive and Strichartz estimates were obtained in [RoScg-p]; the smallness is used to rule out bound states, among other things.In the important case of the charge transfer model (the time-dependent potential that arises in the stability analysis of multisolitons) see Ya1980, Grf1990, Zi1997, [RoScgSf-p], [RoScgSf-p2], where energy, dispersive, and Strichartz estimates are obtained, with application to the asymptotic stability of multisolitons. The nonlinear interactions between the bound states of a Schrodinger operator with potential (which have no dispersion or decay properties in time) and the absolutely continuous portion of the spectrum (where one expects dispersion and Strichartz estimates) is not well understood.A preliminary result in this direction is in [GusNaTsa-p], which shows in R3 that if there is only one bound state, and Strichartz estimates hold in the remaining portion of the spectrum, and the non-linearity does not have too high or too low a power (say $4/3 \le p \le 4\,,$ or a Hartree-type nonlinearity) then every small $H^1\,$ solution will asymptotically decouple into a dispersive part evolving like the linear flow (with potential), plus a nonlinear bound state, with the energy and phase of this bound state eventually stabilizing at a constant.In [SfWs-p] the interaction of a ground state and an excited state is studied, with the generic behavior consisting of collapse to the ground state plus radiation.	2013-05-19 13:28:03	{"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": 67, "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": 0.8962554335594177, "perplexity": 557.4080570071084}, "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-2013-20/segments/1368697552127/warc/CC-MAIN-20130516094552-00010-ip-10-60-113-184.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2918019/simplex-method-why-tableau-row-operation-produces-correct-indicator-vector	# Simplex method: Why tableau row operation produces correct indicator vector? I learned the objective variable can be written as: $$z = c_B^T \bar{b} - \xi^Tx$$ where $\bar b = B^{-1} b$ and $\xi$ is the "indicator vector" which is usually put at the top or bottom of simplex tableau. But I do not understand why each iteration, the row operation to unset (zero) previous $\xi_B$ will produce the correct new $\xi$. In my textbook, the $\xi$ has zeros in its first $m$ element and $\xi_j = c_B^T \bar{A}_j - c_j$ when $j = m+1, m+2 ... n$. I understand initially if $c_B^T = \mathbf{0}$ then $\xi_j = -c_j$ and the row operation to zero previous $\xi_B$ is essentially add $c_B^T \bar A_j$ onto it. So in the first iteration I can successfully reason why row operation produces $\xi_j = c_B^T \bar{A}_j - c_j$. But after the first iteration, $c_B^T \not= \mathbf{0}$ then why the row operation performed on objective row will produce the $\xi_j = c_B^T \bar{A}_j - c_j$?	2019-07-18 23:24:38	{"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": 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": 0.9189929962158203, "perplexity": 347.5117502576059}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525863.49/warc/CC-MAIN-20190718231656-20190719013656-00014.warc.gz"}
https://math.stackexchange.com/questions/366678/bit-permutations-and-collisions-of-compression-function/370888	# Bit permutations and collisions of compression function I'm having trouble finding a good method for solving the following problem: If $n$ is a positive integer, let $S_n$ denote the group of permutations of the set $\{1,2,\dots, n\}$. For a permutation $\pi$ in $S_3$, let $e_\pi$ be the bit permutation of bit strings of length 3. For each $\pi \in S_3$ determine the number of collisions of the compression function $h(x) = e_\pi(x) \oplus x$. Does anyone have any advice on how to approach this? • Anyone have any tips? Thanks! – willow Apr 22 '13 at 15:54 This will be the same for all permutations in the same conjugacy class, so you only have to do it for $(12)$, $(123)$ and the identity. For the identity all bit strings are mapped to $000$. For $(12)$, a bit string is mapped to $000$ if bits $1$ and $2$ are the same, and $011$ if bits $1$ and $2$ are different (with four strings mapped to each of the strings). For $(123)$, a bit string is mapped to $000$ if it's constant, and to one of the strings $011$, $101$ and $110$ if it isn't (with two strings mapped to each of those strings). I'm not sure what exactly the "number of collisions" is, but if you know how it's defined you should be able to determine it from the above. • Thanks for the response, joriki. Why is it that for the identity all bit strings are mapped to 000? For instance, one of the bit strings of length 3 is 111. Wouldn't the identity permutation map this to itself? I'm trying to figure out where exactly my confusion lies. Thanks again! – willow Apr 24 '13 at 16:06 • @willow: Yes, the identity permutation maps this string to itself. But the value of the compression function isn't $e_\pi(111)=111$, but $e_\pi(111) \oplus 111=111\oplus111=000$. – joriki Apr 24 '13 at 16:16 • Ah ok, yes, I hadn't taken that next step and computed the compression function yet. Thanks. – willow Apr 24 '13 at 16:27 • Oh, and just to clarify what "number of collisions" is: it's when the compression function maps two different inputs to the same output, i.e. $h(x) = h(x')$ but $x \neq x'$. So, if $\pi$ is the identity permutation then there is a collision for $h(000)$ and $h(111)$ since $e_\pi(000) \oplus 000 = 000 \oplus 000 = 000 = 111 \oplus 111 = e_\pi(111) \oplus 111$. At least that's my understanding. Thanks again! – willow Apr 24 '13 at 16:34 • @willow: I was aware what a collision is; I just wasn't sure what exactly the number of collisions is. If four strings are mapped to the same string, is that one collision (one multiply achieved result), or three collisions (three times when the strings are mapped one after the other that a string is mapped to a string that had already been obtained), or six collisions (six pairs of strings that are mapped to the same string)? – joriki Apr 24 '13 at 16:51	2019-08-19 05:56:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9190137982368469, "perplexity": 237.96799102150194}, "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/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00014.warc.gz"}
https://testbook.com/question-answer/a-person-rows-30-km-upstream-and-the-same-distance--605dbf097c898180c05b3b0e	# A person rows 30 km upstream and the same distance downstream. The total time taken by him for this whole journey is $$3\dfrac{1}{5}$$ hours. If the speed of the stream is 5 km/h, then in how many hours does the person row a distance of 144 km upstream? This question was previously asked in Dehli Police Constable Previous Paper 22 (Held on: 11 Dec 2020 Shift 2) View all Delhi Police Constable Papers > 1. 9.6 2. 8.4 3. 10 4. 9 Option 1 : 9.6 Free Delhi Police Constable 2020: Full Mock Test 93896 100 Questions 100 Marks 90 Mins ## Detailed Solution Given: 30km upstream and downstream distance. Time taken is 16/5 hours. Speed of stream is 5 km/hr Formula Used: Speed = Distance/Time Speed in upstream direction = x – y Speed in downstream direction = x + y Where, x is speed of a boat in still water y is speed of a stream Calculation: According to the question, ⇒ 16/5 = 30/(x – y) + 30/x + y) ⇒ 16/5 = 30/(x – 5) + 30/(x + 5) ⇒ 16/5 = {30(x + 5) + 30(x – 5)}/(x – 5)(x + 5) ⇒ 16/5 = (30x + 150 + 30x – 150)/(x– 25) ⇒ 16/5 = 60x/(x– 25) ⇒ 4/5 = 15x/(x– 25) ⇒ 4(x– 25) = 5 × 15x ⇒ 4x2 – 100 = 75x ⇒ 4x– 75x – 100 = 0 ⇒ 4x– 80x + 5x – 100 = 0 ⇒ 4x(x – 20) + 5 (x – 20) = 0 ⇒ 4x + 5 = 0 or x – 20 = 0 ⇒ x = –5/4 or x = 20 Speed can not be negative, ⇒ x = 20 Upstream speed = x – y ⇒ Upstream speed = 20 – 5 ⇒ Upstream speed = 15km/hr To cover 144 km distance upstream, ⇒ Time = 144/15 ⇒ Time = 9.6 ∴ Time taken to cover 144 km distance is 9.6 hours.	2021-09-20 01:37:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.6142455339431763, "perplexity": 6435.765630873153}, "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-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00392.warc.gz"}
http://renormalization.com/tag/quantum-gravity/page/2/	## Quantum gravity I study some aspects of the renormalization of quantum field theories with infinitely many couplings in arbitrary space-time dimensions. I prove that when the space-time manifold admits a metric of constant curvature the propagator is not affected by terms with higher derivatives. More generally, certain lagrangian terms are not turned on by renormalization, if they are absent at the tree level. This restricts the form of the action of a non-renormalizable theory, and has applications to quantum gravity. The new action contains infinitely many couplings, but not all of the ones that might have been expected. In quantum gravity, the metric of constant curvature is an extremal, but not a minimum, of the complete action. Nonetheless, it appears to be the right perturbative vacuum, at least when the curvature is negative, suggesting that the quantum vacuum has a negative asymptotically constant curvature. The results of this paper give also a set of rules for a more economical use of effective quantum field theories and suggest that it might be possible to give mathematical sense to theories with infinitely many couplings at high energies, to search for physical predictions. PDF Class.Quant.Grav. 20 (2003) 2355-2378 \| DOI: 10.1088/0264-9381/20/11/326 arXiv:hep-th/0212013 We study a regularization of the Pauli-Villars kind of the one loop gravitational divergences in any dimension. The Pauli-Villars fields are massive particles coupled to gravity in a covariant and nonminimal way, namely one real tensor and one complex vector. The gauge is fixed by means of the unusual gauge-fixing that gives the same effective action as in the context of the background field method. Indeed, with the background field method it is simple to see that the regularization effectively works. On the other hand, we show that in the usual formalism (non background) the regularization cannot work with each gauge-fixing.In particular, it does not work with the usual one. Moreover, we show that, under a suitable choice of the Pauli-Villars coefficients, the terms divergent in the Pauli-Villars masses can be corrected by the Pauli-Villars fields themselves. In dimension four, there is no need to add counterterms quadratic in the curvature tensor to the Einstein action (which would be equivalent to the introduction of new coupling constants). The technique also works when matter is coupled to gravity. We discuss the possible consequences of this approach, in particular the renormalization of Newton’s coupling constant and the appearance of two parameters in the effective action, that seem to have physical implications. PDF Phys.Rev. D48 (1993) 5751-5763 \| DOI: 10.1103/PhysRevD.48.5751 arXiv:hep-th/9307014 ### Search this site Support Renormalization If you want to support Renormalization.com you can spread the word on social media or make a small donation ### Book 14B1 D. Anselmi Renormalization PDF Last update: May 9th 2015, 230 pages Contents: Preface \| 1. Functional integral \| 2. Renormalization \| 3. Renormalization group \| 4. Gauge symmetry \| 5. Canonical formalism \| 6. Quantum electrodynamics \| 7. Non-Abelian gauge field theories \| Notation and useful formulas \| References Course on renormalization, taught in Pisa in 2015. (More chapters will be added later.)	2018-02-22 11:13:45	{"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": 0.8139676451683044, "perplexity": 444.44365101438194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00626.warc.gz"}
http://www.ck12.org/algebra/Graphs-of-Linear-Systems/lesson/Solving-Linear-Systems-by-Graphing/	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Graphs of Linear Systems ## Graph lines to identify intersection points 0% Progress Practice Graphs of Linear Systems Progress 0% Solving Linear Systems by Graphing Two trains leave the station going in the same direction. One train leaves two hours before the other one. If the first train travels at an average speed of 65 mph and the second train travels at an average speed of 90 mph, then how long will it take the second train to catch up to the first train? In this concept, you will learn to solve linear systems by graphing. ### Guidance The solution for a system of two linear equations with two variables is an ordered pair that will make both equations true. If an ordered pair makes one of the linear equations true then the ordered pair represents a point on the line. If the same ordered pair makes the second linear equation true then the ordered pair represents a point on that line. If the same point is on the two lines, then the lines both pass through this same point. This means that the two lines will cross each other or intersect at this point. A system of equations can have one solution, no solution or an infinite number of solutions. Let’s look at an example. Solve the following system of linear equations by graphing. yy==2x42x+8 Notice that both equations are written in slope-intercept form y=mx+b\begin{align}y=mx+b\end{align}. First, write down the information given in each equation. yym===mx+b2x42;b=4 yym===mx+b2x+82;b=8 Next, graph each of the equations on the same Cartesian grid. The two lines intersect at the point (3,2)\begin{align}(3,2)\end{align}. The solution is (xy)=(32)\begin{align}\dbinom{x}{y} = \dbinom{3}{2}\end{align}. Let’s look at another example. Solve the following system of linear equations by graphing. yy==3x13x+2 Notice that both equations are written in slope-intercept form y=mx+b\begin{align}y=mx+b\end{align}. First, write down the information given in each equation. yym===mx+b3x13;b=1 yym===mx+b3x+23;b=2 The equations have the same slopes but different y\begin{align}y\end{align}-intercepts. Next, graph each of the equations on the same Cartesian grid. The lines are parallel which means they will never intersect. There is no solution for this system of equations. Let’s look at another example. Solve the following system of linear equations by graphing. yx+4y==14x14 First, write down the information given in each equation. yym===mx+b14x114;b=1 x+4yxinterceptyintercept===4written in standard form Ax+By=C(4,0)(0,1) The second equation is a multiple (4)\begin{align}(4)\end{align} of the first equation. Next, graph each of the equations on the same Cartesian grid. The lines coincide which means they are actually the graph of the same line. There are an infinite number of solutions for this system of equations. ### Guided Practice Solve the following system of linear equations by graphing: 2x+3yx5y==411 Both equations are written in standard form Ax+By=C\begin{align}Ax+By=C\end{align}. First, write each equation in slope-intercept form y=mx+b\begin{align}y=mx+b\end{align} and write down the value of the slope and the y\begin{align}y\end{align}-intercept. yym===mx+b23x+4323;b=43 yym===mx+b15x+11515;b=115 Then, graph both equations on the same Cartesian grid. The two lines intersect at the point (1,2)\begin{align}(-1,2)\end{align}. The solution for this system of linear equations is: (xy)=(12) ### Examples Answer the following examples using True or False. If the answer is false, explain why. #### Example 1 If two lines intersect at one point, then the coordinates of that point are the solution to the linear system. #### Example 2 If a system of linear equations is such that both equations have the same slope but different y\begin{align}y\end{align}-intercepts then there are an infinite number of solutions for the linear system. If two equations have the same slope then the graph of the equations will display two parallel lines with different y\begin{align}y\end{align}-intercepts. Parallel lines will never intersect. There is no solution for this linear system. #### Example 3 If a system of linear equations is such that the two lines coincide then there is no solution for the linear system. If the graph of the linear system displays two lines that coincide then every ordered pair is on both lines and satisfy both equations. There is an infinite number of solutions for the linear system. Credit: Gwydion M. Williams Source: https://www.flickr.com/photos/45909111@N00/4868309667/in/photolist-8qcmRz-6rNw9E-96uCNG-crLdZb-dhpES3-8qfwe7-7ed27J-okNLH6-c91FM1-7Uzv4G-buG92r-aC7s41-osv4Gw-dDfzn3-nr9Q5Y-eaMgyV-7J6nEa-dhpEYU-48UStC-d8an7d-aLQdzp-eh1yhs-aC4Nkg-bPnkTF-o4AAAF-7e98rD-dk2Gtr-4GSDyt-auixFE-cDKzwf-dNoitL-gwCAPG-daFc8s-cDKyKo-rX2LeW-8YyQye-dbKTKr-78ucr5-nBzPy2-aC4Quz-dhr4KA-dhr7SG-dhr7dP-agxZAF-6WBq85-pjdYjc-bbjFLk-aaN9TJ-kQ1LAM-J72Yf Remember the two trains that left the station at different times? The second train left two hours after the first train but was travelling at a greater speed than the first train. You need to figure out when the second train will catch up to the first train. First, write a system of linear equations to represent the distance travelled by each of the trains. Remember that distance (d)\begin{align}(d)\end{align} is a function of the product of speed (s)\begin{align}(s)\end{align} and time (t)\begin{align}(t)\end{align}. dTrain 1:dd===st65(t+2)65t+130 dTrain 2:d==st90t Next, graph the two equations on the same Cartesian grid. The lines display an intersection point just past five hours. This means the two trains will catch up with each other at this point. ### Explore More Graph the following systems of equations. Identify the solution or write no solution or infinitely many solutions, if appropriate. 1. yy==2x3x1\begin{align}\begin{array}{rcl} y &=& 2x-3 \\ y &=& x-1 \end{array}\end{align} 2. 2x+2yy==1x+12\begin{align}\begin{array}{rcl} 2x+2y &=& 1 \\ y &=& -x+\frac{1}{2} \end{array}\end{align} 3. y3xy==3x+17\begin{align}\begin{array}{rcl} y &=& -3x+1 \\ 3x-y &=& -7 \end{array}\end{align} 4. yy2==2xx52\begin{align}\begin{array}{rcl} y &=& 2x \\ \frac{y}{2} &=& x - \frac{5}{2} \end{array}\end{align} 5. yy==3x+53x+8\begin{align}\begin{array}{rcl} y&=&3x+5 \\ y&=&-3x+8 \end{array}\end{align} 6. yy==2x+13x+3\begin{align}\begin{array}{rcl} y&=&2x+1 \\ y&=&3x+3 \end{array}\end{align} 7. yy==12x22x\begin{align}\begin{array}{rcl} y&=&\frac{1}{2}x-2 \\ y&=&-2x \end{array}\end{align} 8. yy==2x+12x2\begin{align}\begin{array}{rcl} y &=& -2x+1 \\ y&=&-2x-2 \end{array}\end{align} 9. yy==4x12x+2\begin{align}\begin{array}{rcl} y&=&4x-1 \\ y&=&2x+2 \end{array}\end{align} 10. yy==5x35x+1\begin{align}\begin{array}{rcl} y &=&5x-3 \\ y&=&-5x+1 \end{array}\end{align} 11. yy==2x3x5\begin{align}\begin{array}{rcl} y&=&2x \\ y &=&3x-5 \end{array}\end{align} 12. yy==x1x+6\begin{align}\begin{array}{rcl} y&=&-x-1 \\ y&=&-x+6 \end{array}\end{align} Answer each question true or false. 13. Some linear systems do not have a solution. 14. Perpendicular lines have one solution. 15. Lines with the same slope and y\begin{align}y\end{align}-intercept have infinite solutions. ### Vocabulary Language: English Consistent Consistent A system of equations is consistent if it has at least one solution. Dependent Dependent A system of equations is dependent if every solution for one equation is a solution for the other(s). Independent Independent A system of equations is independent if it has exactly one solution. linear equation linear equation A linear equation is an equation between two variables that produces a straight line when graphed. Linear Function Linear Function A linear function is a relation between two variables that produces a straight line when graphed. system of equations system of equations A system of equations is a set of two or more equations.	2015-12-01 05:12:03	{"extraction_info": {"found_math": true, "script_math_tex": 28, "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": 0.9899047017097473, "perplexity": 529.4344240029797}, "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-2015-48/segments/1448398464396.78/warc/CC-MAIN-20151124205424-00190-ip-10-71-132-137.ec2.internal.warc.gz"}
https://socratic.org/questions/how-do-you-solve-s-2-14s-45-0	# How do you solve s^ { 2} - 14s + 45= 0? Mar 13, 2017 $s = 9 , s = 5$ #### Explanation: There are multiple ways of solving this. (The easiest in this case): Find two numbers that when multiplied gives you $45$ and when added gives you $- 14$ This usually a trial and error process but you'll find that these two numbers are $- 9$ and $- 5$ When we multiply $- 9$ and $- 5$ we get $45$ and when we add $- 9$ and $- 5$ we get $- 14$ We can rewrite this information as $\left(s - 9\right) \left(s - 5\right) = 0$ We then have to solve for the variable $s$ We treat this as two separate equations such that: $s - 9 = 0$ and $s - 5 = 0$ $s = 9 , s = 5$ Note: you can check your answer by multiplying out $\left(s - 9\right) \left(s - 5\right)$ and you should get what you started with.	2020-08-06 13:17:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9412412643432617, "perplexity": 180.71925861665258}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00012.warc.gz"}
http://math.stackexchange.com/questions/112408/how-to-deal-with-linear-regression-model-with-some-data-aggregated/121652	# How to deal with Linear Regression model with some data aggregated Lets say I am trying to find a linear regression between Weight and Height of a person. $W=b_0+b_1 H+e$ The data I have gathered from 8 people is like this: # W(kg) H(cm) 1. 68 168 2. 64 170 3. ? 160 4. ? 180 5. ? 145 6. ? 191 7. 69 185 8. 80 191 Where I know that the sum of ?'s is 280, but I do not know exact data of each of them (because, let's say, at the time, I only had scales, which had a minimum scale of $200$. Dumb reason, I know, but that's just for the sake of example). So, my question is this: how do I create my $W$ matrix so I can make computations (to find out $b_0$ and $b_1$ using least squares method)? :) - Since both answers including comments have ended up rather long, here's a summary: Both answers lead to the same result, with different approaches; Gottfried's iteratively uses the fact that the slope can be computed using only the deviations from the mean, whereas mine integrates over all values of the missing weights consistent with the known sum to show that the four measurements should be replaced by four measurements of their averages. – joriki Feb 23 '12 at 23:13 The minimization of the mean square error can be regarded as a result of maximizing the likelihood that the data resulted from normally distributed errors. To deal with the aggregated data, we could integrate this likelihood over all values consistent with the constraint. The likelihood is $$\prod_i\mathrm e^{-\beta(w_i-(b_1h_i+b_0))^2}\;,$$ and the integral over all values consistent with the constraint is $$\iiiint\mathrm dw_1\mathrm dw_2\mathrm dw_3\mathrm dw_4\delta(w_1+w_2+w_3+w_4-280)\prod_i\mathrm e^{-\beta(w_i-(b_1h_i+b_0))^2}$$ (where I've numbered the unknown values $1$ to $4$ for convenience). Integrating out $w_4$ yields $$\prod_{i\gt4}\mathrm e^{-\beta(w_i-(b_1h_i+b_0))^2}\iiiint\mathrm dw_1\mathrm dw_2\mathrm dw_3\mathrm e^{-\beta(w^\top Aw-2s^\top w+c)}$$ with $$\begin{eqnarray} A_{ij}&=&\delta_{ij}+1\;,\\ s_i&=&280+b_1(h_i-h_4)\;,\\ c&=&280^2+\sum_{k=1}^4{(b_1h_k+b_0)^2}-2\cdot280(b_1h_4+b_0)\;, \end{eqnarray}$$ where $i$ and $j$ run from $1$ to $3$. The integral is proportional to $\mathrm e^{\beta(s^\top A^{-1}s-c)}$. With $$\displaystyle A^{-1}=\frac14\pmatrix{3&-1&-1\\-1&3&-1\\-1&-1&3}\;,$$ this is $\mathrm e^{-4\beta(\overline w-(b_1\overline h+b_0))^2}$, with $\overline w=(w_1+w_2+w_3+w_4)/4$ and $\overline h=(h_1+h_2+h_3+h_4)/4$. Thus, you should treat these four measurements as if you had made four measurements of the average weight $\overline w$ at the average height $\overline h$. - It seems as if you may be trying to show that you'll get the same estimated slope by using the average of the three weights and the average of the three heights, and giving that data point multiplicity 4. But there's also the question of whether a confidence interval for the slope, as opposed to a point estimate of the slope, would be different. – Michael Hardy Feb 23 '12 at 13:56 @Michael: I think the confidence intervals should also be the same. We can derive the integration approach by regarding the unknown measurements (except for their average) as nuisance parameters, assuming a flat prior and marginalizing over them. Then the resulting likelihood is exactly the likelihood we'd get for four measurements of the average values. Since the confidence intervals are determined by the likelihood, they should be the same, too, no? (Note that the average is over four (not three) weights and heights, and the calculation yields both the slope $b_1$ and the intercept $b_0$.) – joriki Feb 23 '12 at 14:34 Could you provide an estimate for the regression-coefficients and especially for the sum-of-squares of the residuals? I'd be interested in a comparision with the result in my answer. – Gottfried Helms Feb 23 '12 at 18:00 It appears to me that your answer is wrong. I've done some stuff with regression and anova software, and and fiddled with the matrices. Let $Y=(1,3,2,4,3)$. Let $X=(1,2,3,4,10)$. The least-squares estimate of the slope is $2.04$. Then replace the middle three values of $Y$ with their average so that $Y=(1,3,3,3,3)$ and do the same with $X$, so that it's $X=(1,3,3,3,10)$. Then the least-squares estimate of the slope is $2.1$. The F-statistics for the null hypothesis that the slope is $0$ are also coming out different from each other. – Michael Hardy Feb 23 '12 at 18:04 Suppose the "true" slope is $\beta$ and the least-square estimate is $\hat\beta$. Assume identically normally distributed independent errors with expectation $0$ and variance $\sigma^2$. Then $\hat\beta$ is normally distributed with expected value $\beta$ and variance $\sigma^2/\sum_i (x_i-\bar x)^2$. Notice that the denominator, $\sum_i (x_i-\bar x)^2$, gets smaller if you replace several of the $x$s with their average. Hence the variance of the least-squares estimate of the slope is bigger, so you have more uncertainty in its esimation. Hence longer confidence intervals. – Michael Hardy Feb 23 '12 at 18:13 Here I propose an iterative approach. [Update 2]: Unfortunately I've taken the regression in the wrong direction, but the general idea is unaffected of this. The numerical results where the regression takes the other direction is at the end First we know, that the regression-parameter for the slope can be computed by the deviations from the mean only. In the setup for the problem we have two groups of data: A = 4 (x,y)-measures where both x and y are known and B = 4 (x,y)-measures, where the x-values are not known, but sum up to 280. While their mean is known, their deviations from the mean is arbitrary except that they sum up to zero. Thus we can define that x-deviations having just the same values as their known y-deviations, scaled by any constant factor b, which is then some arbitrary slope in the scatterplot. Next we do a regression based on the data in A only. What we get is the equation for the regression $\small \hat{y}_A = 83.4 + 1.35 x_A$ (here the slope is b~1.35) Since we can determine the x-deviations in B arbitrarily, just to sum up to zero, we can use the deviations of the y-values and rescale them by the factor of $\small {1 \over 1.35}$ What we get then is the table for the B-data $\small \text{ B =} \begin{array} {rr} 63.350& 160\\ 78.127& 180\\ 52.268& 145\\ 86.255& 191 \end{array}$ If we insert that into the original table we get the sum-of-squares of the residues to about 291.01 (which was also the minimum that I could get with some experimenting). This might still be incomplete (and thus suboptimal) because the mean of the x-values in the complete data-set is slightly different from the means of the x-values in A (70.25) and in B (70) and the common optimum must be determined over the complete dataset; so possibly this must be extended to a recursive procedure. So if the above is not completely wrong or misleading, but useful so far, then that recursive procedure might be added later. Update: I did recursion to adapt the solution according to the problem of different means in the A and B x-data. I got a small improvement. B becomes now $\small \text{ B =} \begin{array} {rr} 63.50640& 160\\ 77.93662& 180\\ 52.68374& 145\\ 85.87324& 191 \end{array}$ the equation becomes $\small \hat{y} = 76.55811894 + 1.385980479 x$ and the sum-of-squares of the residues becomes now 290.887311 which is an improvement of about -0.26. After this the recursion becomes stable in the leading six decimals [update 2,3] Upps, I've taken the wrong direction of the regression. If I take the other way I get $\small \text{ B =} \begin{array} {rr} 66.8704363308& 160\\ 73.8250222623& 180\\ 61.6544968822& 145\\ 77.6500445247& 191 \end{array}$ the equation becomes $\small \hat{x} = 9.707035 + 0.34773 y$ and the sum-of-squares of the residues becomes now 72.980855 after a couple of recursions. [Update 4] The excel-generated image shows the regression lines for the complete data (black) for the incomplete/estimated data(red) and the complete data(blue). The solution of Joriki might be explainable by the effect, that the 4-fold imputation of the mean of the incomplete set adds the same "weight" of errors to the complete model as the imputation found by the iterative method, because the slope for the incomplete data can be set arbitrarily. - This is an interesting approach; I like the idea. But I wonder why you say that it's "much simpler" -- how is something that requires recursion (and we don't know whether the recursion would always be stable if the means of the two groups were less similar) much simpler than the prescription to simply replace the four unknown measurements by four measurements of the average values? – joriki Feb 23 '12 at 19:46 @joriki: It should not be so, but still it is that if I come across integral signs, and even stacked, I'm completely blocked and best I can do is becoming very formal; I'm completely losing any "inner imagination" of that problem -and that's why I say that the aproach above is "simpler". But you're right: after I see recursion is needed, I should remove that "simpler".... – Gottfried Helms Feb 23 '12 at 19:52 I guess it depends on whether we're comparing the simplicity of the derivations or of the resulting prescription for the calculation. My derivation isn't simple (though as a theoretical physicist I'm necessarily on speaking terms with integrations over multivariate Gaussians :-), but the resulting prescription to the practitioner couldn't be simpler. – joriki Feb 23 '12 at 19:56 I've done 5 pictures of the iteration process. Initially I insert the mean 70 for each of the missing values and compute the regression for the complete set and for the estimated set. Next I adapt the estimated values according to the regression-slope of the complete set and repeat. I've the zeroth,first, second, tenth and fifitieth iteration. See go.helms-net.de/math/divers/mse/MSE_120223_Regr.htm The optimum is achieved when the two slopes are identical (the four y-distances are equal). – Gottfried Helms Feb 23 '12 at 22:51 Just for the future generations I am going to put a link to a program that I used to diaggregate the data for me. ECOTRIM. Event it is an old program, it is working decently and diaggregated data for me very well. I did compare the real GDP data with diaggregated one, and it was shooting as close as not more than 5% away. Quite a decent tool. -	2015-04-26 13:13:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9067749977111816, "perplexity": 389.6342039415991}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246654467.42/warc/CC-MAIN-20150417045734-00105-ip-10-235-10-82.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/151882-find-length-curve-no-bounds-given.html	Thread: Find the length of the curve? NO bounds are given. 1. Find the length of the curve? NO bounds are given. Find the length of the curve? NO bounds are given. ---------------------------------------------------------------------------------- length of the curve formula using wolframalpha I got this int sqrt( 1+ ((1-2 x)/(2 sqrt(x-x^2))+1/(2 sqrt(1-x) sqrt(x)))^2) dx - Wolfram\|Alpha ================================================== ======== So, Is there any answer for this thing?? Thanks 2. To get the bounds, try looking for the natural domain of the original function. 3. Got it. its 0 to 1 the final answer = 2 4. Hmm. I get the limits as 0 to 1, and the value of the integral as different from yours. Can you post your work?	2016-10-23 01:29:27	{"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": 0.8186654448509216, "perplexity": 1524.2478259201066}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00163-ip-10-171-6-4.ec2.internal.warc.gz"}
https://byjus.com/ncert-solutions-class-11-maths/chapter-11-conic-sections/	# NCERT Solutions For Class 11 Maths Chapter 11 ## NCERT Solutions Class 11 Maths Conic Sections ### Ncert Solutions For Class 11 Maths Chapter 11 PDF Free Download BYJU’S, India’s largest learning app provides free NCERT Solutions for all the classes and are available in chapter wise with an objective to help students to prepare and understand concepts in an easy manner. NCERT solutions for class 11 Maths chapter 11 conic sections provide solutions for all questions present in the class 11 maths textbooks. Those students preparing for their board exams or any other competitive exams can use this study material to have a quick review of all important equations and formulas of conic sections. The conic sections are a kind of nondegenerate curves produced by the crossings of a plane and a cone. If a plane is normal to the axis of the cone, a circle is generated. For a plane that is not normal to the axis and meets only a single nappe, the curve generated is either a parabola or an ellipse. whereas hyperbola is the curve generated by a plane meeting both nappes. Both hyperbola and ellipse are termed as central conics. You can also find the whole list of NCERT solutions for Class 11 maths, to learn different solutions from. Due to this geometric assumption, the conics were examined by the Greeks even before its importance to the orbital inverse square law was recognised. Kepler initially noticed the ellipse shape of the orbits. Accordingly, Newton derived the pattern of orbits by utilizing calculus, by assuming that the gravitational force is inverse square of distance. Based on the energy of the orbiting object, 4 types of conic sections are possible. Solutions for NCERT class 11 Maths chapter 11 is available in PDF format for free and are mainly prepared by the team of our subject experts to assist students in their preparations. Interested students can download the solutions for class 11 Maths chapter 11 pdf files by visiting our website at BYJU’S. ### NCERT Solutions Class 11 Maths Chapter 11 Exercises Exercise 11.1 Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 Given: (h, k) = centre of a circle = (0, 3) and radius r = 3 The equation of a circle is: (a – 0)2 + (b – 3)2 = 32 a2 + b2 – 6b + 9 = 9 a2 + b2 – 6b + 9 – 9 = 0 a2 + b2 – 6b = 0 Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 Given: Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4 The equation of a circle is: (a – (- 1))2 + (b – 2)2 = 42 (a + 1)2 + (b – 2)2 = 42 a2 + 2a + 1 + b2 – 4b + 4 = 16 a2 + 2a + b2 – 4b + 5 = 16 a2 + 2a + b2 – 4b + 5 – 16 = 0 a2 + 2a + b2 – 4b – 11 = 0 Q.3: A circle is given with centre ($\frac{1}{3}$, $\frac{1}{2}$) and radius $\frac{1}{14}$. Find the equation of the circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 Given: Centre of a circle = (h, k) = ($\frac{1}{3}$, $\frac{1}{2}$) and the radius (r) = 4 The equation of a circle is: (a – $\frac{1}{3}$)2 + (b – $\frac{1}{2}$)2 = $\frac{1}{14}$2 (a – $\frac{1}{3}$)2 + (b – $\frac{1}{2}$)2 = $\frac{1}{14}$2 $a^{2} – 2.\frac{1}{3}.a + \frac{1}{9} + b^{2} – 2.\frac{1}{2}.b + \frac{1}{4} = (\frac{1}{14})^{2} \\ a^{2} – \frac{2a}{3} + \frac{1}{9} + b^{2} – b + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{1}{9} + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} – \frac{1}{196} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{637 – 9}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{628}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + 0.35 = 0 \\ 3a^{2} + 3b^{2} – 2a – 3b + 1.068 = 0 \\$ Q.4: A circle is given with centre (2, 2) and radius $\sqrt{5}$. Obtain the equation of the given circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 Given: Centre of a circle = (h, k) = (2, 2) and radius (r) = $\sqrt{5}$ The equation of a circle is: (a – 2)2 + (b – 2)2 = $\sqrt{5}^{2}$ (a – 2)2 + (b – 2)2 = $\sqrt{5}^{2}$ a2 – 4a + 4 + b2 – 4b + 4 = 5 a2 – 4a + b2 – 4b + 8 = 5 a2 – 4a + b2 – 4b + 8 – 5 = 0 a2 – 4a + b2 – 4b + 3 = 0 a2 + b2 – 4a – 4b + 3 = 0 Q.5: A circle is given with centre (- x, – y) and radius $\sqrt{x^{2} – y^{2}}$. Find the equation of the circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 Given: Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ = $\sqrt{x^{2} – y^{2}}$ The equation of a circle is: (a – (- x)2 + (b – (- y)2 = $\sqrt{x^{2} – y^{2}}^{2}$ (a + x)2 + (b + y)2 = $\sqrt{x^{2} – y^{2}}^{2}$ (a + x)2 + (b + y)2 = x2 – y2 a2 + 2ax + x2 + b2 + 2by + y2 = x2 – y2 a2 + 2ax + b2 + 2by + y2 = 0 Q.6: The equation of a given circle is given as (a + 5)2 + (b – 3)2 = 36. Find the centre and radius of the circle. Sol: Given: The equation of a given circle (a + 5)2 + (b – 3)2 = 36 (a + 5)2 + (b – 3)2 = 36 {a – (- 5)}2 + (b – 3) = 62 Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = – 5, k = 3 and r = 6. Hence, the centre of the circle is (-5, 3), and the radius of circle is 6. Q.7: The equation of a given circle is given as a2 + b2 – 4a – 4b + 3 = 0 Find the radius and centre of the circle. Sol: Given: The equation of a given circle a2 + b2 – 4a – 4b + 3 = 0 a2 + b2 – 4a – 4b + 3 = 0 a2 – 4a + b2 – 4b + 3 = 0 {a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} – 22 – 22 + 3 = 0 {a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} – 8 + 3 = 0 {a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} = 5 (a – 2)2 + (b – 2)2 = $\sqrt{5}^{2}$ Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 2 and r = $\sqrt{5}$. Thus, the centre of the circle is (2, 2), and the radius of circle is $\sqrt{5}$. Q.8: The equation of a given circle is given as a2 + b2 – 6a + 8b – 14 = 0 Find the centre and radius of the circle. Sol: Given: The equation of a given circle a2 + b2 – 6a + 8b – 14 = 0 a2 – 6a + b2 + 8b – 14 = 0 {a2 – 2.a.3 + 32} + {b2 + 2.b.4 + 42} – 32 – 42 – 14 = 0 (a – 3) 2 + (b + 4) 2 – 9 – 16 – 14 = 0 (a – 3) 2 + (b + 4) 2 – 39 = 0 (a – 3) 2 + (b + 4) 2 = 39 (a – 3) 2 + (b – (- 4)) 2 = 39 Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 3, k = 4 and r = $\sqrt{39}$. Thus, the centre of the circle is (3, 4), and the radius of circle is $\sqrt{39}$. Q.9: The equation of a given circle is given as 2a2 + 2b2 – 8a = 0 Find the centre and radius of the circle. Sol: Given: The equation of a given circle 2a2 + 2b2 – 8a = 0 2a2 + 2b2 – 8a = 0 2a2 – 8a + 2b2 = 0 2[a2 – 4a + b2] = 0 {a2 – 2.a.2 + 22} – 22 + (b – 0)2 = 0 [a2 – 2.a.2 + 22] + [b – 0]2 – 22 = 0 (a – 2)2 + (b – 0)2 = 22 = 4 Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 0 and r = $\sqrt{4}$ = 2. Thus, the centre of the circle is (3, 4), and the radius of circle is 2. Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle. Sol: The equation of a required circle with centre (h, k) and radius r is given as (a – h)2 + (b – k)2 = r2 As the circle passes through (6, 2) and (4, 3) (6 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1) (4 – h)2 + (3 – k)2 = r2 . . . . . . . . . . . . . . . . (2) The centre (h, k) of the circle lies on the line 2a + y = 8, 2h + k = 8 . . . . . . . . . . . . . . . . . . . . . (3) Now, On Comparing equations 1 and 2, we will get: (6 – h)2 + (2 – k)2 = (4 – h)2 + (3 – k)2 36 – 12h + h2 + 4 – 4k + k2 = 16 – 8h + h2 + 9 – 6k + k2 36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k – 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4 4h – 2k = 15 . . . . . . . . . . . . . . . . . . (4) Now, on Solving Equations (3) and (4), we will get: h = $\frac{31}{8}$ k = $\frac{1}{4}$ Substituting the values of h and k in equation (1), we will get: (6 – $\frac{31}{8}$)2 + (2 – $\frac{1}{4}$)2 = r2 $\\(\frac{17}{8})^{2} + (\frac{7}{4})^{2} = r^{2}$ $\\\Rightarrow$ $\frac{289}{64} + \frac{49}{16} = r^{2}$ $\\\Rightarrow$ $\frac{289 + 196}{64} = r^{2}$ $\\\Rightarrow$ $\frac{485}{64} = r^{2}$ $\Rightarrow$ r = 7.57 The required equation is: (a – $\frac{31}{8}$)2 + (b – $\frac{1}{4}$)2 = r2 (a – $\frac{31}{8}$)2 + (b – $\frac{1}{4}$)2 = 7.572 $(a – \frac{31}{8})^{2} + (b – \frac{1}{4})^{2} = 7.57^{2} \\$ $\\\Rightarrow$ $a^{2} – \frac{31}{4}a + (\frac{31}{8})^{2} + b^{2} – \frac{b}{2} + \frac{1}{16} = 57.30\\$ $\\\Rightarrow$ $a^{2} – \frac{31}{4}a + \frac{961}{64} + b^{2} – \frac{b}{2} + \frac{1}{16} =57.30$ Therefore, the required equation is: 64a2– 496a + 64b2 -32b = 2702.51 Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle. Sol: The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 As the circle passes through (3, 2) and (-2, 1) (3 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1) (-2 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . (2) The centre (h, k) of the circle lies on the line a – 3y – 13 = 0, h – 3k = 13 . . . . . . . . . . . . . . . (3) Now, on comparing equations 1 and 2, we will get: (3 – h)2 + (2 – k)2 = (-2 – h)2 + (2 – k)2 9 – 6h + h2 = 4 + 4h + h2 9 – 6h = 4 + 4h h = 0.5 . . . . . . . . . . . . . . . . . . (4) Now, on solving equation (3), we will get: k = – $\frac{25}{6}$ = – 4.166 Now, on substituting the values of h and k in equation (1), we will get: (3 – 0.5)2 + (2 – (- $\frac{25}{6}$))2 = r2 (2.5) 2 + (2 + $\frac{25}{6}$)2 = r2 (2.5) 2 + (2 + 4.16)2 = r2 (2.5) 2 + (6.16)2 = r2 r2 = 44.1956 r = 6.65 The required equation is: (a – h)2 + (b – k)2 = r2 (a – 0.5)2 + (b – (- 4.16))2 = r2 (a – 0.5)2 + (b + 4.16)2 = r2 a2 – a + 6.25 + b2 + 8.32b + 17.30 = 44.19 a2 – a + b2 + 8.32b = 44.19 – 6.25 – 17.30 a2 + b2 – a + 8.32b – 20.64 = 0 Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation. Sol: Given: The radius of the circle is 6 and passes through the point (3, 2) The General equation of a circle with centre (h, k) and radius r is: (a – h)2 + (b – k)2 = r2 As, the centre lies on x – axis, k = 0, the equation becomes: (a – h) 2 + b 2 = 36 (3 – h) 2 + 22 = 36 (3 – h) 2 = 36 – 4 3 – h = $\sqrt{32}$ h = $3 + 4\sqrt{2}$ or h = $3 – 4\sqrt{2}$ Q.13: The circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Find the equation of the circle. Sol: Suppose, the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2 The centre of the circle passes through (0, 0): (0 – h) 2 + (0 – k) 2 = r 2 h 2 + k 2 = r 2 Therefore, the equation of the circle is: (x – h) 2 + (y – k) 2 = h 2 + k 2 . Given, the circle makes intercepts a and b on the coordinate axes. The circle passes through points (a, 0) and (0, b). Therefore, (a – h) 2 + (0 – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . (1) (0 – h) 2 + (b – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . . . . (2) Now, from equation (1), we obtain: a 2 – 2ah + h 2 + k 2 = h 2 + k 2 a 2 – 2ah = 0 a (a – 2h) = 0 a = 0 or (a – 2h) = 0 We know, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2. Now, from equation (2), we will get: h 2 + b 2 – 2bk + k 2 = h 2 + k 2 b 2 – 2bk = 0 b (b – 2k) = 0 b = 0 or(b – 2k) = 0 However, b ≠ 0; Therefore, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is: $(x – \frac{a}{2})^{2} + (y – \frac{b}{2})^{2} = (\frac{a}{2})^{2} + (\frac{b}{2})^{2} \\ (\frac{2x – a}{2})^{2} + (\frac{2y – b}{2})^{2} = \frac{a^{2} + b^{2}}{2} \\ 4x^{2} – 4ax + a^{2} + 4y^{2} – 4bx + b^{2} = a^{2} + b^{2} \\ 4x^{2} + 4y^{2} – 4ax – 4by = 0 \\ x^{2} + y^{2} – ax – by = 0$ Q.14: The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle? Sol: Given: The centre of the circle (h, k) = (3, 3). As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5). $r = \sqrt{(3 – 6)^{2} + (3 – 5)^{2}} + \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}\\$ Therefore, the equation of the circle is: $\\(a – h)^{2} + (b – k)^{2} = r^{2} \\ (a – 3)^{2} + (b – 3)^{2} = \sqrt{13}^{2} \\ a^{2} – 6a + 9 + b^{2} – 6b + 9 = 13 \\ a^{2} + b^{2} – 6a – 6b + 18 – 13 = 0 \\ a^{2} + b^{2} – 6a – 6b + 5 = 0$ Q.15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x 2 + y 2 = 25? Sol: Given: The equation of the given circle is x 2 + y 2 = 25 x 2 + y 2 = 25 (x – 0)2 + (y – 0)2 = 52 , which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 0, k = 0, and r = 5. Centre = (0, 0) and radius = 5 Distance between point (– 2.0, 3.0) and centre (0, 0) $\sqrt{(- 2 – 0)^{2} + (3 – 0)^{2}}\\$ =$\sqrt{4 + 9} = \sqrt{13} = 3.60 (approx.) < 5$ As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, point (– 2.0, 3.0) lies inside the circle. Exercise 11.2 Q.1: For the equation y2 = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: y2 = 16x As the coefficient of x is positive, the given parabola has the opening on the right side. By comparing this equation with y2 = 4ax, we get: 4a = 16 ⇒ a = 4 Therefore, focus coordinates = (a, 0) = (4, 0) The given equation has y2, the x-axis is the axis of the parabola. Directrix equation, x = – a = – 4 So, x + 4 = 0 Therefore, Length of the latus rectum = 4a = 4 × 4 = 16 Q.2: For the equation x2 = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: x2 = 8y As the coefficient of y is positive, the given parabola has the opening upwards. By comparing this equation with x2 = 4ay, we get 4a = 8 ⇒ a = 2 Therefore, focus coordinates = (0, a) = (0, 2) The given equation has x2, the y – axis is the axis of the parabola. Directrix equation, y = – a = – 2 So, x + 2 = 0 Therefore, Length of the latus rectum = 4a = 4 × 2 = 8 Q.3: For the equation y2 = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: y2 = – 12x As the coefficient of x is negative, the given parabola has the opening on the left side. By comparing this equation with y2 = – 4ax, we get 4a = 12 ⇒ a = 3 Therefore, focus coordinates = (- a, 0) = (- 3, 0) The given equation has y2, the x-axis is the axis of the parabola. Directrix equation, x = a = 2 So, x + 4 = 0 Therefore, Length of the latus rectum = 4a = 4 × 3 = 12 Q.4: For the equation x2 = – 20y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: x2 = – 20y As the coefficient of y is negative, the given parabola has the opening downwards. By comparing this equation with x2 = – 4ay, we get: – 4a = – 20 ⇒ a = 5 Therefore, focus coordinates = (0, a) = (0, 5) The given equation has x2, the y – axis is the axis of the parabola. Directrix equation, y = a = 5 Therefore, Length of the latus rectum = 4a = 4 × 5 = 20 Q.5: For the equation y2 = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: y2 = 24x As the coefficient of x is positive, the given parabola has the opening on the right side. By comparing this equation with y2 = 4ax, we get: 4a = 24 ⇒ a = 6 Therefore, focus coordinates = (a, 0) = (6, 0) The given equation has y2, the x-axis is the axis of the parabola. Directrix equation, x = – a = – 6 So, x + 6 = 0 Therefore, Length of the latus rectum = 4a = 4 × 6 = 24 Q.6: For the equation x2 = – 7y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola. Sol: Given: x2 = – 7y As the coefficient of y is negative, the given parabola has the opening downwards. By comparing this equation with x2 = – 4ay, we get: – 4a = – 7 ⇒ a = $\frac{7}{4}$ Therefore, focus coordinates = (0, a) = (0, $\frac{7}{4}$) The given equation has x2, the y – axis is the axis of the parabola. Directrix equation, y = a = $\frac{7}{4}$ Therefore, Length of the latus rectum = 4a = 4 × $\frac{7}{4}$ = 7 Q7. Obtain the equation of the parabola which satisfies the given conditions below: (i) Focus coordinates = (8, 0) (ii) Directrix (x) = – 8 Sol: Given: (i) Focus coordinates = (8, 0) (ii) Directrix (x) = – 8 The axis of the given parabola is the x – axis as the focus is represented on the x – axis. The required equation can be either y2 = 4ax and y2 = – 4ax As we know, directrix (x) = – 8 As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (8, 0) Therefore, the equation of the parabola = y2 = 4ax, Here, a = 8 Thus, the equation of the parabola = y2 = 32 x. Q.8: Obtain the equation of the parabola which satisfies the given conditions below: (i) Focus coordinates = (0, – 5) (ii) Directrix (y) = 5 Sol: Given: (i) Focus coordinates = (0, – 5) (ii) Directrix (x) = 5 The axis of the given parabola is the y – axis as the focus is represented on the y – axis. The required equation can be either x2 = 4ay and x2 = – 4ay As we know, directrix (x) = 5 As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (0, -5) Therefore, the equation of the parabola = x2 = – 4ay, Here, a = 5 Thus, the equation of the parabola = x2 = – 20y Q.9: Obtain the equation of the parabola which satisfies the given conditions below: (i) Vertex is at origin. (ii) Focus coordinates = (4, 0) Sol: Given: (i) Vertex is at origin i.e., (0, 0) (ii) Focus coordinates = (4, 0) As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis. The required equation is y2 = 4ax As we know, focus is (4, 0) Therefore, the equation of the parabola = y2 = 4ax Here, a = 4 Thus, the equation of the parabola = y2 = 16 x Q.10: Obtain the equation of the parabola which satisfies the given conditions below: (i) Vertex is at origin. (ii) Focus coordinates = (- 3, 0) Sol: Given: (i) Vertex is at origin i.e., (0, 0) (ii) Focus coordinates = (- 3, 0) As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, the axis of the parabola is x – axis. The required equation is y2 = – 4ax Therefore, the equation of the parabola = y2 = – 4ax As we know, focus is (- 3, 0) Here, a = 3 Thus, the equation of the parabola = y2 = – 12 x Q.11: Obtain the equation of the parabola which satisfies the given conditions below: (i) Vertex is at origin. (ii) Passing through coordinates = (3, 5) The axis of the parabola is on x – axis. Sol: Given: (i) Vertex is at origin i.e., (0, 0) (ii) Passing through coordinates = (3, 5) As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5). The required equation is either y2 = 4ax or y2 = – 4ax The coordinates (3, 5) is in first quadrant. Therefore, the equation of the parabola = y2 = 4ax, whereas (3, 5) should satisfy the equation So, 52 = 4a (3) 25 = 12 a a = $\frac{25}{12}$ The parabola’s equation is: y2 = 4 ($\frac{25}{12}$) x 3 y2 = 25 x Q.12: Obtain the equation of the parabola which satisfies the given conditions below: (i) Vertex is at origin. (ii) Passing through coordinates = (6, 4) The equation is symmetric as considered with y – axis. Sol: Given: (i) Vertex is at origin i.e., (0, 0) (ii) Passing through coordinates = (6, 4) As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4). The required equation is either x2 = 4ay or x2 = – 4ay The coordinates (6, 4) is in first quadrant. Therefore, the equation of the parabola = x2 = 4ay, whereas (6, 4) should satisfy the equation So, 62 = 4a (4) 36 = 16 a a = $\frac{36}{16}$ a = $\frac{9}{4}$ The parabola’s equation is: x2 = 4($\frac{9}{4}$)y x2 = 9 y Exercise 11.3 Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$. Sol: Given: $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{25}$ > the denominator of $\frac{y^{2}}{9}$ So, along x – axis consist the major axis and along the y – axis consists the minor axis. $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 9} = \sqrt{16} = 4$ We get: m = 5, n = 3 The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0) The foci’s coordinates are (c, 0) and (- c, 0) = (4, 0) and (- 4, 0) Length of the axis: Major axis = 2m = 10 Minor axis = 2n = 6 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.3^{2}}{5} = \frac{18}{5}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{4}{5}$ Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$. Sol: Given: $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{9}$ < the denominator of $\frac{y^{2}}{16}$ So, along y – axis consist the major axis and along the x – axis consists the minor axis. $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{16 – 9} = \sqrt{7}$ We get,: m = 4, n = 3 The vertices coordinates are (0, m) and (0, – m) = (0, 4), (0, – 4) The foci’s coordinates are (0, c) and (0, – c) = ($\sqrt{7}$, 0) and (-$\sqrt{7}$, 0) Length of the axis: Major axis = 2m = 8 Minor axis = 2n = 6 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.3^{2}}{4} = \frac{9}{2}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{7}}{4}$ Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$. Sol: Given: $\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{25}$ > the denominator of $\frac{y^{2}}{4}$ So, along x – axis consist the major axis and along the y – axis consists the minor axis. $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 4} = \sqrt{21}$ We get: m = 5, n = 2 The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($\sqrt{21}$, 0) and (-$\sqrt{21}$, 0) Length of the axis: Major axis = 2m = 10 Minor axis = 2n = 4 Length of the latus rectum = $\frac{2n^{2}}{m}=\frac{2.2^{2}}{5} = \frac{8}{5}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{21}}{5}$ Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$. Sol: Given: $\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$ . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{36}$ < the denominator of $\frac{y^{2}}{121}$ So, along y – axis consist the major axis and along the x – axis consists the minor axis. $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{121 – 36} = \sqrt{85}$ We get: m = 11, n = 6 The vertices coordinates are (0, m) and (0, – m) = (0, 11), (0, – 11) The foci’s coordinates are (0, c) and (0, – c) = (0, $\sqrt{85}$) and (0, -$\sqrt{85}$) Length of the axis: Major axis = 2m = 22 Minor axis = 2n = 12 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.6^{2}}{11} = \frac{72}{11}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{85}}{11}$ Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$. Sol: Given: $\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$ . . . . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{64}$ > the denominator of $\frac{y^{2}}{25}$ So, along x – axis consist the major axis and along the y – axis consists the minor axis. $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{64 – 25} = \sqrt{39}$ We get: m = 8, n = 5 The vertices coordinates are (m, 0) and (- m, 0) = (8, 0), (- 8, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($\sqrt{39}$, 0) and (-$\sqrt{39}$, 0) Length of the axis: Major axis = 2m = 16 Minor axis = 2n = 10 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.5^{2}}{8} = \frac{25}{4}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{39}}{8}$ Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$. Sol: Given: $\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$ . . . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{400}$ > the denominator of $\frac{y^{2}}{100}$ So, along x – axis consist the major axis and along the y – axis consists the minor axis. $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{400 – 100} = \sqrt{300} = 10\sqrt{3}$ We get: m = 20, n = 10 The vertices coordinates are (m, 0) and (- m, 0) = (20, 0), (- 20, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($10 \sqrt{3}$, 0) and (-$10 \sqrt{3}$, 0) Length of the axis: Major axis = 2m = 40 Minor axis = 2n = 20 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.10^{2}}{20} = 10$ Eccentricity, e = $\frac{c}{m}$ = $\frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2}$ Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x2 + 9y2 = 441 Sol: Given: 49x2 + 9y2 = 441 $\frac{x^{2}}{9} + \frac{y^{2}}{49} = 1$ . . . . . . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{9}$ < the denominator of $\frac{y^{2}}{49}$ So, along y – axis consist the major axis and along the x – axis consists the minor axis. $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{49 – 9} = \sqrt{40} = 2 \sqrt{10}$ We get: m = 7, n = 3 The vertices coordinates are (0, m) and (0, – m) = (0, 7), (0, – 7) The foci’s coordinates are (0, c) and (0, – c) = (0, $2 \sqrt{10}$) and (0, $2 \sqrt{10}$) Length of the axis: Major axis = 2m = 14 Minor axis = 2n = 6 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.3^{2}}{7} = \frac{18}{7}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{2 \sqrt{10}}{7}$ Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + 16y2 = 64 Sol: Given: 4x2 + 16y2 = 64 $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 169$ . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{16}$ > the denominator of $\frac{y^{2}}{4}$ So, along x – axis consist the major axis and along the y – axis consists the minor axis. $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c=\sqrt{m^{2}-n^{2}}=\sqrt{16 – 4}=\sqrt{12}=2\sqrt{3}$ We get: m = 4, n = 2 The vertices coordinates are (m, 0) and (- m, 0) = (4, 0), (- 4, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($2 \sqrt{3}$, 0) and (-$2 \sqrt{3}$, 0) Length of the axis: Major axis = 2m = 8 Minor axis = 2n = 4 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.4^{2}}{4} = 8$ Eccentricity, e = $\frac{c}{m}$ = $\frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + y2 = 4 Sol: Given: 4x2 + y2 = 4 $\frac{x^{2}}{1} + \frac{y^{2}}{4} = 1$ . . . . . . . . . . . . . . . . (1) is the equation of the ellipse. As we know that the denominator of $\frac{x^{2}}{1}$ < the denominator of $\frac{y^{2}}{4}$ So, along y – axis consist the major axis and along the x – axis consists the minor axis. $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} – n^{2}} = \sqrt{4 – 1} = \sqrt{3}$ We get: m = 2, n = 1 The vertices coordinates are (0, m) and (0, – m) = (0, 2), (0, – 2) The foci’s coordinates are (0, c) and (0, – c) = (0, $\sqrt{3}$) and (0, -$\sqrt{3}$) Length of the axis: Major axis = 2m = 4 Minor axis = 2n = 2 Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.1^{2}}{2} = 1$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{3}}{2}$ Q.10: For the given condition obtain the equation of the ellipse. (i) Vertices (±6, 0) (ii) Foci (±3, 0) Sol: Given: (i) Vertices (±6, 0) (ii) Foci (±3, 0) The vertices are represented along x – axis. The required equation of the ellipse is of the form $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1) We get: m = 6 (semi major axis), c = 3 As we know: m2 = n2 + c2 62 = n2 + 32 n2 = 62 – 32 n2 = 36 – 9 = 27 n = $\sqrt{27} = 3\sqrt{3}$ Hence, $\frac{x^{2}}{36} + \frac{y^{2}}{27} = 1$ is the equation of the ellipse. Q.11: For the given condition obtain the equation of the ellipse. (i) Vertices (0, ±8) (ii) Foci ( 0, ±4) Sol: Given: (i) Vertices (0, ±8) (ii) Foci ( 0, ±4) The vertices are represented along y – axis. The required equation of the ellipse is of the form $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . .(1) We get: m = 8 (semi major axis), c = 4 As we know: m2 = n2 + c2 82 = n2 + 42 n2 = 82 – 42 n2 = 64 – 16 = 48 n = $\sqrt{48} = 4\sqrt{3}$ Hence, $\frac{x^{2}}{48} + \frac{y^{2}}{64} = 1$ is the equation of the ellipse. Q.12: For the given condition obtain the equation of the ellipse. (i) Vertices (±5, 0) (ii) Foci (±3, 0) Sol: Given: (i) Vertices (± 5, 0) (ii) Foci (± 3, 0) The vertices are represented along x – axis. The required equation of the ellipse is of the form $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1) We get: m = 5 (semi major axis), c = 3 As we know: m2 = n2 + c2 52 = n2 + 32 n2 = 52 – 32 n2 = 25 – 9 = 16 n = $\sqrt{16} = 4$ Hence, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is the equation of the ellipse. Q.13: For the given condition obtain the equation of the ellipse. (i) Coordinates of major axis (±5, 0) (ii) Coordinates of minor axis (0, ±3) Sol: Given: (i) Coordinates of major axis (±5, 0) (ii) Coordinates of minor axis (0, ±3) The major axis is represented along x – axis. The required equation of the ellipse is of the form $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1) We get: m = 5 (semi major axis), n = 3 Hence, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is the equation of the ellipse. Q.14: For the given condition obtain the equation of the ellipse. (i) Coordinates of major axis (0, ±2) (ii) Coordinates of minor axis (±$\sqrt{3}$, 0) Sol: Given: (i) Coordinates of major axis (0, ±2) (ii) Coordinates of minor axis (±$\sqrt{3}$, 0) The major axis is represented along y – axis. The required equation of the ellipse is of the form $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . . . (1) We get: m = 2 (semi major axis), n = $\sqrt{3}$ Hence, $\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1$ is the equation of the ellipse. Q.15: For the given condition obtain the equation of the ellipse. (i) Length of major axis 30 (ii) Coordinates of foci (±4, 0) Sol: Given: (i) Length of major axis 30 (ii) Coordinates of foci (±4, 0) The major axis are represented along x – axis as foci is along x – axis The required equation of the ellipse is of the form $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . (1) We get: 2m = 30 (semi major axis), m = 15 c = 4 As we know: m2 = n2 + c2 152 = n2 + 42 n2 = 152 – 42 n2 = 225 – 16 = 209 n = $\sqrt{209}$ Hence, $\frac{x^{2}}{225} + \frac{y^{2}}{209} = 1$ is the equation of the ellipse. Q.16: For the given condition obtain the equation of the ellipse. (i) Length of minor axis 26 (ii) Coordinates of foci (0, ±9) Sol: Given: (i) Length of minor axis 26 (ii) Coordinates of foci (0, ±9) The major axis are represented along y – axis as foci is along y – axis The required equation of the ellipse is of the form $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . (1) We get: 2n = 26 (semi major axis), n = 13 c = 9 As we know: m2 = n2 + c2 m2 = 132 + 92 m2 = 169 + 81 m2 = 250 m = $\sqrt{250}$ = $5 \sqrt{10}$ Hence, $\frac{x^{2}}{169} + \frac{y^{2}}{250} = 1$ is the equation of the ellipse. Q.17: For the given condition obtain the equation of the ellipse. (i) Coordinates of foci (±4, 0) (ii) m = 6 Sol: Given: (i) Coordinates of foci (±4, 0) (ii) m = 6 The major axis are represented along x – axis as foci is along x – axis The required equation of the ellipse is of the form $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . (1) We get: c = 4 m = 6 As we know: m2 = n2 + c2 62 = n2 + 42 n2 = 36 – 16 n2 = 20 n = $\sqrt{20}$ = $2 \sqrt{5}$ Hence, $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$ is the equation of the ellipse. Q.18: For the given condition obtain the equation of the ellipse. (i) n = 2, c = 3 (on the x – axis) (ii) Coordinates of centre (0, 0) Sol: Given: (i) n = 2, c = 3 (on the x – axis) (ii) Coordinates of centre (0, 0) i.e., centre is at origin. The major axis are represented along x – axis as foci is along x – axis The required equation of the ellipse is of the form: $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (1) We have: n = 2, c = 3 As we know: m2 = n2 + c2 m2 = 22 + 32 m2 = 4 + 9 m2 = 13 m = $\sqrt{13}$ Hence, $\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1$ is the equation of the ellipse. Q.19: For the given condition obtain the equation of the ellipse. (i) Centre is at origin and major axis is along the y – axis (ii) It passes through the points (1, 6) and (3, 2) Sol: Given: (i) Centre is at origin and major axis is along the y – axis (ii) It passes through the points (1, 6) and (3, 2) The major axis are represented along y – axis as foci is along y – axis The required equation of the ellipse is of the form: $\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$ . . . . . . . . . . . . . . . (1) As the ellipse passes through (2, 1) and (2, 3) points, then $\frac{1^{2}}{n^{2}} + \frac{6^{2}}{m^{2}} = 1$ $\frac{3^{2}}{n^{2}} + \frac{2^{2}}{m^{2}} = 1$ $\frac{1}{n^{2}} + \frac{36}{m^{2}}$ = 1 . . . . . . . . . . . . . . . (a) $\frac{9}{n^{2}} + \frac{4}{m^{2}}$ = 1 . . . . . . . . . . . . . . . . (b) Now, on comparing equation (a) and equation (b) we will get: m2 = 40, n2 = 10 Hence, $\frac{y^{2}}{40} + \frac{x^{2}}{10} = 1$ is the equation of the ellipse. Q.20: For the given condition obtain the equation of the ellipse. (i) Major axis is represented along x – axis (ii) It passes through the points (6, 2) and (4, 3) Sol: Given: (i) Major axis is represented along x – axis (ii) It passes through the points (6, 2) and (4, 3) The major axis are represented along y – axis as foci is along y – axis The required equation of the ellipse is of the form: $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (1) As the ellipse passes through (6, 2) and (4, 3) points, then $\frac{6^{2}}{m^{2}} + \frac{2^{2}}{n^{2}} = 1$ $\frac{4^{2}}{m^{2}} + \frac{3^{2}}{n^{2}} = 1$ $\frac{36}{m^{2}} + \frac{4}{n^{2}}$ = 1 . . . . . . . . . . . . . . . . (a) $\frac{16}{m^{2}} + \frac{9}{n^{2}}$ = 1 . . . . . . . . . . . . . . . . (b) Now, on comparing equation (a) and equation (b) we will get: m2 = 52 and n2 = 13 Hence, $\frac{y^{2}}{13} + \frac{x^{2}}{52} = 1$ is the equation of the ellipse. Exercise 11.4 Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola $\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$. Sol: Given: $\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$ . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . .(2) Now, on comparing equation (1) and equation (2) we will get: $c = \sqrt{m^{2} + n^{2}} = \sqrt{25 + 16} = \sqrt{41}$ We get: m = 5 and n = 4 The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($\sqrt{41}$, 0) and (-$\sqrt{41}$, 0) Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.4^{2}}{5} = \frac{32}{5}$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{41}}{5}$ Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola $\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$. Sol: Given: $\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$ . . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . . . .(2) Now, on comparing equation (1) and equation (2) we will get: m = 1and n = $\sqrt{8}$ As we know: $c = \sqrt{m^{2} + n^{2}} = \sqrt{1 + 8} = \sqrt{9}$ = 3 The vertices coordinates are (0, m) and (0, – m) = (0, 1), (0, – 1) The foci’s coordinates are (0, c) and (0, – c) = (0, 3), (0, – 3) Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{16}{1} = 16$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{3}}{1}$ = 3 Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y2 – 4 x2 = 100 Sol: Given: 25 y2 – 4 x2 = 100 $\frac{y^{2}}{4} – \frac{x^{2}}{25} = 1$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . .(2) Now, on comparing equation (1) and equation (2) we will get: m = 2 and n = 5. $c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 25} = \sqrt{29}$ The vertices coordinates are (0, m) and (0, – m) = (0, 2) and (0, – 2) The foci’s coordinates are (0, c) and (0, – c) = (0, $\sqrt{29}$) and (0, -$\sqrt{29}$) Length of the latus rectum = $\frac{2.n^{2}}{m}$ = $\frac{2.5^{2}}{2} = \frac{50}{2}$ = 25 Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{29}}{2}$ Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x2 – 4 y2 = 36 Sol: Given: 9 x2 – 4 y2 = 36 $\frac{x^{2}}{4} – \frac{y^{2}}{9} = 1$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: m = 2 and n = 3. $c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 3} = \sqrt{13}$ The vertices coordinates are (m, 0) and (- m, 0) = (2, 0), (- 2, 0) The foci’s coordinates are (c, 0) and (- c, 0) = ($\sqrt{13}$, 0) and (-$\sqrt{13}$, 0) Length of the latus rectum = $\frac{2n^{2}}{m}$ = $\frac{2.3^{2}}{2} = 9$ Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{13}}{2}$ Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y2 – 16 x2 = 96 Sol: Given: 6 y2 – 16 x2 = 96 $\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$ . . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . .(2) Now, on comparing equation (1) and equation (2) we will get: m = 4 and n = $\sqrt{6}$ $c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$ The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4) The foci’s coordinates are (0, c) and (0, – c) = (0, $\sqrt{10}$) and (0, -$\sqrt{10}$) Length of the latus rectum = $\frac{2.n^{2}}{m}$ = $\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$ = 3 Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{10}}{4}$ Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y2 – 9 x2 = 96 Sol: Given: 6 y2 – 16 x2 = 96 $\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$ . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola. The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . (2) Now, on comparing equation (1) and equation (2) we will get: m = 4 and n = $\sqrt{6}$ $c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$ The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4) The foci’s coordinates are (0, c) and (0, – c) = (0, $\sqrt{10}$) and (0, -$\sqrt{10}$) Length of the latus rectum = $\frac{2.n^{2}}{m}$ = $\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$ = 3 Eccentricity, e = $\frac{c}{m}$ = $\frac{\sqrt{10}}{4}$ Q.7: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the vertices (±3, 0) (ii) Coordinates of the foci (±4, 0) Sol: Given: (i) Coordinates of the vertices (±3, 0) (ii) Coordinates of the foci (±4, 0) The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (1) As, coordinates of the vertices (±3, 0), m = 3 And, coordinates of the foci (±4, 0), c = 4 We know that: c2 = m2 + n2 42 = 32 + n2 n2 = 16 – 9 = 5 $\frac{x^{2}}{9} – \frac{y^{2}}{5} = 1$ is the equation of the hyperbola. Q.8: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the vertices (0, ±7) (ii) Coordinates of the foci (0, ±9) Sol: Given: (i) Coordinates of the vertices (0, ±7) (ii) Coordinates of the foci (0, ±9) The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . (1) As, coordinates of the vertices (0, ±7), m = 7 And, coordinates of the foci (0, ±9), c = 9 We know that: c2 = m2 + n2 92 = 72 + n2 n2 = 81 – 49 = 32 $\frac{y^{2}}{49} – \frac{x^{2}}{32} = 1$ is the equation of the hyperbola. Q.9: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the vertices (0, ±8) (ii) Coordinates of the foci (0, ±11) Sol: Given: (i) Coordinates of the vertices (0, ±8) (ii) Coordinates of the foci (0, ±11) The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . . . (1) As, coordinates of the vertices (0, ±8), m = 8 And, coordinates of the foci (0, ±11), c = 11 We know that: c2 = m2 + n2 112 = 82 + n2 n2 = 121 – 64 = 57 $\frac{y^{2}}{64} – \frac{x^{2}}{57} = 1$ is the equation of the hyperbola. Q.10: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the foci (±7, 0), (ii) The length of the transverse axis is 12 Sol: Given: (i) Coordinates of the foci (±7, 0), (ii) The length of the transverse axis is 12 The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . (1) And, coordinates of the foci (±7, 0), c = 7 As, the transverse axis is of length 12, 2 m = 12 m = 6 We know that: c2 = m2 + n2 72 = 62 + n2 n2 = 49 – 36 = 13 $\frac{x^{2}}{36} – \frac{y^{2}}{13} = 1$ is the equation of the hyperbola. Q.11: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the foci (0, ±8) (ii) The length of the conjugate axis is 16 Sol: Given: (i) Coordinates of the foci (0, ±9) (ii) The length of the conjugate axis is 16 The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (1) And, coordinates of the foci (0, ±9), c = 9 As, the conjugate axis is of length 16, 2 m = 16 m = 8 Therefore, m2 = 64 We know that: c2 = m2 + n2 92 = 82 + n2 n2 = 81 – 64 = 17 $\frac{y^{2}}{64} – \frac{x^{2}}{17} = 1$ is the equation of the hyperbola. Q.12: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the foci (±$2\sqrt{3}$, 0 ) (ii) The length of the latus rectum is 8 Sol: Given: (i) Coordinates of the foci (±$2\sqrt{3}$, 0 ) (ii) The length of the latus rectum is 8 The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (1) And, the coordinates of the foci (±$2\sqrt{3}$, 0 ), c = $2\sqrt{3}$ As, the latus rectum is of length 8 $\frac{2n^{2}}{m} = 8 \\ 2n^{2} = 8 m \\ n^{2} = 4 m$ We know that: m2 + n2 = c2 m2 + 4m = $(2\sqrt{3})^{2}$ m2 + 4m = 12 m2 + 4m – 12 = 0 m2 + 6m – 2m – 12 = 0 m (m + 6) – 2 (m + 6) = 0 m = – 6 and m = 2 m is non – negative so, m = 2 and n2 = 8 [Since, n2 = 12 – 22 = 8] $\frac{x^{2}}{4} – \frac{y^{2}}{8} = 1$ is the equation of the hyperbola. Q.13: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the foci (±$2\sqrt{6}$, 0) (ii) The length of the latus rectum is 10 Sol: Given: (i) Coordinates of the foci (±$2\sqrt{3}$, 0 ) (ii) The length of the latus rectum is 10 The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1) And, coordinates of the foci (±$2\sqrt{6}$, 0 ), c = $2\sqrt{6}$ As, the latus rectum is of length 10 Therefore, $\frac{2n^{2}}{m} = 10 \\ 2n^{2} = 10 m \\ n^{2} = 5m$ We know that: m2 + n2 = c2 m2 + 5 m = $(2\sqrt{6})^{2}$ m2 + 5 m = 24 m2 + 5 m – 24 = 0 m2 + 8 m – 3 m – 24 = 0 m (m + 8) – 3 (m + 8) = 0 m = – 8 and m = 3 m is non – negative so, m = 3 and n2 = 5m = 15 $\frac{x^{2}}{9} – \frac{y^{2}}{15} = 1$ is the equation of the hyperbola. Q.14: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the vertices (±9, 0) (ii) Eccentricity (e) = $\frac{9}{4}$ Sol: Given: (i) Coordinates of the vertices (±9, 0) (ii) Eccentricity (e) = $\frac{9}{4}$ The specific equation for hyperbola is $\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1) And, Coordinates of the vertices (±9, 0), m = 9 As, the eccentricity (e) = $\frac{9}{4}$, $\frac{c}{m} = \frac{9}{4}$ $\frac{c}{9} = \frac{9}{4}$ c = $\frac{81}{4}$ We know that: m2 + n2 = c2 92 + n2 = $(\frac{81}{4})^{2}$ n2 = $(\frac{81}{4})^{2}$ – 92 n2 = $\frac{6561}{16} – 81$ n2 = $\frac{6561 – 1296}{16} = \frac{5265}{16}$ $\frac{x^{2}}{81} – \frac{16 y^{2}}{5265} = 1$ is the equation of the hyperbola. Q.15: For the given condition obtain the equation of the hyperbola. (i) Coordinates of the foci (0, ± $\sqrt{10}$) (ii) It passes through (2, 3) Sol: Given: (i) Coordinates of the foci (0, ±$\sqrt{10}$) (ii) It passes through (2, 3) The specific equation for hyperbola is $\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . (1) And, coordinates of the foci (0, ±$\sqrt{10}$), c = $\sqrt{10}$ We know that: m2 + n2 = c2 n2 = c2 – m2 n2 = 10 – m2 . . . . . . . . . . . . (2) As the hyperbola passes through (2, 3) $\frac{3^{2}}{m^{2}} – \frac{2^{2}}{n^{2}} = 1$ $\frac{9}{m^{2}} – \frac{4}{n^{2}} = 1$ . . . . . . . . . . . (3) Putting Equation (2) in equation (3), we get: $\frac{9}{m^{2}}-\frac{4}{10 – m^{2}}= 1$ 9 (10 – m2) – 4m2 = m2 (10 – m2) m4 – 23m2 + 90 = 0 m4 – 18m2 – 5m2 + 90 = 0 (m2 – 5) (m2 – 18) = 0 m2 = 5 or 18 As in hyperbola: m2 < c2 m2 = 5 n2 = 10 – 5 [From Equation (2)] = 5 $\frac{y^{2}}{5} – \frac{x^{2}}{5} = 1$ is the equation of the hyperbola.	2019-03-19 23:25:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8338724970817566, "perplexity": 459.3507787241153}, "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-2019-13/segments/1552912202161.73/warc/CC-MAIN-20190319224005-20190320010005-00070.warc.gz"}
https://en.wikipedia.org/wiki/Age_crisis	# Cosmic age problem (Redirected from Age crisis) "Age crisis" redirects here. For life crises, see Quarter-life crisis and Midlife crisis. The cosmic age problem is a historical problem in astronomy concerning the age of the universe. The problem was that at various times in the 20th century, some objects in the universe were estimated to be older than the time elapsed since the Big Bang,[1] as estimated from measurements of the expansion rate of the universe known as the Hubble constant, denoted H0. (This is more correctly called the Hubble parameter, since it generally varies with time). Since around 1997–2003, the problem is believed to be solved by most cosmologists: modern measurements give an accurate age of the universe of 13.8 billion years, and recent age estimates for the oldest objects are either younger than this, or consistent allowing for measurement uncertainties. ## Early years Following theoretical developments of the Friedmann equations by Alexander Friedmann and Georges Lemaitre in the 1920s, and the discovery of the expanding universe by Edwin Hubble in 1929, it was immediately clear that tracing this expansion backwards in time predicts that the universe had almost zero size at a finite time in the past. This concept, initially known as the "Primeval Atom" by Lemaitre, was later elaborated into the modern Big Bang theory. If the universe had expanded at a constant rate in the past, the age of the universe now (i.e. the time since the Big Bang) is simply the inverse of the Hubble constant, often known as the Hubble time. For Big Bang models with zero cosmological constant and positive matter density, the actual age must be somewhat younger than this Hubble time; typically the age would be between 66% and 90% of the Hubble time, depending on the density of matter. Hubble's early estimate of his constant[2] was 550 km/s/Mpc, and the inverse of that is 1.8 billion years. It was believed by many geologists such as Arthur Holmes in the 1920s that the Earth was probably over 2 billion years old, but with large uncertainty.[citation needed] The possible discrepancy between the ages of the Earth and the universe was probably one motivation for the development of the Steady State theory in 1948 as an alternative to the Big Bang;[3] in the (now obsolete) steady state theory, the universe is infinitely old and on average unchanging with time. The steady state theory postulated spontaneous creation of matter to keep the average density constant as the universe expands, and therefore most galaxies still have an age less than 1/H0. However, if H0 had been 550 km/s/Mpc, our Milky Way galaxy would be exceptionally large compared to most other galaxies, so it could well be much older than an average galaxy, therefore eliminating the age problem. ## 1950–1970 In the 1950s, two substantial errors were discovered in Hubble's extragalactic distance scale: first in 1952, Walter Baade discovered there were two classes of Cepheid variable star. Hubble's sample comprised different classes nearby and in other galaxies, and correcting this error made all other galaxies twice as distant as Hubble's values, thus doubling the Hubble time.[4] A second error was discovered by Allan Sandage and coworkers: for galaxies beyond the Local Group, Cepheids were too faint to observe with Hubble's instruments, so Hubble used the brightest stars as distance indicators. Many of Hubble's "brightest stars" were actually HII regions or clusters containing many stars, which caused another underestimation of distances for these more distant galaxies.[citation needed] Thus, in 1958 Sandage[5] published the first reasonably accurate measurement of the Hubble constant, at 75 km/s/Mpc, which is close to modern estimates of 68–74 km/s/Mpc.[6] The age of the Earth (actually the Solar System) was first accurately measured around 1955 by Clair Patterson at 4.55 billion years,[7] essentially identical to the modern value. For H0 ~ 75 km/s/Mpc, the inverse is 13.0 billion years; so after 1958 the Big Bang model age was comfortably older than the Earth. However, in the 1960s and onwards, new developments in the theory of stellar evolution enabled age estimates for large star clusters called globular clusters: these generally gave age estimates of around 15 billion years, with substantial scatter.[citation needed] Further revisions of the Hubble constant by Sandage and Gustav Tammann in the 1970s gave values around 50–60 km/s/Mpc,[8] and an inverse of 16-20 billion years, consistent with globular cluster ages. ## 1975–1990 However, in the late 1970s to early 1990s, the age problem re-appeared: new estimates of the Hubble constant gave higher values, with Gerard de Vaucouleurs estimating values 90–100 km/s/Mpc,[9] while Marc Aaronson and co-workers gave values around 80-90 km/s/Mpc.[10] Sandage and Tammann continued to argue for values 50-60, leading to a period of controversy sometimes called the "Hubble wars".[citation needed] The higher values for H0 appeared to predict a universe younger than the globular cluster ages, and gave rise to some speculations during the 1980s that the Big Bang model was seriously incorrect. ## Late 1990s: probable solution The age problem was eventually thought to be resolved by several developments between 1995-2003: firstly, a large program with the Hubble space telescope measured the Hubble constant at 72 km/s/Mpc with 10 percent uncertainty.[11] Secondly, measurements of parallax by the Hipparcos spacecraft in 1995 revised globular cluster distances upwards by 5-10 percent;[12] this made their stars brighter than previously estimated and therefore younger, shifting their age estimates down to around 12-13 billion years.[13] Finally, from 1998-2003 a number of new cosmological observations including supernovae, cosmic microwave background observations and large galaxy redshift surveys led to the acceptance of dark energy and the establishment of the Lambda-CDM model as the standard model of cosmology. The presence of dark energy implies that the universe was expanding more slowly at around half its present age than today, which makes the universe older for a given value of the Hubble constant. The combination of the three results above essentially removed the discrepancy between estimated globular cluster ages and the age of the universe.[14] More recent measurements from WMAP and the Planck spacecraft have established a very accurate age of the universe of 13.80 billion years[15] with only 0.3 percent uncertainty (based on the standard Lambda-CDM model), and modern age measurements for globular clusters [16] and other objects are currently smaller than this value (within the measurement errors). A substantial majority of cosmologists therefore believe the age problem is now resolved.[17] ## References 1. ^ Evidence for the Big Bang by Björn Feuerbacher and Ryan Scranton. January 25, 2006. Retrieved 16 April 2007. 2. ^ Hubble, E. (15 March 1929). "A relation between distance and radial velocity among extra-galactic nebulae". Proceedings of the National Academy of Sciences. 15 (3): 168–173. Bibcode:1929PNAS...15..168H. doi:10.1073/pnas.15.3.168. PMC 522427. PMID 16577160. 3. ^ Kragh, Helge (1999). Cosmology and Controversy. Princeton Univ. Press. ISBN 978-0691005461. 4. ^ Baade, W. (February 1956). "The Period-Luminosity Relation of the Cepheids". Publications of the Astronomical Society of the Pacific. 68: 5. Bibcode:1956PASP...68....5B. doi:10.1086/126870. 5. ^ Sandage, Allan (1958). "Current Problems in the Extragalactic Distance Scale". Astrophysical Journal. 127: 513. Bibcode:1958ApJ...127..513S. doi:10.1086/146483. 6. ^ Riess, A.; Macri, Lucas; Casertano, Stefano; Lampeitl, Hubert; Ferguson, Henry C.; Filippenko, Alexei V.; Jha, Saurabh W.; Li, Weidong; Chornock, Ryan (2011). "A 3% solution: determination of the Hubble constant with the Hubble Space Telescope". ApJ. 730 (119). arXiv:1103.2976. Bibcode:2011ApJ...730..119R. doi:10.1088/0004-637X/730/2/119. 7. ^ Patterson, C.; Tilton, G.; Inghram, M. (21 January 1955). "Age of the Earth". Science. 121 (3134): 69–75. Bibcode:1955Sci...121...69P. doi:10.1126/science.121.3134.69. 8. ^ Sandage, A.; Tammann, G. A. "Steps toward the Hubble constant. VII - Distances to NGC 2403, M101, and the Virgo cluster using 21 centimeter line widths compared with optical methods: The global value of H sub 0". The Astrophysical Journal. 210: 7. Bibcode:1976ApJ...210....7S. doi:10.1086/154798. 9. ^ de Vaucouleurs, G. (23 September 1982). "Five crucial tests of the cosmic distance scale using the Galaxy as fundamental standard". Nature. 299 (5881): 303–307. Bibcode:1982Natur.299..303D. doi:10.1038/299303a0. 10. ^ Aaronson, M.; Bothun, G.; Mould, J.; Huchra, J.; Schommer, R. A.; Cornell, M. E. "A distance scale from the infrared magnitude/H I velocity-width relations. V - Distance moduli to 10 galaxy clusters, and positive detection of bulk supercluster motion toward the microwave anisotropy". The Astrophysical Journal. 302: 536. Bibcode:1986ApJ...302..536A. doi:10.1086/164014. 11. ^ Madore, Barry F.; Freedman, Wendy L.; Silbermann, N.; Harding, Paul; Huchra, John; Mould, Jeremy R.; Graham, John A.; Ferrarese, Laura; Gibson, Brad K.; Han, Mingsheng; Hoessel, John G.; Hughes, Shaun M.; Illingworth, Garth D.; Phelps, Randy; Sakai, Shoko; Stetson, Peter (10 April 1999). "The Key Project on the Extragalactic Distance Scale. XV. A Cepheid Distance to the Fornax Cluster and Its Implications". The Astrophysical Journal. 515 (1): 29–41. arXiv:astro-ph/9812157. Bibcode:1999ApJ...515...29M. doi:10.1086/307004. 12. ^ Reid, N (1998). "Globular clusters, Hipparcos, and the age of the galaxy". Proceedings of the National Academy of Sciences of the United States of America. 95 (1): 8–12. Bibcode:1998PNAS...95....8R. doi:10.1073/pnas.95.1.8. 13. ^ Chaboyer, Brian; Demarque, P.; Kernan, Peter J.; Krauss, Lawrence M. (10 February 1998). "The Age of Globular Clusters in Light of : Resolving the Age Problem?". The Astrophysical Journal. 494 (1): 96–110. arXiv:astro-ph/9706128. Bibcode:1998ApJ...494...96C. doi:10.1086/305201. 14. ^ Krauss, Lawrence M.; Chaboyer, Brian (3 January 2003). "Age Estimates of Globular Clusters in the Milky Way: Constraints on Cosmology". Science. 299 (5603): 65–69. Bibcode:2003Sci...299...65K. doi:10.1126/science.1075631. PMID 12511641. 15. ^ Planck Collaboration, Planck; Ade, P. A. R.; Aghanim, N.; Armitage-Caplan, C.; Arnaud, M.; Ashdown, M.; Atrio-Barandela, F.; Aumont, J.; Baccigalupi, C.; Banday, A. J.; Barreiro, R. B.; Bartlett, J. G.; Battaner, E.; Benabed, K.; Benoît, A.; Benoit-Lévy, A.; Bernard, J. -P.; Bersanelli, M.; Bielewicz, P.; Bobin, J.; Bock, J. J.; Bonaldi, A.; Bond, J. R.; Borrill, J.; Bouchet, F. R.; Bridges, M.; Bucher, M.; Burigana, C.; Butler, R. C.; et al. (2013). "Planck 2013 results XVI: Cosmological Parameters". arXiv:1303.5076 [astro-ph.CO]. 16. ^ VandenBerg, Don A.; Brogaard, K.; Leaman, R.; Casagrande, L. (1 October 2013). "THE AGES OF 55 GLOBULAR CLUSTERS AS DETERMINED USING AN IMPROVED $\Delta V^{\rm HB}_{\rm TO}$ METHOD ALONG WITH COLOR-MAGNITUDE DIAGRAM CONSTRAINTS, AND THEIR IMPLICATIONS FOR BROADER ISSUES". The Astrophysical Journal. 775 (2): 134. arXiv:1308.2257. Bibcode:2013ApJ...775..134V. doi:10.1088/0004-637X/775/2/134. 17. ^ "Cosmological Parameters" (PDF). Review of Particle Properties 2014. Particle Data Group.	2016-08-24 03:00:43	{"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": 0.8594744205474854, "perplexity": 5117.449148350309}, "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-2016-36/segments/1471982290765.41/warc/CC-MAIN-20160823195810-00063-ip-10-153-172-175.ec2.internal.warc.gz"}
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-10-infinite-sequences-and-series-section-10-6-alternating-series-and-conditional-convergence-exercises-10-6-page-603/46	## Thomas' Calculus 13th Edition Let us apply the Ratio Test to the given series. $\lim\limits_{n \to \infty} \|\dfrac{u_{n+1}}{u_n}\|=\lim\limits_{n \to \infty} \dfrac{e^n-e^{-n}}{e^{n+1}-e^{-n-1}}\\=\lim\limits_{n \to \infty} \dfrac{1-0}{e -0} \\=\dfrac{1}{e} \lt 1$ This implies that the series is Absolutely Convergent by the ratio test.	2019-12-11 08:06:25	{"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": 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": 0.9562088251113892, "perplexity": 115.71455159385346}, "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/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00514.warc.gz"}
http://physics.stackexchange.com/tags/velocity/hot	# Tag Info 10 Put your math aside for a minute, and take a lesson from Robert H. Goddard, in one of my all-time favorite papers. Basically your rocket consists of a payload H, and the rest of the rocket consisting of fuel mass P, plus non-fuel mass (i.e. tank) K. The secret is, as you shed P through combustion, you must also shed K. Otherwise as P gets smaller and ... 6 The answer is right there in your own math. You derived that the delta V that results from using the rocket as a single stage rocket is $$\Delta V_{\text{single stage}} = v_{ex}\ln\Bigl(\frac{M}{M-(m_{fa}+m_{fb})}\Bigr)$$ while the delta V from using the rocket as a two stage rocket is $$\Delta V_{\text{two stage}} = v_{ex} \Biggl( ... 3 Your approach is correct, now simply add indices to everything, i.e.$$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$and note that t_2 = t_1 - 2\,\text{s}. Then solve 15\,\text{m}\stackrel!=y_1 - y_2. 2 The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say,$$ v_{drift}=\eta E $$where \eta ... 1 If you take the time of throwing of the first object as 0 then the second object will start falling at 0+2 second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e0+2 \longrightarrow 0 andt \longrightarrow t-2 So for the second one$$ \frac {dx} {dt} =u+at dx=udt +atdt\int^y_0 ... 1 If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and change in position. $s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$ , where $s$ is displacement and $R$ is position. However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small ... 1 The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - ... 1 There is one error in the derivation, if you want to have$v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that$v_f = v(t_f)$. Once you use all this, you should be able to divide out$t_f-t_i$in the corrected version of your last line and get the result you seek. 1 The$x$and$y$velocities should not add to$V_0$. To understand why, imagine something moving with$V_x = 1 \frac{m}{s}$and$V_y = -1 \frac{m}{s}$. This is something going horizontally and down; there's no reason why its velocity should be zero. The answer is that$V_0$is the length of the velocity vector$\vec{V}$, and so it's calculated using ... 1 I would say, $$\sum \vec{F} = m\,\vec{a}_C$$ where the left hand side are the net forces applied, and$\vec{a}_C\$ is the acceleration of the center of mass. Only top voted, non community-wiki answers of a minimum length are eligible	2014-10-21 07:07:43	{"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": 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": 0.8949012756347656, "perplexity": 936.051705001558}, "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/1413507444209.14/warc/CC-MAIN-20141017005724-00169-ip-10-16-133-185.ec2.internal.warc.gz"}
http://www.researchgate.net/publication/2060716_Generalized_Monopoles_in_Six-dimensional_Non-Abelian_Gauge_Theory	Article Generalized Monopoles in Six-dimensional Non-Abelian Gauge Theory • Muneto Nitta Physical review D: Particles and fields 08/2004; 71(4). DOI: 10.1103/PhysRevD.71.041701 Source: arXiv ABSTRACT A spherically symmetric monopole solution is found in SO(5) gauge theory with Higgs scalar fields in the vector representation in six-dimensional Minkowski spacetime. The action of the Yang-Mills fields is quartic in field strengths. The solution saturates the Bogomolny bound and is stable. 0 Bookmarks · 65 Views • Source Article: A SELF-DUALITY SOLUTION TO A SIX-DIMENSIONAL GAUGE THEORY [Hide abstract] ABSTRACT: We give an exact solution to the generalized self-duality equations suggested by Tchrakian on a six-dimensional twisted space-time and work on the gauge theory $\text{SO}(3,3)$ with a higher-derivative coupling term. The coupling term is considered as geometry as well as interaction dependent. The topological properties are also studied. Far East Journal of Mathematical Sciences 12/2009; 11(2):197-204. • Source Article: A self-duality solution to a six-dimensional gauge theory [Hide abstract] ABSTRACT: We give an exact solution to the generalized self-duality equations suggested by Tchrakian on a six-dimensional twisted space-time and work on the gauge theory SO(3,3) with a higher-derivative coupling term. The coupling term is considered as geometry as well as interaction dependent. The topological properties are also studied. Far East Journal of Dynamical Systems. 01/2009; 11(2). • Source Article: Classical solution in six-dimensional gauge theory with higher derivative coupling [Hide abstract] ABSTRACT: We show that the spin connection of the standard metric on a six-dimensional sphere gives an exact solution to the generalized self-duality equations suggested by Tchrakian some years ago. We work on a SO(6) gauge theory with a higher-derivative coupling term. The model consists of vector fields only. The pseudoenergy is bounded from below by a topological charge, which is proportional to the winding number of spatial S5 around the internal space SO(6). The fifth homotopy group of SO(6) is, indeed, Z. The coupling constant of the higher derivative term is quadratic in the radius of the underlying space S6. Physical review D: Particles and fields 02/2008; 77(4).	2015-01-26 17:01:35	{"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": 0.7345108389854431, "perplexity": 1427.848678849494}, "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-2015-06/segments/1422115864313.15/warc/CC-MAIN-20150124161104-00011-ip-10-180-212-252.ec2.internal.warc.gz"}
https://www.gamedev.net/forums/topic/20329-errors-but-still-runs/	#### Archived This topic is now archived and is closed to further replies. # Errors, but still runs This topic is 6421 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts In all of my projects before when ever i get a compile error with VC5, it won''t run the program, (which is what i want), but now i have a new project, and If i hit the exclamation mark button, that will compile then execute, if there are errors during the compile, it still exectutes. I noticed that the folder it created is called Release instead of my other projects which were called Debug, so I guess i''m in release mode and not debug mode. Is this normal? and why would you ever want to run your program if it had errors in it? t2sherm ô¿ô ##### Share on other sites These errors are probably just warning (which will not stop the creation of the exe file). If they say error and not warnings, there is an option in VC++ that displays warnings as errors, it''s probably turned on. Goto "Settings..." under the project menu, select "C/C++" tab, Select "General" from the Category list box, and you''ll see a check box "Warnings as errors" BTW, this has noting to do with Release and Debug. Debug mode can track runtime problems in your app (ex. if your app crashes with a General Protection Fault and you click on Debug it will show you the exact line of code that caused the crash), while Release is for speed and performance. Use "Set Active Configuration..." under the "Build" menu to change between the two. +AA_970+ ##### Share on other sites These are not just warnings, they are errors like this, i took the semicolon of the end of a line then i hit the execute button, and i get this: Compiling... ddex4.cpp D:\Pong\ddex4.cpp(72) : error C2146: syntax error : missing '';'' before identifier ''BOOL'' D:\Pong\ddex4.cpp(72) : fatal error C1004: unexpected end of file found Error executing cl.exe. Creating browse info file... BSCMAKE: error BK1506 : cannot open file ''.\Release\ddex4.sbr'': No such file or directory Error executing bscmake.exe. Pong.exe - 3 error(s), 0 warning(s) But it STILL runs, very strange. Do you think it could have anything to do with creating the browse info?, is that a bad thing to do? t2sherm ô¿ô ##### Share on other sites the likelyhood is that you''re running an old executable, not a newly compiled one... somewhere, like WinMain, insert this code: MessageBox(NULL, "Time of build was: ", __TIMESTAMP__, 0); this will let you check how new or old the executable is (i''ve just realised you could actually just look at the date in explorer instead ... but real programmers never take such easy options) ##### Share on other sites If something seems unusual when compiling a project - then do a Rebuild All. This will clean the target directory, removing previously created OBJs and more importantly the previous EXE.	2018-02-23 19:07:08	{"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": 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": 0.20008069276809692, "perplexity": 5206.4081160741625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814827.46/warc/CC-MAIN-20180223174348-20180223194348-00382.warc.gz"}
https://arbital.greaterwrong.com/p/bayes_log_odds?l=1zh	# Bayes' rule: Log-odds form The odds form of Bayes’s Rule states that the prior odds times the likelihood ratio equals the posterior odds. We can take the log of both sides of this equation, yielding an equivalent equation which uses addition instead of multiplication. Letting $$H_i$$ and $$H_j$$ denote hypotheses and $$e$$ denote evidence, the log-odds form of Bayes’ rule states: $$\log \left ( \dfrac {\mathbb P(H_i\mid e)} {\mathbb P(H_j\mid e)} \right ) = \log \left ( \dfrac {\mathbb P(H_i)} {\mathbb P(H_j)} \right ) + \log \left ( \dfrac {\mathbb P(e\mid H_i)} {\mathbb P(e\mid H_j)} \right ).$$ This can be numerically efficient for when you’re carrying out lots of updates one after another. But a more important reason to think in log odds is to get a better grasp on the notion of ‘strength of evidence’. # Logarithms of likelihood ratios Suppose you’re visiting your friends Andrew and Betty, who are a couple. They promised that one of them would pick you up from the airport when you arrive. You’re not sure which one is in fact going to pick you up (prior odds of 50:50), but you do know three things: 1. They have both a blue car and a red car. Andrew prefers to drive the blue car, Betty prefers to drive the red car, but the correlation is relatively weak. (Sometimes, which car they drive depends on which one their child is using.) Andrew is 2x as likely to drive the blue car as Betty. 2. Betty tends to honk the horn at you to get your attention. Andrew does this too, but less often. Betty is 4x as likely to honk as Andrew. 3. Andrew tends to run a little late (more often than Betty). Betty is 2x as likely to have the car already at the airport when you arrive. All three observations are independent as far as you know (that is, you don’t think Betty’s any more or less likely to be late if she’s driving the blue car, and so on). Let’s say we see a blue car, already at the airport, which honks. The odds form of this calculation would be a $$(1 : 1)$$ prior for Betty vs. Andrew, times likelihood ratios of $$(1 : 2) \times (4 : 1) \times (2 : 1),$$ yielding posterior odds of $$(1 \times 4 \times 2 : 2 \times 1 \times 1) = (8 : 2) = (4 : 1)$$, so it’s 45 = 80% likely to be Betty. Here’s the log odds form of the same calculation, using 1 bit to denote each factor of $$2$$ in belief or evidence: • Prior belief in Betty of $$\log_2 (\frac{1}{1}) = 0$$ bits. • Evidence of $$\log_2 (\frac{1}{2}) = {-1}$$ bits against Betty. • Evidence of $$\log_2 (\frac{4}{1}) = {+2}$$ bits for Betty. • Evidence of $$\log_2 (\frac{2}{1}) = {+1}$$ bit for Betty. • Posterior belief of $$0 + {^-1} + {^+2} + {^+1} = {^+2}$$ bits that Betty is picking us up. If your posterior belief is +2 bits, then your posterior odds are $$(2^{+2} : 1) = (4 : 1),$$ yielding a posterior probability of 80% that Betty is picking you up. Evidence and belief represented this way is additive, which can make it an easier fit for intuitions about “strength of credence” and “strength of evidence”; we’ll soon develop this point in further depth. # The log-odds line Imagine you start out thinking that the hypothesis $$H$$ is just as likely as $$\lnot H,$$ its negation. Then you get five separate independent $$2 : 1$$ updates in favor of $$H.$$ What happens to your probabilities? Your odds (for $$H$$) go from $$(1 : 1)$$ to $$(2 : 1)$$ to $$(4 : 1)$$ to $$(8 : 1)$$ to $$(16 : 1)$$ to $$(32 : 1).$$ Thus, your probabilities go from $$\frac{1}{2} = 50\%$$ to $$\frac{2}{3} \approx 67\%$$ to $$\frac{4}{5} = 80\%$$ to $$\frac{8}{9} \approx 89\%$$ to $$\frac{16}{17} \approx 94\%$$ to $$\frac{32}{33} \approx 97\%.$$ Graphically representing these changing probabilities on a line that goes from 0 to 1: We observe that the probabilities approach 1 but never get there — they just keep stepping across a fraction of the remaining distance, eventually getting all scrunched up near the right end. If we instead convert the probabilities into log-odds, the story is much nicer. 50% probability becomes 0 bits of credence, and every independent $$(2 : 1)$$ observation in favor of $$H$$ shifts belief by one unit along the line. (As for what happens when we approach the end of the line, there isn’t one! 0% probability becomes $$-\infty$$ bits of credence and 100% probability becomes $$+\infty$$ bits of credence. knows-requisite(Math 2): %%note: This un-scrunching of the interval $$(0,1)$$ into the entire real line is done by an application of the inverse logistic function.%% ) # Intuitions about the log-odds line There are a number of ways in which this infinite log-odds line is a better place to anchor your intuitions about “belief” than the usual 1 probability interval. For example: • Evidence you are twice as likely to see if the hypothesis is true than if it is false is $${+1}$$ bits of evidence and a $${^+1}$$-bit update, regardless of how confident or unconfident you were to start with—the strength of new evidence, and the distance we update, shouldn’t depend on our prior belief. • If your credence in something is 0 bits—neither positive or negative belief—then you think the odds are 1:1. • The distance between $$0.01$$ and $$0.000001$$ is much greater than the distance between $$0.11$$ and $$0.100001.$$ To expand on the final point: on the 0-1 probability line, the difference between 0.01 (a 1% chance) and 0.000001 (a 1 in a million chance) is roughly the same as the distance between 11% and 10%. This doesn’t match our sense for the intuitive strength of a claim: The difference between “1 in 100!” and “1 in a million!” feels like a far bigger jump than the difference between “11% probability” and “a hair over 10% probabiility.” On the log-odds line, a 1 in 100 credibility is $${^-2}$$ orders of magnitude, and a “1 in a million” credibility is $${^-6}$$ orders of magnitude. The distance between them is minus 4 orders of magnitude, that is, $$\log_{10}(10^{-6}) - \log_{10}(10^{-2})$$ yields $${^-4}$$ magnitudes, or roughly $${^-13.3}$$ bits. On the other hand, 11% to 10% is $$\log_{10}(\frac{0.10}{0.90}) - \log_{10}(\frac{0.11}{0.89}) \approx {^-0.954}-{^-0.907} \approx {^-0.046}$$ magnitudes, or $${^-0.153}$$ bits. The log-odds line doesn’t compress the vast differences available near the ends of the probability spectrum. Instead, it exhibits a “belief bar” carrying on indefinitely in both directions—every time you see evidence with a likelihood ratio of $$2 : 1,$$ it adds one more bit of credibility. The Weber-Fechner law says that most human sensory perceptions are logarithmic, in the sense that a factor-of-2 intensity change feels like around the same amount of increase no matter where you are on the scale. Doubling the physical intensity of a sound feels to a human like around the same amount of change in that sound whether the initial sound was 40 decibels or 60 decibels. That’s why there’s an exponential decibel scale of sound intensities in the first place! Thus the log-odds form should be, in a certain sense, the most intuitive variant of Bayes’ rule to use: Just add the evidence-strength to the belief-strength! If you can make your feelings of evidence-strength and belief-strength be proportional to the logarithms of ratios, that is. Finally, the log-odds representation gives us an even easier way to see how extraordinary claims require extraordinary evidence: If your prior belief in $$H$$ is −30 bits, and you see evidence on the order of +5 bits for $$H$$, then you’re going to wind up with −25 bits of belief in $$H$$, which means you still think it’s far less likely than the alternatives. # Example: Blue oysters Consider the blue oyster example problem: You’re collecting exotic oysters in Nantucket, and there are two different bays from which you could harvest oysters. • In both bays, 11% of the oysters contain valuable pearls and 89% are empty. • In the first bay, 4% of the pearl-containing oysters are blue, and 8% of the non-pearl-containing oysters are blue. • In the second bay, 13% of the pearl-containing oysters are blue, and 26% of the non-pearl-containing oysters are blue. Would you rather have a blue oyster from the first bay or the second bay? Well, we first note that the likelihood ratio from “blue oyster” to “full vs. empty” is $$1 : 2$$ in either case, so both kinds of blue oyster are equally valuable. (Take a moment to reflect on how obvious this would not seem before learning about Bayes’ rule!) But what’s the chance of (either kind of) a blue oyster containing a pearl? Hint: this would be a good time to convert your credences into bits (factors of 2). 89% is around 8 times as much as 11%, so we start out with $${^-3}$$ bits of belief that a random oyster contains a pearl. Full oysters are 12 as likely to be blue as empty oysters, so seeing that an oyster is blue is $${^-1}$$ bits of evidence against it containing a pearl. Posterior belief should be around $${^-4}$$ bits or $$(1 : 16)$$ against, or a probability of 1/17… so a bit more than 5% (1/20) maybe? (Actually 5.88%.) <div><div> # Real-life example: HIV test Find & cite the referenced study A study of Chinese blood donors noteCitation needed found that roughly 1 in 100,000 of them had HIV (as determined by a very reliable gold-standard test). The non-gold-standard test used for initial screening had a sensitivity of 99.7% and a specificity of 99.8%, meaning respectively that $$\mathbb P({positive}\mid {HIV}) = .997$$ and $$\mathbb P({negative}\mid \neg {HIV}) = .998$$, i.e., $$\mathbb P({positive} \mid \neg {HIV}) = .002.$$ That is: the prior odds are $$1 : 100,000$$ against HIV, and a positive result in an initial screening favors HIV with a likelihood ratio of $$500 : 1.$$ Using log base 10 (because those are easier to do in your head): • The prior belief in HIV was about −5 magnitudes. • The evidence was a tad less than +3 magnitudes strong, since 500 is less than 1,000. ($\log_{10}(500) \approx 2.7$). So the posterior belief in HIV is a tad underneath −2 magnitudes, i.e., less than a 1 in 100 chance of HIV. Even though the screening had a $$500 : 1$$ likelihood ratio in favor of HIV, someone with a positive screening result really should not panic! Admittedly, this setup had people being screened randomly, in a relatively non-AIDS-stricken country. You’d need separate statistics for people who are getting tested for HIV because of specific worries or concerns, or in countries where HIV is highly prevalent. Nonetheless, the points that “only a tiny fraction of people have illness X” and that “preliminary observations Y may not have correspondingly tiny false positive rates” are worth remembering for many illnesses X and observations Y. # Exposing infinite credences The log-odds representation exposes the degree to which $$0$$ and $$1$$ are very unusual among the classical probabilities. For example, if you ever assign probability absolutely 0 or 1 to a hypothesis, then no amount of evidence can change your mind about it, ever. On the log-odds line, credences range from $$-\infty$$ to $$+\infty,$$ with the infinite extremes corresponding to probability $$0$$ and $$1$$ which can thereby be seen as “infinite credences”. That’s not to say that $$0$$ and $$1$$ probabilities should never be used. For an ideal reasoner, the probability $$\mathbb P(X) + \mathbb P(\lnot X)$$ should be 1 (where $$\lnot X$$ is the logical negation of $$X$$).noteFor us mere mortals, consider avoiding extreme probabilities even then. Nevertheless, these infinite credences of 0 and 1 behave like ‘special objects’ with a qualitatively different behavior from the ordinary credence spectrum. Statements like “After seeing a piece of strong evidence, my belief should never be exactly what it was previously” are false for extreme credences, just as statements like “subtracting 1 from a number produces a lower number” are false if you insist on regarding knows-requisite(Math 2): $$\aleph_0$$ !knows-requisite(Math 2): infinity as a number. # Evidence in decibels E.T. Jaynes, in Probability Theory: The Logic of Science (section 4.2), reports that using decibels of evidence makes them easier to grasp and use by humans. If an hypothesis has a likelihood ratio of $$o$$, then its evidence in decibels is given by the formula $$e = 10\log_{10}(o)$$. In this scheme, multiplying the likelihood ratio by 2 means approximately adding 3dB. Multiplying by 10 means adding 10dB. Jayne reports having used decimal logarithm first, for their ease of calculation and having tried to switch to natural logarithms with the advent of pocket calculators. But decimal logarithms were found to be easier to grasp. Parents: • Bayes' rule Bayes’ rule is the core theorem of probability theory saying how to revise our beliefs when we make a new observation. • I recommend rethinking the magnet metaphor, on the grounds that it is physically wrong. If you have two magnets on either end of a ruler, and one is twice as strong as the other, then an iron ball at the center of the ruler is going to roll all the way to the larger magnet (accelerating as it goes, because inverse square law), unless I’m missing something. Perhaps a better physical metaphor would be something like rubber bands, with each bit of evidence adding another rubber band from the belief level to pins at the ends of the ruler? • I don’t understand this sentence. • odds ratios? The thing inside the log(this part) is an odds ratio, right? • I would expect this sentence only after another telling me that the observations were red car, honking, and punctuality. I think the next sentence should be broken apart and this should be inserted inside. • It would be nice to show how to go from 99.8% to the 500:1 ratio. • I don’t think these terms have been defined yet. The difference between “strength of credence” and “strength of evidence” isn’t obvious to me, but it seems like it’s assumed throughout the rest of the article that the reader knows what they mean. • Is “-1 against” the same as “+1 for”? Expressing the first practical example entirely in terms of negative numbers seems like a poor pedagogical choice. Phrasing as “3 bits against” and then “a further 1 bit against” may help. Adding that the blue ones are not a great pick if you want pearls may help people understand the direction of “against”. • “Extreme credences” here should likely be “infinite credences”. Even so, previous page made the exact counterpoint: While strong evidence may not change your view of things, things, extreme evidence absolutely should make you revisit your estimate of even an infinite credence level. • One of these does log( prob/ 1 - prob) the other does log( prob) … I get your point about orders of magnitude difference, but for me this ends up more confusing then anything. • Wrong, they are exactly the same distances. I read the next paragraph so I get where you were going with this, but I find it confusing to start off with a blatantly wrong claim, especially when the next line compares 0.11 to 0.1 (11% to 10%) -- not to 0.100001 -- in order to describe how the significance of 0.00001 gets “lost in translation” when speaking in probabilities and not in bits. • But that really gives a different magnitude to the evidence. Why not be consistent with the log base? For example, if we were to use log base 2, the prior would be ~16.6 magnitudes strong and the evidence ~8. This means that the evidence would alter the prior by (slightly) less than half the order of magnitudes, where’s in the case of log base 10 the alteration is (slightly) more than half the order of magnitudes (5 vs 2.7). Also, imagine the absurd choice of log base 100k. The prior would remain practically intact in terms of this kind of order of magnitudes. • It is really confusing to apply one of the initial steps of a study as evidence to a prior which is the result (last step) of the same study. • Easier to grasp perhaps, but dangerously misleading. Increasing the likelihood of an event from 10^-100 to 10^-99 is very different and much less significant than increasing it from 10^-2 (1%) to 10^-1 (10%). I hope this is covered later in this guide.	2020-07-06 04:41:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "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": 0.6750023365020752, "perplexity": 11713.528598150404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2020-29/segments/1593655890105.39/warc/CC-MAIN-20200706042111-20200706072111-00142.warc.gz"}
https://math.stackexchange.com/questions/1325360/how-can-i-plot-this-function-that-has-a-fraction-that-has-an-absolute-value-in-t/1325378	# How can I plot this function that has a fraction that has an absolute value in the denominator? I have this piecewise function: $$f(x)=\begin{cases} \dfrac{x^2-x-2}{\|x-2\|}, & x \neq 2 \\ 0, & x = 2\text{.} \end{cases}$$ I can't figure out how to graph it. I punched these numbers into my calculator, and it created a parabola, but I haven't been able to get there on my own without the calculator. So far I have plotted the point $(2,0)$ on my graph, and I factored the first function to $\dfrac{(x-2)(x+1)}{\|x-2\|}$ and now I am stuck. I tried multiplying both the top and bottom by $x+2$, but after simplifying I ended up with $x+1$, which I am pretty positive is not correct. Where did I go wrong? Hint: recall (ignoring the $x = 2$ case): $$\|x-2\| = \begin{cases} x-2, & x-2 > 0 \\ -(x-2), & x-2 < 0 \end{cases} = \begin{cases} x-2, & x > 2 \\ -(x-2), & x < 2\text{.} \end{cases}$$ So $$\dfrac{x^2-x-2}{\|x-2\|} = \begin{cases} \dfrac{x^2-x-2}{x-2} = \dfrac{(x-2)(x+1)}{x-2} = x+1, & x > 2 \\ \dfrac{x^2 - x - 2}{-(x-2)} = -(x+1), & x < 2\text{.} \end{cases}$$ • Do I ignore the x=2 case when I graph this function? – matryoshka Jun 14 '15 at 20:08 • @Grace When $x = 2$, $f(x) = 0$. This is by definition in the problem. – Clarinetist Jun 14 '15 at 20:09 Dividing the two cases: $$x-2>0 \iff x>2 \Rightarrow \|x-2\|=x-2$$ and $$x-2<0 \iff x<2 \Rightarrow \|x-2\|=2-x$$ your function become: $$f(x)= \begin {cases} y=-x-1 \quad , \quad x<2\\ y=0 \quad,\quad x=2\\ y=x+1\quad,\quad x>2 \end{cases}$$ the graph is done by two open half-lines ( decreasing for $x<2$ and crescent for $x>2$) and an isolated point $(2,0)$. Note that $\frac{x}{\|x\|}=1$ for $x>$ and $\frac{x}{\|x\|}=-1$ for $x<0$. Thus, $$\bbox[5px,border:2px solid #C0A000]{\frac{(x-2)(x+1)}{\|x-2\|}=\text{sgn}(x-2)(x+1)}$$ where $\text{sgn}(x-2)=1$ for $x>2$ and $\text{sgn}(x-2)=-1$ for $x<2$. We can define the sign function as $0$ when its argument is $0$. And we're done!	2019-09-23 15:46:58	{"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": 0.7939123511314392, "perplexity": 156.01395088686755}, "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/1568514577363.98/warc/CC-MAIN-20190923150847-20190923172847-00477.warc.gz"}
http://math.stackexchange.com/questions/101939/proving-an-integral-equality	# Proving an integral equality I have a Homework question which is: Let $f(x)$ be an integrable function in $[0,1]$ and there exists a value $M>0$ such that for $x\in[0,1]$ then $f(x)>M$. Let $a=\frac{1}{2}\int_{0}^{1}{f(x)dx}.$ Prove that there is a value $c\in[0,1]$ such that $a=\int_{0}^{c}{f(x)dx}$, and that there is only one number $c$ which satisfies that condition. I am not really sure how to solve this, I can prove that $\frac{1}{2}\int_{0}^{1}{f(x)dx}\ge M/2$. But I don't think that is the right direction. Can someone please help me out? Thanks :) - Put $F(x)=\int_0^xf(t)dt$. Then $F$ is continuous and $F(0)=0$, $F(1)=2a>a$ since $a>0$. – Davide Giraudo Jan 24 '12 at 10:46 ## 1 Answer As Davide suggested, put $F(x)=\displaystyle\int_0^x f(t)dt$. Then $F(0)=\displaystyle\int_0^0 f(t)dt=0$ and $F(1)=\displaystyle\int_0^1 f(t)dt=2a$ since $a=\displaystyle\frac{1}{2}\int_0^1 f(t)dt$, which shows that $F(0)=0\leq a\leq 2a=F(1)$. Note that $F$ is continuous since $f$ is integrable, by Intermediate value theorem, there exists $c\in [0,1]$ such that $$\int_0^c f(t)dt=F(c)=a.$$ To prove that $c$ is unique, suppose by contradiction that there exists $c'\neq c$ such that $F(c')=a$. By definiton of $F$, we have $$\int_0^{c'} f(t)dt=F(c')=a=F(c)=\int_0^{c} f(t)dt.$$ Without loss of generality, we assume $c<c'$. Hence, we have $\displaystyle\int_c^{c'} f(t)dt=0$. By this is a contradiction, because by assumption $f(x)>M$ we have $\displaystyle\int_c^{c'} f(t)dt\geq M(c'-c)>0$. Note: In my previous answer, I made a mistake by assuming that $f$ is continuous. Thank you for Didier pointing out the mistake. Note added: Proof of $F$ being continuous: $$\|F(x)-F(y)\|=\left\|\int_0^{x} f(t)dt-\int_0^{y} f(t)dt\right\|\leq\int_{[x,y]}\|f(t)\|dt \rightarrow 0\mbox{ as }x\rightarrow y$$ since $f$ is integrable. - Great answer :) Thank you very much – Jason Jan 24 '12 at 11:54	2015-04-19 05:23:14	{"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": 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": 0.9809529185295105, "perplexity": 71.75966427677798}, "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-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00283-ip-10-235-10-82.ec2.internal.warc.gz"}
http://mathhelpforum.com/math-puzzles/218300-gulliver-king-print.html	# Gulliver and the King • April 28th 2013, 01:28 AM baxteryay Sigma What does sigma mean I maths • April 28th 2013, 05:11 AM Soroban Re: Gulliver and the King Hello, baxteryay! This is based on a classic problem. He could cut one link like this: . $\subset\!\supset\!\!\!\subset\!\!\supset \;\;\subset\!\supset \;\;\subset\!\!\supset \!\!\!\subset\!\supset \!\!\! \subset\!\supset \!\!\!\subset\!\supset$ Then he can present from 1 to 7 links. . . $\begin{array}{cc} \text{Number} & \text{Combination} \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 1+2 \\ 4 & 4 \\ 5 & 1+4 \\ 6 & 2+4 \\ 7 & 1+2+4 \end{array}$ Since he needed two cuts, he had more than 7 links, . . up to a maximum of 15 links.	2014-09-18 08:05:08	{"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": 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": 0.7374455332756042, "perplexity": 5530.438799306348}, "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-2014-41/segments/1410657126053.45/warc/CC-MAIN-20140914011206-00071-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"}
https://math.stackexchange.com/questions/94821/biased-lower-bounded-random-walk	# Biased lower bounded random walk I have a random walk with the following rules: • It starts at 2 • At each step it goes up by 1 with chance .4, down by one with chance .4 and up by 2 with chance .2 • The walk ends if it reaches 0 I want to know two things... 1. Firstly is this guaranteed to terminate at some point - I believe it is but confirmation would be nice. If not can a probability be given to it not finishing? 2. What is the probability of the walk having a given maximum value (eg what is the probability of it getting to 10 but no higher before terminating at zero). And just so you know I used to study maths at university but that was a while ago now so best to assume that I know basic stuff but try not to assume too much knowledge on my part in an answer. :) I also was never great at statistics. ;-) I tried solving the problem using a difference equation but I got stuck trying that. I'm not sure that was the right way to go really... • Regarding your first question: What do you now about the time-to-ruin of a "standard" biased random walk with probability of an up move of $p > 1/2$? Can you make a simple comparison of your walk to that? Conclusion? – cardinal Dec 28 '11 at 23:30 • Hmm... Since there is a higher probability of going up than down (0.6 to 0.4), it seems that in the long run, it will tend to increase. That's all I can figure out. Nice question! Also, what do you mean by lower bounded here? – itdoesntwork Dec 28 '11 at 23:30 • He means absorption at 0. – gnometorule Dec 28 '11 at 23:34 • @gnometorule: I did mean that. Its a while since I last did this stuff and I count myself lucky to have remembered the name "random walk" at all. ;-) – Chris Dec 30 '11 at 9:01 I can handle part 1. I'm thinking about part 2, but it looks quite difficult in general - I'll edit this answer if I figure it out. For part 1, look at the probability of eventually going one space left. Call this P. Then the probability of eventually going two spaces left is just going to be the probability of going one space left, and then going another space left: $P^2$. Similarly, the probability of going three squares left is $P^3$, et cetera... Now, directly from the random walk's parameters, we have: $P=.4+.4P^2+.2P^3$ Why? Well, we have a .4 chance of going left immediately, in which case we're finished ($.4$). We have a .4 chance of starting by going right, in which case we have to work our way left two spaces eventually ($.4P^2$). And we have a .2 chance of starting by going right two spaces, in which case we have to go left three spaces eventually ($.2P^3$). WolframAlpha tells us that of the three solutions to this equation, one is negative and the other two are P=1 and P=~0.561553. The negative solution is obviously wrong. The P=1 solution is non-obviously wrong, but since your random walk tends to increase, there's a positive probability that it will just spiral towards infinity. Therefore P is about 0.561553, and the probability of finishing from the starting point of 2 is $0.561553^2$ which is about $0.315341$. • The equation has higher order terms. This gets more convoluted looking first hit times later. I can't give the proper solution myself, but am fairly certain this neglects terms we can't neglect. This is just the probability to go 1 down in no more than 3 steps. – gnometorule Dec 29 '11 at 0:26 • Unless I made a mistake somewhere, I'm fairly certain that the formula I give is the probability of going down 1 step eventually, not just in the next 3 steps. I've just edited to clarify how I got the cubic equation in the middle. – Lopsy Dec 29 '11 at 2:01 • @gnometorule, the argument leading to the third degree equation in $P$ is right. The key word is eventually. – Did Dec 29 '11 at 10:13 • @Didier: You're probably right, and it's embarrassing for personal reasons. – gnometorule Dec 29 '11 at 15:26 • Thanks. That's the sort of thing I was wanting. :) I was just going about it all wrong. Part 2 is of less interest to me if there is such a high chance of never reaching zero anyway so don't worry too much about part 2. :) – Chris Dec 30 '11 at 9:00	2019-04-23 22:47:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8491783738136292, "perplexity": 392.9503450545305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00558.warc.gz"}
https://iwaponline.com/wst/article/75/4/833-846/19802	In this study, process parameters in chemical oxygen demand (COD) and turbidity removal from metal working industry (MWI) wastewater were optimized by electrocoagulation (EC) using aluminum, iron and steel electrodes. The effects of process variables on COD and turbidity were investigated by developing a mathematical model using central composite design method, which is one of the response surface methodologies. Variance analysis was conducted to identify the interaction between process variables and model responses and the optimum conditions for the COD and turbidity removal. Second-order regression models were developed via the Statgraphics Centurion XVI.I software program to predict COD and turbidity removal efficiencies. Under the optimum conditions, removal efficiencies obtained from aluminum electrodes were found to be 76.72% for COD and 99.97% for turbidity, while the removal efficiencies obtained from iron electrodes were found to be 76.55% for COD and 99.9% for turbidity and the removal efficiencies obtained from steel electrodes were found to be 65.75% for COD and 99.25% for turbidity. Operational costs at optimum conditions were found to be 4.83, 1.91 and 2.91 €/m3 for aluminum, iron and steel electrodes, respectively. Iron electrode was found to be more suitable for MWI wastewater treatment in terms of operational cost and treatment efficiency. ## NOMENCLATURE • MWI metal working industry • • EC electrocoagulation • • J current density, mA/cm2 • • COD chemical oxygen demand, mg/L • • RSM response surface methodology • • CCD central composite design • • ANOVA variance analyses • • ENC electrical energy consumption, kWh/m3 • • ELC electrode material consumption, kg/m3 • • MW molecular weight, g/mole ## INTRODUCTION Metal working industry (MWI) wastewater is generated as a result of bath washing and infiltration of the metal material produced at the end of procedures such as material lubrication, corrosion prevention, surface cleaning, sandpapering, acid cleaning, cooling and polishing (Canizares et al. 2004; Machado et al. 2016). Wastewaters emerging at the end of the manufacturing processes can contain chemically complex and specially formulated components, metal components, permanent and highly resoluble organics, detergents, heavy metals, cyanide and other chemicals (Gast et al. 2004; Gabaldon et al. 2007; Machado et al. 2016). As a result of serious environmental problems owing to their high pollutant potential (Cheng et al. 2005; Grijalbo et al. 2016), the discharge of these wastewaters to the sewer system without treatment has been prohibited according to the Water Pollution Control Regulation (Ministry of Environment and Urbanisation 2004). Physical, chemical and biological treatment processes are being widely used for the treatment of MWI wastewater (Schreyer & Coughlin 1999; Hu et al. 2002; Muszyński & Łebkowska 2005; Gabaldon et al. 2007; Gutierrez et al. 2007; Kobya et al. 2008; Tir & Moulai-Mostefa 2009; Machado et al. 2016). Electrocoagulation (EC) is a suitable method for the removal of high chemical oxygen demand (COD) and turbidity of these wastewaters (Tir & Moulai-Mostefa 2009; Kobya et al. 2011). EC is a complex process including multiple mechanisms such as adsorption, co-precipitation, chemical oxidation-reduction and flotation, and the EC process uses electrodes to coagulate ions in wastewater (Arslan-Alaton et al. 2008). Recently, EC has been commonly preferred in wastewater treatment due to simplicity, low energy consumption, short operation time, no additional reagent, low sludge production and high pollutant removal (Varank et al. 2016; Zodi et al. 2010). In the EC process, an electric current related to the chemical coagulant is applied to the anode and cathode. While electrodes are dissolved via electrolysis and coagulant types and metal hydroxides are formed on the one hand, dissolved pollutants within wastewater and colloidal particles become destabilized and aggregates emerge (Thirugnanasambandham et al. 2015) on the other hand. While the wastewater is treated via the EC process, the following four successful reactions take place. • (i) During the electrolytic reactions of electrodes, as M, where n = 2 or 3 (aluminum, iron or steel), coagulants are formed: 1 • (ii) Electrolysis of water occurs on the anode and the cathode: 2 3 • (iii) Destabilization of pollutants, particle suspensions and emulsions: 4 5 • (iv) Aggregation of destabilized phase for the formation of flocs. Various multi-variate statistical models have been used to optimize wastewater treatment processes in recent years (Karichappan et al. 2014; Varank et al. 2014; Pakravan et al. 2015; Akarsu et al. 2016; Šereš et al. 2016). Among these multi-variate statistical models, response surface methodology (RSM) is a useful and a highly preferred technique to minimize the amount of time needed and the cost of experimental sets. This mathematical model is a technique that utilizes modeling and analysis of a problem and is dependent on numerous variables, provides estimated answers to this problem, and checks the accuracy of the model (Thirugnanasambandham et al. 2013). RSM designs an optimum multi-factor model for EC processes by evaluating the interactions between multiple explanatory variables and one or more response variables, thus reducing the experimental set numbers (Chavalparit & Ongwandee 2009). Central composite design (CCD) is the most commonly used sub-design model of RSM. CCD is a flexible method demonstrating the interaction between variables using a minimized experimental set. In this study, EC experiments were performed for MWI wastewater treatment, using aluminum, iron and steel electrodes. The main objective of the study was to investigate and optimize the variable parameters such as pH, current density (J) and electrolysis time (t) for the removal of COD and turbidity from MWI wastewater via RSM. EC process operational cost (OC) analyses were also conducted under optimum experimental conditions. ## EXPERIMENTAL METHOD ### MWI wastewater The MWI wastewater used in the study was supplied from a facility in Gebze-Kocaeli (Turkey). The wastewater samples obtained from the input of the wastewater treatment plant were preserved at +4 °C in order to prevent microbial activity before the study was conducted. The characterization of the real wastewater used in the study is given in Table 1. All the analyses performed on the wastewater before and after the EC process were conducted according to Standard Methods (APHA 2005). Table 1 Characteristics of MWI wastewater Parameter Average value pH 6.89 COD (mg/L) 1,270 Chloride (mg/L) 190 Conductivity (μS/cm) 1,485 Turbidity (NTU) 1,820 Copper (mg/L) 0.120 Chromium (mg/L) <0.01 Nickel (mg/L) <0.01 Zinc (mg/L) 0.1200 Parameter Average value pH 6.89 COD (mg/L) 1,270 Chloride (mg/L) 190 Conductivity (μS/cm) 1,485 Turbidity (NTU) 1,820 Copper (mg/L) 0.120 Chromium (mg/L) <0.01 Nickel (mg/L) <0.01 Zinc (mg/L) 0.1200 ### Experimental setup and procedure Experimental studies were carried out in a laboratory-scale plexiglass reactor with 9 cm diameter and 13 cm length. A schematic view of the reactor used in the experimental studies is shown in Figure 1. Electrode sets consisting of anode and cathode electrodes are made up of two single-pole parallel plates (6 cm width × 11.5 cm length and 0.1 cm thickness) and the effective area of each plate is 42.5 cm2. Electrodes are placed together at 3 cm intervals. A 500 mL wastewater sample was used for each test. Due to the low conductivity of the wastewater content, 2,000 mg/L sodium chloride solution was added as electrolyte solution before each set. Before each experimental set, electrodes were washed with acetone and the impurities on the electrodes were removed by dipping in the solution prepared by mixing 100 mL hydrochloric acid solution (35% (v/v)) and 200 mL hexamethylenetetramine aqueous solution (2.80% (v/v)) for 5 minutes. EC experiments were performed at the ranges of pH 6–10, 10–30 minutes electrolysis time and 11–55 mA/cm2 current density provided by the power supply. Analyses were conducted by taking samples from the supernatant at the end of the precipitation time (60 minutes) after each experimental step. Figure 1 Schematic diagram of the EC process apparatus. Figure 1 Schematic diagram of the EC process apparatus. ### Experimental design and model development In this study, EC process optimization for COD and turbidity removal from MWI wastewater was achieved in three analytical steps (i.e. determination of factors and variables with the help of preliminary studies, variance analysis and drawing of response surface graphs, realization of optimization of model suitability). For statistical calculations, levels of three parameters as X1 (pH), X2 (J), X3 (t) were coded as Xi depending on Equation (6). The graphical perspective of the mathematical model resulted in the creation of the response surface method term. The relationship between answers and variables is illustrated in Equation (6). For the statistical design of the experiments and data analysis, Statgraphics Centurion XVI.I software was used. In this study, a full factorial CCD model with three independent variables and three different levels was employed and a total of 15 experimental sets were used. Operation parameters of pH: 6–10, J: 11–55 mA/cm2, t: 10–30 minutes were the independent variables; COD and turbidity removal rates were determined as system responses (Y). Independent variables and their levels are given in Table 2, while the experimental design matrix is shown in Table 3. 6 Table 2 Coded and actual values of variables of the experimental design matrix for EC Coded factors Factors Original factor (X) −1 +1 pH X1 10 Current density (mA/cm2X2 11 33 55 Time (min) X3 10 20 30 Coded factors Factors Original factor (X) −1 +1 pH X1 10 Current density (mA/cm2X2 11 33 55 Time (min) X3 10 20 30 Table 3 CCD experimental design with experimental data for the EC by aluminum, iron and steel electrodes Sludge volume mL/0.5 L of wastewater Final pH Run X1 X2 X3 X1 X2 X3 Aluminum Iron Steel Aluminum Iron Steel −1 −1 11 20 53 56 60 6.23 7.40 6.40 −1 10 11 20 200 68 76 8.50 8.13 8.77 −1 55 20 200 90 160 6.02 6.56 6.18 10 55 20 140 100 100 6.78 8.64 8.70 −1 −1 33 10 100 92 55 6.11 7.30 7.10 −1 10 33 10 60 70 52 8.14 9.43 7.25 −1 33 30 250 150 90 9.01 9.30 8.80 10 33 30 140 110 94 7.04 9.25 8.93 −1 −1 11 10 35 125 17 8.50 8.90 7.87 10 −1 55 10 105 200 76 8.34 8.61 7.60 11 −1 11 30 91 200 64 8.23 7.75 8.08 12 55 30 110 74 92 7.52 8.50 7.49 13 33 20 115 105 100 8.32 8.32 8.19 14 33 20 130 100 100 8.12 8.35 8.40 15 33 20 125 100 100 8.10 8.40 8.25 Sludge volume mL/0.5 L of wastewater Final pH Run X1 X2 X3 X1 X2 X3 Aluminum Iron Steel Aluminum Iron Steel −1 −1 11 20 53 56 60 6.23 7.40 6.40 −1 10 11 20 200 68 76 8.50 8.13 8.77 −1 55 20 200 90 160 6.02 6.56 6.18 10 55 20 140 100 100 6.78 8.64 8.70 −1 −1 33 10 100 92 55 6.11 7.30 7.10 −1 10 33 10 60 70 52 8.14 9.43 7.25 −1 33 30 250 150 90 9.01 9.30 8.80 10 33 30 140 110 94 7.04 9.25 8.93 −1 −1 11 10 35 125 17 8.50 8.90 7.87 10 −1 55 10 105 200 76 8.34 8.61 7.60 11 −1 11 30 91 200 64 8.23 7.75 8.08 12 55 30 110 74 92 7.52 8.50 7.49 13 33 20 115 105 100 8.32 8.32 8.19 14 33 20 130 100 100 8.12 8.35 8.40 15 33 20 125 100 100 8.10 8.40 8.25 In Equation (6), y represents the observable response variable, f represents the function response, x1, x2, x3, … xn represent independent variables, n represents the number of the independent variables and ɛ represents statistical errors. After the selection of the design, the model equation becomes definable, and model equation coefficients become predictable. Selected independent variables are coded according to Equation (7). 7 In this equation, xi, represents the real value of factor i, xavg represents the mean of high and low values of factor i, Δx represents the value of change (Bajpai et al. 2012). To correlate the relationship between independent variables and responses, the second-order polynomial model was selected for further analysis. The generalized mathematical form of the second-order polynomial equation is shown below: 8 where Y is the response in coded units, β0 is a constant, and (β1, β2, β3), (β12, β13, β23), and (β11, β22, β33) are the linear, interactive, and quadratic coefficients respectively. The adequacy of the developed mathematical model was tested by variance analysis (ANOVA). The effectiveness of the fit of the model was expressed by the determination coefficient (R2) and its statistical significance was checked by the Fisher F-test. Model terms were determined by the P-value and the F-value. ## RESULTS AND DISCUSSION ### Statistical analysis A second-order polynomial response surface model was applied in order to analyze the match of the experimental results obtained via CCD with the predicted values. In order to explain COD and turbidity removal from MWI wastewater via the EC process, which is conducted by using aluminum, iron and steel electrodes, the regression equation, which includes coded variables created based on the experimental results provided in Table 3, is given below: when aluminum electrode is used: 9 10 when iron electrode is used: 11 12 when steel electrode is used: 13 14 In Equations (9)–(14), a positive sign of the coefficients represents synergic impact and a negative sign is representative of antagonistic effect (Bajpai et al. 2012). When the independent operation variables are taken into account, it is seen that for Equation (9), there is a significantly positive impact of pH of solution, current density and electrolysis time on COD removal. For Equation (10), however, pH of solution and electrolysis time had significant negative effect on turbidity removal but current density had a positive effect. In Equation (11), pH of solution and electrolysis time had a significantly negative effect on COD removal efficiency, whereas current density had a significantly positive effect. In Equation (12), while pH of solution and electrolysis time had a positive effect on turbidity removal, current density had a negative effect. In Equations (13) and (14), it is seen that all three variables had significantly positive effects on COD and turbidity removal. The higher the coefficient values of the independent operation parameters with positive coefficient signs, the higher the COD and turbidity removal efficiency; and the smaller the coefficient values of the independent operation parameters with negative coefficient signs, the smaller the removal efficiency. Removal efficiency rates obtained using aluminum, iron and steel electrodes were predicted via Equations (9)–(14). The statistical analysis of the model was conducted via ANOVA and the ANOVA results of model are given in Table 4. High value of F represents the significance of the relevant term. The larger the F-value, the more significant is the corresponding term. Furthermore, the P-value related to the F-value could be used to show whether the F-value is large enough or not. Table 4 ANOVA results of regression parameters of the predicted response surface quadratic model Model R2 Sum of squares Mean square F-value Prob > F Aluminum electrodes COD 0.94 1,205.580 133.9533 8.288877 0.015698 Turbidity 0.93 0.626425 0.069603 8.412312 0.015198 Iron electrodes COD 0.93 354.9039 39.43377 7.943327 0.017229 Turbidity 0.98 72.52459 8.058288 30.15499 0.000783 Steel electrodes COD 0.93 65.35961 7.262179 7.910522 0.017385 Turbidity 0.98 99.61293 11.06810 28.72289 0.000881 Model R2 Sum of squares Mean square F-value Prob > F Aluminum electrodes COD 0.94 1,205.580 133.9533 8.288877 0.015698 Turbidity 0.93 0.626425 0.069603 8.412312 0.015198 Iron electrodes COD 0.93 354.9039 39.43377 7.943327 0.017229 Turbidity 0.98 72.52459 8.058288 30.15499 0.000783 Steel electrodes COD 0.93 65.35961 7.262179 7.910522 0.017385 Turbidity 0.98 99.61293 11.06810 28.72289 0.000881 P-values smaller than 0.05 prove that the regression model is statistically significant (Kumar et al. 2009; Amani-Ghadima et al. 2013). Additionally, besides the evaluation of whether or not the variable is significant, the sum of squares must be taken into account. As the sum of squares value grows, the significance of the variables grows at the same time (Ravikumar et al. 2007; Jing et al. 2011). In the case that the ‘Prob > F’ value is smaller than 0.0001, it can be accepted that the model is statistically highly significant and model terms are significant at the percentage of 95% probability. In the case that ‘Prob > F’ values are smaller than 0.05, model terms are significant (Zhang et al. 2010). In ANOVA results of the EC study conducted with aluminum electrode (Table 4), 8.288877 and 8.412312 F-values of the model and P-values 0.015698 and 0.015198, respectively, for COD and turbidity removal indicate that the results obtained are significant and the relationship between variables and targets can be explained by the model for COD and turbidity removal. As can be seen in Table 4, when iron electrode was used, for COD and turbidity removal, F-values 7.943327 and 30.15499 and P-values 0.017229 and 0.000783, respectively, indicate that the results obtained are significant and the relationship between variables and targets can be explained by the model for COD and turbidity removal. According to the ANOVA results obtained via the EC process carried out using steel electrodes, in Table 4 the F- and P-values of the effect of the model on the COD removal efficiency were found to be 7.910522 and 0.017385, respectively, whereas the F- and P-values of the effect of the model on the turbidity removal efficiency were found to be 28.72289 and 0.000881, respectively. These values demonstrate that the results are significant and the relationship between the variables and targets for COD and turbidity removal can be explained via this model. According to the ANOVA results in Table 4, for the COD removal in EC processes in which aluminum, iron and steel electrodes are used, model correlation coefficients (R2) were found to be 0.94, 0.93 and 0.93, respectively. These values demonstrate that only 6.28%, 6.54% and 6.57% of the total variance cannot be explained by the empirical model for the EC process in which aluminum, iron and steel are used. In order to achieve compliance with the model, it is enough to have a correlation coefficient above 0.80 (Olmez 2009). For turbidity removal in EC processes in which aluminum, iron and steel electrodes are used, model correlation coefficients were found as 0.93, 0.98 and 0.98, respectively. For turbidity removal, 6.2%, 1.81% and 1.9% of the total variance cannot be explained by the empirical model in EC processes in which aluminum, iron and steel are used. That all R2 values are above 0.80 indicates that the process can be explained via the regression model. Table 5 shows the predicted response surface model regression parameters variance analysis results obtained using experimental results for COD removal via the EC process using aluminum electrodes. According to the ANOVA results in Table 5, while current density and electrolysis time independent variables X3X3 quadratic coefficients had a significant effect on COD removal efficiency, it is seen that solution pH, all interactive coefficients, X1X1 and X2X2 quadratic coefficients did not have a significant effect. Table 5 ANOVA results of the response surface quadratic model of COD removal using aluminum electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 1,205.58 133.9533 8.288 0.0157 X1 20.6238 20.62380 1.280 0.3099 NS X2 725.432 725.4320 44.89 0.0011 X3 238.107 238.1070 14.73 0.0121 X1X1 0.36321 0.363206 0.020 0.8867 NS X1X2 10.6945 10.69450 0.660 0.4529 NS X1X3 6.11339 6.113390 0.380 0.5654 NS X2X2 4.29307 4.293070 0.270 0.6282 NS X2X3 88.4495 88.44950 5.470 0.0664 NS X3X3 114.493 114.4930 7.080 0.0448 Total error 80.803 16.16060 Total (corr.) 1,286.38 14 R2 93.72% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 1,205.58 133.9533 8.288 0.0157 X1 20.6238 20.62380 1.280 0.3099 NS X2 725.432 725.4320 44.89 0.0011 X3 238.107 238.1070 14.73 0.0121 X1X1 0.36321 0.363206 0.020 0.8867 NS X1X2 10.6945 10.69450 0.660 0.4529 NS X1X3 6.11339 6.113390 0.380 0.5654 NS X2X2 4.29307 4.293070 0.270 0.6282 NS X2X3 88.4495 88.44950 5.470 0.0664 NS X3X3 114.493 114.4930 7.080 0.0448 Total error 80.803 16.16060 Total (corr.) 1,286.38 14 R2 93.72% S: Significant. NS: Not significant. The ANOVA results of the predicted response surface model regression parameters for COD removal by the EC process using iron electrodes are given in Table 6. According to the ANOVA results in Table 6, pH of solution and current density, which are independent variables, had significant effects on COD removal efficiency. It is seen that X2X2 and X3X3 quadratic coefficients and X1X2 and X1X3 interactive coefficients had significant effects on COD removal efficiency. Also, the electrolysis time, an independent variable, X1X1 quadratic coefficient and X2X3 interactive coefficient did not have significant effects on COD removal efficiency. Table 6 ANOVA results of the response surface quadratic model of COD removal using iron electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 354.904 39.43377 7.943 0.01722 X1 33.4370 33.43700 6.740 0.04850 X2 107.878 107.8780 21.73 0.00550 X3 20.2532 20.25320 4.080 0.09940 NS X1X1 11.8647 11.86470 2.390 0.18280 NS X1X2 56.3874 56.38740 11.36 0.01990 X1X3 39.1198 39.11980 7.880 0.03770 X2X2 36.5879 36.58790 7.370 0.04200 X2X3 6.11339 6.113390 1.230 0.31760 NS X3X3 37.0149 37.01490 7.460 0.04130 Total error 24.8213 4.964260 Total (corr.) 379.726 14 R2 93.46% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 354.904 39.43377 7.943 0.01722 X1 33.4370 33.43700 6.740 0.04850 X2 107.878 107.8780 21.73 0.00550 X3 20.2532 20.25320 4.080 0.09940 NS X1X1 11.8647 11.86470 2.390 0.18280 NS X1X2 56.3874 56.38740 11.36 0.01990 X1X3 39.1198 39.11980 7.880 0.03770 X2X2 36.5879 36.58790 7.370 0.04200 X2X3 6.11339 6.113390 1.230 0.31760 NS X3X3 37.0149 37.01490 7.460 0.04130 Total error 24.8213 4.964260 Total (corr.) 379.726 14 R2 93.46% S: Significant. NS: Not significant. Table 7 shows the ANOVA results of the predicted response surface model regression parameters for COD removal by the EC process using steel electrodes. As can be seen from Table 7, the pH of solution, electrolysis time linear coefficients and X3X3 quadratic coefficients had a significant effect on COD removal, whereas current density, all interactive coefficients and X1X1 and X2X2 coefficients did not. Table 7 ANOVA results of the response surface quadratic model of COD removal using steel electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 65.3596 7.262179 7.910 0.01738 X1 24.8409 24.84090 27.06 0.00350 X2 2.22829 2.228290 2.430 0.18000 NS X3 12.1016 12.10160 13.18 0.01500 X1X1 4.81936 4.819360 5.250 0.07050 NS X1X2 0.83860 0.838600 0.910 0.38310 NS X1X3 5.22868 5.228680 5.700 0.06270 NS X2X2 2.63457 2.634570 2.870 0.15100 NS X2X3 0.00582 0.005822 0.010 0.93960 NS X3X3 15.0672 15.06720 16.41 0.00980 Total error 4.5902 0.918040 Total (corr.) 69.9498 14 R2 93.43% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 65.3596 7.262179 7.910 0.01738 X1 24.8409 24.84090 27.06 0.00350 X2 2.22829 2.228290 2.430 0.18000 NS X3 12.1016 12.10160 13.18 0.01500 X1X1 4.81936 4.819360 5.250 0.07050 NS X1X2 0.83860 0.838600 0.910 0.38310 NS X1X3 5.22868 5.228680 5.700 0.06270 NS X2X2 2.63457 2.634570 2.870 0.15100 NS X2X3 0.00582 0.005822 0.010 0.93960 NS X3X3 15.0672 15.06720 16.41 0.00980 Total error 4.5902 0.918040 Total (corr.) 69.9498 14 R2 93.43% S: Significant. NS: Not significant. Table 8 shows the predicted response surface model regression parameters variance analysis results obtained using experimental results for turbidity removal via the EC process using aluminum electrodes. In Table 8, it is seen that solution pH and current density independent variables and X2X2 and X3X3 quadratic coefficients had a significant effect on turbidity removal efficiency, whereas electrolysis time, all interactive coefficients and X1X1 quadratic coefficient did not. Since the coefficients found to be non-significant, depending on the F- and P-values as a result of the ANOVA analysis, do not have a significant impact on removal efficiency, they are ignored in the RSM equation. Removal of these coefficients from the equation makes the operations simpler and makes it easier according to obtained results. Table 8 ANOVA results of the response surface quadratic model of turbidity removal using aluminum electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 0.626425 0.069603 8.412 0.01519 X1 0.094210 0.094210 11.39 0.01980 X2 0.131574 0.131574 15.90 0.01040 X3 0.000003 0.000003 0.000 0.98450 NS X1X1 0.002771 0.002771 0.330 0.58780 NS X1X2 0.033819 0.033819 4.090 0.09910 NS X1X3 0.005478 0.005478 0.660 0.45280 NS X2X2 0.074469 0.074469 9.000 0.03010 X2X3 0.000055 0.000055 0.010 0.93780 NS X3X3 0.307930 0.307930 37.22 0.00170 Total error 0.0413696 0.008273 Total (corr.) 0.667794 14 R2 93.8% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 0.626425 0.069603 8.412 0.01519 X1 0.094210 0.094210 11.39 0.01980 X2 0.131574 0.131574 15.90 0.01040 X3 0.000003 0.000003 0.000 0.98450 NS X1X1 0.002771 0.002771 0.330 0.58780 NS X1X2 0.033819 0.033819 4.090 0.09910 NS X1X3 0.005478 0.005478 0.660 0.45280 NS X2X2 0.074469 0.074469 9.000 0.03010 X2X3 0.000055 0.000055 0.010 0.93780 NS X3X3 0.307930 0.307930 37.22 0.00170 Total error 0.0413696 0.008273 Total (corr.) 0.667794 14 R2 93.8% S: Significant. NS: Not significant. The ANOVA results of the predicted response surface model regression parameters for turbidity removal by the EC process using iron electrodes are given in Table 9. The results in Table 9 demonstrate that among the independent variables, pH of solution had the greatest effect on turbidity removal. Electrolysis time, X1X1 and X2X2 quadratic coefficients and X1X2 and X1X3 interactive coefficients had significant effects on turbidity removal. However, the current density independent variable and X2X3 interactive coefficient and X3X3 quadratic coefficient did not have a significant effect on turbidity removal. Table 9 ANOVA results of the response surface quadratic model of turbidity removal using iron electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 72.5245 8.058288 30.15 0.00078 X1 37.5254 37.52540 140.4 <0.0001 HS X2 0.02989 0.029891 0.110 0.75160 NS X3 2.58469 2.584690 9.670 0.02660 X1X1 17.5501 17.55010 65.67 0.00050 X1X2 5.23706 5.237060 19.60 0.00680 X1X3 7.01959 7.019590 26.27 0.00370 X2X2 3.27442 3.274420 12.25 0.01730 X2X3 0.26675 0.266755 1.000 0.36360 NS X3X3 0.01895 0.018959 0.070 0.80060 NS Total error 1.33618 0.267235 Total (corr.) 73.8604 14 R2 98.19% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 72.5245 8.058288 30.15 0.00078 X1 37.5254 37.52540 140.4 <0.0001 HS X2 0.02989 0.029891 0.110 0.75160 NS X3 2.58469 2.584690 9.670 0.02660 X1X1 17.5501 17.55010 65.67 0.00050 X1X2 5.23706 5.237060 19.60 0.00680 X1X3 7.01959 7.019590 26.27 0.00370 X2X2 3.27442 3.274420 12.25 0.01730 X2X3 0.26675 0.266755 1.000 0.36360 NS X3X3 0.01895 0.018959 0.070 0.80060 NS Total error 1.33618 0.267235 Total (corr.) 73.8604 14 R2 98.19% S: Significant. NS: Not significant. HS: Highly significant. Table 10 shows the ANOVA results of the predicted response surface model regression parameters for turbidity removal by the EC process using steel electrodes. The results presented in Table 10 indicate that pH of solution, current density and electrolysis time independent variables had significant effects on turbidity removal. It is seen that X1X1 and X2X2 quadratic coefficients had significant effects on turbidity removal, whereas X3X3 coefficients did not. It is clear from the results in the table that while X1X2 and X2X3 interactive coefficients had significant effects on turbidity removal, X1X3 coefficient did not. Table 10 ANOVA results of the response surface quadratic model of turbidity removal using steel electrodes Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 99.6129 11.0681 28.72 0.00088 X1 35.3539 35.3539 91.75 0.00020 X2 16.3391 16.3391 42.40 0.00130 X3 9.64377 9.64377 25.03 0.00410 X1X1 0.63421 0.63421 1.650 0.25580 NS X1X2 13.7205 13.7205 35.61 0.00190 X1X3 2.43754 2.43754 6.330 0.05350 NS X2X2 3.64092 3.64092 9.450 0.02770 X2X3 17.1382 17.1382 44.48 0.00110 X3X3 1.30932 1.30932 3.400 0.12460 NS Total error 1.92664 0.38532 Total (corr.) 101.54 14 R2 98.1% Source Sum of squares Degrees of freedom Mean square F-ratio P-value Remark Model 99.6129 11.0681 28.72 0.00088 X1 35.3539 35.3539 91.75 0.00020 X2 16.3391 16.3391 42.40 0.00130 X3 9.64377 9.64377 25.03 0.00410 X1X1 0.63421 0.63421 1.650 0.25580 NS X1X2 13.7205 13.7205 35.61 0.00190 X1X3 2.43754 2.43754 6.330 0.05350 NS X2X2 3.64092 3.64092 9.450 0.02770 X2X3 17.1382 17.1382 44.48 0.00110 X3X3 1.30932 1.30932 3.400 0.12460 NS Total error 1.92664 0.38532 Total (corr.) 101.54 14 R2 98.1% S: Significant. NS: Not significant. As can be seen from Figure 2, there is a high level of match between the data obtained experimentally in COD and turbidity removal and the values that are predicted via the model for all three processes applied. Table 11 presents the optimized process conditions. Under the optimized conditions determined with the help of the model, for the EC processes in which aluminum, iron and steel electrodes are used, maximum COD removal efficiency values were found to be 76.72%, 76.55% and 65.75%, respectively, and turbidity removal efficiency values were found to be 99.97%, 99.99% and 99.99%, respectively. Table 11 Optimum operating conditions of the process variables Aluminum electrodes Iron electrodes Steel electrodes Factor COD Turbidity COD Turbidity COD Turbidity pH 6.000 6.000 6.000 9.970 6.620 10.00 Current density (mA/cm255.00 55.00 37.00 48.45 30.18 11.00 Time (min) 21.78 10.00 10.00 10.00 18.89 30.00 Sludge volume (mL) 185.0 135.0 100.0 86.00 75.00 88.00 Aluminum electrodes Iron electrodes Steel electrodes Factor COD Turbidity COD Turbidity COD Turbidity pH 6.000 6.000 6.000 9.970 6.620 10.00 Current density (mA/cm255.00 55.00 37.00 48.45 30.18 11.00 Time (min) 21.78 10.00 10.00 10.00 18.89 30.00 Sludge volume (mL) 185.0 135.0 100.0 86.00 75.00 88.00 Figure 2 Predicted versus actual plots for the removal of COD and turbidity (a) and (b) for aluminum electrodes; (c) and (d) for iron electrodes; (e) and (f) for steel electrodes. Figure 2 Predicted versus actual plots for the removal of COD and turbidity (a) and (b) for aluminum electrodes; (c) and (d) for iron electrodes; (e) and (f) for steel electrodes. In response surface model graphs given in Figures 35, while a variable is held fixed in the center, the other two variables get values between the contours determined. The response surface and contour graph is a function of the variable held fixed and two variables that have values between the contours. Figure 3 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (aluminum electrodes). Figure 3 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (aluminum electrodes). Figure 4 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (iron electrodes). Figure 4 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (iron electrodes). Figure 5 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (steel electrodes). Figure 5 Three-dimensional response surface graphs for the effects of variables on the COD and turbidity removals (steel electrodes). ### Cost analysis The energy consumption and the amount of electrode material are two important parameters in the OC analysis of the EC process. The OC at optimum conditions was calculated by the following equation: 15 where aENC (kWh/m3) denotes the electrical energy consumed and bELC (kg/m3) estimates the material cost. The electrical energy consumption was calculated using the following equation (El-Ashtoukhy et al. 2009): 16 where U is the applied voltage (V), i is the current (A), tEC is the operating time (s) and v is the volume (m3) of the wastewater. Electrode material consumption was calculated using the following equation: 17 where i is the current (A), tEC is the operating time (s) and v is the volume (m3) of the wastewater, MW is the molecular mass of aluminium (26.98 g/mol) or iron (55.84 g/mol), z is the number of electrons transferred (z = 3 for aluminium and iron electrodes) and F is the Faraday constant (96,487 C/mol). Cost of chemicals for adjustment of a desired pH is ignored. In Equation (15), SDC is the sludge disposal cost. According to Turkish regulations, the generated sludge by EC is classified as hazardous. Therefore, the incineration process must be chosen as the sludge disposal method. The hazardous sludge disposal cost is about 240 €/ton (http://www.istac.istanbul/tr/hizmetlerimiz/endustriyel-atik-yonetimi). The sludge disposal cost is calculated as 1.8, 0.96 and 0.72 €/m3 and 2, 1.1 and 0.85 $/m3 for sludge volume generated in the EC process using aluminum, iron and steel electrodes at the optimum conditions, respectively. At the optimum conditions, OC of the EC process using aluminum, iron and steel electrodes applied on MWI wastewater for COD removal was found to be 4.83, 1.91 and 2.91 €/m3 and 5.40, 2.14 and 3.26$/m3 respectively. In comparison with the OC of the EC process using aluminum, iron and steel electrodes, the OC of the EC process using iron electrodes was lower than the others. ## CONCLUSIONS In the present study, treatment of MWI wastewaters via the EC process using aluminum, iron and steel electrodes was investigated. The results of the study indicated that the EC process is a suitable treatment alternative in order to achieve high pollutant removal from MWI wastewater and low energy consumption. Pollutant removal efficiency is a function of solution pH, current density and electrolysis time. Response surface method has been successfully conducted in applying the EC process in MWI wastewater. The impacts of solution pH, current density and electrolysis time on COD and turbidity removal were investigated. Three operational parameters were determined, pH: 6–10, J: 11–55 mA/cm2, t: t 10–30 min intervals, as independent parameters, while COD and turbidity removal rates were identified as system targets. The model applied in the study indicated that there was a high level of correlation (R2 > 0.80) between the experimental data and the values predicted via the model. High correlation coefficient values obtained from variance analysis indicate that the second-order model sufficiently matches with experimental data. Under optimum conditions, removal efficiency rates achieved using iron electrodes were found to be 76.55% for COD and 99.9% for turbidity, removal efficiency rates obtained using aluminum electrodes were found to be 76.72% for COD and 99.97% for turbidity, and removal efficiency rates achieved using steel electrodes were found to be 65.75% for COD and 99.9% for turbidity. Under optimum conditions, operational cost that includes both electricity and electrode consumption was found to be 1.91 €/m3 for iron electrode, 4.83 €/m3 for aluminum electrode and 2.91 €/m3 for steel electrode. Response surface graphs and low error rates in experimental and predictable values are the indicators of the match between the values obtained from the models and the real data. The results demonstrate that response surface method is an efficient method for optimizing the operational conditions of MWI wastewater treatment by the EC process. ## REFERENCES REFERENCES Akarsu C. Ozay Y. Dizge N. Gulsen H. E. 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https://puzzling.stackexchange.com/questions/5231/generalized-path-guards?noredirect=1	# Generalized Path Guards Suppose you are traveling and come to a fork in the road. One of them leads to something good like a cat shelter and a company handing out free ice cream and money, while the other leads to something terrible like a series of pasta parties you can't join. The path is guarded by $n>1$ guards. Each one either always tells the truth or tells lies. They all have perfect knowledge of which of them is honest or dishonest and which path goes to which destination. You know that at least one guard is honest and at least one is dishonest, but you don't know which or how the others answer questions. As a function of $n$, what is the smallest number of questions you must ask to be guaranteed to discover which path to take? To keep the problem mathematically formal, questions should be able to be expressed as Boolean logic expressions, which the guards will then answer with true or false, depending on their dispositions. You solve the puzzle when you can use the guards' answers to know which path to take. You can decide which questions to ask depending on the responses you get, but if an approach has an upper and lower bound for how many questions you must ask, you want to minimize the upper bound. Also, the guards know how to evaluate the statement $H(G)$, which is true if guard $G$ is honest, and $I(P)$, which is true if the path $P$ takes you to your destination. If you can think of other statements that make sense for the context of the puzzle, feel free to use them. • Hint: what is the answer when $n=2$? – Kevin Nov 20 '14 at 15:33 $1$ question will suffice - for whatever $n$ (given the other constraints). Take guard 1, $g_1$, given that we have $2$ paths $p_1 ,p_2$ and we disregard the way we came from. Ask $g_1$ "(you are a liar $\land$ $p_1$ leads to the cat shelter) $\lor$ (you are a truthteller $\land$ $p_1$ leads to the pasta party)?" Explanation: There are 4 possible conditions: Case 1: $g_1$ is a liar AND $p_1$ leads to the cat shelter. You will get the answer: FALSE Case 2: $g_1$ is a truthteller AND $p_1$ leads to the cat shelter. You will get the answer: FALSE. Case 3: $g_1$ is a liar AND $p_1$ leads to the pasta party. You will get the answer: TRUE Case 4: $g_1$ is a truthteller AND $p_1$ leads to the pasta party. You will get the answer: TRUE Thus, if you get the answer TRUE you take $p_2$, if you get FALSE, you take $p_1$. • @AeJey But he's a liar... so he'll say FALSE – d'alar'cop Nov 20 '14 at 8:46 • Ya ya. Now got it :) – AeJey Nov 20 '14 at 8:47 • @AeJey Awesome :D – d'alar'cop Nov 20 '14 at 8:47 The classic 1-guard solution works for any number of guards. Pick a guard $G$, and ask him $H(G) \iff I(P_1)$: "Are you honest XNOR is path 1 good?" Then, take path 1 if he says "True" and path 2 if he says "False". Liars lying and honest guards not lying is equivalent to every guard $G$ XNORing the answers to questions with $H(G)$. By building an XNOR with $H(G)$ into the question we ask, we cancel this out, and the answer we get will always be $I(P_1)$. • isn't xoring two questions together still two questions? I mean, you are asking "Are you honest" (which, btw, both knaves and knights will answer yes) and then xoring that value with "is path one good". Technically it is two questions, and technically the first question will always be true. So you are indeed asking "is path 1 not good". – rodolphito Nov 20 '14 at 8:05 • @Rodolvertice: No. I'm not asking the guard two questions and XORing the results; I'm asking the guard to compute the value of a single boolean expression and give me a single true/false answer. – user2357112 supports Monica Nov 20 '14 at 8:43 • @user is correct. The XOR statement can be expressed in a single Boolean expression, so by the rules of the puzzle, it counts as a single question. If you had to ask two guards, or ask the guard a different question depending on the results of a first question, that would count as a second question. – Kevin Nov 20 '14 at 15:56 • Am I misunderstanding how this works? In the case that p1 is good, the liar will say (yes XOR no) = true, but the honest will say (yes XOR yes) = false, so it doesn't tell you anything. – Nattgew Nov 20 '14 at 16:55 • @Nattgew: The liar doesn't lie about both halves separately and XOR the lies; he lies about the question as a whole. In other words, he computes "not (no XOR yes)", not "(yes XOR no)". (If he did XOR two lies, you could instead ask "Is the sky blue XOR is path 1 good".) – user2357112 supports Monica Nov 20 '14 at 17:35 The minimum number of questions required is 3. Ask the first guard "If I ask you and the second guard which is the correct path, will your answers be the same?", then go to the second guard and ask him "If I ask you and the first guard which is the correct path, will your answers be the same?" If the responses by both the guards are different, you can confirm that one of them is a truth telling guard (knight) and other is a liar (knave). Now you can ask the normal knight/knave question to get the right path: Ask either of them, "If I ask the other guard which path will lead to the cat shelter, which path will he show me?" Then go down the other path than the one the guard pointed to. Now if both the guards answered the same, check what they have answered. If they both answered yes, they both are knights. If they both answered no, both of them are knaves. Now if both are knights, ask one of them which is the path to cat shelter and go down the path he points to. If both are knaves, ask one of the which is the path to cat shelter take the other path. Still 3 questions. So the final answer is that whatever the value of n may be, the total number of questions needed to find the right path will be 3. • I'm excited to see your explanation! – Kevin Nov 20 '14 at 5:48 • I would like to point out that there's a trivial case when $n=2$: that's the standard version of the problem, for which you only need one question. Are you sure that doesn't affect your solution for small values of $n$? – Kevin Nov 20 '14 at 5:52 • No it won't, even if there is just 2 guards, we need to figure out whether they both are knight, or both are knave or one is knight and the other is knave – AeJey Nov 20 '14 at 6:00 • I have updated the answer to 3 since I haven't included the normal single question used to find the path in knight knave problem. – AeJey Nov 20 '14 at 6:01 • You don't need to figure out anything about which guards tell the truth or lie; you can just ask questions that truth-tellers and liars will give the same answers to. – user2357112 supports Monica Nov 20 '14 at 6:17 You only have to ask one question. Ask either Guard "If I were to ask the other Guard which is the best path to take what would be his answer?" You then follow the opposite path. The logic is one lies one tells the truth. So if you ask the question to the liar he will tell you the opposite of the true answer from the other guard, and if you ask the true guard he will tell you the answer from the liar which will be the wrong path. • But what if n>2 ? – A E Nov 20 '14 at 15:09	2020-08-13 21:30:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.620354175567627, "perplexity": 718.3996743833958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00021.warc.gz"}
http://math.stackexchange.com/questions/125157/how-to-find-the-limit-lim-limits-x-to-infty-left-sqrt-x23x-sq	# How to find the limit $\lim\limits _{ x\to \infty } \left( \sqrt { x^2+3x } -\sqrt { x^2+x } \right)$? When I try to do this type of indeterminations I reach to this point: $\lim\limits_{ x\to \infty } \dfrac { 2x }{ \sqrt { x^ 2 +3x } +\sqrt { x^2 +x } }$ but I don't know how to continue. Thanks. - Divide numerator and denominator by $x$ and let $x\to \infty$ then. – martini Mar 27 '12 at 18:57 yes, that is the simplest way to do it. the answer should be 1. – chango Mar 27 '12 at 18:58 @martini A brief explanation? Can you answer solving it? Thanks. – Garmen1778 Mar 27 '12 at 19:07 Mathlover did :) – martini Mar 27 '12 at 19:12 – Aryabhata Mar 27 '12 at 19:34 $\lim\limits_{ x\to \infty }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x } } }$ $\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3}{x}}+\sqrt{ 1+\frac{1}{x}}}}=\frac {2}{\large{\sqrt{ 1+0}+\sqrt{ 1+0}}}=1$	2016-05-28 16:41:12	{"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": 0.8801789283752441, "perplexity": 689.3023331292285}, "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-2016-22/segments/1464049278042.30/warc/CC-MAIN-20160524002118-00203-ip-10-185-217-139.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/275542/modularity-of-theta-functions-attached-to-hecke-characters	# modularity of Theta functions attached to Hecke characters Let $K/\mathbb{Q}$ be a quadratic imaginary field, and let $\chi$ be a Hecke character on $K$. Using Poisson summation, one can show that the theta function $$\theta(z):=\sum_{I\subseteq \mathcal{O}_K} \chi(I) q^{\|I\|}, q = e^{2\pi i z}$$ satisfies a functional equation relating $\theta(z)$ to $\theta(-1/Nz)$, where $N$ is equal to the discriminant of $K$ times the norm of the conductor of $\chi$. But from this equation, how can I deduce that $\theta$ is a modular form for $\Gamma_1(N)$ (or even for $\Gamma_0(N)$, with central character)? What I have tried: I thought it was easy, because the functional equation roughly says that $\theta$ is invariant under the action of the matrix $\begin{pmatrix} & 1 \\ -N & \end{pmatrix}$, and it is obvious that $\theta$ is invariant under the action of the matrix $\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}$. Hence using a conjugation, we get the invariance by the matrix $\begin{pmatrix}1 & \\ N & 1\end{pmatrix}$. Now the problem is solved, if one can show that the group $\Gamma_1(N)$ is generated by the two matrices $\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}$ and $\begin{pmatrix}1 & \\ N & 1\end{pmatrix}$. But is this true? I tried to prove it with elementary methods, but failed ... Or do I need other ingredients to prove the modularity? EDIT: According to the comment of Will Sawin, the above approach may be too naive. So are there any reference (in classical language, i.e. no adelization or Weil representations) on how to prove this modularity? The character $\chi$ that I'm interested in is not necessarily of finite order: it could have some infinity type like $z \mapsto z^{-k}$. • The rank of the abelianization of $\Gamma_1(N)$ grows with $N$, and whenever it's larger than $2$ it can't be generated by two elements, so i think you need other ingredients. – Will Sawin Jul 16 '17 at 8:05 • Weil's converse theorem may help - because the set of functions you are considering is closed under twisting by Dirichlet characters. – Jeremy Rouse Jul 16 '17 at 17:44 • It should be possible, because Hecke did this. For a proof of modularity of (positive-definite, even-dimensional) theta series generally, see R. Gunning's book on modular forms. However, upon closer examination of his argument (which maybe goest back to Schoenberg, I forget), it is far better done on an adele group. Indeed, this is one of the things that persuaded me. (See also my book on Hilbert modular forms.) But/and, also, in Mumford's Tata lectures on thetas generators for congruence subgroups are given, even in larger symplectic groups. Again, not so illuminating, in my opinion. – paul garrett Jul 16 '17 at 19:06	2019-07-15 22:58:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9525848031044006, "perplexity": 219.66870434792045}, "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/1563195524254.28/warc/CC-MAIN-20190715215144-20190716001144-00114.warc.gz"}
https://codereview.stackexchange.com/questions/91367/unique-combinations-of-sets-with-n-min-selections-of-overall-set	# Unique combinations of sets with N min selections of overall set I'm in the process of attempting to write a parser compiler. In this, sets play a major role. I'm in the 'lexical ambiguity' isolation phase, and to that, I need to yield a set of every possible permutation of a given set of items which represent an ambiguity. An ambiguity is reached when there are at two or more items in the active context that are lexically identical, but differ by their defined identity (The contextual keyword 'from' versus an Identifier of 'from'.) That being said, I first detect the ambiguities of the language, and generate a set of sets which represent the full extent of each ambiguity. Upon each set I then need to yield all permutations of two or greater. Below is the initial result that appears, upon testing, to do what I want, but I went overboard and have a method that operates on multiple sets, so if given: { 'a', 'b' } and { 'c', 'd' } it would yield { 'a', 'c' }, { 'a', 'd' }, { 'b', 'c' }, and { 'b', 'd' }, which I solved with the 'Splay' method. public static IEnumerable<IEnumerable<T>> Splay<T>(this IEnumerable<T> series) { /* When you want to yield an enumerable over each element in the series. / foreach (var element in series) yield return new T[1] { element }; } public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, params T[][] series) { return GetAllPermutations(minSetLength, (IEnumerable<IEnumerable<T>>)series); } public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, IEnumerable<IEnumerable<T>> series) { var jaggedVariation = series.Select(set => set.ToArray()).ToArray(); for (int minDepth = minSetLength; minDepth <= jaggedVariation.Length; minDepth++) foreach (var set in GetPermutationsOfLength<T>(minDepth, jaggedVariation)) yield return set; } private static IEnumerable<IEnumerable<T>> GetPermutationsOfLength<T>(int elementsPerSet, T[][] series) { for (int subsetIndex = 0; subsetIndex < series.Length - (elementsPerSet - 1); subsetIndex++) foreach (var subset in GetPermutationsOfLength<T>(elementsPerSet, subsetIndex, 0, series).Select(k=>k.ToArray())) if (subset.Length == elementsPerSet) / Keeps the logic below very simple. / yield return subset; } private static IEnumerable<IEnumerable<T>> GetPermutationsOfLength<T>(int elementsPerSet, int startingAt, int currentLength, T[][] series) { if (startingAt >= series.Length \|\| currentLength >= elementsPerSet) yield break; var currentFrontSet = series[startingAt]; foreach (var forefront in currentFrontSet) { var forefrontSet = new T[1] { forefront }; / Continue expanding recursively until the above constraints cause it to short circuit. / for (int i = startingAt + 1; i < series.Length; i++) { var subsets = GetPermutationsOfLength<T>(elementsPerSet, i, currentLength + 1, series); foreach (var subset in subsets) yield return forefrontSet.Concat(subset); } yield return forefrontSet; } } I'll be calling this with a single array that I call Splay on due to my over eagerness, but the general idea is I need every possible combination with two or greater elements, with no repeats. So if an ambiguity represents 5 separate identities I would yield 26 different sets, two through five items long. Each permutation within these ambiguity sets would become unique identities themselves, and have a bit mask I could check against unambiguous transition tables to identify the specific ambiguity, unify that ambiguity to determine the proper follow set and how to differentiate which identity it once was. Does anyone have any suggestions/insight on the approach? I tried to keep the approach simple: I used iterators due to the simplicity they provide. They will have a slight memory footprint due to the allocation of the iterator objects and the lifting of the locals; however, the environment this runs in is already utilizing 6+GB of memory to handle unbound look-ahead ambiguity resolution, so this is just another step. public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, IEnumerable<IEnumerable<T>> series) { var jaggedVariation = series.Select(set => set.ToArray()).ToArray(); for (int minDepth = minSetLength; minDepth <= jaggedVariation.Length; minDepth++) foreach (var set in GetPermutationsOfLength<T>(minDepth, jaggedVariation)) yield return set; } Firstly, I don't like the absence of braces in your loops. I know it's technically correct, but it makes it a lot harder to parse in your head, and if you ever want to add an extra statement in there, you're going to have to go in and surround it with braces first, or you'll get some really weird errors. public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, IEnumerable<IEnumerable<T>> series) { var jaggedVariation = series.Select(set => set.ToArray()).ToArray(); for (int minDepth = minSetLength; minDepth <= jaggedVariation.Length; minDepth++) { foreach (var set in GetPermutationsOfLength<T>(minDepth, jaggedVariation)) { yield return set; } } } Secondly, your calls to ToArray don't make much sense to me. All your functions are returning IEnumerable<IEnumerable<T>>, but they all take T[][]. Why not simplify the whole affair by accepting IEnumerable<IEnumerable<T>> too? But let's look closer. For your methods, when do you actually need an array? public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, params T[][] series) { return GetAllPermutations(minSetLength, (IEnumerable<IEnumerable<T>>)series); } Not here, you're literally casting it to IEnumerable<IEnumerable<T>> and that's it. public static IEnumerable<IEnumerable<T>> GetAllPermutations<T>(int minSetLength, IEnumerable<IEnumerable<T>> series) { var jaggedVariation = series.Select(set => set.ToArray()).ToArray(); for (int minDepth = minSetLength; minDepth <= jaggedVariation.Length; minDepth++) foreach (var set in GetPermutationsOfLength<T>(minDepth, jaggedVariation)) yield return set; } Certainly not here, and cutting out all those ToArray() calls will cut down on execution time. private static IEnumerable<IEnumerable<T>> GetPermutationsOfLength<T>(int elementsPerSet, T[][] series) { for (int subsetIndex = 0; subsetIndex < series.Length - (elementsPerSet - 1); subsetIndex++) foreach (var subset in GetPermutationsOfLength<T>(elementsPerSet, subsetIndex, 0, series).Select(k=>k.ToArray())) if (subset.Length == elementsPerSet) / Keeps the logic below very simple. / yield return subset; } Not here, either, Length can be substituted for a call to Count(). That'll cut out both a .ToArray() and a Select() call. private static IEnumerable<IEnumerable<T>> GetPermutationsOfLength<T>(int elementsPerSet, int startingAt, int currentLength, T[][] series) { if (startingAt >= series.Length \|\| currentLength >= elementsPerSet) yield break; var currentFrontSet = series[startingAt]; foreach (var forefront in currentFrontSet) { var forefrontSet = new T[1] { forefront }; / Continue expanding recursively until the above constraints cause it to short circuit. */ for (int i = startingAt + 1; i < series.Length; i++) { var subsets = GetPermutationsOfLength<T>(elementsPerSet, i, currentLength + 1, series); foreach (var subset in subsets) yield return forefrontSet.Concat(subset); } yield return forefrontSet; } } And not here, either. Length can be replaced by Count() and the one time you're getting something by index, you can simply call ElementAt(x) or, if you must, convert it to a List or an Array here. • #1. Lack of braces is style choice, the lack of them doesn't impact my coding. #2. I translate everything into an array because: 2.a. When you use Element at, mentioned by you later, you're spinning up an enumerator for each request, I can't see any performance gain here, in fact the opposite. 2.b. I use length later in the results of the iterator .ToArray() to simplify the logic in the later method, cheap, yes, but I made a tradeoff for maintainability #3. Count(), ElementAt(x), and so on don't have advantages. #4. The version accepting an array of array is to allow me to test it quickly. – Alexander Morou May 21 '15 at 23:44 The approach works. After further evaluating the needs of the use case, I determined that there on average very few actual permutations, even on larger languages. A majority of this was obviated due to token precedences. 'Performance' matters, but only when you're dealing with very large sets, the results I was getting were... under a hundred subsets. Once I handled unbound look-ahead and follow projection, I reduced that set of ambiguities to those which actually surfaced in the result language and only about 16 actually appear as used. Worrying about the small set of items that result in the scope of the other 110 million+ computations that take 6GB+ ended up being quite unnecessary.	2019-05-21 03:58:10	{"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": 0.34007057547569275, "perplexity": 7513.694405910606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00435.warc.gz"}
http://mathhelpforum.com/algebra/62803-partial-fractions.html	Math Help - Partial Fractions 1. Partial Fractions Express 1/r(r+1) in partial fractions and use the result to find the sum of the following series as a single fraction in terms of n: 1/1x2 + 1/2x3 + 1/3x2 + ............... + 1/n(n+1) I can express it as partial fractions to 1/r - 1/r-1 but from there, I don't know, I don't think it's an arithmetic sequence as there is no common difference. 2. Originally Posted by Leodinas Express 1/r(r+1) in partial fractions and use the result to find the sum of the following series as a single fraction in terms of n: 1/1x2 + 1/2x3 + 1/3x2 + ............... + 1/n(n+1) I can express it as partial fractions to 1/r - 1/r-1 but from there, I don't know, I don't think it's an arithmetic sequence as there is no common difference. The partial fraction decomposition is $\frac{1}{r} - \frac{1}{r{\color{red}+}1}$. Then the series can be written: $\left(\frac{1}{1} - \frac{1}{1+1}\right) + \left(\frac{1}{2} - \frac{1}{2+1}\right) + \left(\frac{1}{3} - \frac{1}{3+1}\right) + \, .... \, + \left(\frac{1}{n} - \frac{1}{n+1}\right)$ $= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \, .... \, + \left(\frac{1}{n} - \frac{1}{n+1}\right)$ and the rest should be clear. 3. Thank you for your help, I still don't see where to go with it after that, could you give me another clue? Similarly, $\frac{a^2-b^2}{(ax+b)(bx+a)}$ I end up with two seemingly unhelpful simultaneous equations, any advice or direction for this question would be appreciated also. 4. Originally Posted by Leodinas Thank you for your help, I still don't see where to go with it after that, could you give me another clue? [snip] Perhaps the answer will make it clear what you have to do: $1 - \frac{1}{n+1}$. Originally Posted by Leodinas [snip] $\frac{a^2-b^2}{(ax+b)(bx+a)}$ I end up with two seemingly unhelpful simultaneous equations, any advice or direction for this question would be appreciated also. It will be easier to help if you show all of the work you've done. 5. $\frac{a^2-b^2}{(ax+b)(bx+a)} = \frac{A}{ax+b} + \frac{B}{bx+a}$ $a^2-b^2 = A(bx+a) + B(ax+b)$ Comparing $x$: $0 = Ab + Ba$ Comparing constants: $a^2-b^2 = Aa + Bb$ I also recognise that $a^2-b^2$ can be $(a+b)(a-b)$. 6. Originally Posted by Leodinas $\frac{a^2-b^2}{(ax+b)(bx+a)} = \frac{A}{ax+b} + \frac{B}{bx+a}$ $a^2-b^2 = A(bx+a) + B(ax+b)$ Comparing $x$: $0 = Ab + Ba$ Comparing constants: $a^2-b^2 = Aa + Bb$ I also recognise that $a^2-b^2$ can be $(a+b)(a-b)$. $0 = Ab + Ba$ .... (1) $a^2-b^2 = Aa + Bb$ .... (2) Multiply equation (1) by -a and equation (2) by b. Add the results: $b(a^2 - b^2) = b^2 B - a^2 B \Rightarrow b(a^2 - b^2) = -B (a^2 - b^2) \Rightarrow B = -b$. Use this value to solve for A. 7. Thank you for that, I never would have thought to do it that way, I'll bear that in mind. Though, I'm still no further with the original question. I've added some of the terms together and recognise that the answer you provided does also produce that, but I don't see how? 8. Originally Posted by Leodinas Thank you for that, I never would have thought to do it that way, I'll bear that in mind. Though, I'm still no further with the original question. I've added some of the terms together and recognise that the answer you provided does also produce that, but I don't see how? Do the arithmetic and substract the like terms. There will only be two terms remaining after you do this. 9. $(\frac{1}{n}-\frac{1}{n+1})+(\frac{1}{n+1}-\frac{1}{n+2})+(\frac{1}{n+2}-\frac{1}{n+3})$ I see that these cancel, well most of them, but I don't see where the 1 or the 1/n+1 would come from. Thanks for your continued, rapid response; I realise it must be a pain. 10. Apologies, the obvious has finally dawned on me! Thanks again	2016-07-27 10:58:25	{"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": 24, "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": 0.88998943567276, "perplexity": 493.076253056039}, "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-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00020-ip-10-185-27-174.ec2.internal.warc.gz"}
http://www.livephysics.com/problems-and-answers/classical-mechanics/kinetic-energy-single-particle-constant-mass/	# Kinetic energy for single particle with constant mass Show that for a single particle with constant mass the equation of motionimplies the following differential equation for the kinetic energy: while if the mass varies with time the corresponding equation is (Source: Herbert Goldstein, Classical Mechanics - Chapter 01) with time variable mass,	2017-01-20 09:56:55	{"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": 0.8567237854003906, "perplexity": 1056.1619835146635}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280825.87/warc/CC-MAIN-20170116095120-00497-ip-10-171-10-70.ec2.internal.warc.gz"}
http://jvera.rbind.io/post/2017/07/15/some-essential-r-packages/	Some essential R packages \| Swimming the Data Lake # Some essential R packages ## 2017/17/07 For me, there’s a bunch of packages considered as “essential” ‘cause in the end, sooner or later I use them in any project that involved opening the RStudio regardless of the type of issue that I’m trying to address. At some point in the process I need to present preliminary data (knitr, rmarkdown, janitor), display graphs (ggplot2, ggthemes, ggThemeAssist, gridBase, grid, corrplot) or simply manipulate data (dplyr, tidyr, stringr). So I use the same snippet to load them at the beginning of any script. [code snippet at the end of post] I start loading pacman, that saves me the trouble of checking if the mentioned package is installed or not, and installs it and/or loads it in the environment as needed. I will briefly comment on some of them: • tidyverse: try install.packages (“tidyverse”) , it is the shortest way of installing following packages in a single call: ggplot2, dplyr, tidyr, readr, purr, tibble, hms, stringr, lubridate, forcats, DBI, haven, http, jsonlite, readxl, rvest , Xml2, modelr, broom. Notice that loading through command library(tidyverse) only loads the essentials, namely: ggplot2, tibble, tidyr, readr, purrr, and dplyr. • corrplot: my favorite option when plotting all the existing correlations in the data set. rticles, rmdformats, tufte: Various additions and different themes to increase the available output formats. Especially for HTML and PDF. • RDocumentation: adds the help documentation from this site inside the RStudio environment. janitor: I love the way it helps, combined with magrittr, the exploratory analysis applied to any data set. I’ll talk more on this package in the future. Promise. • formatR: Easy code formatting improves readability. • ggThemeAssist: An excellent wizard to modify the parameters for your graphic objects generated with ggplot2. • rio: Import and export data the easy way. Many different formats with an extremely simple syntax. It is a meta-package actually, that incorporates some of the best in this area. if (!require(pacman)) install.packages(“pacman”) “gridBase”, “grid”, “plyr” )	2019-02-21 04:04:15	{"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": 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": 0.21212489902973175, "perplexity": 8281.62303626647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247499009.48/warc/CC-MAIN-20190221031117-20190221053117-00097.warc.gz"}
https://stat.ethz.ch/R-manual/R-devel/library/stats/html/prop.trend.test.html	prop.trend.test {stats} R Documentation Test for trend in proportions Description Performs chi-squared test for trend in proportions, i.e., a test asymptotically optimal for local alternatives where the log odds vary in proportion with score. By default, score is chosen as the group numbers. Usage prop.trend.test(x, n, score = seq_along(x)) Arguments x Number of events n Number of trials score Group score Value An object of class "htest" with title, test statistic, p-value, etc. Note This really should get integrated with prop.test Author(s) Peter Dalgaard prop.test smokers <- c( 83, 90, 129, 70 )	2022-10-06 21:30:58	{"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": 0.8173564672470093, "perplexity": 6814.75913592466}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337855.83/warc/CC-MAIN-20221006191305-20221006221305-00726.warc.gz"}
https://spark.apache.org/docs/1.3.0/mllib-clustering.html	# MLlib - Clustering Clustering is an unsupervised learning problem whereby we aim to group subsets of entities with one another based on some notion of similarity. Clustering is often used for exploratory analysis and/or as a component of a hierarchical supervised learning pipeline (in which distinct classifiers or regression models are trained for each cluster). MLlib supports the following models: ## K-means k-means is one of the most commonly used clustering algorithms that clusters the data points into a predefined number of clusters. The MLlib implementation includes a parallelized variant of the k-means++ method called kmeans\|\|. The implementation in MLlib has the following parameters: • k is the number of desired clusters. • maxIterations is the maximum number of iterations to run. • initializationMode specifies either random initialization or initialization via k-means\|\|. • runs is the number of times to run the k-means algorithm (k-means is not guaranteed to find a globally optimal solution, and when run multiple times on a given dataset, the algorithm returns the best clustering result). • initializationSteps determines the number of steps in the k-means\|\| algorithm. • epsilon determines the distance threshold within which we consider k-means to have converged. Examples The following code snippets can be executed in spark-shell. In the following example after loading and parsing data, we use the KMeans object to cluster the data into two clusters. The number of desired clusters is passed to the algorithm. We then compute Within Set Sum of Squared Error (WSSSE). You can reduce this error measure by increasing k. In fact the optimal k is usually one where there is an “elbow” in the WSSSE graph. import org.apache.spark.mllib.clustering.KMeans import org.apache.spark.mllib.linalg.Vectors // Load and parse the data val data = sc.textFile("data/mllib/kmeans_data.txt") val parsedData = data.map(s => Vectors.dense(s.split(' ').map(_.toDouble))).cache() // Cluster the data into two classes using KMeans val numClusters = 2 val numIterations = 20 val clusters = KMeans.train(parsedData, numClusters, numIterations) // Evaluate clustering by computing Within Set Sum of Squared Errors val WSSSE = clusters.computeCost(parsedData) println("Within Set Sum of Squared Errors = " + WSSSE) All of MLlib’s methods use Java-friendly types, so you can import and call them there the same way you do in Scala. The only caveat is that the methods take Scala RDD objects, while the Spark Java API uses a separate JavaRDD class. You can convert a Java RDD to a Scala one by calling .rdd() on your JavaRDD object. A self-contained application example that is equivalent to the provided example in Scala is given below: import org.apache.spark.api.java.; import org.apache.spark.api.java.function.Function; import org.apache.spark.mllib.clustering.KMeans; import org.apache.spark.mllib.clustering.KMeansModel; import org.apache.spark.mllib.linalg.Vector; import org.apache.spark.mllib.linalg.Vectors; import org.apache.spark.SparkConf; public class KMeansExample { public static void main(String[] args) { SparkConf conf = new SparkConf().setAppName("K-means Example"); JavaSparkContext sc = new JavaSparkContext(conf); String path = "data/mllib/kmeans_data.txt"; JavaRDD<String> data = sc.textFile(path); JavaRDD<Vector> parsedData = data.map( new Function<String, Vector>() { public Vector call(String s) { String[] sarray = s.split(" "); double[] values = new double[sarray.length]; for (int i = 0; i < sarray.length; i++) values[i] = Double.parseDouble(sarray[i]); return Vectors.dense(values); } } ); parsedData.cache(); // Cluster the data into two classes using KMeans int numClusters = 2; int numIterations = 20; KMeansModel clusters = KMeans.train(parsedData.rdd(), numClusters, numIterations); // Evaluate clustering by computing Within Set Sum of Squared Errors double WSSSE = clusters.computeCost(parsedData.rdd()); System.out.println("Within Set Sum of Squared Errors = " + WSSSE); } } The following examples can be tested in the PySpark shell. In the following example after loading and parsing data, we use the KMeans object to cluster the data into two clusters. The number of desired clusters is passed to the algorithm. We then compute Within Set Sum of Squared Error (WSSSE). You can reduce this error measure by increasing k. In fact the optimal k is usually one where there is an “elbow” in the WSSSE graph. from pyspark.mllib.clustering import KMeans from numpy import array from math import sqrt # Load and parse the data data = sc.textFile("data/mllib/kmeans_data.txt") parsedData = data.map(lambda line: array([float(x) for x in line.split(' ')])) # Build the model (cluster the data) clusters = KMeans.train(parsedData, 2, maxIterations=10, runs=10, initializationMode="random") # Evaluate clustering by computing Within Set Sum of Squared Errors def error(point): center = clusters.centers[clusters.predict(point)] return sqrt(sum([x2 for x in (point - center)])) WSSSE = parsedData.map(lambda point: error(point)).reduce(lambda x, y: x + y) print("Within Set Sum of Squared Error = " + str(WSSSE)) ## Gaussian mixture A Gaussian Mixture Model represents a composite distribution whereby points are drawn from one of k Gaussian sub-distributions, each with its own probability. The MLlib implementation uses the expectation-maximization algorithm to induce the maximum-likelihood model given a set of samples. The implementation has the following parameters: • k is the number of desired clusters. • convergenceTol is the maximum change in log-likelihood at which we consider convergence achieved. • maxIterations is the maximum number of iterations to perform without reaching convergence. • initialModel is an optional starting point from which to start the EM algorithm. If this parameter is omitted, a random starting point will be constructed from the data. Examples In the following example after loading and parsing data, we use a GaussianMixture object to cluster the data into two clusters. The number of desired clusters is passed to the algorithm. We then output the parameters of the mixture model. import org.apache.spark.mllib.clustering.GaussianMixture import org.apache.spark.mllib.linalg.Vectors // Load and parse the data val data = sc.textFile("data/mllib/gmm_data.txt") val parsedData = data.map(s => Vectors.dense(s.trim.split(' ').map(_.toDouble))).cache() // Cluster the data into two classes using GaussianMixture val gmm = new GaussianMixture().setK(2).run(parsedData) // output parameters of max-likelihood model for (i <- 0 until gmm.k) { println("weight=%f\nmu=%s\nsigma=\n%s\n" format (gmm.weights(i), gmm.gaussians(i).mu, gmm.gaussians(i).sigma)) } All of MLlib’s methods use Java-friendly types, so you can import and call them there the same way you do in Scala. The only caveat is that the methods take Scala RDD objects, while the Spark Java API uses a separate JavaRDD class. You can convert a Java RDD to a Scala one by calling .rdd() on your JavaRDD object. A self-contained application example that is equivalent to the provided example in Scala is given below: import org.apache.spark.api.java.; import org.apache.spark.api.java.function.Function; import org.apache.spark.mllib.clustering.GaussianMixture; import org.apache.spark.mllib.clustering.GaussianMixtureModel; import org.apache.spark.mllib.linalg.Vector; import org.apache.spark.mllib.linalg.Vectors; import org.apache.spark.SparkConf; public class GaussianMixtureExample { public static void main(String[] args) { SparkConf conf = new SparkConf().setAppName("GaussianMixture Example"); JavaSparkContext sc = new JavaSparkContext(conf); String path = "data/mllib/gmm_data.txt"; JavaRDD<String> data = sc.textFile(path); JavaRDD<Vector> parsedData = data.map( new Function<String, Vector>() { public Vector call(String s) { String[] sarray = s.trim().split(" "); double[] values = new double[sarray.length]; for (int i = 0; i < sarray.length; i++) values[i] = Double.parseDouble(sarray[i]); return Vectors.dense(values); } } ); parsedData.cache(); // Cluster the data into two classes using GaussianMixture GaussianMixtureModel gmm = new GaussianMixture().setK(2).run(parsedData.rdd()); // Output the parameters of the mixture model for(int j=0; j<gmm.k(); j++) { System.out.println("weight=%f\nmu=%s\nsigma=\n%s\n", gmm.weights()[j], gmm.gaussians()[j].mu(), gmm.gaussians()[j].sigma()); } } } In the following example after loading and parsing data, we use a GaussianMixture object to cluster the data into two clusters. The number of desired clusters is passed to the algorithm. We then output the parameters of the mixture model. from pyspark.mllib.clustering import GaussianMixture from numpy import array # Load and parse the data data = sc.textFile("data/mllib/gmm_data.txt") parsedData = data.map(lambda line: array([float(x) for x in line.strip().split(' ')])) # Build the model (cluster the data) gmm = GaussianMixture.train(parsedData, 2) # output parameters of model for i in range(2): print ("weight = ", gmm.weights[i], "mu = ", gmm.gaussians[i].mu, "sigma = ", gmm.gaussians[i].sigma.toArray()) ## Power iteration clustering (PIC) Power iteration clustering (PIC) is a scalable and efficient algorithm for clustering vertices of a graph given pairwise similarties as edge properties, described in Lin and Cohen, Power Iteration Clustering. It computes a pseudo-eigenvector of the normalized affinity matrix of the graph via power iteration and uses it to cluster vertices. MLlib includes an implementation of PIC using GraphX as its backend. It takes an RDD of (srcId, dstId, similarity) tuples and outputs a model with the clustering assignments. The similarities must be nonnegative. PIC assumes that the similarity measure is symmetric. A pair (srcId, dstId) regardless of the ordering should appear at most once in the input data. If a pair is missing from input, their similarity is treated as zero. MLlib’s PIC implementation takes the following (hyper-)parameters: • k: number of clusters • maxIterations: maximum number of power iterations • initializationMode: initialization model. This can be either “random”, which is the default, to use a random vector as vertex properties, or “degree” to use normalized sum similarities. Examples In the following, we show code snippets to demonstrate how to use PIC in MLlib. PowerIterationClustering implements the PIC algorithm. It takes an RDD of (srcId: Long, dstId: Long, similarity: Double) tuples representing the affinity matrix. Calling PowerIterationClustering.run returns a PowerIterationClusteringModel, which contains the computed clustering assignments. import org.apache.spark.mllib.clustering.PowerIterationClustering import org.apache.spark.mllib.linalg.Vectors val similarities: RDD[(Long, Long, Double)] = ... val pic = new PowerIteartionClustering() .setK(3) .setMaxIterations(20) val model = pic.run(similarities) model.assignments.foreach { a => println(s"${a.id} ->${a.cluster}") } A full example that produces the experiment described in the PIC paper can be found under examples/. PowerIterationClustering implements the PIC algorithm. It takes an JavaRDD of (srcId: Long, dstId: Long, similarity: Double) tuples representing the affinity matrix. Calling PowerIterationClustering.run returns a PowerIterationClusteringModel which contains the computed clustering assignments. import scala.Tuple2; import scala.Tuple3; import org.apache.spark.api.java.JavaRDD; import org.apache.spark.mllib.clustering.PowerIterationClustering; import org.apache.spark.mllib.clustering.PowerIterationClusteringModel; JavaRDD<Tuple3<Long, Long, Double>> similarities = ... PowerIterationClustering pic = new PowerIterationClustering() .setK(2) .setMaxIterations(10); PowerIterationClusteringModel model = pic.run(similarities); for (PowerIterationClustering.Assignment a: model.assignments().toJavaRDD().collect()) { System.out.println(a.id() + " -> " + a.cluster()); } ## Latent Dirichlet allocation (LDA) Latent Dirichlet allocation (LDA) is a topic model which infers topics from a collection of text documents. LDA can be thought of as a clustering algorithm as follows: • Topics correspond to cluster centers, and documents correspond to examples (rows) in a dataset. • Topics and documents both exist in a feature space, where feature vectors are vectors of word counts. • Rather than estimating a clustering using a traditional distance, LDA uses a function based on a statistical model of how text documents are generated. LDA takes in a collection of documents as vectors of word counts. It learns clustering using expectation-maximization on the likelihood function. After fitting on the documents, LDA provides: • Topics: Inferred topics, each of which is a probability distribution over terms (words). • Topic distributions for documents: For each document in the training set, LDA gives a probability distribution over topics. LDA takes the following parameters: • k: Number of topics (i.e., cluster centers) • maxIterations: Limit on the number of iterations of EM used for learning • docConcentration: Hyperparameter for prior over documents’ distributions over topics. Currently must be > 1, where larger values encourage smoother inferred distributions. • topicConcentration: Hyperparameter for prior over topics’ distributions over terms (words). Currently must be > 1, where larger values encourage smoother inferred distributions. • checkpointInterval: If using checkpointing (set in the Spark configuration), this parameter specifies the frequency with which checkpoints will be created. If maxIterations is large, using checkpointing can help reduce shuffle file sizes on disk and help with failure recovery. Note: LDA is a new feature with some missing functionality. In particular, it does not yet support prediction on new documents, and it does not have a Python API. These will be added in the future. Examples In the following example, we load word count vectors representing a corpus of documents. We then use LDA to infer three topics from the documents. The number of desired clusters is passed to the algorithm. We then output the topics, represented as probability distributions over words. import org.apache.spark.mllib.clustering.LDA import org.apache.spark.mllib.linalg.Vectors // Load and parse the data val data = sc.textFile("data/mllib/sample_lda_data.txt") val parsedData = data.map(s => Vectors.dense(s.trim.split(' ').map(_.toDouble))) // Index documents with unique IDs val corpus = parsedData.zipWithIndex.map(_.swap).cache() // Cluster the documents into three topics using LDA val ldaModel = new LDA().setK(3).run(corpus) // Output topics. Each is a distribution over words (matching word count vectors) println("Learned topics (as distributions over vocab of " + ldaModel.vocabSize + " words):") val topics = ldaModel.topicsMatrix for (topic <- Range(0, 3)) { print("Topic " + topic + ":") for (word <- Range(0, ldaModel.vocabSize)) { print(" " + topics(word, topic)); } println() } import scala.Tuple2; import org.apache.spark.api.java.*; import org.apache.spark.api.java.function.Function; import org.apache.spark.mllib.clustering.DistributedLDAModel; import org.apache.spark.mllib.clustering.LDA; import org.apache.spark.mllib.linalg.Matrix; import org.apache.spark.mllib.linalg.Vector; import org.apache.spark.mllib.linalg.Vectors; import org.apache.spark.SparkConf; public class JavaLDAExample { public static void main(String[] args) { SparkConf conf = new SparkConf().setAppName("LDA Example"); JavaSparkContext sc = new JavaSparkContext(conf); // Load and parse the data String path = "data/mllib/sample_lda_data.txt"; JavaRDD<String> data = sc.textFile(path); JavaRDD<Vector> parsedData = data.map( new Function<String, Vector>() { public Vector call(String s) { String[] sarray = s.trim().split(" "); double[] values = new double[sarray.length]; for (int i = 0; i < sarray.length; i++) values[i] = Double.parseDouble(sarray[i]); return Vectors.dense(values); } } ); // Index documents with unique IDs JavaPairRDD<Long, Vector> corpus = JavaPairRDD.fromJavaRDD(parsedData.zipWithIndex().map( new Function<Tuple2<Vector, Long>, Tuple2<Long, Vector>>() { public Tuple2<Long, Vector> call(Tuple2<Vector, Long> doc_id) { return doc_id.swap(); } } )); corpus.cache(); // Cluster the documents into three topics using LDA DistributedLDAModel ldaModel = new LDA().setK(3).run(corpus); // Output topics. Each is a distribution over words (matching word count vectors) System.out.println("Learned topics (as distributions over vocab of " + ldaModel.vocabSize() + " words):"); Matrix topics = ldaModel.topicsMatrix(); for (int topic = 0; topic < 3; topic++) { System.out.print("Topic " + topic + ":"); for (int word = 0; word < ldaModel.vocabSize(); word++) { System.out.print(" " + topics.apply(word, topic)); } System.out.println(); } } } ## Streaming k-means When data arrive in a stream, we may want to estimate clusters dynamically, updating them as new data arrive. MLlib provides support for streaming k-means clustering, with parameters to control the decay (or “forgetfulness”) of the estimates. The algorithm uses a generalization of the mini-batch k-means update rule. For each batch of data, we assign all points to their nearest cluster, compute new cluster centers, then update each cluster using: $$c_{t+1} = \frac{c_tn_t\alpha + x_tm_t}{n_t\alpha+m_t}$$ $$n_{t+1} = n_t + m_t$$ Where $c_t$ is the previous center for the cluster, $n_t$ is the number of points assigned to the cluster thus far, $x_t$ is the new cluster center from the current batch, and $m_t$ is the number of points added to the cluster in the current batch. The decay factor $\alpha$ can be used to ignore the past: with $\alpha$=1 all data will be used from the beginning; with $\alpha$=0 only the most recent data will be used. This is analogous to an exponentially-weighted moving average. The decay can be specified using a halfLife parameter, which determines the correct decay factor a such that, for data acquired at time t, its contribution by time t + halfLife will have dropped to 0.5. The unit of time can be specified either as batches or points and the update rule will be adjusted accordingly. Examples This example shows how to estimate clusters on streaming data. First we import the neccessary classes. import org.apache.spark.mllib.linalg.Vectors import org.apache.spark.mllib.regression.LabeledPoint import org.apache.spark.mllib.clustering.StreamingKMeans Then we make an input stream of vectors for training, as well as a stream of labeled data points for testing. We assume a StreamingContext ssc has been created, see Spark Streaming Programming Guide for more info. val trainingData = ssc.textFileStream("/training/data/dir").map(Vectors.parse) val testData = ssc.textFileStream("/testing/data/dir").map(LabeledPoint.parse) We create a model with random clusters and specify the number of clusters to find val numDimensions = 3 val numClusters = 2 val model = new StreamingKMeans() .setK(numClusters) .setDecayFactor(1.0) .setRandomCenters(numDimensions, 0.0) Now register the streams for training and testing and start the job, printing the predicted cluster assignments on new data points as they arrive. model.trainOn(trainingData) model.predictOnValues(testData.map(lp => (lp.label, lp.features))).print() ssc.start() ssc.awaitTermination() As you add new text files with data the cluster centers will update. Each training point should be formatted as [x1, x2, x3], and each test data point should be formatted as (y, [x1, x2, x3]), where y is some useful label or identifier (e.g. a true category assignment). Anytime a text file is placed in /training/data/dir the model will update. Anytime a text file is placed in /testing/data/dir you will see predictions. With new data, the cluster centers will change!	2022-08-18 19:53:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.327840656042099, "perplexity": 4947.344148041199}, "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/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00201.warc.gz"}
https://brilliant.org/practice/basic-properties-of-integrals/	Calculus # Basic Properties of Integrals If $\int_0^1 f(x) \, dx = 4$, what is the value of $\int_0^1 2 f(x) \, dx ?$ If $\int_0^1 f(x) \, dx = 2$ and $\int_0^1 g(x) \, dx = 6$, what is the value of $\int_0^1 \left( 6 f(x) + 2 g(x) \right) \, dx ?$ If $\int_0 ^ {4} f(x) \, dx = 3$, what is the value of $\int_0^{4} \left( 3 - f(x) \right) \, dx ?$ If $\int_0^{10} f(x) \, dx = 27$ and $\int_0^5 f(x) \, dx = 7$, then what is the value of $\int_5^{10} f(x) \, dx ?$ Suppose $f(x)$ is an odd function and $g(x)$ is an even function such that $\int_0 ^ {5} f(x) \, dx = 7 \hspace{.6cm} \int_{5}^{15} f(x) \, dx = 8 \\ \int_{-5}^{0} g(x) \, dx = 1 \hspace{.6cm} \int_{5}^{15} g(x) \, dx = 2$ What is the value of $\int_{-5}^{15} \left( f(x) + g(x) \right) \, dx ?$ ×	2022-05-22 15:03:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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": 0.8153185248374939, "perplexity": 148.1027209862092}, "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/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00037.warc.gz"}
https://www.sarthaks.com/2816633/certain-amounts-years-simple-interest-the-same-sum-compounded-half-yearly-10-annum-frac	# A certain sum amounts to Rs. 15,500 in 2 years at 12% p.a. simple interest. If the same sum is compounded half-yearly at 10% per annum for $1 \frac{1 0 votes 235 views closed A certain sum amounts to Rs. 15,500 in 2 years at 12% p.a. simple interest. If the same sum is compounded half-yearly at 10% per annum for \(1 \frac{1}{2}$ years, what will be the amount received? 1. Rs. 14,470 2. Rs. 15,125 3. Rs. 14,360 4. Rs. 13,460 by (59.9k points) selected Correct Answer - Option 1 : Rs. 14,470 22Given: A certain sum amounts to Rs. 15,500 in 2 years at 12% p.a. simple interest. Next time is 1.5 year and the rate of interest is 10% p.a. and interest compounded half yearly. Concept Used: Interest = Principal - amount If p be the principal and the rate of interest is r then simple interest in t year is ptr/100 If p be the principal, r be the rate of interest, the interest compounded half yearly then amount after t year will be p(1 + r/200)2t Calculation: Let the principal be p Simple interest get in 2 years at 12% p.a. simple interest is (2 × 12p)/100 ⇒ 24p/100 Amount = 24p/100 + p ⇒ 124p/100 Accordingly, 124p/100 = 15500 ⇒ p = 12500 The principal is 12500 Total time is 1.5 year The rate of interest is 10% p.a. Interest compounded half yearly Amount after 1.5 years will be 12500 × (1 + 10/200)1.5 × 2 ⇒ 12500 × (21/20)3 ⇒ 14470.31 ⇒ 14470 ∴ The amount will be 14470 after 1.5 year	2022-11-27 12:05:08	{"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": 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": 0.8842865228652954, "perplexity": 6043.817395670235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00637.warc.gz"}
https://socratic.org/questions/what-should-the-ph-of-salt-water-be	# What should the pH of salt water be? Of aqueous sodium chloride? $p H = 7$. Sodium chloride is the salt of a strong base, $N a O H$, and a strong acid, $H C l$. Because this is a strong base/strong acid pair, by definition, their counterions, $N {a}^{+}$, and $C {l}^{-}$, do not undergo water hydrolysis. Contrast this behaviour with an aqueous solution of $\text{sodium fluoride}$. The $p H$ of this solution is $> 7$. Given that $H F \left(a q\right)$, unlike $H C l , H B r , H I$, is a WEAK ACID, how can this solution behaviour be rationalized?	2021-10-19 03:39:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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": 0.731776773929596, "perplexity": 1610.2114532059313}, "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/1634323585231.62/warc/CC-MAIN-20211019012407-20211019042407-00463.warc.gz"}
http://list.seqfan.eu/pipermail/seqfan/2013-May/011123.html	[seqfan] Re: proof of finiteness of A073396 Max Alekseyev maxale at gmail.com Tue May 7 16:12:14 CEST 2013 No. Prime factorization is not defined for 0 (and neither are A008472 and A001414). Max On Tue, May 7, 2013 at 7:02 AM, Andrew Weimholt <andrew.weimholt at gmail.com> wrote: > Should 0 be in the sequence? > > > > On Sat, May 4, 2013 at 3:25 PM, Max Alekseyev <maxale at gmail.com> wrote: > >> Let m = \prod_{i=1}^n p_i^{k_i}, where p_1 < ... < p_n are primes and >> k_i >= 1, be an element of A073396, i.e., >> >> m = (\sum_{i=1}^n p_i) * (\sum_{i=1}^n p_ik_i). >> >> First notice that if n=1, then m = p^k = kp^2, implying that m = 2^4 or >> 3^3. >> For the rest assume that n >= 2. >> >> Denote >> K = \sumi=1^n k_i = Omega(m) = A001222(m) >> s_t = \sum_{i=1}^t p_i >> S_t = \sum_{i=1}^t p_ik_i >> >> Then m = s_n S_n and thus >> 0 == m == s_{n-1} * S_{n-1} (mod p_n), >> implying that p_n <= S_{n-1}. >> >> Consider two cases, depending on whether k_n = 1 or k_n > 1. >> >> 1) if k_n > 1, then s_n <= S_n < Kp_n (the last inequality is strict >> since n>=2) >> >> m = s_n S_n < K^2 * p_n^2 >> >> Let m' = m / p_n^2, which is an integer with Omega(m') = K-2. >> Then >> 2^(K-2) <= m' < K^2 >> implying that K <= 7 and m' <= 49. >> >> Now, we go over all possible values of m' and solve the equation >> A008472(m') * (A001414(m') + p) = m' * p^2 (this corresponds to the >> case k_n >= 3) >> or the equation >> (A008472(m') + p) * (A001414(m') + p) = m' * p^2 (this corresponds to >> the case k_n = 2) >> with respect to prime p = p_n. >> In either case, p must divide A008472(m')A001414(m') and we can >> simply let p go over prime divisors of A008472(m')A001414(m') and try >> m'p as m. >> >> 2) If k_n = 1, then s_n <= S_n = S_{n-1} + p_n <= 2 S_{n-1} <= 2 * >> (K-1) * p_{n-1}, implying that >> >> m = s_n * S_n <= 4 * (K-1)^2 * p_{n-1}^2 >> >> Let m' = m / (p_n * p_{n-1}), which is an integer with Omega(m') = K-2. >> Then >> 2^(K-2) <= m' = m / (p_n * p_{n-1}) <= 4 * (K-1)^2 * p_{n-1}/p_n < 4 * >> (K-1)^2, >> implying that K <= 10 and m' < 324. >> >> Now, we go over all possible values of m' and look for m of the form m >> = m' * p * q where p < q are primes (in fact, p = p_{n-1} and q = >> p_n). >> Consider two cases: >> >> 2.1) if p divides m' (this corresponds to k_{n-1} >= 2), we let p go >> over all prime divisors of m'. Similarly to the case 1), we further >> have that q must divide A008472(m'p)A001414(m'p), which is a finite >> number of candidates to consider. >> >> 2.2) if p does not divide m' (this corresponds to the case k_{n-1} = >> 1), we solve the equation >> (A008472(m') + p + q) (A001414(m') + p + q) = m' * p * q >> with respect to primes p < q. >> We notice that q divides (A008472(m') + p) * (A001414(m') + p), which >> >> 2.2.1) q = A008472(m') + p or q = A001414(m') + p. In either case, p >> must divide >> 2A008472(m')A001414(m')(A008472(m')+A001414(m')) >> and we let p go over the prime divisors of this number. >> >> 2.2.2) q is a proper divisor of A008472(m') + p or A001414(m') + p. Then >> p < q <= (A001414(m') + p)/2, >> implying that p < A001414(m'). >> So we let p go over the primes below A001414(m'), and let q go over >> the prime divisors of (A008472(m') + p) (A001414(m') + p). >> >> In summary, we have a finite number of cases to consider. I carefully >> went over them all and found that there are no other elements of >> A073396 besides currently known 16, 27, and 150. >> >> Regards, >> Max >> >> _______________________________________________ >> >> Seqfan Mailing list - http://list.seqfan.eu/ >> > > _______________________________________________ > > Seqfan Mailing list - http://list.seqfan.eu/	2022-01-17 10:55:03	{"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": 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": 0.864445149898529, "perplexity": 5416.006778404069}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00707.warc.gz"}
https://testbook.com/question-answer/a-certain-sum-is-lent-at-6-p-a-for-2-years-4-5--60406a08056c4adb0238d66c	# A certain sum is lent at 6% p.a for 2 years, 4.5% p.a for the next 4 years, and 12% p.a beyond 6 years. If for a period of 10 years, The simple interest obtained is Rs. 49140 then the sum is? 1. Rs. 60000 2. Rs. 63000 3. Rs. 61000 4. Rs. 65000 Option 2 : Rs. 63000 ## Detailed Solution GIVEN: rate for 2 years = 6% Rate for next 4 years = 4.5% Rate after 6 years = 12% T = 10 years Interest = Rs. 49140 EXPLANATION: Interest in 2 years = 6 × 2 = 12% Interest in next 4 years = 4.5 × 4 = 18% Interest in next 4 years = 12 × 4 = 48% Total interest = 12 + 18 + 48 = 78% A.T.Q 78% of P = 49140 P = (49140/78) × 100 = Rs. 63000 ∴ The sum is Rs. 63000	2021-10-19 06:07:10	{"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": 0.8481767177581787, "perplexity": 3033.8440456432677}, "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/1634323585242.44/warc/CC-MAIN-20211019043325-20211019073325-00343.warc.gz"}
https://www.ias.ac.in/listing/bibliography/boms/KOWSAR_MAJID	• KOWSAR MAJID Articles written in Bulletin of Materials Science • Synthesis, characterization, thermal and electrical properties of composite of polyaniline with cobaltmonoethanolamine complex The present paper involves the synthesis of polyaniline (PANI) composite with cobaltmonoethanolamine [Co(mea)2(H2O)2Cl2] complex via in situ oxidative polymerization by ammonium persulphate. The complex has been synthesized by refluxing method. The composite has been subjected to UV–Visible spectra, FT–IR, X-ray diffraction, SEM and electrical conductivity characterization techniques. Thermal analysis has been done by using TG and DSC techniques. FT–IR absorption peaks confirm the insertion of complex in the backbone of PANI. SEM of the composite also supports its successful synthesis. The XRD of composite also shows crystalline structure hence, proving the successful synthesis of PANI. Thermal analysis shows enhanced thermal stability of polyaniline. In the present composite system, the polymerization of PANI with [Co(mea)2(H2O)2Cl2] complex causes strong interfacial interactions between PANI and [Co(mea)2(H2O)2Cl2] complex crystallites, also suggested by the FT–IR and XRD studies, thereby changing the molecular conformation of PANI from compact coil structure to an expanded coil-like structure. As a consequence, there is an enhancement in the conductivity of composite of PANI up to certain dopant concentration. The anticorrosive property of a coating of PANI/[Co(mea)2(H2O)2Cl2] composite on mild steel coupon in 3 M HNO3 was evaluated using weight loss measurement and compared with pure polyaniline coating. The said composite has shown anticorrosive property and can thus, act as a potent dopant for enhancing corrosion resistance of PANI coatings. • Spectroscopic, morphological, thermal and dielectrical analysis of composite of polythiophene with photoactive transition metal complex of W(IV) The present work involves the synthesis of polythiophene–potassium octacyanotungstate(IV) dihydrate compositevia in-situ oxidative chemical polymerization method using FeCl$_3$ as an oxidant. The resulting composite hasbeen subjected to Fourier transform infrared, X-ray diffraction (XRD) and scanning electron microscopy characterizationtechniques, which confirm the successful synthesis of the composite. XRD shows that the crystalline structure ofK$_4$[W(CN)$_8$] · 2H$_2$O has been retained in the composite. Thermogravimetric analysis data confirm the higher thermal stability of the composite in comparison with pure polythiophene, thus allowing it to be used as a promising material for high-temperature application purposes. Dielectric studies reveal that the dielectric constant and ac-conductivity of the compositeincreased by several orders of magnitude as compared with pure polythiophene at all frequencies, thus showing that thematerial can be used for various applications in the fields of charge storage devices and high-frequency device applications,and can also serve as a potential candidate for solar cell applications. • Synthesis, characterization and optical absorption studies of Na[Fe(CN)$_4$(C$_3$H$_4$N$_2$)NO]2H$_2$O crystals Crystals of Na[Fe(CN)$_4$(C$_3$H$_4$N$_2$)NO]2H$_2$O (SNP-d) have been prepared by photochemical route using sodium nitroprusside (SNP) as precursor complex. The crystals were subjected to various characterization techniques such as energy dispersive X-ray analysis, Fourier transform infrared, UV–visible spectroscopy and X-ray diffraction. From X-ray diffraction, it has been observed that SNP-d belongs to orthorhombic crystal system with primitive lattice. Crystallite size has been found to be 79.2 nm. Molecular structure of SNP-d depicts its non-centrosymmetric nature. As observed from UV–visible spectroscopy, SNP-d crystals form a double bandgap material, between which transitions have been found to be direct allowed. High value of refractive index and low value of reflectance, as obtained from Reddy, Anjaneyulu and Fresnel’s equations, indicate transparency of material to visible radiation. This is clearly justified from UV–visible spectra as the wavelength cutoff is around 300 nm. • # Bulletin of Materials Science Volume 44, 2021 All articles Continuous Article Publishing mode • # Dr Shanti Swarup Bhatnagar for Science and Technology Posted on October 12, 2020 Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru Chemical Sciences 2020	2021-09-19 01:59:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5742942690849304, "perplexity": 6903.021920341267}, "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/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00085.warc.gz"}
https://pavelnaumov.com/395/2017/2/27/midterm-questions	# Midterm Questions Midterm will include 10 out the following 14 questions. Each out of the 10 questions will have the same weight. 1. Find all pure Nash equilibria in a given finite two-player game. 2. Find all pure Nash equilibria in a given finite multiplayer game. 3. Find all pure strict Nash equilibria in a given finite two-player game. 4. Find all (pure and mixed) Nash equilibria in a given two-player game where each player has only two strategies. 5. List all strictly dominating strategies in a given game. 6. List all weakly dominating strategies in a given game. 7. List all strictly dominated strategies in a given game. 8. List all weakly dominated strategies in a given game. 9. Find all maxmin pure strategies of a given two-player game. 10. Find all maxmin mixed strategies of a given two-player game. 11. Find all minmax strategies of a given two-player game. 12. Find all minmax mixed strategies of a given two-player game. 13. Find all minmax regret strategies of a given two-player game. 14. Find all epsilon-Nash equilibria of a given two-player game.	2018-03-19 04:33:04	{"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": 0.9053345918655396, "perplexity": 2200.1851689290575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646375.29/warc/CC-MAIN-20180319042634-20180319062634-00292.warc.gz"}
http://openstudy.com/updates/4f1785d4e4b0aeb795f58027	## anonymous 4 years ago Use the Law of Sines to solve for c to the nearest tenth. 1. anonymous 2. anonymous Please show work so I can understand. Thanks. 3. Hero In general, when solving for the unknown sides or angles of triangles, you can use the Law Sines formula: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$, where Angle A is opposite side a, Angle B is opposite side b, and Angle C is opposite side c. When using Law Of Sines, you only use two fractions at a time in the form of a proportion with one unknown component to solve for. In this case, Angle A (44°), side a (36), Angle C(83°) are given, and side c is unknown, therefore you use the proportion: $\frac{\sin A}{a} = \frac{\sin C}{c}$ to solve for c. Do you believe you can continue to solve from here? 4. anonymous Yes, I got C=51.4 Thank you very much! 5. Hero Good job.	2016-10-24 14:36:51	{"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": 0.8580676913261414, "perplexity": 809.5954449465522}, "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/1476988719638.55/warc/CC-MAIN-20161020183839-00240-ip-10-171-6-4.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/294882/isoperimetric-inequality-and-geometric-measure-theory	# Isoperimetric inequality and geometric measure theory The following version of the isoperimetric inequality can be easily deduced from the Brunn-Minkowski inequality: Theorem. If $$K\subset\mathbb{R}^n$$ is compact, then $$\|K\|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}\mu_+(K),$$ where $$\omega_n$$ is the volume of the unit ball and $$\mu_+(K)=\liminf_{h\to 0} \frac{\|\{x:\, 0<{\rm dist}\, (x,K)\leq h\}\|}{h}$$ is the Minkowski content. If $$K$$ is the closure of a bounded set with $$C^2$$ boundary, then $$\mu_+(K)=H^{n-1}(\partial K)$$ (Hausdorff measure) so we have $$\|K\|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K).$$ However, the above inequality is true for any compact set without assuming anything about regularity of the boundary. My question is: ## Is there a simple proof of the following result? Theorem. If $$K\subset\mathbb{R}^n$$ is compact, then $$\|K\|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K).$$ This result can be proved using the machinery of the geometric measure theory. If $$H^{n-1}(\partial K)=\infty$$, the inequality is obvious. If $$H^{n-1}(\partial K)<\infty$$, then $$K$$ has finite perimeter and the isoperimetric inequality for sets of finite perimeter yields $$\|K\|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}P(K)= n^{-1}\omega_n^{-1/n} H^{n-1}(\partial^* K)$$ where $$P(K)$$ is the perimeter of $$K$$ and $$\partial^*K\subset\partial K$$ is the reduced boundary. For details see [1]. Unfortunately, this argument is very far from being elementary. [1] Ambrosio, L., Fusco, N., Pallara, D.: Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000.	2019-05-24 03:56:50	{"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9114518165588379, "perplexity": 95.93651895479772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257497.4/warc/CC-MAIN-20190524024253-20190524050253-00470.warc.gz"}
https://reviewersph.com/mathematics-third?namadamadan=1$a8ecbb1a06677a6172765d9ede0d08264c8f19f81c13904522e23ba9b2c2f0bf73ce1c58$0	### Math Notes Subjects #### Algebra Solutions ##### Topics \|\| Problems Calculation of Clock Problems In clock problems you usually calculate time and angles. Clock Hands • Minute Hand (Base) • Hour Hand • Seconds Hand As described by the minute hand Minute Hand Second Hand Minute Hand Hour Hand 1 min 60 min 1 min 1/12 min 2 min 120 min 2 min 1/6 min 30 min 1800 min 30 min 5/2 min ... ... ... ... n min 60n min n min n/12 min This simply means that, if the minute hand moves n minutes the second hand moves 60n minutes and the hour hand moves n/12 minutes. Conversion factor between minutes to degrees Minutes Angle (degree) from 12 15 90 30 180 60 360 Use this as a conversion unit, so if you like to calculate how many degrees is 12 minutes. 12 x 90 / 15 = 72 degrees What is the angle between the hour hand and the minute hand at 3:30? Solution: sum forward (blue line) = $$15 + 30/12 + \theta$$ sum backwards (black line) = 30 $$15 +30/12 +\theta$$ = 30 $$\theta$$ = 25 minutes = 75 degrees Using the hour hand, rewrite it as described by minute hand. From 12 to 3 that's 15 minutes, from 3 to where it is now is 30/12 minutes(refer to the table above) and the rest up to 6 is unknown. The minute hand just moves 30 minutes. Another Solution: Convert all minutes to degrees 15 minutes = 90 degrees 30/12 minutes = 15 degrees 30 minutes = 180 degrees sum forward (blue line) = $$90 + 15 + \theta$$ sum backwards (black line) = 180 $$90 + 15 + \theta$$ = 180 $$\theta$$ = 75 degrees	2023-03-31 19:09:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.375977098941803, "perplexity": 4053.626548621767}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00342.warc.gz"}
https://www.nature.com/articles/s41598-020-78714-3?error=cookies_not_supported&code=af85f882-02a2-4ead-ac9c-36506e6f6712	# The critical role of pore size on depth-dependent microbial cell counts in sediments ## Abstract Cell counts decrease with sediment depth. Typical explanations consider limiting factors such as water availability and chemistry, carbon source, nutrients, energy and temperature, and overlook the role of pore size. Our analyses consider sediment self-compaction, the evolution of pore size with depth, and the probability of pores larger than the microbial size to compute the volume fraction of life-compatible pores. We evaluate cell counts vs. depth profiles gathered at 116 sites worldwide. Results confirm the critical role of pore size on cell counts in the subsurface and explain much of the data spread (from ~ 9 orders of magnitude range in cell counts to ~ 2 orders). Cells colonize pores often forming dense biofilms, thus, cell counts in pores are orders of magnitude higher than in the water column. Similar arguments apply to rocks. ## Introduction Bioactivity has been found at sediment depths in excess of 1000 m1,2,3. Overall, cell counts decrease with depth2,4,5,6,7; typical explanations include limiting factors such as water availability and chemistry, carbon source, nutrients, energy and temperature8,9,10,11,12,13,14,15. Some studies have also recognized the critical role of pore size on microbial life and the need for interconnected and traversable habitable spaces16,17,18. A previous experimental study and related analyses identified three distinct regions defined by particle size and sediment depth: “active and motile” for coarse silts and sands at all depths, and either “trapped” or “membrane punctured” for clay-controlled sediments19. Still, the role of pore size as a limiting factor to microbial activity in sediments remains unnoticed or overlooked. Here, we analyze cell counts versus depth data gathered at 116 sites as part of the global Ocean Drilling Program ODP, Integrated Ocean Drilling Program IODP, and other expeditions. Our goal is to assess the role of pore size on microbial cell counts in marine sediments. ## Analyses A detailed description of the analytical approach follows. Additional details, the complete dataset and references can be found in the Supplementary Information associated to this manuscript. ### From sediment self-compaction to cell counts The equilibrium analysis of a sediment slice of thickness dz at depth z predicts that the effective stress gradient dσ'z/dz is a function of the local void ratio ez (Note: the void ratio ez is the volume of voids Vv normalized by the volume of the solid mineral phase Vm): $$\frac{{d\sigma_{z}^{^{\prime}} }}{dz} = \rho_{w} \left[ {\frac{{G_{s} - 1}}{{1 + e_{z} }}} \right]g$$ (1) where gravitational acceleration is g = 9.81 m/s2 and the specific gravity Gs = ρm/ρw is the ratio between the mineral density ρm and water density ρw. The sediment void ratio ez decreases with depth z as the vertical effective stress σ'z increases. We adopt an asymptotically-correct exponential compaction model, where the void ratio decreases from the asymptotic maximum value eL at the water-seabed interface where σ'z = 0, to the limiting void ratio eH at very high effective stress20: $$e_{z} = e_{H} + \left( {e_{L} - e_{H} } \right)\exp \left[ { - \left( {\frac{{\sigma_{z}^{^{\prime}} }}{{\sigma_{c}^{^{\prime}} }}} \right)^{\eta } } \right]$$ (2) The characteristic effective stress is typically between σ'c = 500 and 3000 kPa, and the model parameter η reflects the sensitivity of the void ratio to effective stress (Note: most sediments exhibit η = 1/3—See related analysis21). The solution of the differential Eq. (1) with the constitutive model in Eq. (2) results in the following implicit equation that relates the void ratio ez and the effective stress σ'z at depth z (for η = 1/3—See related example21,22): $$z = \frac{{\left( {1 + e_{H} } \right)}}{{\left( {G_{s} - 1} \right)\rho_{w} g}}\sigma_{z}^{^{\prime}} + 3\frac{{\left( {e_{L} - e_{H} } \right)}}{{\left( {G_{s} - 1} \right)\rho_{w} g}}\sigma_{c}^{^{\prime}} \left\{ {\left[ {\left( {\frac{{\sigma_{z}^{^{\prime}} }}{{\sigma_{c}^{^{\prime}} }}} \right)^{\frac{2}{3}} + 2\left( {\frac{{\sigma_{z}^{^{\prime}} }}{{\sigma_{c}^{^{\prime}} }}} \right)^{\frac{1}{3}} + 2} \right] \cdot \exp \left[ { - \left( {\frac{{\sigma_{z}^{^{\prime}} }}{{\sigma_{c}^{^{\prime}} }}} \right)^{\frac{1}{3}} } \right] - 2} \right\}$$ (3) Finally, we obtain the void ratio profile ez = f (z) with depth z by replacing a selected effective stress σ'z in both Eqs. (3) and (2). ### Mean pore size The geometrical analysis of various sediment fabrics shows that the mean pore size μd [m] is a function of the void ratio ez and the specific surface Ss defined as the ratio between the surface area of particles As and their mass, Ss = As/(ρmVm) [m2/g]18,23, $$\mu_{d} = k\frac{{e_{z} }}{{S_{s} \rho_{m} }}$$ (4) where the k-factor reflects the soil fabric (see geometric analyses and values for various fabrics in the Supplementary Table S1). Furthermore, our database shows a strong correlation between the specific surface Ss and the asymptotic void ratio eL (see Supplementary Fig. S1): $$e_{L} = 0.9 + 0.03\frac{{S_{s} }}{{[{\text{m}}^{2} /{\text{g}}]}}$$ (5) For large rotund particles, the specific surface Ss → 0 while the asymptotic void ratio tends to eL → 0.9 which corresponds to the loose, simple cubic packing of monosize spherical particles. ### Pore size distribution (Soils and rocks) We compiled a large database of pore size distributions measured from a wide range of soils (39 specimens) and intact rocks (44 specimens), and fitted each dataset with a log-normal distribution (Supplementary Table S2 and Figs. S2S6). Figure 1 shows the standard deviation σd [ln(d/µm)] plotted against the mean pore size μd [ln(d/µm)]. The trend reveals a surprisingly strong relationship across all sediments and intact rocks: σd/μd ≈ 0.4, from nm-size pores in shales to mm-size pores in sandy sediments (previously observed for a small set of sediments18). ### Cell count in sediments The pore size d must be larger than the cell size b [m]. Consequently, cell counts per sediment unit volume must relate to the probability of pores P(d ≥ b). The probability P(d ≥ b) for a log-normal distribution assuming σd/μd ≈ 0.4 simplifies to (see inset in Fig. 1): $$P(d \ge b) = \int\limits_{\ln b}^{\infty } {\exp \left\{ { - \frac{2.5}{4}\left[ {\ln \left( {\frac{d}{{\mu_{d} }}} \right) + 0.05} \right]^{2} } \right\}{\text{d}} d}$$ (6) Finally, the cell count c [cells/cm3] in a sediment at depth z depends on: (1) the cell concentration cfl [cells/cm3] in the pore fluid within the pores, (2) the sediment void ratio ez (Eqs. 2 and 3), and (3) the probability P(d ≥ b) (Eq. 6) for a given mean pore size μd (Eq. 4) and standard deviation σd/μd = 0.4 (Fig. 1): $$c = c_{fl} \cdot \left( {\frac{{e_{z} }}{{1 + e_{z} }}} \right) \cdot P(d \ge b)$$ (7) where porosity n = e/(1 + e). We adopt a nominal cell size b = 1 µm for all analyses presented in this manuscript. ### Implementation We use the effective stress dependent, asymptotically correct self-compaction model to match the reported void ratio versus depth ez-z profiles (Eqs. 13). The fitted asymptotic void ratio eL allows the estimation of the specific surface Ss (Eq. 5). Then, the mean pore size μd(z) at depth z is computed from the local void ratio ez and the sediment specific surface Ss (Eq. 4, Supplementary Table S1). We complete the probabilistic pore size analysis by invoking the strong correlation σd/μd = 0.4 between the mean pore size μd and the standard deviation σd (Fig. 1, Supplementary Table S2 and Figs. S2S6). While there is variability, we adopt a single σd/μd ≈ 0.4 for all analyses to avoid additional degrees of freedom (The Supplementary Fig. S7 shows the effect of σd/μd on predicted cell counts—all other parameters are kept constant). Finally, we estimate cell count profiles c(z) [cells/cm3] using a single value of cell concentration in the pore fluid cfl for the full profile at each site (Eq. 7). ## Results Results in Fig. 2 show the fitted compaction model and predicted cell counts for various marine sediments. Computed trends fit the compiled data well. In particular, there is a significant reduction in cell count with depth for high specific surface sediments (yellow and red data points). By contrast, pore size is not the limiting factor for microbial cell counts in silty or sandy sediments (low specific surface—blue data points). In fact, the cell count in the pore fluid cfl and the sediment porosity n determine the cell counts in coarse-grained sediments c = cfl·[ez/(1 + ez)] = cfl·nz [refer to Eqs. (6) and (7)]. We followed the same methodology to analyze all 116 profiles in the database (a total of 2696 measurements—Supplementary Table S3 and Figs. S8S121). Figure 3a presents the complete dataset. We use the fitted asymptotic void ratio eL to discriminate cell count profiles and cluster bio-habitats into three distinct groups: (1) green data points correspond to sandy and silty sediments (eL < 2, Ss < 1.1 m2/g, and LL < 30), (2) black data points show intermediate plasticity sediments (eL = 2–5, Ss = 10–70 m2/g, and LL ≈ 50-to-120), and (3) red data points represent very high plasticity clayey sediments (eL > 5, Ss > 120 m2/g, and LL > 140). For completeness, the values in parentheses include the estimated specific surface Ss and liquid limit LL, where the liquid limit LL is a gravimetric water content of a water–sediment mixture at the paste-slurry transition. Data clustering by sediment type highlights the role of sediment texture and effective stress-dependent pore size on microbial cell counts in the subsurface (For clarity, cell count data for the different void ratio categories are presented in the Supplementary Fig. S122). Cell counts in sediments vary across > 8 orders of magnitude (Fig. 3a), and the overall depth distribution deviates from previously suggested trends4,5,24 (black line24: log10[cell counts] = 8.05–0.68·log10[depth/m]). Figure 3b shows the ratio between measured and predicted (Eq. 7) cell counts versus depth. Data points collapse onto a single trend within ± one log cycle (standard deviation σ = 0.52). The contraction in the spread from Fig. 3a to Fig. 3b reflects the extent to which observed cell counts can be justified by pore size as a limiting factor. The remaining spread reflects physical factors (e.g., sediment layering and heterogeneity, non-constant cell concentration in the pore fluid cfl with depth due to nutrient availability and environmental conditions such as temperature), experimental difficulties (e.g., cell counts and void ratio measurements), inherent uncertainties in the analysis and material parameters (e.g., validity of correlations, adopted nominal cell size, and correlation between eL and Ss—Eq. 5). Our depth dependent cell count analysis identifies two parameters of particular significance: the cell concentration in the pore fluid cfl and the asymptotic void ratio eL, i.e., sediment type. Figure 4 presents cumulative distributions for the asymptotic void ratio eL and the cell concentration in the pore fluid cfl obtained by fitting the analytical model to void ratio and cell count profiles at each of the 116 sites. The fitted asymptotic void ratio values eL fall between 2.4 ≤ eL ≤ 4.8 for 68% of data (mean value eL = 3.6—Fig. 4a). This suggests a prevalence of intermediate plasticity sediments at the studied sites. Intermediate plasticity sediments with asymptotic void ratios in the range of 2 ≤ eL ≤ 5 tend to host life with the highest cell volume density cfl. In these sediments, the inferred cell concentrations in the pore fluid varies between cfl = 107.8 and 1010.2 cells/cm3 for 68% of the data (mean value cfl = 109 cells/cm3; Fig. 4b), and can reach volume saturation levels found in dense biofilms at ~ 1011 cells/cm3. (Note: biofilm concentrations are typically reported in areal density; a high biofilm density of 107 cells/cm2 corresponds to 1011 cells/cm3 for biofilm layers separated at 1 μm; for comparison, the packing of micron-size spheres in simple cubic configuration corresponds to the ratio cm3/μm3 that is 1012 cells/cm3). These cell counts are orders of magnitude higher than in the water column in most oceans, which ranges between 2 × 104 and 5 × 105 [cells/cm3]9,25,26,27. ## Discussion and implications Active microorganisms require traversable pore throats larger than the nominal b ≈ 1 μm size28. Pores and pore throats also limit the advective nutrient transport. In fact, the 1 μm size correlates with a hydraulic conductivity kh ≈ 0.1-to-10 cm/day29. We can anticipate low flow velocities v = kh·i given the typically low hydraulic gradients in nature i < 1.0; therefore, the ensuing advective-reactive regime hinders nutrient transport and bio-activity, as reported by others2,30. A small fraction of fines can be sufficient to fill the pore space between coarse grains, control the pore size and eventually limit microbial cell counts. The Revised Soil Classification System RSCS recognizes the critical role of fines on mechanical and fluid flow properties in sediments31,32,33,34. For example, the fines fraction required to fill the pores in a sandy sediment is within the range of 12% for kaolinite, 7% for illite, and 2% for bentonite. Consequently, the detailed analysis of bioactivity in sediments must carefully consider the presence of fines and their mineralogy. Data compiled in this study and the mecho-geometrical probabilistic analyses provide strong evidence for the critical role of pore size on microbial cell counts in sediments (together with other limiting factors such as water, carbon source, nutrients and temperature). The sediment type and effective-stress dependent pore size analysis adequately capture the decreasing cell counts with depth, and highlight the controlling role of the sediment specific surface Ss. Similarly, pore size emerges as a critical limiting factor for life in rocks as well; for example, it is unlikely that active life will take place in the small pores of intact shales (Fig. 1), however, life may indeed thrive in large carbonate vugs. Furthermore, we expect to find microbial activity in most fractures, even at depth35,36. In fact, fractures can be active bio-reactors within rock masses. ## References 1. 1. Moore, G. F., Taira, A. & Klaus, A. Shipboard scientific party. In Proceedings of the Ocean Drilling Program, Initial Reports, Vol. 190 (2001). 2. 2. Wellsbury, P., Mather, I. & Parkes, R. J. Geomicrobiology of deep, low organic carbon sediments in the Woodlark Basin, Pacific Ocean. FEMS. Microbiol. Ecol. 42, 59–70 (2002). 3. 3. Inagaki, F. et al. Exploring deep microbial life in coal-bearing sediment down to~ 2.5 km below the ocean floor. Science 349, 420–424 (2015). 4. 4. Parkes, R. J. et al. Deep bacterial biosphere in Pacific Ocean sediments. Nature 371, 410–413 (1994). 5. 5. Parkes, R. J., Cragg, B. A. & Wellsbury, P. Recent studies on bacterial populations and processes in subseafloor sediments: a review. Hydrogeol. J. 8, 11–28 (2000). 6. 6. Schippers, A. et al. Prokaryotic cells of the deep sub-seafloor biosphere identified as living bacteria. Nature 433, 861–864 (2005). 7. 7. Jørgensen, B. B. & D’Hondt, S. A starving majority deep beneath the seafloor. Science 314, 932–934 (2006). 8. 8. Jannasch, H. W. & Wirsen, C. O. Deep-sea microorganisms: in situ response to nutrient enrichment. Science 180, 641–643 (1973). 9. 9. Whitman, W. B., Coleman, D. C. & Wiebe, W. J. Prokaryotes: the unseen majority. Proc. Natl. Acad. Sci. USA 95, 6578–6583 (1998). 10. 10. D’Hondt, S. et al. Distributions of microbial activities in deep subseafloor sediments. Science 306, 2216–2221 (2004). 11. 11. Mitchell, J. K. & Santamarina, J. C. Biological considerations in geotechnical engineering. J. Geotech. Geoenviron. 131, 1222–1233 (2005). 12. 12. Pernthaler, J. Predation on prokaryotes in the water column and its ecological implications. Nat. Rev. Microbiol. 3, 537–546 (2005). 13. 13. Jørgensen, B. B. & Boetius, A. Feast and famine—microbial life in the deep-sea bed. Nat. Rev. Microbiol. 5, 770–781 (2007). 14. 14. Fry, J. C., Parkes, R. J., Cragg, B. A., Weightman, A. J. & Webster, G. Prokaryotic biodiversity and activity in the deep subseafloor biosphere. FEMS. Microbiol. Ecol. 66, 181–196 (2008). 15. 15. Røy, H. et al. Aerobic microbial respiration in 86-million-year-old deep-sea red clay. Science 336, 922–925 (2012). 16. 16. Fredrickson, J. K. et al. Pore size constraints on the activity and survival of subsurface bacteria in a late cretaceous shale sandstone sequence, northwestern New Mexico. Geomicrobiol. J. 14, 183–202 (1997). 17. 17. Bartlett, R. et al. Lithological controls on biological activity and groundwater chemistry in Quaternary sediments. Hydrol. Process. 24, 726–735 (2010). 18. 18. Phadnis, H. S. & Santamarina, J. C. Bacteria in sediments: pore size effects. Geotech. Lett. 1, 91–93 (2011). 19. 19. Rebata-Landa, V. & Santamarina, J. C. Mechanical limits to microbial activity in deep sediments. Geochem. Geophys. Geosyst. 7, Q11006 (2006). 20. 20. Chong, S. H. & Santamarina, J. C. Soil compressibility models for a wide stress range. J. Geotech. Geoenviron. Eng. 142, 06016003 (2016). 21. 21. Lyu, C., Park, J. & Santamarina, J. C. Depth-dependent seabed properties: geoacoustic assessment. J. Geotech. Geoenviron. Eng. 147, 04020151 (2021). 22. 22. Terzariol, M., Park, J., Castro, G. M. & Santamarina, J. C. Methane hydrate-bearing sediments: Pore habit and implications. Mar. Petrol. Geol. 116, 104302 (2020). 23. 23. Santamarina, J. C., Klein, K. A. & Fam, M. A. Characterization of Particles and Particulate Media in Soils and waves: Particulate Materials Behavior, Characterization and Process Monitoring (Wiley, New York, 2001). 24. 24. Parkes, R. J. et al. A review of prokaryotic populations and processes in sub-seafloor sediments, including biosphere: geosphere interactions. Mar. Geol. 352, 409–425 (2014). 25. 25. Arístegui, J., Gasol, J. M., Duarte, C. M. & Herndld, G. J. Microbial oceanography of the dark ocean’s pelagic realm. Limnol. Oceanogr. 54, 1501–1529 (2009). 26. 26. De Corte, D., Sintes, E., Yokokawa, T., Reinthaler, T. & Herndl, G. J. Links between viruses and prokaryotes throughout the water column along a North Atlantic latitudinal transect. ISME. J. 6, 1566–1577 (2012). 27. 27. Mapelli, F. et al. Biogeography of planktonic bacterial communities across the whole Mediterranean Sea. Ocean Sci. 9, 585–595 (2013). 28. 28. NAE (National Academy of Engineering), "Flow and transport - Underlying processes" in Characterization, Modeling, Monitoring, and Remediation of Fractured Rock 31–52 (The National Academies Press, Washington, DC, 2015). 29. 29. Ren, X. W. & Santamarina, J. C. The hydraulic conductivity of sediments: a pore size perspective. Eng. Geol. 233, 48–54 (2018). 30. 30. Boivin-Jahns, V., Ruimy, R., Bianchi, A., Daumas, S. & Christen, R. Bacterial diversity in a deep-subsurface clay environment. Appl. Environ. Microbiol. 62, 3405–3412 (1996). 31. 31. Jang, J. & Santamarina, J. C. Fines classification based on sensitivity to pore-fluid chemistry. J. Geotech. Geoenviron. Eng. 142, 06015018 (2016). 32. 32. Jang, J. & Santamarina, J. C. Closure to “Fines classification based on sensitivity to pore-fluid chemistry” by Junbong Jang and J. Carlos Santamarina. J. Geotech. Geoenviron. Eng. 143, 07017013 (2017). 33. 33. Park, J. & Santamarina, J. C. Revised soil classification system for coarse-fine mixtures. J. Geotech. Geoenviron. Eng. 143, 04017039 (2017). 34. 34. Park, J., Castro, G. M. & Santamarina, J. C. Closure to “Revised soil classification system for coarse-fine mixtures” by Junghee Park and J. Carlos Santamarina. J. Geotech. Geoenviron. Eng. 144, 07018019 (2018). 35. 35. Brodsky, E. E., Kirkpatrick, J. D. & Candela, T. Constraints from fault roughness on the scale-dependent strength of rocks. Geology 44, 19–22 (2016). 36. 36. Al-Fahmi, M. M., Ozkaya, S. I. & Cartwright, J. A. New insights on fracture roughness and wall mismatch in carbonate reservoir rocks. Geosphere. 14, 1851–1859 (2018). ## Acknowledgements Gabrielle. E. Abelskamp edited the manuscript. Support for this research was provided by the KAUST Endowment at King Abdullah University of Science and Technology. ## Author information Authors ### Contributions J.P. and J.C.S. have contributed equally to all parts of the study and manuscript. ### Corresponding author Correspondence to J. Carlos Santamarina. ## Ethics declarations ### Competing interests The authors declare no competing interests. ### Publisher's note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Park, J., Santamarina, J.C. The critical role of pore size on depth-dependent microbial cell counts in sediments. Sci Rep 10, 21692 (2020). https://doi.org/10.1038/s41598-020-78714-3 • Accepted: • Published:	2021-02-25 02:38:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5924920439720154, "perplexity": 7961.437899134377}, "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/1614178350706.6/warc/CC-MAIN-20210225012257-20210225042257-00293.warc.gz"}
http://www.encyclopediaofmath.org/index.php/Complete_metric_space	# Complete metric space 2010 Mathematics Subject Classification: Primary: 54E50 [MSN][ZBL] A metric space in which each Cauchy sequence converges. A complete metric space is a particular case of a complete uniform space. A closed subset $A$ of a complete metric $(X,d)$ space is itself a complete metric space (with the distance which is restiction of $d$ to $A$). The converse is true in a general metric space: if $(X,d)$ is a metric space, not necessarily complete, and $A\subset X$ is such that $(A,d)$ is complete, then $A$ is necessarily a closed subset. Given any metric space $(X,d)$ there exists a unique completion of $X$, that is a triple $(Y,\rho,i)$ such that: • $(Y, \rho)$ is a complete metric space; • $i: X \to Y$ is an isometric embedding, namely a map such that $d(x,y) = \rho (i(x), i(y))$ for any pair of points $x,y\in X$; • $i(X)$ is dense in $Y$. Often people refer to the metric space $(Y, \rho)$ as the completion. Both the space and the isometric embedding are unique up to isometries. How to Cite This Entry: Complete metric space. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Complete_metric_space&oldid=30896 This article was adapted from an original article by M.I. Voitsekhovskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	2014-08-31 00:18:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9289622902870178, "perplexity": 215.28054912565656}, "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-2014-35/segments/1408500835844.19/warc/CC-MAIN-20140820021355-00116-ip-10-180-136-8.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/171428/are-there-additive-subgroups-of-reals-of-dimension-1-with-no-subgroups-of-dimens	# Are there additive subgroups of reals of dimension 1 with no subgroups of dimension strictly between 0 and 1? I will use $dimA$ to denote the Hausdorff dimension of a set $A \subseteq \mathbb{R}$. Being a null set means having Lebesgue measure zero. In the 1966 paper "Additive gruppen mit vorgegebener hausdorffscher dimension" by Paul Erdös and Bodo Volkmann, it was shown that under the continuum hypothesis $(2^{\aleph_0} = \aleph_1)$ there is an additive subgroup $G$ of $\mathbb{R}$ such that (i) $G$ is non-null. In particular, $dimG = 1$. (ii) If $A$ is a null set, then $G \cap A$ is countable. In particular, there does not exist a subgroup $H$ of $G$ with $dimH \in (0,1)$. My question is as follows: Assuming only ZFC, is there an additive subgroup $G$ of $\mathbb{R}$ such that (i) dim $G$ = 1 and (ii) there does not exist a subgroup $H$ of $G$ with $dimH \in (0,1)$. The proof of Erdös and Volkmann starts by enumerating the collection of null $G_\delta$ sets, and their argument relies heavily on the continuum hypothesis. However, their result appears to be stronger than the question I am asking. The 1984 paper "A Very Sparse Set of Dimension 1" by Anatole Beck seems to be a step in the right direction, but I've found little information otherwise.	2019-06-27 11:21:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9188294410705566, "perplexity": 131.93256546186078}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001089.83/warc/CC-MAIN-20190627095649-20190627121649-00335.warc.gz"}
https://math.stackexchange.com/questions/1159016/choose-any-false-statements-regarding-the-graph	# Choose any false statements regarding the graph. Choose any false statements regarding the graph. Select all that apply. (a) The graph is a function that has an inverse function. (b) The graph is a function. (c) The graph is a function that does not have an inverse function. (d) The graph passes the horizontal line test. (e) The graph is a one-to-one function. (f) The graph passes the vertical line test. I assumed that The false statements are: The graph is a one-to-one function. The graph is a function that has an inverse function. The graph passes the horizontal line test. but not sure Yes, the false statements are a,d, and e. The others are true. You need two facts here: 1) A graph passes the vertical line test (every vertical line intersects the graph in at most one point) if and only if the graph represents a function. 2) If we know that a graph represents a function, these three statements are equivalent (any one is true if and only if another one is true): a) The graph passes the horizontal line test (every horizontal line intersects the graph in at most one point); b) The function is one-to-one; c) The function has an inverse function. Looking at your graph, we see that every vertical line intersects the graph exactly once, so it passes the vertical line test. Therefore by the first fact we know the graph represents a function. So (b) and (f) are true. We see that the horizontal line $y=6$ intersects the graph seven times that we can see and perhaps infinitely many times that we do not see, so it fails the horizontal line test. That means that by the second fact, (a), (d), and (e) are false. Item (c) is the opposite of (a), so (c) is true. • yes please more details thanks – user155971 Feb 21 '15 at 18:38 • @user155971: See the additions to my answer. – Rory Daulton Feb 21 '15 at 19:49 • thank you sir for helping me – user155971 Feb 21 '15 at 19:56	2019-05-27 13:06:19	{"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": 0.6428184509277344, "perplexity": 430.38385742517096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232262600.90/warc/CC-MAIN-20190527125825-20190527151825-00153.warc.gz"}
https://deepai.org/publication/a-nonsmooth-nonconvex-descent-algorithm	# A nonsmooth nonconvex descent algorithm The paper presents a new descent algorithm for locally Lipschitz continuous functions f:X→R. The selection of a descent direction at some iteration point x combines an approximation of the set-valued gradient of f on a suitable neighborhood of x (recently introduced by Mankau Schuricht) with an Armijo type step control. The algorithm is analytically justified and it is shown that accumulation points of iteration points are critical points of f. Finally the algorithm is tested for numerous benchmark problems and the results are compared with simulations found in the literature. 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For the minimization of a nonsmooth function f(x)→min(x∈X) (1.1) numerous algorithms based on quite different methods have been proposed in the literature. Let us mention, without being complete, bundle-type methods (cf. Alt [3], Frangioni [11], Gaudioso & Monaco [12], Hiriat-Urruty [17], Kiwiel [18], Makela & Neittaanmaki [22], Mifflin [24], Schramm [28], Wolfe [32], Zowe [34]), proximal point and splitting methods as e.g. the Fista or the primal dual method (cf. Beck [5], Eckstein & Bertsekas [10], Chambolle & Pock [7]), gradient sampling algorithms (cf. Burke, Lewis & Overton [6], Kiwiel [19]), algorithms based on smoothing techniques (cf. Polak & Royset [26]) and the discrete gradient method (cf. Bagirov & Karasozen [4]). Bundle-type methods, proximal point methods, and splitting methods require to be convex or to have some other special structure. Many algorithms for locally Lipschitz continuous functions as the discrete gradient method need to know the entire generalized gradient of at given points. Stochastic methods like the gradient sampling algorithm are robust without the knowledge of the entire generalized gradient, but at the cost of high computational effort. Therefore they are limited to minimization problems on low dimensional spaces. Recall that the derivative indicates a direction of descent for near . However if the direction of descent changes rapidly in a small neighborhood of , which is typical for functions having large second derivatives or that are even nonsmooth, then some knowledge of on a whole neighborhood of is necessary for the determination of a suitable direction of descent near . For a new robust and fast algorithm we combine ideas of bundle-methods and gradient sampling methods. We use the concept of gradients of on sets as introduced in Mankau & Schuricht [23] (which extends ideas from Goldstein [15]). Here, similar to gradient sampling methods, generalized gradients of on a whole neighborhood of are considered for the determination of a suitable descent direction near point . But, in contrast to e.g. Burke, Lewis & Overton [6] and Kiwiel [19], the set-valued gradient on a neighborhood of point is not approximated stochastically. We rather use an elaborate recursive inner approximation coupled with the computation of related descent directions until a generalized Armijo condition is satisfied (a condition similar to that used in Alt [3] and Schramm [28] in connection with the -subdifferential). Finally a line search along a direction of sufficient descent gives the next iteration point (cf. Pytlak [27]). For better performance we may also adapt the norm of in each step. It turns out that our algorithm requires substantially less gradient computations than in [6] and [19]. Therefore it is also applicable on high dimensional spaces as needed for variational problems. For a locally Lipschitz continuous function our methods merely demand that, at any point , both the value and at least one element of the generalized gradient (in the sense of Clarke [9]) can be computed. Notice that this mild requirement is assumed in any of the above mentioned gradient based algorithms and that it is typically met in applications. In an upcoming paper an extended algorithm is presented where quasi-Newton methods and preconditioning methods are included by a suitable change of norm in each iteration step. Section 2 gives a brief overview about gradients on sets as needed for our treatment. The algorithm and some convergence results are given in Section 3. After the formulation of the condition of sufficient descent and of several general assumptions, Section 3.1 provides the main Algorithm 3.8 and its properties. Algorithm 3.8 calls Algorithm 3.14 for the computation of a suitable inner approximation of the set-valued gradient on a neighborhood of the current iteration point and the computation of a related descent direction while Step 3 of Algorithm 3.14 calls Algorithm 3.17 for some subiteration. Figure 1 gives an overview of the whole algorithm and several statements justify essential steps of it. Theorem 3.24 shows that every accumulation point of iteration points produced by Algorithm 3.8 is a critical point in the sense of Clarke. The proofs are collected in Section 3.2. Comprehensive numerical tests of our algorithm for classical benchmark problems can be found in Section 4. Here the simulations are also compared with results from Burke, Lewis & Overton [6], Kiwiel [19], Alt [3], Schramm [28] and the BFGS algorithm. Notation: is a Hilbert space111Notice that any Hilbert space is uniformly convex and reflexive. with scalar product where the dual is always identified with . For a set we write for its closure, for its convex hull and for its closed convex hull. and are the open -neighborhood of point and set , respectively. stands for the open segment between and, in particular, for the open interval. denotes the positive real numbers. For a locally Lipschitz continuous function we write for Clarke’s generalized directional derivative of at in direction and for Clarke’s generalized gradient of at (cf. Clarke [9]). ## 2 Gradient on sets Let be a locally Lipschitz continuous function on a Hilbert space . Clarke’s generalized gradient of at and the corresponding generalized directional derivative of at in direction somehow express the behavior of at point (cf. Clarke [9]). However, for the construction of a descent step in a numerical scheme, some information about the behavior of on a whole neighborhood of is useful in general. In particular, for describing the behavior of on the whole -ball , we use some set-valued gradient of and some corresponding generalized directional derivative as introduced in Mankau & Schuricht [23] by using Clarke’s pointwise quantities. For the convenience of the reader we present some brief specialized summary of that material as needed for our treatment. For we define the gradient of on by ∂εf(x):=¯¯¯¯¯¯¯¯¯¯¯conv⋃y∈¯¯¯¯¯¯¯¯¯¯¯¯Bε(x)∂f(y) (2.1) (notice that the closed convex hull agrees with the weakly closed convex hull) and the directional derivative of on in direction by f0ε(x;h):=supy∈¯¯¯¯¯¯¯¯¯¯¯¯Bε(x)f0(y;h). (2.2) We have the following basic properties (cf. Propositon 2.3 and Corollary 2.10 in [23]). ###### Proposition 2.3. Let be Lipschitz continuous of rank on a neighborhood of with and . Then • is nonempty, convex, weak-compact and bounded by . • is finite, positively homogeneous, subadditive, and Lipschitz continuous of rank . Moreover it is the support function of , i.e. f0ε(x;h)=maxa∈∂εf(x)⟨a,h⟩for allh∈X. (2.4) • We have ∂εf(x)={a∈X∣∣⟨a,h⟩≤f0ε(x;h)for allh∈X}. (2.5) • Let with and let with . Then f(x+th)≤f(x)+tf0ε(x;h) • Let with and let . Then limk→∞f0εk(x;h)=f0(x;h)and⋂k∈N∂εkf(x)=∂f(x). Regularity of at , i.e. , implies regularity of on some by Proposition 2.16 in [23]. ###### Lemma 2.6. Let be locally Lipschitz continuous and let for some . Then there exist and with such that −∥a∥≤⟨a,h⟩≤f0ε(x;h)<0for alla∈∂εf(x). Moreover, by (2.5). Motivated by Proposition 2.3 (4) we say that is a descent direction of on if . We call steepest or optimal descent direction of on (with respect to ) if ∥~h∥=1andf0ε(x;~h)=min∥h∥≤1f0ε(x;h)<0. (2.7) Theorem 3.10 of [23] ensures the existence of optimal descent directions and of norm-minimal elements in . ###### Proposition 2.8. Let be Lipschitz continuous on a neighborhood of for some , . Then there is a unique with ∥~a∥=mina∈∂εf(x)∥a∥. (2.9) If or, equivalently, for some (cf. (2.5)), then there is a unique optimal descent direction on . In particular ~h=−~a∥~a∥,f0ε(x;~h)=min∥h∥≤1f0ε(x;h)=−∥~a∥. (2.10) Corollary 3.15 and Corollary 3.16 in [23] state some stability of descent directions. ###### Corollary 2.11. Let be Lipschitz continuous of rank on a neighborhood of for some , , let , and let , be as in Proposition 2.8. Then every with is a descent direction on . This allows to get descent directions by suitable approximations of , which is important for our numerical algorithms. ###### Corollary 2.12. Let be Lipschitz continuous on a neighborhood of for some , , let , and let , be as in Proposition 2.8. Then for any there is some such that for every with ∥a′∥≤mina∈∂εf(x)∥a∥+τ=∥~a∥+τ we have that is a descent direction on and satisfies f0ε(x;h′)(???)=maxa∈∂εf(x)⟨a,h′⟩<−δ∥~a∥. ## 3 Descent algorithm We now introduce some descent algorithm for locally Lipschitz continuous functions on a Hilbert space . At each iteration point we determine an approximation of the norm-minimal element (cf. (2.9)) with respect to some suitable radius . We are interested in pairs satisfying a condition of sufficient descent in the sense of a generalized Armijo step of the form f(x−εh)−f(x)≤−δε∥a∥withh=a∥a∥ (3.1) where is fixed for the whole scheme. As new iteration point we then select for some such that (3.1) still holds with instead of . If , the norm will be very small and the null step condition ∥a∥ (with a suitable control function that is fixed for the whole scheme) indicates that situation. Here we cannot expect (3.1) in general and we have two possibilities. If is on the desired level of accuracy for the minimizer (or critical point), we can stop the algorithm. Otherwise the used ball is too large for an iteration step with sufficient descent. Therefore we decrease and look for sufficient descent with a new pair . Our approximation of combined with the analytically justified step size control ensures that we always get sufficient descent for some small enough (cf. Lemma 2.6 and also the proof of Theorem 3.24). That we finally end up with the null step condition on the desired scale, has to become sufficiently small during the algorithm, which is ensured by control functions and . But, that the algorithm doesn’t get stuck in a small ball without critical point, shouldn’t approach zero too fast, which is ensured by control functions and . Thus a careful selection of the step size, that is related to , plays a very important role. The algorithm can be improved by choosing suitable equivalent norms at each iteration step. Let us start with general requirements for the control functions , , and . ###### Assumption 3.3. Suppose that: • are non-decreasing functions such that limt→0T1(t)=0andlimn→∞Tn2(t)=0for allt>0 where is given inductively by and for all . Notice that this implies T2(t)0 (otherwise for some and, since is non-decreasing, induction would give ). • is a function having at least one of the following properties: • implies for any sequences and . • For any there is some such that for all and . Since the conditions for and are quite technical, we provide some typical examples. ###### Example 3.4 (examples for T2 and G). • for . • where, in particular, . • with a constant satisfies (a). • with satisfies (b). • with non-decreasing and satisfies (a). • and satisfy (a) or (b) if , satisfy both (a) or (b), respectively. • satisfies (a) and satisfies (b) for any . As already mentioned, it might be useful to adapt the norm in every iteration (recall that the Newton method can be considered as descent algorithm with changing norm at each step). In our algorithm we allow a change of norm in every step as long as we have some uniform equivalence. ###### Assumption 3.5. The norms and on are uniformly equivalent, i.e. there is some such that 1C∥⋅∥≤∥⋅∥k≤C∥⋅∥% for allk∈N. (3.6) In practice is related to the Hessian of some smooth function at iteration point and is some (usually not explicitly known) bound of that Hessian. Notice that the definition of and of as subset of merely uses convergence in and, thus, does not depend on equivalent norms on . However the Riesz mapping identifying with the Hilbert space depends on the norm. Therefore depends on the norm if it is considered as subset of , which we usually do for simplification of notation. ###### Remark 3.7. The gradient based on the norm is understood as subset of equipped with where, in particular, is taken with respect to . ### 3.1 Algorithm Now we introduce the main algorithm based on two subalgorithms presented afterwards. We formulate several results that justify the single steps and that finally show convergence of the algorithm (cf. Figure 1 below for a rough overview). The proofs are collected in Section 3.2. ###### Algorithm 3.8 (Main Algorithm). • Initialization: Choose and satisfying Assumption 3.3, δ∈]0,1[,x0∈X,ε0>0 and set . • Choose some norm subject to Assumption 3.5, some (w.r.t. ), and some . • Determine (w.r.t. ) by Algorithm 3.14 such that the null step condition ∥ak,i∥k or the condition of sufficient descent f(xk−εk,ihk,i)−f(xk)≤−δ∥ak,i∥kεk,iwithhk,i:=ak,i∥ak,i∥k (3.10) is satisfied (recall Remark 3.7 for the meaning of related to and notice that (3.9) and (3.10) can be satisfied simultaneously). • In case (3.9) set , increment by one, and go to Step 3. • If (3.9) is not true, choose such that the condition of sufficient descent f(xk−σkhk,i)−f(xk)≤−δ∥ak,i∥kσk (3.11) is satisfied (notice that is always possible, since (3.10) is satisfied in this case). Then fix the new iteration point xk+1:=xk−σkhk,i, (3.12) set , increment by one, set , and go to Step 2. ###### Remark 3.13. • Instead of in Step 2 one could also choose εk,0≥G(∥ak−1∥k−1,εk)fork>0. • The selection of in Step 5 can be done by some line search in direction (cf. Pytlak [27]). • One can easily ensure that by requiring that e.g. εk≤σk=∥xk−xk+1∥≤T3(εk) for some with and since the proof of Theorem 3.24 shows that . But in practice this is usually not necessary. The essential point in Algorithm 3.8 is Step 3 with the computation of a suitable approximation of the norm-smallest element (cf. (2.9)) such that the null step condition or the condition of sufficient descent is satisfied for given . Let us briefly discuss the main idea before we formulate the corresponding subalgorithm. Usually the sets defining are not known explicitely. For the algorithm we merely suppose that always at least one element can be determined numerically (cf. Remark 3.19 below for a brief discussion of that point). On this basis we select step by step elements for suitable and determine certain such that, roughly speaking, the convex hull of all , with is an approximating subset of and is a norm-minimal element in . In doing so we still manage that is decreasing sufficiently. Therefore we reach for and large that the null step condition (3.9) is satisfied if or, otherwise, that approximates sufficiently well in the sense of Corollary 2.12. In the last case is a descent direction on and, by Proposition 2.3 (4), f(xk−εk,ihk,i)−f(xk)≤εk,if0εk,i(x;−h)≤−δεk,i∥~ak,i∥ with from Algorithm 3.8, i.e. condition (3.10) of sufficient descent is satisfied with the standard norm. Clearly the quality of the algorithm is closely related to the quality of the approximating set and, in some applications, we can improve the quality substantially by choosing a suitable equivalent norm in each step. Let us now provide the precise algorithm where quantities determined here are marked by . ###### Algorithm 3.14. Let , , , , and be as in Step 3 of Algorithm 3.8 for some . • Choose some (w.r.t. ) and some and set . (Typically, but not necessarily, agrees with from Algorithm 3.8.) • Set and . If satisfies the null step condition (3.9) or condition (3.10) of sufficient descent, stop and return . • Otherwise compute some (w.r.t. ) for some by Algorithm 3.17 such that ⟨a′j,b′j⟩k≤δ′∥∥a′j∥∥2k. (3.15) • Choose some subset such that and set A′j:=convB′j. • Compute a′j+1:=argmin{∥∥a′∥∥k∣∣a′∈A′j}, increment by one, and go to Step 2. Notice that for all by induction and that A′j=convB′j⊆∂εk,if(xk). Hence can be considered as some inner approximation of and the norm-smallest element is an approximation of the norm-smallest element . Algorithm 3.14 ensures with (3.15) that decreases sufficiently, i.e. we have for some as long as the null step condition (3.9) is not fulfilled (cf. the proof of Lemma 3.30). Hence, the null step condition (3.9) has to be satisfied for some after finitely many steps if we do not meet condition (3.10) of sufficient descent before. In practice we usually take B′j = {a′0}∪{a′l∣∣j−m≤l≤j}∪{b′j}or B′j = with . ###### Remark 3.16. Note that the computation of is equivalent to the minimization of a quadratic function defined on some -simplex. This can be easily done with SQP or semi smooth Newton methods (cf. [2, 25, 30]). Since for typical applications, we can neglect the computational time for compared to that needed for the computation of a gradient of . We complete our algorithm with the precise scheme about the selection of in Step 3 of Algorithm 3.14 by some nesting procedure for the segment . New quantities determined in the subalgorithm are marked by . ###### Algorithm 3.17. Let , , , , , and be as in Step 3 of Algorithm 3.14. (Notice that both the null step condition (3.9) and condition (3.10) of sufficient descent are violated for .) • Set , , and (notice that ). • Choose some (w.r.t. ). • If satisfies (3.15) stop and return . • Otherwise choose and such that ∥∥y′′l+1−x′′l+1∥∥k = 12∥∥y′′l−x′′l∥∥kand f(y′′l+1)−f(x′′l+1) > −δ∥∥a′j∥∥k∥∥y′′l+1−x′′l+1∥∥k (3.18) where we take if (this way the condition of sufficient descent is violated on segment with ). • Increment by and go to Step 2. A slightly simplified survey about the complete algorithm is given in Figure 1. ###### Remark 3.19. While the implementation of the most steps in Algorithm 3.8 and its subalgorithms should be quite clear, let us briefly discuss how to choose some element . In our applications we usually have a representation of that allows the numerical computation of some element . More precisely, in many cases is continuously differentiable on an open set such that has zero Lebesgue measure. Here we can use Proposition 2.1.5 or Theorem 2.5.1 from Clarke [9] to get single elements . If is defined to be the pointwise maximum or minimum of smooth functions, Proposition 2.3.12 or Proposition 2.7.3 in [9] can be used to determine some . Moreover we can combine this with other calculus rules as e.g. the chain rule [9, Theorem 2.3.9]. Beyond these methods, that are sufficient for the benchmark problems considered in Section 4, also discrete approximations of elements of like e.g. in [4] can be used. Let us finally state that the presented algorithm assumes the possibility to compute at least one element of . Let us now justify the essential steps of the algorithm, i.e. that the required conditions can be reached and that the iterations typically terminate after finitely many steps. We start with Algorithm 3.17 and consider in particular Step 4. ###### Proposition 3.20 (properties of Algorithm 3.17). Let the assumptions of Algorithm 3.17 be satisfied. Then: • The choice in Step 4 of Algorithm 3.17 is possible for every . • The set Λl:={λ∈[0,1]∣∣there is someb′j∈∂f(λx′′l+(1−λ)y′′l)satisfying {\rm(???)}\,} has positive Lebesgue measure for every . • If Algorithm 3.17 does not terminate and, therefore, produces sequences and converging to some , then there is some satisfying (3.15) and is not strictly differentiable at . • If is convex on a neighborhood of , then Algorithm 3.17 terminates in Step 3 already for . Though it is not trusted that we find some satisfying (3.15) after finitely many steps, there is an extremely good chance according to Proposition 3.20 (2). In practice Algorithm 3.17 always terminated, also for rather complex simulations presented here. Nevertheless there are examples where the algorithm does not terminate (at least theoretically) as a simple induction argument shows, e.g., for with and . ###### Remark 3.21. Typically it is much cheaper for time consumption to compute merely the scalar in (3.15 ) instead of the complete vector . Therefore we compute only if (3.15) is satisfied. ###### Proposition 3.22 (properties of Algorithm 3.14). Let the assumptions of Algorithm 3.14 be satisfied, let be Lipschitz continuous on some neighborhood of , and suppose that Algorithm 3.17 always terminates. Then Algorithm 3.14 stops after finitely many steps and returns some satisfying (3.9) or (3.10). ###### Proposition 3.23 (properties of Algorithm 3.8). Let the assumptions of Algorithm 3.8 be satisfied and let be an iteration point from that algorithm. Then: • If is related to for some , then there exists satisfying the null step condition (3.9) or condition (3.10) of sufficient descent. • If , then there are only finitely many such that (3.9) is satisfied. Though Proposition 3.23 (1) already follows from Propositions 3.22 we will still give a brief independent proof of it in the next section. Summarizing we can say that, in principle, the presented algorithm always works and cannot get stuck, i.e. at most finitely many subiterations are necessary to find a new iteration point . The only point is that Algorithm 3.17 might not terminate which, however, is quite unlikely according to Proposition 3.20 (2) and which never happened in our simulations. Let us finally confirm that the presented descent algorithm can reach both minimizers and critical points of . ###### Theorem 3.24 (accumulation points are critical points). Let the assumptions of Algorithm 3.8 be satisfied and let be a corresponding sequence of iteration points. Then is strictly decreasing. Moreover, if is an accumulation point of , then and . As consequence we can formulate some more precise statement. ###### Proposition 3.25. Let the assumptions of Algorithm 3.8 be satisfied, let be a corresponding sequence of iteration points with step sizes , and suppose that is relatively compact. Then each accumulation point of is a critical point of and, if contains only finitely many critical points, converges to a critical point of . Moreover, if is not convergent, then has no isolated accumulation point. ###### Remark 3.26. If and is bounded, then is relatively compact. ### 3.2 Proofs Proof of Proposition 3.20. Since does not satisfy (3.9), we have . By construction . (1) Since (3.10) is not fulfilled, we have (3.18) for with . Assume that (3.18) holds for . Then f(x′′l)−f(x′′l+y′′l2)+f(x′′l+y′′l2)−f(x′′l)>−δ∥∥a′j∥∥k∥∥x′′l−y′′l∥∥k. (3.27) If neither and	2021-10-22 17:13:40	{"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": 0.9200581908226013, "perplexity": 851.7070896826827}, "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-2021-43/segments/1634323585516.51/warc/CC-MAIN-20211022145907-20211022175907-00157.warc.gz"}
http://pe-cn.github.io/199/	0% # Problem 199 Iterative Circle Packing Three circles of equal radius are placed inside a larger circle such that each pair of circles is tangent to one another and the inner circles do not overlap. There are four uncovered “gaps” which are to be filled iteratively with more tangent circles. At each iteration, a maximally sized circle is placed in each gap, which creates more gaps for the next iteration. After 3 iterations (pictured), there are 108 gaps and the fraction of the area which is not covered by circles is 0.06790342, rounded to eight decimal places. What fraction of the area is not covered by circles after 10 iterations? Give your answer rounded to eight decimal places using the format x.xxxxxxxx .	2020-10-23 05:54:10	{"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": 0.87286776304245, "perplexity": 568.058052930226}, "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-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00296.warc.gz"}
https://devzone.nordicsemi.com/questions/scope:all/sort:activity-desc/page:5/	# 22,463 questions 139 views 11 2 ## Nordic Tech Tour target audience? I'm an intern in my second year for a bachelor in computer engineering. I've been working with the nRF52832 development kit as part of my assignment. I was looking through Nordic's events when the tech tour caught ... (more) 47 views no 2 ## I've connected nRF52 board to custom nRF52 board with serial communication chip. How to setup serial communication now? What changes are to be made in example dtm code so that I set up a serial communication link. 25 views no no ## nrfprog returns wrong family error when programming nrf52840 on a custom board [closed] Hello, I get the "ERROR: The --family option given with the command (or the default from nrfjprog. ini) ERROR: does not match the device connected." message after entering pretty much any command, erase, program etc., for example, this one: nrfjprog ... (more) 36 views no 2 ## nRF52840 USB MSD Slow Write Speed Hi, I have recently acquire a nRF52840 preview development board to prototype a USB MSD. With the example provided in the SDK V13, I tested the write speed and it was far from usable. I am not sure if it ... (more) 12 views no no ## Store data in flash with NRF52832 and MBED and GCC Hi, I'm trying to use an existing NRF51822 mbed codebase on a new NRF52832 product and am having some issues. First of all, it seems that I can't use mbedlib, but need to update to mbed-os because BLE ... (more) 47 views no 1 ## Why can't I keep using S110/120 devices on nrf52840? It appears that I must use S134 and S140 etc. Is it backward compatible? By that I mean I keep using S110/120 at the cost of not being able to utilize some of the latest feature? 34 views no no ## what is result err code 3 in app_twi_callback_t I am getting result code 3 after app_twi_schedule in the call back. What does that mean? Thanks Paul 42 views no 2 ## After DFU Application at DFU_BANK_0_REGION_START is not valid [closed] Hi, I have a device with nRF52832 and firmware is composed of Bootloader (BL), softdevice (SD) S132 v2.0.0, application and I am using SDK 11.0. I have created different firmware updates (BL, SD, App, BL+SD, etc ... (more) 24 views no no ## ble_app_uart packet not sent more than 278 packets [closed] I am testing ble_app_uart. Softdevice is S130 and SDK is nRF5_SDK_12.2.0_f012efa. My development board is Waveshare's NRF51822 Eval Kit. When I test the ble_app_uart firmware, I cannot send more than 278 packets to ... (more) 19 views no no ## DFU Security. I do not know the BLE Event Hi; I do not know the BLE Event(DFU bootLoader/SDK Sample) so I will ask you a question. The environment using is below. SDK v10.0.0 (Sample code/dfu_transport_ble.c) SoftDevice S110&S120(v8.0.0?&v2.1 ... (more) 47 views no 1 ## Building sdk12 examples Hi So far I have been using the examples from Keil 5 pack installer. Now I tried loading some from the SDK example folder and I have issues with includes. What do I need to adjust to have those includes ... (more) 36 views 9 1 ## Unable to Send Messages NRF52 BLE Central Custom Device Currently I am using the NRF52 device to connect with a custom peripheral. The current goal I am trying to reach is toggling an LED by sending the command 0x8900 through Service UUID 0x0001 with Characterisic ID 0x0002. This is ... (more) 182 views 4 2 ## Restartig app_timer does not work with nRF52832 and SDK12.0.0 I am having an app_timer (nRF52832 / SDK12.0.0) that's interval must be changed during its lifetime. I am using following code to change the interval: app_timer_stop(m_timer_id); uint32_t interval = APP_TIMER_TICKS(interval_ms, APP_TIMER_PRESCALER); app_timer_start(m_timer_id, interval, NULL); So far ... (more) 38 views -1 no ## How to prgram nRF beacon kit as peripheral with KEIL and nRF DK52. I cannot see the beacon kit to be programmed. KEIL says DLL error How to prgram nRF beacon kit as peripheral with KEIL and nRF DK52. I cannot see the beacon kit to be programmed. KEIL says DLL error. Please let me know thw KEIL settings to see the beacon as a device 43 views no 1 ## Where can I get example for transfer data(file) (nRF52840) [closed] I have 2 board nrf52840 DK, I now I want to send file ( 1~2 Mbyte) from board_1 to board_2 without use UART. where can I get example to do that ? PS: I also need example with different kit or ... (more) 27 views no no ## nRF5 SDK 12.3.0 Eddystone on nRF51822 beacon Hi, I've seen that SDK 12.3.0 contains a new version of the application sample using the Eddystone protocol for BLE. I would like to know if this version is compatible with nRF51 chips as it was said ... (more) 25 views no no ## command line nrfutil dfu ble I am attempting to do a BLE firmware update via command line tools in an attempt to automate programming in a script file. I have zero problems performing an over the air dfu when using nRF Connect, but I can ... (more) 37 views 1 vote 1 ## nrf52 reboot and VDD overvoltage Hi, I am trying to figure out the reboot reason on an NRF52 in order to solve it in the best way. We have a design with a dual supply 3.6V/5V going through a 3V LDO. Problem is ... (more) 56 views 2 1 ## Interrupt Vectors REMAP How can I relocate or manage the transition between the bootloaders interrupt vector map and the applications interrupt vector map which will be at different addresses in FLASH. On the M3/M4, you can remap this but presumably on the ... (more) 43 views 2 1 ## Reduce connection establish time Hello, i have a BLE Peripheral that has a service that we use to simulate UART over BLE. Very similar to the nRF UART. I consider the time to establish a link that allows to send asynchronous data to be ... (more) 66 views 1 vote 1 ## Can I get two NFC 13.56MHz antennas from Nordic? I have four of the nrf52 DK's. I lost two of the NFC 13.56MHz antennas. Can I get/buy two from Nordic? Richard Ferraro 287 views no 3 ## AES CCM encryption/decryption examples Hello, Are there examples of AES CCM encryption/decription? Is there a recommended procedure for generating nounces? Are there recommended procedures for using CCM? Thank you. Update: I forgot to mention I am using a nrf52832. 30 views no 1 ## AirFuel SDK 2.2.0 IROM and IRAM settings I download "nRF5_AirFuel_SDK_2.2.0_7b89a57" and use "ptu_app_beacon" development, I am currently using "nrf51822-QFAA" and "nrf51822-QFAC", how do I set up IROM and IRAM? 155 views 5 1 ## nrf52840 802.15.4 start command Hi, I'm trying to use the command "mlme_start_req" to setup the device as a pan coordinator to send beacons to the others devices in the network, but after submit the command the callback is not called. For instance, if ... (more) 40 views 2 1 ## Connecting Jlink EDU-SWD mode to Nrf52832 wqewqeqwe.PNGI want to program the nrf52 chip using J link Edu -SWD mode. I actually connected the SWDCLK, SDI, and Vtref and GND of Jlink EDU programer to SWDCLK,SDI,3.3V& GND of nrf52 in my Self ... (more) 41 views no 1 ## CCM module resets the mcu I have written a simple implementation for using the ccm module with mesh. It encrypts and decrypts as it should but when enabled the processor makes a soft reset. It happens quiet often but not on every sent or received ... (more) 56 views no 1 ## Capacitance reading circuit for nRF51 Currently, we are trying to have the nRF51 chip working with a type of capacitive pressure sensor. We did quite a lot of work on searching proper circuits of reading the capacitance of the sensor, including capacitance voltage divider, charging ... (more) 102 views 3 no ## 32-bit overflow in nrf_drv_wdt_init() In nRF5 SDK 12.1.0 a bug was fixed in nrf_drv_wdt_init(). Watchdog timeout could not be larger than 131 seconds because of (x * 32768) multiplication before. This bug is again present in SDK 12.2.0, 12.3.0 ... (more) 16 views no 1 ## Send and receive interrupt lines from nBLE51822 to any MCU I have not gone into much details of 51822. while interfacing BLE to any MCU, Is it possible to add interrupt lines from BLE module to MCU to receive interrupts in MCU. Like, if we added one RX INTR line ... (more) 64 views no 1 ## Timer not going off for 7-50 seconds Hi, We're working on the connection params negotiation. Therefore we need a timer to go off 2 seconds after connecting to send a conn params change request. Most of the time this works as expected. We toggle some pins ... (more) #### Statistics • Total users: 20387 • Latest user: Debdasora • Resolved questions: 9454 • Total questions: 22463 ## Recent blog posts • ### The complete tutorial for developing and debugging nRF52 applications on a Mac Posted 2017-05-11 04:13:06 by Mohammad Afaneh • ### Bluetooth 5 2Mbps Demo with nRF52 Series and Samsung Galaxy S8 Posted 2017-05-09 14:20:22 by John Leonard • ### Setting up Segger on a mac or pc and debug - step by step for humans Posted 2017-05-05 14:03:35 by Lola Posted 2017-04-11 11:49:17 by Radosław Koppel • ### Taking a deeper dive into Bluetooth 5 Posted 2017-04-07 14:53:51 by John Leonard ## Recent questions • ### BLE GATT_INTERNAL_ERROR 129 Posted 2017-05-30 14:53:16 by Mrunali • ### Data storing in flash based on timer. Posted 2017-05-30 13:54:41 by Prashant Singh • ### Testing coded PHY with UART? Posted 2017-05-30 13:45:39 by slowhand	2017-05-30 13:04:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.23853719234466553, "perplexity": 10519.42681510567}, "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/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00003.warc.gz"}
https://crypto.stackexchange.com/questions/19242/for-calculating-the-index-of-coincidence-for-each-sequence	For calculating the index of coincidence for each sequence I was learning about the finding the key length reading the following web site... http://practicalcryptography.com/cryptanalysis/stochastic-searching/cryptanalysis-vigenere-cipher/ and I really don't understand some part of the example shown in the web. They assume the key length is 2 and extract the two sequence 1,3,5,7.... and 2,4,6,8... and from these two sequences they calculate the I.C. I wasn't sure how they calculate the I.C. for each sequence. Do they find the frequency count of each letter from a to z, and calculate the I.C using the following formula? As explained on the link you posted, the Vigenere cipher with a key on length $n$ encrypts every $n$-th symbol with the same key under the Caesar cipher. So to calculate the IC you should take all the $n$ sub-sequences separately: $\{1, 1+n, \dots, 1+kn, \dots\}$, $\{2, 2+n, \dots, 2+kn, \dots\}$ and so on and compute the IC for every sub-sequence.	2020-09-26 09:58:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7783312797546387, "perplexity": 566.4907483637951}, "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-2020-40/segments/1600400238038.76/warc/CC-MAIN-20200926071311-20200926101311-00058.warc.gz"}
https://nus.kattis.com/problems/tournament	OpenKattis National University of Singapore # Tournament Every September, the Kingdom of Loowater holds a jousting tournament. In each of a series of events, a pair of knights attempt to knock each other from their respective horses. The winning knight is paired with another, while the loser is eliminated. This process continues until all but one knight is eliminated; this knight is declared champion. The tournament schedule is organized so that no knight needs to compete in more than $e$ events to be champion, for the minimum possible $e$ given $k$, the number of knights. In order to construct the schedule, it may be necessary to identify several knights who compete in fewer than $e$ events; these knights are said to be awarded a bye and are excluded from the first round of competition. The first round of competition involves pairing as many knights as possible among those who are not awarded a bye. The competition is more interesting if the knights in each pair are as evenly matched in ability as possible. You are to determine which knights should be awarded a bye so as to make the first round as interesting as possible. ## Input Standard input consists of several test cases (at most $10$) followed by a line containing $0$. Each test case begins with an integer $2 \le k \le 2500$, the number of knights. $k$ lines follow, each giving the name and ability of a knight. The name is a nonempty string of lower case letters not longer than $20$; the ability is an integer between $0$ and $10\, 000$. No two names in a test case are equal. The mismatch between knights with abilities $a$ and $b$ respectively is defined to be $(a-b)^2$. ## Output For each test case, output the number of knights to be given a bye, followed by their names. The knights to be given a bye should be chosen such that the sum of all mismatch values for pairs of knights competing in the first round is minimized (assuming these are paired up in an optimal way). If there are several solutions, any will do. Sample Input 1 Sample Output 1 3 1	2021-06-19 22:03:23	{"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": 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": 0.5646876096725464, "perplexity": 683.7991590080765}, "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/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00282.warc.gz"}
http://www.ck12.org/book/Geometry---Second-Edition/r1/section/10.5/anchor-content	<meta http-equiv="refresh" content="1; url=/nojavascript/"> Areas of Circles and Sectors \| CK-12 Foundation You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version. 10.5: Areas of Circles and Sectors Created by: CK-12 Learning Objectives • Find the area of circles, sectors, and segments. Review Queue Find the area of the shaded region in the following figures. 1. Both figures are squares. 2. Each vertex of the rhombus is 1.5 in from midpoints of the sides of the rectangle. 3. The figure is an equilateral triangle. (find the altitude) 4. Find the area of an equilateral triangle with side $s$. Know What? Back to the pizza. In the previous section, we found the length of the crust for a 14 in pizza. However, crust typically takes up some area on a pizza. Leave your answers in terms of $\pi$ and reduced improper fractions. a) Find the area of the crust of a deep-dish 16 in pizza. A typical deep-dish pizza has 1 in of crust around the toppings. b) A thin crust pizza has $\frac{1}{2}$ - in of crust around the edge of the pizza. Find the area of a thin crust 16 in pizza. c) Which piece of pizza has more crust? A twelfth of the deep dish pizza or a fourth of the thin crust pizza? Area of a Circle Recall in the previous section we derived $\pi$ as the ratio between the circumference of a circle and its diameter. We are going to use the formula for circumference to derive the formula for area. First, take a circle and divide it up into several wedges, or sectors. Then, unfold the wedges so they are all on one line, with the points at the top. Notice that the height of the wedges is $r$, the radius, and the length is the circumference of the circle. Now, we need to take half of these wedges and flip them upside-down and place them in the other half so they all fit together. Now our circle looks like a parallelogram. The area of this parallelogram is $A=bh=\pi r \cdot r=\pi r^2$. To see an animation of this derivation, see http://www.rkm.com.au/ANIMATIONS/animation-Circle-Area-Derivation.html, by Russell Knightley. Area of a Circle: If $r$ is the radius of a circle, then $A=\pi r^2$. Example 1: Find the area of a circle with a diameter of 12 cm. Solution: If the diameter is 12 cm, then the radius is 6 cm. The area is $A=\pi (6^2)=36 \pi \ cm^2$. Example 2: If the area of a circle is $20 \pi$, what is the radius? Solution: Work backwards on this problem. Plug in the area and solve for the radius. $20 \pi &= \pi r^2\\20 &= r^2\\r &= \sqrt{20}=2 \sqrt{5}$ Just like the circumference, we will leave our answers in terms of $\pi$, unless otherwise specified. In Example 2, the radius could be $\pm 2 \sqrt{5}$, however the radius is always positive, so we do not need the negative answer. Example 3: A circle is inscribed in a square. Each side of the square is 10 cm long. What is the area of the circle? Solution: The diameter of the circle is the same as the length of a side of the square. Therefore, the radius is half the length of the side, or 5 cm. $A=\pi 5^2=25 \pi \ cm$ Example 4: Find the area of the shaded region. Solution: The area of the shaded region would be the area of the square minus the area of the circle. $A=10^2-25 \pi =100-25 \pi \approx 21.46 \ cm^2$ Area of a Sector Sector of a Circle: The area bounded by two radii and the arc between the endpoints of the radii. The area of a sector is a fractional part of the area of the circle, just like arc length is a fractional portion of the circumference. Area of a Sector: If $r$ is the radius and $\widehat{AB}$ is the arc bounding a sector, then $A= \frac{m\widehat{AB}}{360^\circ} \cdot \pi r^2$. Example 5: Find the area of the blue sector. Leave your answer in terms of $\pi$. Solution: In the picture, the central angle that corresponds with the sector is $60^\circ$. $60^\circ$ would be $\frac{1}{6}$ of $360^\circ$, so this sector is $\frac{1}{6}$ of the total area. $area \ of \ blue \ sector=\frac{1}{6} \cdot \pi 8^2=\frac{32}{3} \pi$ Another way to write the sector formula is $A=\frac{central \ angle}{360^\circ} \cdot \pi r^2$. Example 6: The area of a sector is $8 \pi$ and the radius of the circle is 12. What is the central angle? Solution: Plug in what you know to the sector area formula and then solve for the central angle, we will call it $x$. $8 \pi &= \frac{x}{360^\circ} \cdot \pi 12^2\\8 \pi &= \frac{x}{360^\circ} \cdot 144 \pi\\8 &= \frac{2x}{5^\circ}\\x &= 8 \cdot \frac{5^\circ}{2}=20^\circ$ Example 7: The area of a sector of circle is $50 \pi$ and its arc length is $5 \pi$. Find the radius of the circle. Solution: First plug in what you know to both the sector formula and the arc length formula. In both equations we will call the central angle, “$CA$.” $50 \pi &= \frac{CA}{360} \pi r^2 && \quad \ 5 \pi =\frac{CA}{360} 2 \pi r\\50 \cdot 360 &= CA \cdot r^2 && 5 \cdot 180=CA \cdot r\\18000 &= CA \cdot r^2 && \quad 900=CA \cdot r$ Now, we can use substitution to solve for either the central angle or the radius. Because the problem is asking for the radius we should solve the second equation for the central angle and substitute that into the first equation for the central angle. Then, we can solve for the radius. Solving the second equation for $CA$, we have: $CA=\frac{900}{r}$. Plug this into the first equation. $18000 &= \frac{900}{r} \cdot r^2\\18000 &= 900r\\r &= 20$ We could have also solved for the central angle in Example 7 once $r$ was found. The central angle is $\frac{900}{20}=45^\circ$. Segments of a Circle The last part of a circle that we can find the area of is called a segment, not to be confused with a line segment. Segment of a Circle: The area of a circle that is bounded by a chord and the arc with the same endpoints as the chord. Example 8: Find the area of the blue segment below. Solution: As you can see from the picture, the area of the segment is the area of the sector minus the area of the isosceles triangle made by the radii. If we split the isosceles triangle in half, we see that each half is a 30-60-90 triangle, where the radius is the hypotenuse. Therefore, the height of $\triangle ABC$ is 12 and the base would be $2 \left( 12 \sqrt{3} \right)=24 \sqrt{3}$. $A_{sector} &= \frac{120}{360} \pi \cdot 24^2 && A_{\triangle} =\frac{1}{2} \left( 24 \sqrt{3}\right)(12)\\&= 192 \pi && \quad \ = 144\sqrt{3}$ The area of the segment is $A=192 \pi - 144 \sqrt{3} \approx 353.8$. In the review questions, make sure you know how the answer is wanted. If the directions say “leave in terms of $\pi$ and simplest radical form,” your answer would be the first one above. If it says “give an approximation,” your answer would be the second. It is helpful to leave your answer in simplest radical form and in terms of $\pi$ because that is the most accurate answer. However, it is also nice to see what the approximation of the answer is, to see how many square units something is. Know What? Revisited The area of the crust for a deep-dish pizza is $8^2 \pi - 7^2 \pi =15 \pi$. The area of the crust of the thin crust pizza is $8^2 \pi - 7.5^2 \pi =\frac{31}{4} \pi$. One-twelfth of the deep dish pizza has $\frac{15}{12} \pi$ or $\frac{5}{4} \pi \ in^2$ of crust. One-fourth of the thin crust pizza has $\frac{31}{16} \pi \ in^2$. To compare the two measurements, it might be easier to put them both into decimals. $\frac{5}{4} \pi \approx 3.93 \ in^2$ and $\frac{31}{16} \pi \approx 6.09 \ in^2$. From this, we see that one-fourth of the thin-crust pizza has more crust than one-twelfth of the deep dish pizza. Review Questions Fill in the following table. Leave all answers in terms of $\pi$. 1. 2 2. $16 \pi$ 3. $10 \pi$ 4. $24 \pi$ 5. 9 6. $90 \pi$ 7. $35 \pi$ 8. $\frac{7}{\pi}$ 9. 60 10. 36 Find the area of the blue sector or segment in $\bigodot A$. Leave your answers in terms of $\pi$. You may use decimals or fractions in your answers, but do not round. Find the central angle of each blue sector. Round any decimal answers to the nearest tenth. 1. The quadrilateral is a square. 2. Carlos has 400 ft of fencing to completely enclose an area on his farm for an animal pen. He could make the area a square or a circle. If he uses the entire 400 ft of fencing, how much area is contained in the square and the circle? Which shape will yield the greatest area? 3. The area of a sector of a circle is $54 \pi$ and its arc length is $6\pi$. Find the radius of the circle. 4. The area of a sector of a circle is $2304 \pi$ and its arc length is $32 \pi$. Find the central angle of the sector. 1. $8^2 - 4^2 = 64 - 16 = 48$ 2. $6(10) - \frac{1}{2} (7)(3) = 60-10.5 = 49.5$ 3. $\frac{1}{2} (6) \left(3 \sqrt{3} \right) = 9 \sqrt{3}$ 4. $\frac{1}{2} (s) \left( \frac{1}{2} s \sqrt{3} \right) = \frac{1}{4} s^2 \sqrt{3}$ Feb 23, 2012 Dec 23, 2014 Reviews Image Detail Sizes: Medium \| Original CK.MAT.ENG.SE.2.Geometry.10.5	2015-01-29 21:50:55	{"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": 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": 68, "texerror": 0, "math_score": 0.8726798892021179, "perplexity": 265.50723657960464}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2015-06/segments/1422115868812.73/warc/CC-MAIN-20150124161108-00135-ip-10-180-212-252.ec2.internal.warc.gz"}
https://astarmathsandphysics.com/ib-maths-notes/complex-numbers/4954-cartesian-form-of-an-equation-from-the-complex-form.html	## Cartesian Form of an Equation From the Complex Form The equation $\| z - w \\| = \\|z-u \\|$ is the complex form of a line. To get the Cartesian form of the line let $z=x+iy, \: w=a+bi, \; u=c+di$ then $\| (x+iy)-(a+bi) \\| = \\|(x+iy)-(c+di) \\|$ $\| (x-a)+(y-b)i \\| = \\|(x-c)+(y-d)i \\|$ $(x-a)^2+(y-b)^2 = (x-c)^2+(y-d)^2$ $x^2-2ax+a^2+y^2-2by+b^2 = x^2-2cx+c^2+y^2-2yd+d^2$ $-2ax+a^2-2by+b^2 =-2cx+c^2-2yd+d^2$ $a^2+2yd-2by+b^2 =2ax-2cx+c^2-2yd+d^2$ $2yd-2by =2ax-2cx-2yd-a^2-b^2+c^2+d^2$ $2y(d-b) =2(a-c)x-a^2-b^2+c^2+d^2$ $y =\frac{a-c}{d-b}x+ \frac{-a^2-b^2+c^2+d^2}{2(d-b)}$	2018-05-28 08:01:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9126558303833008, "perplexity": 662.4380617115571}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00019.warc.gz"}
http://codeforces.com/blog/entry/4207	### PavelKunyavskiy's blog By PavelKunyavskiy, 10 years ago, translation, Hello! We are happy to invite you to participate in today's round. I hope everyone will find interesting problems for him. And most of the round participants will like it as well as the previous one. Today's contest was prepared for you by SPb SU 4 team (Alex-Gran (Alexandr Granovskiy), Dmitry_Egorov (Dmitry Egorov), PavelKunyavskiy (Pavel Kunyavskiy)). After long thoughts you can notice that we present St.Petersburg State University. Thanks very much for help in the tasks preparation to Artem Rakhov(RAD), Gerald Agapov (Gerald) and Maria Belova (Delinur) for translation statements. Also thanks to Petr Kalinin (KAP) for statements proofreading. In today's contest there will be seven problems (5 in each division) about the country where wizards live, and this causes many interesting events. Today you have to take part in local meeting, understand the intricacies of writing spells, ride on the magic transport, try to take away magic prizes, play in popular wizards game, help magic government in ruling the country and resolve financial dispute of two famous wizards. Scoring is standard is standard in both divisions today (500-1000-1500-2000-2500). UPD: analyse was published. Good luck! • +207 » 10 years ago, # \| +21 After a long time both there is a contest for both divisions. I really liked the previous one which had rated the problems dynamically. » 10 years ago, # \| -46 so,just fight! » 10 years ago, # \| -58 One more comment. » 10 years ago, # \| 0 This may be a silly question at this point but — how do I ask for clarifications? (if there is such thing) • » » 10 years ago, # ^ \| +2 Ask a question link on problems tab » 10 years ago, # \| -23 The problems were as if I was sitting in my English examination to read comprehensions... » 10 years ago, # \| -7 It should be 500 1500 1500 2000 2500 in Div 2 • » » 10 years ago, # ^ \| -11 maybe 500 1000 1000 2500 2500?)) • » » » 10 years ago, # ^ \| -7 the 4rd problem is too hard to read>_< i can not analyse the samples >_< • » » » » 10 years ago, # ^ \| ← Rev. 2 → +9 4th, not 4rd. • » » 10 years ago, # ^ \| ← Rev. 3 → -12 maybe 2500 2500 500 500 500? why not? //sarcasm » 10 years ago, # \| +29 My screencast will be here shortlyNot very eventful through as I spent most of my time with pen and paper ;) • » » 10 years ago, # ^ \| 0 Why can you test the sample testcases so fast? That's amazing! • » » » 10 years ago, # ^ \| +3 they are parsed and auto tested by CHelper • » » » 10 years ago, # ^ \| 0 For Java check thisThere are some tools to parse tests for other languages, cannot easily locate them through » 10 years ago, # \| ← Rev. 2 → 0 what will happen if try to access the index which is not allocated. By default what is stored in it ( here in C++ ) ? http://www.codeforces.com/contest/168/submission/1429535 » 10 years ago, # \| +25 When are the system test starting??? » 10 years ago, # \| -14 Problem B div1, is a simple dynamic programming, but the Hardest part of problem was undrestanding of that (at least for me). However with help of writers I got that finally. • » » 10 years ago, # ^ \| -17 at last. i also cant not understand the problem。。。。。。it is too ......... » 10 years ago, # \| -27 GREAT PROBLEMS » 10 years ago, # \| +10 Today I learned that in Java StringBuffer is much faster than String .In Div-2 ,Prob B My solution in which i concatenate two Strings give TLE while just replacing the String with StringBuffer gives accepted in this code,and that too in 330 ms. :) • » » 10 years ago, # ^ \| +1 and StingBuilder is even faster coz StringBuffer uses StringBulder. • » » 10 years ago, # ^ \| 0 It's necessary to know all language features if you write in it. Code s += t; , after compilation and decompilation, transforms into s = new StringBuilder(s).append(t).toString(); which works in O(length(s)+length(t)) » 10 years ago, # \| +31 I wonder why updating ratings takes so long. • » » 10 years ago, # ^ \| +27 I'm too. I'm waiting for RAD or Gerald. • » » 10 years ago, # ^ \| +8 update the ratings is longer than the contest it self • » » 10 years ago, # ^ \| +20 Honestly, I think the rating formula should be modified after seeing tourist got -1 rating for his 2nd place. • » » » 10 years ago, # ^ \| 0 However, his rating is much higher than anyone who participated the contest. • » » » » 10 years ago, # ^ \| ← Rev. 3 → 0 Despite that fact (and 5 of top 10 coders didn't compete), I suppose his rating would gain 5-10. It seems tough to participate in a win-or-minus-rating round, even if he is the best. Take a look at Topcoder, I find that there are about 600 coders with rating 1800 or higher, while that number in Codeforces is 1000. I know that it isn't necessary for Codeforces to be compared with Topcoder or something else, but above numbers probably show some "rating inflation". • » » » 10 years ago, # ^ \| 0 Moreover, i'll like to add one more thing......the rating limit separating Div 2 and Div 1(at Present it is 1700) Should be Decreased to 1600 or 1650 at least....bcoz a coder at 1600+ is capable enough....and for him solving Div2 A and B solving is not fun, as they really very simple and implementing them is really a time waste for him(approx 30-40 mins wasted for solving first 2 problems on average)......and hence he able to give comparatively less time for problem C and D.Instead it would be better if he can start from Div2 C which is Div 1 A.....i think this will avoid time waste for that user and he will evolve faster while competing in Div !.Other Suggestions are welcomed.. • » » » » 10 years ago, # ^ \| +11 > they really very simple> approx 30-40 mins wasted for solving first 2 problems on average » 10 years ago, # \| 0 It was a very interesting round with good problems......I think the author is very fond of mathematical problems(my rating goes up yahooooooooo).......however the problem statement of B(Div 1) was not very clear to me and I think for many others........I guessed that participant can take the tours in any order and coded accordingly and got accepted....still I m confused what it says......anyway hoping for a better problem description in future contests..... » 8 years ago, # \| ← Rev. 3 → 0 Please can anyone tell me why this happened ? I could remove TLE in Div1 A after I replaced cout.precision(10) ; for(int i = 1 ; i < n ; ++i){cout << ans << "\n" ; }withfor(int i = 1 ; i < n ; ++i){printf("%.6lf\n",ans) ; }Is printing after using cout.precision slow ? • » » 8 years ago, # ^ \| +1 it is well-known fact that printf is much faster than cout. • » » 8 years ago, # ^ \| +1 Also, try ios_base::sync_with_stdio(false), it makes cout go much faster than usual. • » » 8 years ago, # ^ \| +1 Moreover, you have different precision in cout and printf! Program with small output size is faster than the same program with large output size =) » 8 years ago, # \| 0 In case anybody is looking for the tutorial, it can be found here: http://codeforces.com/blog/entry/4214	2022-07-02 20:43:50	{"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": 0.28366026282310486, "perplexity": 3784.1538036859943}, "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/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00665.warc.gz"}
https://www.gamedev.net/forums/topic/13641-opengl-in-solaris/	#### Archived This topic is now archived and is closed to further replies. # openGL in solaris This topic is 6923 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Hi there, I had all my codes written in Windows, now I am trying to port it to Solaris. I think I have the GL libraries in the /usr/lib already, such as libGL.so, libGL.so.1, libGLU.so and libGLU.so.1@, but I''m afraid the solaris box doesn''t have the glut libraries and the headers. However, I obtained these files (libMesaGL.a, libMesaGL.so, libMesaGLU.a, libMesaGLU.so) and put them in the same directory as my other codes. I also put the glut.h, gl.h, etc in the same folder, and change the #include header to "glut.h". I don''t have root access so I can''t put them in the right place. I always get symbol reference errors on all the GL and glut functions. I wonder if I have the correct libraries in the right place. Does anybody know how to compile? thanks a lot ##### Share on other sites You don''t have the glut libraries from what I can see. I recommend either downloading the latest Mesa (which has glut 3.7) or download and build it directly from: http://reality.sgi.com/mjk/glut3/glut3.html A sample Makefile would then be: CXX = c++ INCLUDES = -I/usr/local/include -I/usr/openwin/include LDFLAGS = -L/usr/local/lib -L/usr/openwin/lib ######## Note the "-lglut" LDLIBS = -lMesaGLU -lMesaGL -lglut ######## LDLIBS += -lm -lX11 -lXext -lXmu -lXi -mt CPPFLAGS = -Wall $(INCLUDES) OBJECTS = lesson1.o .cpp.o :$(CXX) $(CPPFLAGS) -c$< all: lesson1 lesson1: lesson1.o $(CC) -o$@ $(LDFLAGS) lesson1.o$(LDLIBS) Other examples which work on Solaris can be found at: http://opengl.koolhost.com --Andy • 15 • 13 • 35 • 39	2019-04-26 07:56:17	{"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": 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": 0.3917522728443146, "perplexity": 9368.847072893135}, "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-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00499.warc.gz"}
https://www.beatthegmat.com/percent-ds-can-someone-show-me-where-my-logic-is-wrong-t249187.html	• 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Get 300+ Practice Questions Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to $200 Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0 Available with Beat the GMAT members only code Percent DS. can someone show me where my logic is wrong tagged by: hutch27 This topic has 1 expert reply and 0 member replies Percent DS. can someone show me where my logic is wrong In 1994, Company X recorded profits that were 10% greater than in 1993, and in 1993 the company’s profits were 20% greater than they were in 1992. What were the company’s profits in 1992? (1) In 1994, the company’s profits were $100,000 greater than in 1993. (2) For every$3.00 in profits earned in 1992, Company X earned $3.96 in 1994. OA is A but can someone please explain statement 2? Please look at my rephrasing of the target question because that's how i prefer to set up questions like this I did it like this.. .. .. .. 1992 1993 1994 p p(1+20/100) p(1+20/100)(1+10/100) GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 14738 messages Followed by: 1849 members Upvotes: 13060 GMAT Score: 790 hutch27 wrote: In 1994, Company X recorded profits that were 10% greater than in 1993, and in 1993 the company’s profits were 20% greater than they were in 1992. What were the company’s profits in 1992? (1) In 1994, the company’s profits were$100,000 greater than in 1993. (2) For every $3.00 in profits earned in 1992, Company X earned$3.96 in 1994. OA is A but can someone please explain statement 2? Please look at my rephrasing of the target question because that's how i prefer to set up questions like this I did it like this.. .. .. .. 1992 1993 1994 p p(1+20/100) p(1+20/100)(1+10/100) You're overcomplicating the problem. The question stem simply gives the RATIO of the three years. Let 1992 = 100. Since 1993 is 20% greater than 1992, 1993 = 100 + .2(100) = 120. Since 1994 is 10% greater than 1993, 1994 = 120 + .1(120) = 132. Thus: 1994 : 1993 : 1992 = 132 : 120 : 100. Given this ratio, if we know the profit in any of the 3 years -- or the DIFFERENCE between any two years -- we can determine the profit in EACH of the 3 years (and thus the profit in 1992). Statement 1: In 1994, the company’s profits were $100,000 greater than in 1993. SUFFICIENT. Statement 2: For every$3.00 in profits earned in 1992, Company X earned \$3.96 in 1994. In other words: 1994 : 1992 = 396 : 300 = 132 : 100. This statement only confirms the ratio given in the question stem. INSUFFICIENT. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Top First Responders* 1 Jay@ManhattanReview 81 first replies 2 Brent@GMATPrepNow 67 first replies 3 fskilnik 54 first replies 4 GMATGuruNY 37 first replies 5 Rich.C@EMPOWERgma... 13 first replies * Only counts replies to topics started in last 30 days See More Top Beat The GMAT Members Most Active Experts 1 fskilnik GMAT Teacher 201 posts 2 Brent@GMATPrepNow GMAT Prep Now Teacher 155 posts 3 Scott@TargetTestPrep Target Test Prep 105 posts 4 Jay@ManhattanReview Manhattan Review 97 posts 5 GMATGuruNY The Princeton Review Teacher 91 posts See More Top Beat The GMAT Experts	2018-09-24 16:15:59	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3077049255371094, "perplexity": 8388.933458951222}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160568.87/warc/CC-MAIN-20180924145620-20180924170020-00473.warc.gz"}
http://kyle.mindtobe.com/robot/calculate_my_pitch_angle.html	$$\definecolor{color1}{RGB}{0,0,255} \definecolor{color2}{RGB}{255,128,0} \definecolor{color3}{RGB}{102,0,102} \definecolor{delete}{RGB}{255,0,0} \definecolor{answer}{RGB}{0,150,0}$$ # My blog page : BaseketBot ## NXT Pitch Angle I made a mistake at time slot 19:45 where I squared distance $$d$$, that is $$tan\theta=\frac{v^2}{g\ d^\cancel{\textcolor{delete}{2}}} \pm \ \sqrt{...}$$ In order to calculate my pitch angle $$\theta$$, I need to use the projectile motion formula of: $$y - y_o = v_o \times t + \frac{1}{2} a t^2$$ $$y-y_o$$ is the difference between the target height and the tip of the cannon barrel. The tip of the barrel is $$\textcolor{answer}{\delta}$$ the height of the robot ($$h_r$$) plus the length $$l$$ of the barrel for the given angle theta; in another words, this is: $$l \ sin\theta + h_r$$. I already calculated my Lego Mindstorm cannon velocity $$v_o$$, which is constant. The projectile flight time is $$t$$, and the vertical acceleration is $$g$$ for gravity, which is $$9.81 \frac{m}{s^2}$$. The vertical velocity for the angle theta is $$v \ \sin \theta$$. With some substitutions we drive to the following equation. $$h_t - l \ \sin \theta - h_r = v \ \sin \theta \ \times \ \textcolor{color1}{t} - \frac{1}{2} g \textcolor{color1}{t}^2 \tag{1}$$ The above equation is for the vertical motion. We can now do the same for the horizontal motion. Again, we start with the same projectile equation: $$x - x_o = v_o \times t + \frac{1}{2} a t^2$$ $$x-x_o$$ is the horizontal distance that the projectile must travel. $$d_o$$ is the distance from the base of the cannon to the middle of the target. Because the angle of the barrel changes, we must account for the change in the horizontal distance, which is : $$l \ cos \theta$$. Therefore, the total horizontal distance travelled is $$d_o - l \ cos \theta$$. Also, the horizontal acceleration is zero because there isn't any force working with or against the ball.Lastly, the horizontal velocity is $$v\ cos\theta$$. We now have the horizontal projectile motion equation: $$d_o \ - \ l \ cos\theta = v\ cos\theta\ t + 0 \tag{2}$$ We now solve for time $$t$$ in our horizontal equation (2) and then and then plug it in to our vertical equation (1). $$\frac{d_o \ - \ l \ cos\theta}{v\ cos\theta} = \frac{v\ cos\theta \ t}{v\ cos\theta} \\ t = \frac{d_o \ - \ l \ cos\theta}{v\ cos\theta}$$ Now that I have solved for time $$t$$, I'll substitute time into the equation (1), which yields to the following: $$h_t - l \ \textcolor{delete}{\sin \theta}- h_r = v \ \sin \theta \ \bigg( \textcolor{color1}{ \frac {d \textcolor{delete}{- l \ cos \theta} } {v \ cos \theta}\bigg) } - \frac{1}{2} g \bigg( \textcolor{color1}{ \frac {d \textcolor{delete}{- l \ cos \theta}} {v \ cos \theta}\bigg) } ^2 \tag{3}$$ If I continue solving this problem, I'll get to a 4th degree polynomial equation which is very hard to solve; so, I can try and do some estimates. I'm shooting at above a 45$$^{\circ}$$ degree angle because I want the ball to go into the basket, not hit it. As I increase the angle, $$\textcolor{delete}{l \ sin \theta}$$ quickly approaches $$l$$ . Similarly, $$\textcolor{delete}{l \ cos \theta }$$ approaches zero as I get closer to 90$$^{\circ}$$ degrees. Therefore, I'll change those values to $$l$$ and zero, respectively. Later, I'll show the program pseudocode for a more accurate estimate. $$h_t - l- h_r \ = v \ \sin \theta \ \bigg(\frac {d} {v \ cos \theta}\bigg) - \frac{g}{2} \bigg(\frac {d} {v \ cos \theta}\bigg)^2 \tag{4}$$ Now I will simplify the above equation. $$h_t - l- h_r \ = \ \textcolor{delete}{\frac {d\ v \ \sin\theta} {v \ cos \theta}}\ - \frac {g\ d^2} {2\ v^2}\ \textcolor{color1}{\frac{1}{cos^2 \theta} } \tag{5}$$ We know the trigonometry identities $$\frac{ sin \theta }{cos \theta} = \tan \theta$$, $$\frac{1}{cos^2 \theta} = \sec^2 \theta$$, and $$\sec^2 \theta = ( 1 + tan^2 \theta )$$, which leads to the following equations: $$h_t - l- h_r \ = \ d\ \tan\theta\ - \frac {g\ d^2} {2\ v^2}\ \textcolor{color1}{ sec^2 \theta } \tag{6}$$ $$h_t - l - h_r = \ d\ \tan\theta\ - \frac {g\ d^2} {2\ v^2}\ \textcolor{color1}{(1+tan^2 \theta)} \tag{7}$$ $$h_t - l- h_r \ = \ d\ \tan\theta\ - \frac {g\ d^2} {2\ v^2}-\ \frac {g\ d^2} {2\ v^2}tan^2\theta \tag{8}$$ I'm going to reorder this equation so that I can use quadratic equation: $$\textcolor{color1}{ \frac {-g\ d^2} {2\ v^2} } \ tan^2\theta+ \ \textcolor{color2}{d} \ \tan\theta + \textcolor{color3}{ \bigg(\frac {-g\ d^2} {2\ v^2}-h_t + l+ h_r \bigg) } =0 \tag{9}$$ Solving for $$tan \theta$$ using quadratic equation we get: $$tan \theta = \frac {\textcolor{color2}{-d} \pm \sqrt{\textcolor{color2}{d^2}- 4 \textcolor{color1}{ \frac {-g\ d^2} {2v^2} } \textcolor{color3}{ \bigg( \frac {-g\ d^2} {2\ v^2} -h_t + l+ h_r \bigg) } } } {2 \textcolor{color1}{ \frac {-g\ d^2} {2v^2} } } \tag{10}$$ We can simplify the above equation with some algebra: $$tan \theta = \Bigg(-d \pm \sqrt{d^2+ \frac {2\ g\ d^2} {v^2} \bigg( \frac {-g\ d^2} {2\ v^2} -h_t + l+ h_r \bigg) }\ \ \Bigg)\Big( \frac {v^2} {-g\ d^2} \Big) \tag{11}$$ The next part of the equation being simplified is in blue. $$tan \theta = \frac {v^2} {g\ d} \pm \sqrt{\frac {v^4} {g^2\ d^2}+ \Bigg( \textcolor{color1}{ \Big(\frac {v^2} {-g\ d^2} \Big)^2 \times \frac {2\ g\ d^2} {v^2} \bigg( \frac {-g\ d^2} {2\ v^2} -h_t + l+ h_r \bigg) } }\ \ \Bigg) \tag{12}$$ $$tan \theta = \frac {v^2} {g\ d} \pm \sqrt{\frac {v^4} {g^2\ d^2}+\Bigg( \textcolor{color1}{ \frac {2\ v^2} {g\ d^2} \bigg( \frac {-g\ d^2} {2\ v^2} -h_t + l+ h_r \bigg) } }\ \ \Bigg) \tag{13}$$ $$tan \theta = \frac {v^2} {g\ d} \pm \sqrt{ \frac {v^4} {g^2\ d^2} + \textcolor{color1}{ \frac {2\ v^2} {g\ d^2} \bigg( -h_t + l+ h_r\bigg) -1 } }\ \ \tag{14}$$ $$tan \theta = \frac {v^2} {g\ d} \pm \sqrt{ \frac {v^2} {g\ d^2} \bigg( \frac{v^2}{g} - 2 \ h_t + 2 \ l+ h_r \bigg) -1 }\ \ \tag{15}$$ Now that this problem is simplified, I must take $$arctan$$ of both sides and I have my answer in simplest form. $$\enclose{box}{ \textcolor{answer} { \theta = arctan \Bigg( \frac {v^2} {g\ d} \pm \sqrt{ \frac {v^2} {g\ d^2} \bigg( \frac{v^2}{g} + 2 \ h_t - 2 \ l+ h_r \bigg) -1 }\ \ \Bigg) } } \tag{16}$$ This is the calculation for the vertical angle theta. Recall that equation $$(3)$$ led to fourth degree polynomial and I used some estimates to get to equation $$(4)$$. We now use algorithm to get closer to our angle. After the first calculation of theta, I can determine the new vertical offset of $$h_t - l \ \sin \theta - h_r$$ and the horizontal offset $$d_o - l \ cos \theta$$ and then recalculate theta. We repeat this process until the change between each iteration of theta yields to a small difference. In below pseudocode we look for a 0.5 degree difference. theta = 90 do { h = L * sin( theta ) // ReCalc height d = initial_d - L * cos( theta ) // ReCalc distance lastTheta = theta theta = ReCalc( d, h ) } while ( AbsoluteValue(theta - lastTheta) >0.5 )	2019-05-19 14:55:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "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": 0.8912144303321838, "perplexity": 367.4561350142299}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00340.warc.gz"}
https://www.semanticscholar.org/paper/Mid-Range-Wireless-Power-Transfer-at-100-MHz-Using-Roberts-Clements/14a3d5a02ccde9c1a15e8404209b1cc13ea015e8	Mid-Range Wireless Power Transfer at 100 MHz Using Magnetically Coupled Loop-Gap Resonators @article{Roberts2021MidRangeWP, title={Mid-Range Wireless Power Transfer at 100 MHz Using Magnetically Coupled Loop-Gap Resonators}, author={David M. Roberts and Aaron P. Clements and Rowan McDonald and Jake S. Bobowski and Thomas Johnson}, journal={IEEE Transactions on Microwave Theory and Techniques}, year={2021}, volume={69}, pages={3510-3527} } • Published 27 March 2021 • Physics • IEEE Transactions on Microwave Theory and Techniques We describe efficient four-coil inductive power transfer (IPT) systems that operate at 100 MHz. The magnetically coupled transmitter and receiver were made from electrically small and high- ${Q}$ loop-gap resonators (LGRs). In contrast to the commonly used helical and spiral resonators, the LGR design has the distinct advantage that electric fields are strongly confined to the capacitive gap of the resonator. With negligible fringing electric fields in the surrounding space, the IPT system is… 2 Citations Figures and Tables from this paper Inductive Power Transfer Through Saltwater • Physics 2021 13th International Conference on Electromagnetic Wave Interaction with Water and Moist Substances (ISEMA) • 2021 We investigated inductive power transfer (IPT) through a rectangular slab of saltwater. Our inductively-coupled transmitters and receivers were made from loop-gap resonators (LGRs) having resonant A Low-Profile Quasi-Loop Magneto-Electric Dipole Antenna Featuring a Wide Bandwidth and Circular Polarization for 5G mmWave Device-to-Device Communication • Physics Journal of Electromagnetic Engineering and Science • 2022 The deployment of the millimeter (mmWave) frequency spectrum by fifth-generation (5G) device-to-device (D2D) wireless networks is anticipated to meet the growing demands for increased capacity. The References SHOWING 1-10 OF 50 REFERENCES Using split-ring resonators to measure the electromagnetic properties of materials : An experiment for senior physics undergraduates A spilt-ring resonator experiment suitable for senior physics undergraduates is described and demonstrated in detail. The apparatus consists of a conducting hollow cylinder with a narrow slit along Wireless Power Transfer via Strongly Coupled Magnetic Resonances • Physics Science • 2007 A quantitative model is presented describing the power transfer of self-resonant coils in a strongly coupled regime, which matches the experimental results to within 5%. Inductive Power Transfer Through Saltwater • Physics 2021 13th International Conference on Electromagnetic Wave Interaction with Water and Moist Substances (ISEMA) • 2021 We investigated inductive power transfer (IPT) through a rectangular slab of saltwater. Our inductively-coupled transmitters and receivers were made from loop-gap resonators (LGRs) having resonant Optimization of “I” Type Shielding for Low Air-Gap Magnetic and Electric Fields Inductive Wireless Power Transfer • Engineering 2020 IEEE Transportation Electrification Conference & Expo (ITEC) • 2020 In an inductive wireless power transfer (IWPT) system, an “I” type magnetic shielding structure was proposed to simultaneously reduce the air-gap center region magnetic and electric fields without The Complex Permeability of Split-Ring Resonator Arrays Measured at Microwave Frequencies • Physics IEEE Transactions on Microwave Theory and Techniques • 2020 We have measured the relative permeability of split-ring resonator (SRR) arrays used in metamaterials designed to have <inline-formula> <tex-math notation="LaTeX">${\mu ^\prime < 0}$ Guidelines for Limiting Exposure to Electromagnetic Fields (100 kHz to 300 GHz). • Physics Health physics • 2020 ICNIRP has updated the radiofrequency EMF part of the 1998 Guidelines, which provide protection for humans from exposure to EMFs from 100 kHz to 300 GHz. Wireless power transfer based on dielectric resonators with colossal permittivity • Physics • 2016 Magnetic resonant wireless power transfer system based on dielectric disk resonators made of colossal permittivity (e = 1000) and low loss (tan δ = 2.5 × 10–4) microwave ceramic is experimentally Split-ring resonator for use in magnetic resonance from 200-2000 MHz • Physics • 1981 A novel high Q resonator is described that is suitable for magnetic resonance in the frequency region 200–2000 MHz, where ordinary cavities are generally too bulky and solenoidal coils impractical. Q‐band loop‐gap resonator • Physics • 1986 The upper frequency for loop‐gap resonators intended for use in electron‐spin‐resonance spectroscopy has been extended to Q band (35 GHz). A practical structure is described containing sample Split-ring resonators for measuring microwave surface resistance of oxide superconductors • Physics • 1991 A cavity perturbation technique using split‐ring resonators has been developed for measuring the surface resistance of metals in the frequency range 0.3–5.0 GHz. The system is designed particularly	2022-10-05 05:57:48	{"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": 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": 0.27287498116493225, "perplexity": 7715.151916506991}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337537.25/warc/CC-MAIN-20221005042446-20221005072446-00024.warc.gz"}
https://textbooks.cs.ksu.edu/cis526/a-html/03-html-document-structure/	# HTML Document Structure When authoring an HTML page, HTML elements should be organized into an HTML Document. This format is defined in the HTML standard. HTML that does not follow this format are technically invalid, and may not be interpreted and rendered correctly by all browsers. Accordingly, it is important to follow the standard. The basic structure of a valid HTML5 document is: <!doctype HTML> <html lang="en"> <title>Page Title Goes Here</title> <body> <p>Page body and tags go here...</p> </body> </html> We’ll walk through each section of the page in detail. ## Doctype The SGML standard that HTML is based on requires a !doctype tag to appear as the first tag on the page. The doctype indicates what kind of document the file represents. For HTML5, the doctype is simply HTML. Note the doctype is not an element - it has no closing tag and is not self-closing. ## HTML Element The next element should be an <html> element. It should include all other elements in the document, and its closing tag should be the last tag on the page. It is best practice to include a lang attribute to indicate what language is used in the document - here we used "en" for English. The <html> element should only contain two children - a <head> and <body> tag in that order. The next element is the <head> element. A valid HTML document will only have one head element, and it will always be the first child of the <html> element. The head section contains metadata about the document - information about the document that is not rendered in the document itself. This typically consists of meta and link elements, as well as a <title>. Traditionally, <script> elements would also appear here, though current best practice places them as the last children of the <body> tag. The <head> element should always have exactly one child <title> element, which contains the title of the page (as text; the <title> element should never contain other HTML elements). This title is typically displayed in the browser tab. The next element is the <body> element. A valid HTML document will only have one body element, and it will always be the second child of the <html> element. The <body> tag contains all the HTML elements that make up the page. It can be empty, though that makes for a very boring page.	2022-06-26 11:32:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.2413948029279709, "perplexity": 2030.3746245144555}, "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/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00010.warc.gz"}
http://tex.stackexchange.com/questions/122840/how-to-draw-arbitrary-closed-surface-with-tikz?answertab=oldest	# How to draw arbitrary closed surface with TikZ? I want to draw an arbitrary closed surface like following figure. How can i do it? Here is the MWE: \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture} \shade[line width=2pt, color=blue] (0, 0) .. controls(1,2) .. (3, 0) ; \end{tikzpicture} \end{document} In my code, color option doesn't work and I can't close the \controls. - Welcome at Tex.sx! Your request is kind of vague. Are you interested in surface plots defined by data matrizes and/or functional expressions in general? Then you may want to browse the questions under the "Related" section of your question. Or are you interested in schematic figures were the only requirement is that it should look "like a cloud", possibly enriched with annotations? Your figure appears to belong to a schematic explanation (and not a precily defined surface plot). – Christian Feuersänger Jul 6 '13 at 22:28 I want to schematic figures, like a cloud. Arbitrary surface, not defined surface plots. I know, it can be done with using "controls" but i can not closed the lines or bends. Thank you for your interesting. – ferahfeza Jul 6 '13 at 22:35 Welcome to TeX.SX! Generally, we can't do much with 'just a picture' -- this figure was quite apparently TeX'd; do you have the source to this figure? It will help us generalize it into what you need. Would you add a minimal working example (MWE) (from \documentclass to \end{document}? :) – Sean Allred Jul 7 '13 at 1:51 I have updated my question. – ferahfeza Jul 7 '13 at 10:15 For arbitrary closed shapes you might find useful the [hobby](www.ctan.org/pkg/hobby) package. You find an couple of examples here and here. – Claudio Fiandrino Jul 8 '13 at 14:57 ## 1 Answer When you try to shade or fill a path, TikZ closes it indeed, but the last segment which closes it is a straight line. If you want a "rounded" closed path, you have to make the path end at the same point it started, and control the curvature so that it is "smooth" at that point. Doing so with control points is difficult. I would use the hobby package, which provides metapost-like syntax to specify smooth curved paths (both open and closed), and that it is highly recommendable if you want control over the final shape. If you don't need too much control, but instead any "rounded shape" would be enough, you can use the construct to [in=alpha, out=beta] to specify the angles alpha and beta at which the curved path leaves its start and enters its end, respectively. If you use the same construct to conect each pair of points, and ensure that the curve leaves the point at the same angle (+180) that it entered it, you'll get smooth curve also at those points. For example: \begin{tikzpicture} \shade[line width=2pt, top color=blue] (0, 0) to [out=20, in=70] (3,0) to [out=250, in=200] (0,0) ; \fill[red] (0,0) circle(2pt) (3,0) circle(2pt); % Show points \end{tikzpicture} - @ JLDiaz, Thank you very muck. This will be my starting point. – ferahfeza Jul 7 '13 at 11:54	2014-09-22 18:25:25	{"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": 0.8914291262626648, "perplexity": 1772.1659442725806}, "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-2014-41/segments/1410657137145.1/warc/CC-MAIN-20140914011217-00070-ip-10-234-18-248.ec2.internal.warc.gz"}
https://github.com/thorchain/Resources/blob/master/Whitepapers/THORChain/whitepaper-en.md	Skip to content 38130d9 Nov 8, 2018 5 contributors ### Users who have contributed to this file 1057 lines (732 sloc) 90.7 KB # THORChain ## A lightning fast decentralised exchange protocol devs@thorchain.org V0.1 July 2018 ### Abstract THORChain is a highly optimised multi-chain using pBFT consensus to achieve sub-second block finality. Tokens are traded on single chains, known as tokenChains with discrete address spaces. Multi-set sharding is proposed to allow byzantine resistant scaling. The native protocol facilitates on-chain trading and order matching at the protocol level, supporting both limit and market orders. Continuous liquidity pools ensure liquidity is always available for any token pair, and double as the source of trustless on-chain price feeds, a cornerstone of the protocol. Fee incentives are designed to continually attract on-chain liquidity. On-chain token generation is built-in for both fixed and variable supply tokens, with the potential for auditable collateralized stablecoins. Two-way pegs with UTXO, account and contract-based cryptocurrencies allow most existing cryptocurrencies to seamlessly move on and off THORChain. Transparent developer incentivisation strategies allow wallet and exchange developers to be funded at the protocol level which will encourage a vibrant developer community. On-chain governance and smart updates with enforced voting and quadratic voting ensures THORChain is an adaptive and iterative platform that grows with its needs without contentious hard forks. THORChain is built to be compatible with the Flash Network; a layer 2 payment channel network. ### Document Set The following whitepapers should be read in conjunction: ## Overview Introduction • Centralized and Decentralized Exchanges • THORChain and ASGARDEX THORChain Architecture • Overview • TokenChains • MerkleChain • Consensus • Validator Set • Multiple Validator Sets On-chain Governance • Overview • Key Aspects Tokens • Rune Characteristics • Block Rewards and Emission • Fees • TokenChain • Rune Generation • THORChain Bridges Continuous liquidity pools • Solving Liquidity • CLPs • CLP Transactions • Liquidity Fees • Trustless On-chain Price Feeds On-chain Trading • Account Types • Transaction Types • Limit and Market Orders User Experience • Wallets • Trading • Exchanges • Exchange and Wallet Funding • Cross-token Trading • On-chain Initial Exchange Offerings • Validator Staking Advanced Transaction Features • THORChain Name Service (TNS) • Account Recovery • Multi-sig Accounts • Transaction Receipts • Transaction Requests • Recurring Transactions • Escrow Accounts • Automatic Disbursements • Combined Payment Features Other Features • Collectibles • Messaging Protocol • Smart contracts • Additional On-Chain Commands • StableCoins • Token Baskets & Indexes • Anonymity • FIX 4.4 Protocol Conclusion References ## Definitions Item Definition ASGARDEX A Decentralised Exchange built on THORChain Bifröst Protocol Bridges built to other networks (two-way-pegs). CLP Continuous Liquidity Pool DEX Decentralised Exchange ECDSA Elliptical Curve Digital Signature Algorithm ERC-20, NEP-5, QRC-20 Ethereum, NEO, QTUM tokens Flash Network Layer 2 Payment Channel Network FIX4.4 Protocol Foreign Information Exchange 4.4 Protocol for Trading Desks HFT High Frequency Trading Mjölnir Liquidity Nodes. RNG Random Number Generation RUNE THORChain Native token, Rune token, Rune, “RUNE” pBFT Practical Byzantine Fault Tolerance Runes/Tokens Tokens on THORChain TokenChain/RuneChain Chain for each Token TPS Transactions Per Second UTXO Unspent Transaction Output VRF Verifiably Random Function ## Introduction ### Centralized and Decentralized Exchanges Exchanges. One of the most important advents of the cryptocurrency economy is the crypto exchange. While digital assets are useful on their own, exchanges are needed so that creators, consumers, investors, and traders can transfer these assets to one other openly and freely. As digital assets reach market equilibrium prices, all market participants benefit from clearing markets that are otherwise either unavailable or not at true equilibrium due to interference from governments, regulators, or logistical barriers. Venture capitalists, service providers, employees, founders, and other company stakeholders benefit from freely available liquidity that is not locked up in arcane or encumbering pre-IPO restrictions. This is true for tokens regardless of whether they are asset-backed, service-backed, or infrastructure-backed. In the asset-backed case, cryptocurrency exchanges allow for easier transactions of securities that are currently more costly in terms of time and money due to reconciliation requirements that are natively on the blockchain and otherwise handled by back-office workers. Cryptocurrency exchanges allow people increased flexibility in using the cryptocurrency of their choice for each individual transaction they undertake. They can focus on a few key markets that capture their consumption needs. For example, a user who considers data storage a primary consumption need can store much of their wealth in Filecoin or Sia while another who relies heavily on shipping goods can store wealth in Shipchain or VeChain. As specialized cryptocurrencies gain traction, users can segment their wealth in specific token ecosystems. They can reward these ecosystems with more of their wealth for positive governance and pro-social activity, forcing all token ecosystems to compete with each other to provide the best economy for users. This competition is only feasible with the proliferation of exchanges that enable fast, secure, reliable transfers of wealth from one token to another. Finally, exchanges allow traders to profit from predicting the future behavior of markets. As global wealth becomes increasingly concentrated in the hands of the few, so too go the resources needed to predict the future of markets and thereby gain additional wealth. Because cryptocurrency markets are new and lightly regulated, large incumbents have less of an overwhelming advantage over lay-traders in forming accurate, profitable hypotheses. Skilled traders can participate more directly in the rewards of insightful analysis without tethering themselves to major institutions, which can disperse market gains. Decentralized Exchanges. Of course, the above is only true to the extent that other sources of power are also decentralized. One clear target for decentralization is the exchange itself. Centralized exchanges can hold undue power over users. This allow them to engage in many of the regressive practices that cryptocurrency enthusiasts dislike in the fiat world. Some hold users’ funds without explanation or recourse. A decentralized alternative will allow users to maintain self-sovereign rights to their assets at all times. Some centralized exchanges bar users, currencies, or transactions because of local law in their domicile. A decentralized alternative has no central operator upon which a nation state can impose restrictions. A decentralized alternative with public code can benefit from perpetual public security audits and publicly driven code changes. All things being equal, decentralized exchanges (DEXes) more adequately capture the ethos of cryptocurrencies than centralized ones. All things are not yet equal, however. Decentralized exchanges need feature and performance parity with centralized exchanges to successfully compete. Binance, one of the most popular centralized exchanges, processes over 1 million transactions per second. Forkdelta, a popular decentralized exchange, routinely has transactions that take minutes. Bittrex, a popular centralized exchange, has a well-designed UI honed by design specialists and a simple-to-use UX. Barterdex, a DEX, has an obtuse UI and requires several cumbersome steps in its UX. Centralized cryptocurrency exchanges pride themselves on listing a myriad of different token types. Most modern DEXes operate within a single cryptocurrency’s ecosystem (for example, stellar for Stellardex and ether for Etherdelta) allowing users to trade only tokens in one protocol. To avoid some of these constraints, some DEXes take partial steps to toward decentralization but centralize key aspects of the system (like order books and matching engines). This is a nice attempt but these services do not provide censorship resistance and so do not fulfill the promise of decentralization. The market needs a decentralized exchange that adheres to the core tenets of decentralization and still provides a world-class exchange experience that meets the feature and performance standards set by centralized exchanges. THORChain. Thorchain addresses these issues. With an out-of-box performance of 10k TPS and a goal of 1m TPS, Thorchain can power the same traffic as a centralized exchange. Thorchain decentralizes all aspects of a trade, including order matching and order book population, ensuring censorship resistance. Most DEXes have less than $10M in daily trading volume but Thorchain will be seeded with several times that amount in liquidity directly from the Odin Foundation. Moreover, Thorchain’s continuous liquidity pool mechanism (borrowed from Bancor) provides fees to users who provide further liquidity, incentivizing even more capacity. Thorchain’s use of cryptocurrency bridges enable it to natively support transactions across cryptocurrency ecosystems. Users can exchange Bitcoin for Ether or an ERC-20 token for an NEP-5 token directly on a Thorchain DEX. Bridges are addressed in more detail separately. ASGARDEX (the first DEX that will be built on top of THORChain) will have deep liquidity, a publicly available order book, omni-token support for cryptocurrencies compatible with bridges, a world-class UI, and refined UX. ASGARDEX solves the major problems with decentralized exchanges, allowing users to experience the full promise of a secure, transparent, censorship-resistant and fault-tolerant cryptocurrency exchange. ## THORChain Architecture ### Overview THORChain is a multi-blockchain protocol comprised of side chains “TokenChains” and the master “MerkleChain”. The MerkleChain tracks the entire state of the network, storing the latest hash of the Merkle Tree to prevent double-spending and syncing the state of network. All side chains in the protocol are “first-class citizens” with no delineation in security, technology or preference in the Validator Set. TokenChains have a genesis account that describe the characteristics of the token, and solely track transactions for that token. Each TokenChain maintains a discrete address space allowing deconfliction of transaction mempools. TokenChains are account-based in a similar manner to Ethereum (as opposed to UTXO-based where unspent transactions are tracked such as Bitcoin). A nonce tracks the latest state of each account, invalidating all previous account states. The native token of the ecosystem is Rune and is created at the genesis of the protocol, with all Rune transactions tracked on the first side chain, T0. The Rune is unique as it is the settlement currency of the ecosystem and is stored in all on-chain liquidity pools. The genesis block of every sidechain has a continuous liquidity pool (CLP) that defines a price ratio for Rune and that token. CLPs are explained in more detail later in this paper. For now, consider CLPs as providing a counterparty to anyone wishing to make a Rune-denominated trade. If two users trade between two non-Rune tokens, one of the tokens is automatically converted to Rune before the transaction is complete. Figure 1: Each token is tracked on a discrete TokenChain. Genesis accounts track Token information and hold on-chain liquidity. ### TokenChains Each TokenChain has a discrete address space. Figure 2: The THORChain Address Space; here for Rune (T0). The prefix deconflicts the address space with all other protocols. The TokenIndex deconflicts the token from other on-chain tokens, where Rune is T0, the first on-chain token is T1 etc. The Separator separates the TokenIndex (which may be variable length) and the public address, derived from a SHA256 hash of a private key. Alice, who has a public address for her Rune, T0xAlice, will also have a matched address on every other TokenChain that she has tokens for. Her single private key unlocks all addresses. If Alice is sent a new token; her new address is automatically created. T0xAlice -> Rune address for Alice T1xAlice -> T1 address for Alice T5xAlice -> T5 address for Alice TnxAlice -> Tn address for Alice Deconflicting address spaces in multiple chains have the following advantages: • Discrete transactions for each token, with discrete mem-pools. • Each token has a clear and distinct lineage. • Each TokenChain inherits a genesis account, which contains and performs cornerstone functionality for the network. Tokenchains can represent tokens internal to Thorchain or external tokens (like Bitcoin or Ethereum). These tokens can be traded within Thorchain but represent actual external tokens stored within a Thorchain bridge. The UX for trading BTC on ASGARDEX involves user selecting a bridge, transferring the BTC to a wallet controlled by that bridge, and specifying in the data field for that BTC transaction which Thorchain wallet they want the corresponding Thorchain proxy bitcoin tBTC stored in. Bitcoin sent to a bridge without a corresponding tBTC wallet address should be sent back automatically. Once this happens, on-chain trading can occur. To withdraw BTC, users specify a bridge and a BTC wallet and then initiate a tBTC transfer from their wallet to the bridge wallet. The bridge then moves the BTC to the corresponding BTC wallet. Likewise, tBTC sent without the right data fields and formatting will be returned. ### MerkleChain The MerkleChain tracks and syncs the network state by hashing and merkelizing the latest transactions for each TokenChain in the network. The following is the overview of this mechanism: The Validator Set collects Merkle Roots from each tokenchain, assembled into a Merkle Tree, and hashed as a single Merkle Root for block n for each TokenChain. The Token Merkle Roots are then collected into another Merkle Tree and a final Master Merkle Root produced. This is then inserted into the MerkleChain block n + 1 by the Validator Set. The Validator Set at the same time then inserts the same Master Merkle Root into every tokenchain at block n + 1. The MerkleChain also stores network wide information such as the TokenIndex counter that specifies and deconflicts TokenChains on the network. ### Consensus THORChain implements Tendermint, a byzantine fault tolerant state machine replication algorithm. Tendermint has the necessary maturity, performance and security to serve THORChain’s requirements. Tendermint is most useful in that it supports instant finality, which is an indispensible property for a blockchain. Tendermint achieves this by only committing transactions that have already reached super-majority consensus, rather than waiting for consensus to be achieved after a transaction is committed (such as Bitcoin and Ethereum V1). As such, Tendermint can support sub-second block times and process up to 10,000 TPS. Tendermint performance will be sufficient for THORChain initially, but modifications will be required to support the end state of achieving performance parity with centralised exchanges at over 1 million TPS. To achieve this performance, THORChain will need to research and implement various implementations of sharding. Validator Set Tendermint requires full-nodes as Validators or block producers, and each Validator must have a weight on the network. The weight is determined by their staking, so Tendermint can support Proof-of-Stake out of the box. The first implementation of THORChain has a single Validator Set drawn from all available Validators through an auction; the 100 highest staked Validators form the Validator Set. Validators propose blocks, agree and commit them. To join the Validator Set, a Validator must stake higher than the lowest Validator, and by this process ensures that the Validators with the greatest economic investment secure the network. Staking pools are possible to allow wider participation, and anyone can delegate their stake to a chosen Validator. Validators are paid from the block reward, and are paid evenly regardless of stake held. Proxy stakers are paid pro-rata to the total stake of a Validator; and so by design and self-interest; should choose the Validator with the lowest current stake, but in the top 100. This encourages proxy stakers to distribute their stake where they have the most earning potential, and thus consequently will ensure a very flat and competitive group of 100, where the minority can easily influence Validator participation in the Validator Set. Figure 3: Validator Set An in-protocol penalty is required to discourage validators from misbehaving on the network. This is heavily researched in both Casper and Cosmos. Misbehaviour may be double-signing blocks, being off-line or lack of participation in on-chain voting. The amount that a validator is slashed depends on the severity of the incident and will likely be finalised with on-chain voting prior to mainnet. If a Validator is slashed to a level below the highest waiting Validator, the dishonest Validator is evicted and the waiting Validator automatically joins the Validator Set. As a result of the flat group, the heavy level of competition and the likelihood that dishonest participation will cause both economic loss to stakes and eviction from the Validator Set, an attacking Validator is heavily discouraged. Network security is a consideration for THORChain. Historically on-chain voting has extremely low turn-out, as shown by EOS, CarbonVote and Steemit voting events. Cosmos is attempting to solve this by increasing the rate of inflation of the staked token to increase the incentives to bond tokens to validator stakes (directly or via pools). Inflation pivots between a floor of 7% and a ceiling of 20% to drive holders to bond tokens at a desired bond state of 2/3 of the entire circulating supply. With ⅔ bonded, the network is optimally secure and resistant to cartels and plutocracy. A downside to this that the staked token loses ideal currency properties, and as such Cosmos have introduced a dual-token system, with a secondary token being used to pay for transaction fees. It is not known if Casper V2 will implement such a mechanism and if staking for collators will cause Ether to lose ideal currency properties. THORChain will adopt a single token for simplicity but research the real-world economics of other Proof-of-Stake chains until mainnet. If a secondary token is implemented, it can be done using THORChain on-chain governance. ### Multiple Validator Sets THORChain’s primary performance metric of TPS is most dependent on the bottlenecks of the single Validator Set. Adding more validators to a Validator Set will not improve performance, instead it will decrease it. There is an opportunity given the design of THORChain to investigate a Multiple Validator Set architecture, whereby transactions are sharded to each tokenChain and maintained by separate Validator Sets. By processing the transactions on each tokenChain as separate, the network can perform at a potentially unbounded level. Sharding is possible in THORChain as separate mempools for each TokenChain are maintained by unique address spaces. This allows pending transactions to easily be segregated. The GasLimit is capped for THORChain to allow the network to track saturation and signal for splitting Validator Sets. If the GasLimit approaches upper limits (above 90% for 100 blocks), then a separate Validator Set will form to meet the scaling demands. In this case, the next 100 highest staked Validators will be selected to form another Validator Set. The primary Validator Set (Validators 1-100) will evict all TokenChains bar the TokenChain with the highest gas demands and the MerkleChain from their mempool. The secondary Validator Set (101-200) will then maintain all other TokenChains. Further, if the secondary Validator Set approaches its GasLimit, it will also then evict all but the highest TokenChain, and nominate the Tertiary TokenChain, and so on. The MerkleChain stores and tracks multiple Validator Sets. In this way a hierarchy of Validator Sets are established, each tracking their immediate child. At any stage if a Validator Set’s GasLimit reduces below 10% saturation for longer than 100 blocks then the set is instantly dissolved into its parent set. If a parent set signals to dissolve before its child, then the Master VS re-shuffles all child VSs to re-establish an unbroken hierarchy and prevent disruption. In this way THORChain can scale so that it can always cater for the demands of the network. Seq Event Action 1 >90% Saturation Validator Set (VS1) start signalling to split 2 100 Blocks 3 101st Block A new Validator Set (VS2) commissioned 4 <10% Saturation on both Chains VS1 and VS2 signal to merge 5 100 Blocks 6 101st Block VS2 decommissioned and VS1 returns to process both chains. Table: Order for splitting and merging The Master VS, using the functions of the MerkleChain and processing subservient tokenChain MerkleRoots, will continue to ensure that the network is synced and interoperable. Subservient tokenChains will be less secure than the Master VS’s MerkleChain and RuneChain as the Validators who service their mempools will have less stake. Figure 3: Segregated Validator Sets There are a number of unanswered questions around network security, performance and the characteristics of having multiple Validator Sets that will require further research and development, however the fundamentals of how it could be achieved are outlined here. Additional research to our plan for implementing multiple validator sets via sharding is in a separate paper. ## On-chain Governance ### Overview On-chain governance on THORChain is known as “Validator Signalling” as it is a continual process performed by validators, and that votes are more aptly referred to as signals. Validator Signalling is covered extensively in the Æsir-Protocol Whitepaper. An overview is provided here. Any block producer can propose a change in the core software and consensus rules structured as a data packet: {description, newCode, diffPatch} All other validators vote to accept the change and if it reaches supermajority consensus the updated code can be immediately brought in to operation. The proposing and agreeing validators run the compiled core software on standby, so that once approved, the core software is live to produce the very next block. The types of updates that can be rolled out are essentially unlimited and could be: 1. Consensus rules such as supermajority thresholds, or voting rules (relating to on-chain governance itself) 2. Protocol architecture such as a change to consensus algorithms, integration of sharding, change to the blockchain structure or signature schemes. 3. Native on-chain commands as discussed, whereby additional trading rules can be integrated at the protocol level. 4. Changes to the token structure such as supply or inflation. 5. Changes to state, such as amending exploited or unused accounts. ### Key Aspects Economically Enforced Participation. Voter participation is enforced by in-protocol slashing rules. Not voting on a proposed update or poll will result in a Validator’s stake being slashed and redistributed to other Validators who do vote. There is a grace period of n blocks allowing Validators time to poll the community (or their staking pool) and take up a position before casting a vote. Empowered Minorities. Quadratic voting is implemented inside a validator’s staking pool. Each member that stakes with a Validator can cast votes quadratically proportional to their stake, where each vote is a single whole unit, and increases with the number of subsequent votes. 1 vote costs 1 token, 2 votes cost 4 tokens (2 * 2), 3 votes cost 9 tokens (3 * 3) etc. This appropriately removes the biases that large holders can have on voting, preventing a plutocracy. It also enables and empowers the minority to know their vote is meaningful and represented. A Validator can submit votes on behalf of non-voting stakers in their pool but the vote is quadratically weighted as above. Thus non-voting stakers are empowered to easily swing their Validator’s final vote as their individual vote has more comparative weight than the Validator who representatively voted for them using their tokens. Flexibility. Any staker can change their vote at any time to signal differently. Delegators who switch staking pools to swing vote a different Validator are bound by bonding periods and can effectively only switch once. A carefully designed on-chain governance solution (which itself can update) will see THORChain become a forkless self-amending ledger and increase the likelihood of persistence well into the future. ## Tokens ### Rune Characteristics The Rune is the token of the ecosystem and resides on Address Space T0. Initial supply will be an arbitrary but finite number. The following are the uses of Rune: 1. Validators are required to stake Rune to be part of the Validator Sets. Once staked the Rune are bonded for a period of time to prevent nothing-at-stake attacks. 2. All network transaction fees (gas) are paid in Rune. Fees may be transaction fees, trading fees, bridge fees and liquidity fees, imposed by the different elements of the ecosystem. 3. Liquidity is always backed by Rune in the Continuous Liquidity Pools; so the Rune functions as ecosystem settlement currency. 4. The Flash Network requires Rune as liquidity to join Liquidity Hubs and fees are paid in Rune. 5. Block rewards for Layer 1 Validator Sets and Layer 2 Liquidity Nodes are paid in Rune. ### Block Rewards and Emission Rune ticker RUNE is a predictable inflationary currency that aims to create a price-competitive ecosystem, driving transaction fees to zero. Inflation is designed to reward those that stake to either secure the network or add either Layer 1 or Layer 2 liquidity. Noting the current inflation rates of Ethereum, Bitcoin, EOS and Cosmos, an acceptable starting inflation of 2-5% is proposed. Block Rewards are issued with 50% to Validator Sets and 50% to the Layer 2 Liquidity Nodes. Each Validator in the Validator Set will receive: Liquidity Nodes will receive rewards in accordance with the Layer 2 Incentivisation Plan which rewards for liquidity and reliability. If Layer 2 is not launched alongside the mainnet, then Layer 2 rewards may be hard-coded to pay back to the Foundation to prevent a stalemate whereby validators refuse to launch the Layer 2 Incentivisation Plan for the incorrect perception that they are being diluted or penalised. ### Fees THORChain retains the concept of gas fees and a gas limit for each TokenChain’s blocks. The GasLimit is designed to solicit Validator Set splitting when required to scale. Gas fees are paid for each transaction type; of which there are several distinct types. Each transaction type is hard-coded in the protocol so gas estimates can be made. Layer 1 Liquidity Fees are collected when utilising Continuous Liquidity Pools (CLPs), and are a function of liquidity in the CLP. The higher the slips incurred in CLP’s, the more fees are charged, thereby attracting more users to offer liquidity to the pool. Adding liquidity depth reduces slip and thus reduces the fees collected. Liquidity makers can withdraw their liquidity at any stage, including their collected fees. Self-interested liquidity makers will likely bootstrap new CLPs, collect fees, and then withdraw just their initial stake to continue to earn fees in the CLP indefinitely. Layer 2 Trading Fees are paid to Liquidity Nodes for providing liquidity to Liquidity Hubs. The fee for using each Hub is the average of each fee nominated by each Node as they join. There may be multiple Hubs for each Rune Pair, thereby encouraging a price competitive network, where traders will use the Hub with the lowest average fees. Nodes are free to join any Hub they wish, or create their own. All Liquidity Nodes are paid from the Block Reward, but are weighted to their pro-rata liquidity and time online: By paying trading fees and block rewards to Liquidity Hubs, nodes are incentivised to be liquid, online, reliable and there is no limit to the number of nodes; thus the network becomes quickly decentralised with many channels. This is outlined in more detail in the Flash Network paper. ### TokenChain Each TokenChain is created in a special genesis transaction GenTX on the primary Rune chain T0. Once created, the TokenChain is initiated in the next block with a Genesis Account GenAcc. The GenACC is both the on-chain specification for the characteristics of the token, as well as the account that hosts the token’s Continuous Liquidity Pool (defined later). GenTXs require a fee to be paid in Rune; which is an effective anti-sybil measure to spamming the network with new tokens. Figure 5: The Genesis Account for Token1 The GenAcc stores the following information about the Token which can be publicly queried and displayed on order books, wallets and exchanges: Ticker TKN1 Ticker to display Name Token1 Name to display Supply 100,000,000 Total Supply (100m) Decimals 18 Decimals Reserve 1.0 A parameter to specify the fractional reserve of the CLP (default is 1.0). Owner Self If self; the details above are immutable as there is no private key for the genesis account. Table: Fixed Supply Token If the Account Owner is SELF then the token details are immutable and cannot be changed, as is expected in most digital assets. The GenAcc doubles as an easily identifiable continuous liquidity pool (CLP) for that token and with protocol-level security. The CLP can be interacted with by sending transactions to it from any other wallet. In this case the transaction will execute the liquidity transactions discussed further. If there is an external Account Owner specified, then that Account Owner can change any characteristic of the token, including Supply by proxy. They cannot change the TokenIndex which is permanent and identifies the TokenChain. These tokens are known as variable supply tokens as the account owner is permitted to change the supply of the token. This is necessary for tokenised assets, security tokens and even pegged tokens, whereby the account owner can be a single owner or multi-sig owner that is permitted to change the token characteristics in line with external non-THORChain logic. Account owners can even hand over control of the GenAcc to other accounts, including accounts without a private key; effectively turning the token into a fixed supply immutable token. Ticker TKN1 Ticker to display Name Token1 Name to display Supply 100,000,000 Total Supply (100m) Decimals 18 Decimals Reserve 1.0 A parameter to specify the fractional reserve of the CLP (default is 1.0). Owner T1xa1b2c3d4e5f6 Owner can be another address which will allow mutable token details, necessary for Asset/Security/Pegged Tokens. Table: Variable Supply Token A minimum GenTX initiation fee can be set (100 * GasPrice) to prevent token spam, but stopping runaway token initiation cost. ### Rune Generation THORChain allows a unique mechanism of generating both the full supply of a token, as well as its on-chain CLP at the same time. The special GenTX which creates a new TokenChain requires a non-zero amount of Rune to be transacted. This non-zero amount pays for the GenTX transaction fee, as well as funding the CLP for the first time by trapping the Rune inside the GenACC alongside the newly minted tokens; this sets the initial price of the token. The token owner will subsequently transact in a specific amount of Rune to emit the full token amount to their custody. The following will happen in the case that 100 Rune was the GenTX, and 1mn tokens were created (decimals not included): Genesis Emission TX in (Rune) Rune Locked Tokens Locked Price (RUNE) Tokens Emitted GenTX (100 Rune) 100 1,000,000 0.0001 Subsequent CLP Transaction TX in (Rune) Rune Locked Tokens Locked Price (RUNE) Tokens Emitted 90 190 100,000 0.001 900,000 99 199 10,000 0.01 990,000 99.9 199.9 1,000 0.1 999,000 99.99 199.99 100 1.0999,900 99.999 199.999 10 10 999,990 99.9999 199.9999 1 100 999,999 Table: Creating and emitting a new token. There are two transactions; (1) the GenTx and (2) the subsequent CLPTX that emits a specified amount of tokens. They can be performed in one request. The locked Rune and tokens that remain in the GenAcc can never be fully emitted, thus the pricing of the tokens are set from the first block. If initial pricing or liquidity is unsatisfactory, special one-way Liquidity Transactions can be performed to correct pricing and add liquidity. It is imagined that the developer community will build software that allows creating tokens to be performed with an easy to understand user experience. The following is an example of how the creation of the token can be done in one client-side order: Parameter Field Notes Token Parameter {Ticker: TKN1, Name: Token, Supply: 100m, Decimals: 18, Owner: SELF} Create a data packet to attach to GenTX. CLP Parameter {Reserve: 1.0, InitialPrice: 0.001, InitialLiquidity: 1000 Rune, InitialEmission: 99%} Set and calculate the initial price, liquidity and emission of the token in the CLPTX. Table: Parameters to create a new token ### THORChain Bridges THORChain will have native cross-compatibility with UTXO, Account and Contract-based cryptocurrencies. The heart of this is are Trustless Two-Way-Pegs (2WP) known as the Layer 1 Bridges, and multi-signature accounts. The core Validator Set (100 staked Validators) are signatories to 67/100 multi-sig accounts to external chains, and a 2 / 3 signature threshold on THORChain. This allows Bitcoin, Ethereum (and their forks) as well as ERC-20 tokens to be seamlessly moved onto the THORChain ecosystem and off again. The external coin is moved into a multi-sig account, signed by the Validator Set. After observed finality, the Validator Set create a signature threshold to mint new tokens in THORChain via CLPs. BLS signature thresholds are used to prevent issues during Validator Set re-orgs, as only a threshold of signatures is required to be achieved. The reverse occurs to move the coin off the ecosystem. BLS signature threshold theory has been extensively researched by DFinity. The benefits of signature thresholds is that it only requires a threshold of signatures from a super-set, and not specifically a certain sub-set from that super-set. This translates to flexibility in who can be part of a sub-set, and tolerates validators leaving and re-entering the Validator Set, which could be a frequent occurrence in a healthy and competitive environment of validators. Cryptonote Coins such as Monero and Loki do not support m of n multi-signature, but can support n of n or n-1 of n. It is possible to support these coins, however there is a risk in a Validator Set re-org that more than one of the signatories is replaced. If the evicted Validators do not stay online, then the Cryptonote coins that require their signature will remain locked indefinitely on THORChain. There is no risk of theft, and indeed the user can trade from tXMR to tBTC and exit on the Bitcoin bridge safely, putting this inconvenience on a low significance. A mitigation plan to this is to re-shuffle validators immediately if just 1 validator is lost. The BiFrost paper contains additional research on Thorchain’s bridge protocol research. ## Continuous liquidity pools ### Solving Liquidity Decentralised exchanges are notorious for having low or zero liquidity to the point where they can be unusable for low liquid pairs. This can also be the case for low-tier centralised exchanges. Solving liquidity has been a large focus for Bancor Network and arguably a successful endeavour. Bancor introduced the concept of continuous liquidity by using smart tokens and connectors built on smart contracts deployed on Ethereum. Tokens and Ether are held in smart contracts in such a way that they become bonded and are priced according to the ratio at which they are held. Users can send either tokens or ether to these smart contracts and the other asset is emitted according to a slip-factored price, which takes into account the liquidity depth of the pool. As a result, liquidity is “always available” for these tokens. THORChain integrates two on-chain liquidity strategies; an adaption of Bancor’s continuous liquidity strategy, the CLP, and on-chain order-book liquidity multiplication adapted from the Komodo blockchain, discussed later. ### CLPs The CLP is arguably one of the most important features of THORChain. By building in on-chain liquidity the ecosystem receives the following benefits: • Provides “always-on” trustless liquidity to all tokens in the ecosystem. • Improves the user experience for low-liquidity tokens. • Functions as source of trustless on-chain price feeds to power all aspects of the ecosystem, including advanced trading types. • Generates arbitrage opportunities, further increasing token liquidity. • Allows users to trade tokens at trustless prices, without relying on centralised third-parties. • Provides trustless price-anchors to the Layer 2 Flash Network, allowing instant trading at the layer 2 level. Figure 6: The CLP is the genesis account. ### CLP Transactions The GenAcc is the CLP for each TokenChain, holding both Rune and the full supply of the created tokens as locked liquidity. The account owner can not move the tokens; they can only interact with the GenAcc on-chain command. There are six valid transaction types: Rune In. Anyone can send Rune to the GenAcc. The GenAcc will emit TKN1 (minus a fee), sent to the sender’s TKN1 address TKN1 In. Anyone can send the specific token to the GenAcc. The GenAcc will emit Rune (minus a fee), sent to the sender’s Rune address. Figure 7: CLP Transactions Rune LiquidityTx. Anyone can add Rune to the locked liquidity in the GenAcc, and no TKN1 will be emitted. This is a special transaction that registers the sender’s address as well as the balance sent in order to track and pay them pro-rata liquidity fees. TKN1 LiquidityTx. Anyone can add TKN1 to the locked liquidity in the GenAcc, and no Rune will be emitted. Again, the sender’s address and balance will be registered for payment of liquidity fees. Figure 8: Adding liquidity to a CLP Rune LiquidityWithdrawTx. Anyone that added liquidity to the CLP is permitted to withdraw up to the maximum of their initial liquidity and earned fees. TKN1 LiquidityWithdrawTx. Again, anyone can withdraw their staked Token1 liquidity up to the maximum. Figure 9: Rune FeeWithdrawTx. Anyone that added liquidity to the CLP is permitted to withdraw their earned fees. TKN1 FeeWithdrawTx. Again, anyone can withdraw their earned fees. Importantly, the Locked Liquidity will only emit tokens at the internal price; inferred by the locked RUNE:TKN1 ratio. If 10 RUNE is locked alongside 1000 TKN1; then the price of 1 TKN1 is 0.01 RUNE, and tokens will be emitted at this internal price. The emission rate factors in slip which is the price of the token after the emission occurs, governed by the following equation: Reserve Ratio can be adjusted to be lower than 100% to prevent large slip rates and is akin to fractional reserves. ### Liquidity Fees The Liquidity Fee is a function of slip; the rate at which the price will change in a transaction that emits Rune or Tokens. Importantly, there will always be a slip, as every transaction (no matter how small) will change the ratio between locked Rune and Tokens, so the subsequent change in ratio defines a new price. To couple fees with slip and thereby encourage more self-interested users to add liquidity in CLPs that need it most (high volume and low liquidity CLPs), fees are paid out as: Liquidity Depth Transaction Slip Liquidity Fee 1m Rune 100 Rune 0.01% 0.001 Rune 100,000 Rune 1000 Rune 1% 1 Rune 1000 Rune 500 Rune 50% 25 Rune Table: Example Liquidity Fees ### Trustless On-chain Price Feeds On-chain Arbitrage. Any interaction with the GenAcc will emit the opposite pair at a slip-factored internal price. The slip that is caused by a large emission may move the GenAcc pricing away from fair market price. In this case, self-interested arbitrageurs will immediately correct the price by performing a reverse transaction. Figure 10: On-chain Arbitrage Price Oracle. Token pricing in the GenAcc should always the fair market price, else self-interested arbitrageurs would have taken action. After large price movements there may be a delay, but large price movements cause large Liquidity Fees which self-corrects by attracting more liquidity. Thus the THORChain network can employ token pricing trustlessly throughout the ecosystem based on GenAcc pricing. Trustless token pricing could be employed by future versions of the protocol in the following ways: • To facilitate advanced order types such as margin trading and stop loss/take profits. • Used across THORDApps to allow advanced dApp features around token prices. • Used to transmit fair market prices to the Layer 2 Payment Network, facilitating instant trades across supported assets. Price Manipulation. The cost of price manipulation is high in THORChain, due to extant economics. The CLPs that are mostly likely to attract manipulation are the most used pairs, however the act of manipulation immediately exposes the manipulator to on-chain arbitrage and high liquidity fees. Sustained manipulation in a CLP will increase volumes across that CLP, increasing the amount of liquidity staked by self-interested liquidity stakers, thereby reducing the effect of manipulation in the long run. ## On-chain Trading ### Account Types THORChain integrates on-chain trading at the protocol level, fulfilling all aspects of the desired trade experience. THORChain makes use of account-based architecture which offers flexibility over the UTXO model. There are three account types in the ecosystem, created with the appropriate native on-chain command: Wallet Account. Wallet Accounts have the following characteristic: - Store a balance that can only be transferred by the Account Owner. - Store a nonce. Balances are updated with a unique nonce, and old balances are invalidated if a later nonce is published. Figure 11: The THORChain Wallet for Rune (T0). Alice’s public address is [aaa...aaa] Trading Account. Trading Accounts have the following characteristic: - Store a balance that can only be transferred by: - The Account Owner - A valid incoming trade, and only after performing the Trade Execution. - Store a Price; that generates a Trade Execution, such that: - Store an Expiry time in blocks. - Store a Host Fee; such that user-facing clients can add an optional fee. - Store a Host Address to send Host Fees to. - Store a nonce. Figure 12: The THORChain Trading Account for Rune (T0). Alice’s public address for the pair (T1) [aaa....aaa] is automatically inserted, but can be specified. CLP Account. A CLP is a special account that resides on genesis account of each tokenChain: - Store an array of stakers (address and commitment amount) to collect fees for. - Store balances of Rune and Token. - Store fees for Rune and Token. - Store Token Data. - Store CLP Data. - Store a nonce. Figure 13: The THORChain CLP Account for TKN1 (T1). ### Transaction Types Alongside Accounts, THORChain adds unique Transaction types that cover transactions and trades. These are executed as native on-chain commands that perform math on each trade and update account nonces to save state. TX. Send a Balance to a recipient’s wallet. If sent to a CLP, corresponding tokens are emitted to back to the sender’s token wallet. - Transfer Rune or TKN. If the recipient does not have the required wallet, it is created using the same publicAddress. Figure 14: Alice paying Bob on a single chain. Figure 15: Alice paying Bob across chains. Alice can pay to T0x(bob) or T1x(bob) or T2x(bob); but balance only updated at T1x(bob). Figure 16: Alice interacting with TKN1 CLP with T0. She receives TKN1. LiqTX. Send a balance to a CLP to add liquidity. Sender’s address is added to collect earned fees. - Transfer Rune or TKN. If the incorrect TKN is sent, it will instead be sent to the matching token CLP. This may not be desired, but easily recovered. Figure 17: Alice adding liquidity to CLP1. LiqWdTX. Withdraw liquidity from a CLP. - Call Withdraw on the CLP. If signatures match with Stakers Array, all liquidity is emitted to sender’s address, including earned fees. Figure 18: Alice withdrawing her liquidity + fees. FeeWdTX. Withdraw fee from a CLP. - Call WithdrawFees on the Account. If signatures match, all fees emitted to sender’s address. Figure 19: Alice withdrawing her fees. GenTX. The Genesis Transaction is used to create a new Token and TokenChain. - Transfer RUNE. - Set Ticker and Name. - Set Supply. Leave 0 for variable supply. - Set Decimals (18 default) - Set Reserve Ratio. - Set Account Owner, default is Self. Figure: Alice creating Token1 TokenChain. TradeTx. The Account Transaction is used to create a Trade Account. - Transfer Rune or TKN. - Set Price. In Rune or TKN. - Set Expiry (in blocks). - Set destination account (default is Alice’s token address). - Set Host Fee. Optional. - Set Host Account. Optional. Figure: Alice setting a T1 Sell Order for T0 WalletTx. The Account Transaction is used to return a Trade Account to a Wallet Account. - Transfer Rune or TKN. Figure: Alice cancelling her trade. TradeSolicitTx. The Trade Transaction is used to solicit a Trade. - Transfer Rune or TKN. - Set Price. Desired Price to fill at. - Set Limit. Furthest deviation from Price allowed. Leave blank for Market Order. - Set Expiry (in blocks). Used if the Order is not immediately filled, and becomes a WalletTX. - Set Host Fee. Optional. - Set Host Account. Optional. Figure: Bob buying Alice’s T1 Sell Order. The transaction is sent across chains from TO to T1 TradeExecuteTx. The successful outcome of TradeSolicitTx. - Transfer TKN to Buying Party Destination Account. - Refund spare Rune back to Buying Party Account. - Transfer Rune to Selling Party. - Update nonces. Figure: Bob buys Alice’s T1 Sell Order, is refunded the extra. Alice gets the Rune. ### Limit and Market Orders Refund. As transaction fees are non-deterministic and may not be known ahead of time, the final balance sent to an order may create “dust” or excess if the trade amount exceeds the target order. Thus the concept of a Refund is introduced. This Refund becomes a useful mechanism to support other trade types such as Market Orders and Advanced Trading by relaying instead of refunding. Relaying Refunds from one order to another may be seen as inefficient as it generates additional consequential transactions that rely on the previous being filled, however the benefits are twofold. The first is the trader can “fire and forget” their trade, knowing that it will be executed as intended prior to the expiry specified. The second is that the trade remains trustless as it does not involve client software splitting up a trader’s orders on behalf of them which can be vulnerable to interception attacks. Ultimately it is a better user experience as the trade can be performed in one user-facing action instead of multiple. Limit Order. A trader wishes to spend their entire balance on open orders within a price limit. They specify their starting price as well as a limit, and collect all open orders inside that range, before the order expires. The mechanism of this is to relay the Refund from one order to another until all is fulfilled (or not). Figure: Bob issues a limit order. Market Order. A trader may just wish to clear their trade at any market price available. This is done by relaying the Refund from one order to another until all is fulfilled (or not). Figure: Bob issues a market order. ## User Experience ### Wallets The following is the user experience of interacting with THORChain-based wallets and exchanges. Instantiating Wallets. Users create a new wallet by generating some form of off-line entropy that is used to create the private/public key pairs. These key pairs can be represented and secured by seed-words (BIP39), key-stores or even memorable brain wallets. An innovation proposed by FairLayer is to use [email + password] as the source of entropy as it is both unique, memorable and portable. A single private key controls all of a user’s token addresses (one for each tokenChain). Wallets can support Hierarchical Deterministic key generation which allow users to generate n number of private/public key pairs from the same entropy. Maintaining Assets. All of a user’s assets can be viewed inside a single wallet interface. Asset details (name, ticker) can be trustlessly queried from the GenAcc whilst prices for assets can be queried from CLPs. Bridge accounts can also be queried from GenAccounts for tokenised assets (tokenised Bitcoin), so users can easily send external assets across the ecosystem. ### Trading Trading a CLP. Users can swap assets from inside their wallets by simply sending to CLPs for each assets. Pricing is shown ahead of time, as well as any expected liquidity fees. The advantage of trading a CLP instead of a public order book is that the swap can be performed immediately, with transparent fees and trustless pricing. Executing a Trade. Users can import their key pairs or entropy to a decentralised exchange built on THORChain. They have immediate access to all of their assets for public trading. Creating & Updating an Order. Users can seamlessly move Rune or any other token into a trade account and create a public Buy or Sell order (depending on the side). The Order can be updated at any time. If performed in a Wallet or on an Exchange front-end, the developer may insert an optional fee for development revenue. This fee is public and encourages price competition. Large Orders. Traders and developers have two options for completing large orders. Traders can split up their assets manually and pick off individual orders, sending a small amount of asset to each discrete order to complete trades. Developers can also build client-side software to split up orders for traders and send these trades to multiple orders but this is not preferred. ### Exchanges Traders will have access to exchange interfaces that are familiar in experience and features to current exchanges. All exchanges are accessing the same order book and assets, so they all share liquidity. The ASGARDEX exchange is one such exchange to be launched for mainnet but will be open-source and able to be forked easily. This will drive developers to continually innovate instead of building closed-source moats. Viewing Trades. Once orders are served, a block explorer can index trading accounts and display on a public order book. Trade sides are always paired to RUNE; but TKN:TKN markets are also available if specified. In this case the pair is ordered respective of the underlying TokenIndex. The following would be displayed on a client-side exchange, and can be displayed on any client side interface: • Liquidity Depth. • Candlestick and Volume charts. • Previous Closed Trades. • CLP Accounts • Token Information • Bridge Accounts Figure: Reading and displaying the trades. Trading. Traders can access all features of the protocol to perform trades on their assets and leave at any time. Private keys are held on client side and signatures may be queried from hardware wallets. ### Exchange and Wallet Funding Exchanges and Wallets have client-side access to insert a transparent Hosting Fee and Account to each AccountTX. As each trade is made (maker, taker, CLP trades) the fee is immediately paid to the host account. Figure: Exchange inserting their maker fee. This will support the development of excellent user-experiences in exchanges and wallets that use the THORChain protocol. The pricing mechanism only gathers fees from trades originating from each Wallet and Exchange, so it will naturally support apps that are great experiences. The fee mechanism is also price competitive and public, ensuring that users get the best fees. ### Cross-token Trading For Users. The Rune is the settlement currency of THORChain and is paired to all tokens in their CLPs, (no token can exist on THORChain without a CLP being created). Users who wish to trade from one token to another such as T1 to T2 do not have a CLP available that matches T1:T2. Instead they can perform two transactions that settle with Rune; T1:Rune:T2. Which uses the T1:Rune and T2:Rune CLP. In this case the emitted Rune is immediately re-routed to the next CLP. The user can be assured that both prices are trustless and represent fair market price. Fees can also be transparent and known ahead of time, and the entire trade will take 2 blocks to complete. Figure: Routing via two CLPs. For Traders. With the mechanism described above, all tokens now have liquidity with each other. This may be sufficient for the needs of traders, but advanced traders may prefer direct order-book markets from one token to another, such as for tBTC:tETH. This is up for users to create by simply specifying a pair that is not in Rune when creating orders. The THORChain order-book explorer will collect orders and pair them appropriately on any DEX built for THORChain. Figure: A T1:T2 sell order that is not paired to Rune T0. ### On-chain Initial Exchange Offerings THORChain has all the mechanisms to support native on-chain IEOs. The benefits to this instead of a contract-based token is that the token standard is immutable, auditable and no back doors can be built. Additionally, the token would gain instant liquidity with an immediate market. Lastly, users can participate directly from their wallets which minimise interception attacks. The following would be the process for any team to create their own crypto-asset: • The TokenChain is minted with token details. • The project team then emit their token to their own account: liquidity is created and pricing is set. • The project then create a number of large public Sell Orders (may be with different price points “bonuses”. • Contributors then send Rune to each Sell Order. Projects may also choose to pair to non-Rune tokens (like tBTC or tETH). • Contributors receive the new token in their wallet immediately. This will herald a new generation of liquid, auditable and trustless cryptoassets. ### Validator Staking A key part of the protocol is that Validators stake or “bond” Rune to be part of the 100 in the Validator Set and agree to be bound by slashing rules. The economic cost of being slashed defines the security of the protocol. With 67% requirement for consensus, the entire protocol’s security can be objectively observed to be: Thus THORChain attempts to create the right mechanisms to encourage Rune holders to stake as much of the circulating supply. The following are aspects of this: A proposed 5% Annual Inflation encourages Rune holders to reduce the cost of inflation and stake as Validators. An auction-based entry to the Validator Set encourages Rune holders to outbid each other in order to enter. A bonding period of 14 days to lock tokens once accepted as a Validator. This prevents long-range nothing-at-stake attacks. An unbonding period of 14 days to prevent a rogue Validator attempting a quick getaway with no repercussions after malicious activity. A method to allow delegation of stake to a Validator without the cost of running validator infrastructure. This empowers minority Rune holders to participate in the security of the network. The mechanisms and UX to staking is simple in THORChain making use of the T0 Genesis Account. Ordinarily GenAccounts are CLPs, but in the special case for T0, there is no CLP required as the currency pair is Rune itself. T0 GenAccount stores the token information for Rune, such as current circulating supply, decimals, as well as the name and ticker. Instead of Liquidity Stakers staking to earn Liquidity Fees, Validators Stake in T0 to earn block rewards. The transaction types for this use the same base logic as CLP staking transactions. Ticker RUNE Ticker to display Name Rune Name to display Supply 100,050,000 Total Supply after 1 year at 5% inflation. Decimals 18 Decimals Reserve 1.0 A parameter to specify the fractional reserve of the CLP (default is 1.0). Owner Self Owned by the Protocol Table: Rune Information. Total Supply updated every block by the Validator Set. The first 100 Validators Auction their way into the Validator Set by staking into T0. The Top 100 are assigned to be Validators, assessed every block. Staking is simply a transaction to T0, where the incoming address is saved in the Staking Array, alongside balance. Figure: A prospective Validator stakes. The Top 100 Validators in the Staking Array Stakers are Validators that secure the protocol, and this is tracked in the Validators array. If a new validator stakes more than the 100th Validator, the old Validator is removed from the Validators array and the new one is added. This can be observed by anyone and is updated every block. A bondPeriod specifies the minimum amount of blocks before a Validator can withdraw stake (14 days). Block Rewards accumulate in T0. When a block reward is issued, the tokenData field is updated to increase total Rune supply commensurately. Staked Validators can withdraw their split share of Block Rewards at anytime (1/100th of the accumulated rewards). Reward entitlements are tracked in the Validators field. Delegating tokens to an existing Validator (or Staker who wishes to become a Validator) is performed by the Rune holder simply performing a transaction that specifies an address that is already in the Staking Array. Not specifying a Staking Address, or an address that is invalid will result in them contending to be a Validator instead of a Delegator: Figure: A rune holder delegates to a Staker. The Delegator’s address is nested against a Staker, storing balance and their address. The Staker has no authority to spend a Delegator’s balance, but it counts towards their total balance. If Delegators add enough balance to a Staker to become a Validator, the Staker enters at the next block. Delegators are entitled to withdraw their share of rewards at any time, pro-rata against what has been accumulated towards their Validator’s address. Delegators are bound to the bonding period and unbonding period of their parent Validator. Withdrawing Stake and Rewards is via the same logic that allows Liquidity Stakers to withdraw Liquidity and Fees in CLPs, covered in § 6.2. Voting and Validator Signalling is covered in the Validator Signalling Whitepaper. ## Advanced Transaction Features *These features require more research and is not in the primary development pathway* ### THORChain Name Service (TNS) Users who wish to assign short human-readable names to their accounts to improve the user experience around payments, such as @satoshi, can use the TNS. Short names are unique, stored on-chain and indexed in the MerkleChain. • The user names their account by performing a naming transaction. • Their chosen name alice is checked against the MerkleChain for uniqueness and stored. • A single name can be used for all of their asset accounts, but can be specified, such as alice.rune, alice.btc, alice.0 by setting either the index or the ticker as a suffix. • Sending RUNE to alice.btc will forward it to alice.rune regardless. Figure: Alice’s Rune Account is alice.rune THORChain has the opportunity to rethink how a name service can be operated such that it achieves both effective allocative and investment efficiency. Names will end up being redistributed to parties who can derive the most value from it, and each name can become an investment to the owner. The principle behind the TNS is a hybrid between Harbinger taxes and staking auctions to prevent name-squatting. If a “squatter” owns a name that another user wishes to purchase, the buyer can simply stake Rune with the squatter’s account, known as “Name Staking” in a special transaction. If the squatter does not hold more than the buyer in their own account, the buyer can purchase the name trustlessly after m blocks. If the squatter does not wish to sell, or cannot afford to hold more in their account, they begin paying a fee to anyone who Name Stakes in their account. The squatter can sell at any time to the highest bidder by simply withdrawing the highest bidders’ stake and the name is trustlessly swapped. A very keen buyer must either be patient to acquire the name, or stake a higher amount to increase the fees the squatter has to pay. There can be multiple Name Squatters in an account. Account Owner Name Staker Fee 10 Rune 20 Rune 20 * every m blocks 12 Rune 12 * every m blocks 5 Rune 5 * every m blocks 1 Rune 1 * every m blocks Total: 38 Rune 0.1 Rune every m blocks Table: If out-staked, the Owner pays a Fee based on the difference between their stake and the highest bidder. They can sell the name any time and if they don’t pay the fees the name is open to be acquired. Figure: Name staking to acquire a desired name. A benefit to this approach is that squatters are always eventually coerced out of the name they are squatting, but at an appropriate price. If the squatter is in fact a disused account, then eventually the account will be emptied via fees and the name can be acquired. ### Account Recovery Account Permissioning was first described in the EOS whitepaper. While more granular controls over an account can be helpful, the key aspect is account recovery. Account recovery processes are important for mainstream adoption. A user specifies two other types of accounts on their primary account: Recovery and Guardian. Recovery accounts are accounts that can move all funds and TNS name to their account, but are always deactivated. A Guardian account activates Recovery accounts. The recovery process is as follows, relying on primarily out-of-band communication and existing social or trust networks: • A user creates an account. • The user nominates a Guardian account (or multiple). The Guardian is notified and signs the request. • The user nominates a Recovery account (or multiple). Recovery accounts are notified and sign the request. They are deactivated. • The user loses access to their private key. • The user contacts their Guardian to activate one of their Recovery accounts. • The user can then move their funds from their lost account to their activated recovery account, signed from their Recovery address. This mechanism is safe as it requires both their Guardian and Recovery accounts being compromised. Guardian accounts can do nothing but activate pre-selected Recovery accounts. Additionally, Recovery accounts are specified before the fact, so funds can only move to an address specified by the original owner. Figure: Guardian and Recovery addresses are stored on-chain. Figure: A Guardian activates a Recovery Address. The Recovery Address withdraws the balance. ### Multi-sig Accounts Multi-signature accounts are supported at the protocol level as a key part of managing assets. The account owner simply adds external accounts in special transactions. Once added all future transactions require all parties to sign. The signature is aggregated in the account and the last party performs the transaction. The signature is cleared once the transaction is performed. The default is n of n but this can easily be changed at any time by the parties. Figure: A THORChain multi-sig. ### Transaction Receipts Sending payments on a public blockchain is problematic for the point of privacy; but invaluable for recording an immutable history of transactions. For public adoption, transactions need to be tracked and reconciled and these are often done through third party software. THORChain has an opportunity to re-think how transactions can be tracked and reconciled for users on-chain with no third party software, but still done in the private fashion using the assymetric key cryptography. Each transaction is sent with a hash of a transaction description: type Tx struct { balance int64 // Balance to transfer from Sender token int64 // TokenIdentifier to_address string // Address of the Receiver data_hash string // Optional. hash.SenderPrivKey(description) or // hash.ReceiverPubKey(hash.SenderPrivKey(description)) } The description is hashed and stored on-chain. There are two options: 1. Only the Sender private keys are used to hash the description. In this case only the Sender can decrypt. 2. Both the Sender private keys as well as the Receiver public keys are used to hash. In this case both (and only) the Sender and Receiver can decrypt and read the message, yet the hash is stored publicly on-chain. Considerations. There are a number of considerations that need to be factored into this functionality and its implementation: 1. Transaction descriptions will add an extra 256 bits to each transaction; which can add up and increase block sizes. As transaction descriptions are not critical information and present no risk to assets if a collision is detected, smaller SHA digests or other hashing functions can be considered such as SHA-224 (224 bits) SHA-1 (160 bits) or MD5 (142 bits). A user who wants full privacy will not use on-chain descriptions anyway. 2. For full two-way decryption, the Sender must know the Receiver Public Key. As THORChain addresses are trimmed hashes of the Public Key, it is not possible to extract the PubKey from the address, so the PubKey must be shared using off-chain channels, or by using the THORChain Messaging Protocol. Exposing a PubKey may increase the probability of an address being compromised in future by advanced key breaking algorithms. Users should be communicated the risks so they do not expose the PubKeys on high-value addresses. With this lightweight functionality built on-chain any wallet for THORChain can decrypt and show transaction information universally. Third-party services can use the data_hash field to insert payment references allowing automated reconciliation and payment tagging. With the correct cryptography, THORChain transactions are private yet useable. ### Transaction Requests Using similiar functionality to Transaction Receipts, payments can be requested on-chain from the Sender. With the correct implementation the Sender will be notified and can simply pay the request. The Requester may or may not be the final Receiver. Each transaction is sent with a hash of a transaction description: type TxRequest struct { balance int64 // Balance Requested from Sender token int64 // TokenIdentifier to_address string // Address of the Receiver data_hash string // Optional. hash.RequesterPrivKey(description) or // hash.SenderPubKey(hash.RequesterPrivKey(description)) } Once the TxRequest is processed on-chain the Sender is notified by an Events Service from supported wallets, or simply shows up in their wallet. The Sender can then simply pay the request. This functionality has the same considerations as Transaction Receipts. ### Recurring Transactions By advancing the functionality of Transaction Requests, psuedo recurring payments can be built on-chain. Requesters send thru batches transaction requests (such as 12 requests for a 12 month subscription), which are pre-approved by the Sender. Once approved, the Requester can execute the payment with the pre-signed authorisation. Start and expiry block times are inserted to prevent the transaction being executed before or after pre-authorised time window. type TxRequest struct { balance int64 // Balance Requested from Sender token int64 // TokenIdentifier to_address string // Address of the Receiver auth_hash string // Initially empty. The Sender inserts their signature here once they approve. start_blocks int64 // The earliest the TX can be made. expiry_blocks int64 // The latest the TX can be made. data_hash string // Optional. hash.RequesterPrivKey(description) or // hash.SenderPubKey(hash.RequesterPrivKey(description)) } Once the TxRequest is processed on-chain the Sender is notified by an Events Service from supported wallets, or simply shows up in their wallet. The Sender simply authorises the payments one-by-one, but they can't be executed by the Requester until the window of time between start_blocks and expiry_blocks. This functionality has the same considerations as Transaction Receipts. The Sender can remove their auth_hash to cancel the payment. ### Escrow Accounts Escrows are a vital part of a payment protocol. Alice pays into an escrow, and nominates Bob to receive, and Charlie to escrow. Alice can only release the payment to Bob. Bob can only refund the payment to Alice. Charlie can refund or release, or do a partial refund. All transactions can be done in range 0% : 100% allowing partial refunds or releases. The implementation is trivial. Alice makes an EscrowCreate transaction and observes the newly created escrow address. She can include an optional transaction message. type txEscrowCreate struct { balance int64 // Balance to transfer from Sender token int64 // TokenIdentifier to_address string // Address of the Receiver agent_address string // Address of the Escrow Agent. Set Null for Protocol-level escrow data_hash string // Optional. hash.SenderPrivKey(description) or // hash.ReceiverPubKey(hash.SenderPrivKey(description)) } To release, Alice makes an EscrowRelease transaction, which checks her address signature and releases the balance (fully or partially) to Bob. type EscrowRelease struct { escrow_address string // Address of the Escrow amount int64 // Amount to release } To refund, Bob makes an EscrowRefund transaction, which checks his address signature and refunds the balance (fully or partially) to Alice. type EscrowRelease struct { escrow_address string // Address of the Escrow. amount int64 // Amount to refund } To refund or release, Charle makes an EscrowRelease or EscrowRefund transaction. ### Automatic Disbursements Trustlessly disseminating to a number of accounts can be easily done through THORChain's Automatic Disbursement feature. Alice creates a Disbursement Account and adds third-party addresses to the account. Any tokens she sends to this account will now disburse to the addresses listed, in specified proportions. Alice can also specify how to treat different tokens on disbersement. type txCreateDisbursement struct { DisburseRecipient array // Array of [addr, rate] } type DisburseRecipient array { to_address string // Address of the Receiver rate int64 // Rate to send to Receiver token int64 // Token type to disburse. Null to send all } ### Combined Payment Features Any one of THORChain's Payment features can be combined to create powerful payment facilities, such as: • A bounty escrow being created, with the protocol being the escrow agent. The recipient address is a list of community contributors who are paid from the bounty. • Payroll being disbursed to staff from a recurring payment from an escrow account. • A transaction request to an escrow account that is signed by a multi-signature account. • Account recovery being affected by a multi-signature account. • A THORChain name being bought and owned by a multi-signature account. Through the Æsir Protocol more payment features can be easily added as uses for them arise. ## Other Features ### Collectibles The ERC-721 standard of collectibles for Ethereum was the dawn of a new era of digital assets that can be collected and traded in marketplaces. A collectible, also known as a non-fungible token token, is a token that (typically) is not divided below 1 and each token has meta-data that can uniquely add value to the individual token. Each token is give an index and owners move individual tokens around. Some of the metadata that can be attached are names, descriptions and even resource links. { "name": "ODIN", "description": "Odin, ruler of the Æsir, father of Thor.", "image": "https://ipfs.infura.io/ipfs/QmPS7SbxcQoKAcFmTBxqm1c43sWztnFaPrCfgGDy3FhskN", } Collectibles are different to tokens in that collectible trades happen much more sparsely and they normally have far few owners. Additionally, collectibles do not require continuous liquidity as individual items cannot be divided. As such it would be prudent to track all collectibles on a single discrete chain with address space TCx<address>. This will reduce blockchain bloat and allow the collectibleChain to maintain separate features to tokenChain. On the creation of a collectible; the collectible is indexed inside the TCx0 account and the initial details are set: type collectible struct { name string // String of the Collectible Collection "Æsir Gods" symbol string // Symbol of the Collectible "ÆGODS" supply int64 // Number of different Collectibles in the Collection tokenURI uint256 // A distinct Uniform Resource Identifier (URI) for a given Collectible owner T0xBob // Allows the owner (Bob) to details in future, such as name or URIs. } Once created all collectibles are emitted to the creator's address, allowing them to transfer further. The simple transfer of a collectible is: Transfer(address _from, address _to, uint256 indexed _tokenId); All collectible and collection details can be read from TCx0 account at any time. Further features such as minting, burning, pausing or blacklisting can be included in further interations of the standard. Collectibles can also be bundled together and transfered in very large quantities. The modified transfer function includes an array to detail a batch transfer: batchTransferFrom(address _from, address _to, uint256[] _ids, uint256[] _values); ### Messaging Protocol Sending transaction receipts using assymetric key cryptography across THORChain allows private and end-end encrypted data transfer. This functionality allows THORChain to experiment with fully private and peer-peer on-chain messaging. Senders can send messages to the Receiver's address. type msg struct { from_address string // Address of the Sender, such as @alice or T0xaaa...aaa to_address string // Address of the Receiver, such as @bob or T0xbbb...bbb msg_hash string // hash.ReceiverPubKey(hash.SenderPrivKey(description)) } ### Smart contracts THORChain may support smart contracts to enhance the ecosystem, but this is not in the primary development pathway. It will most likely integrated by on-chain governance. Any potential smart contract that benefits the protocol as a whole should be added as a native on-chain command instead. THORChain is not intended to be built to be a dApp platform; rather it is focussed on digital assets and the trading of those assets. ### Additional On-Chain Commands THORChain employs on-chain commands which are lightweight methods hard-coded into the protocol, and called by transactions. Compared with smart contracts, these native on-chain commands do not need Virtual Machine support and are part of the underlying protocol. Whilst smart contracts are generalist and need to be compiled to bytecode, these function as advanced transaction outputs/inputs. Each time a one is executed, the state of an account is updated in accordance with the script. On-chain commands are built into the protocol as standardised methods; however once a standard is agreed upon, it is added into the protocol layer through on-chain governance. The following on-chain commands are tabled to be built into THORChain: • Security Tokens • Stablecoins • Collectibles (ERC-721) • Identity (ERC-725, 735) • Recurring Payments • Account Permissioning • Escrows ### StableCoins THORChain’s macro vision is to create a highly liquid payment network built on a completely trustless structure. An imperative feature for mainstream adoption are stablecoins that are pegged to existing fiat currencies such as USD, YEN, EURO and AUD. The first way to achieve this is to simply tokenise existing stablecoins, while the second is to support the development of stable coins on the platform. The first will naturally be easy if the bridges in § 8 are built. THORChain has all the required features to create StableCoins with auditable supply and collateral using a Variable Supply token and its CLP and full Validator Set participation. The following would be the process: • A tokenChain tUSD is created by the Validator Set. A Rune:tUSD CLP is created that is variable supply, requiring 2 / 3 signatures from Validators, or by a on-chain smart contract. • At each round, Validators (on smart contract) propose a price $ that the Rune is valued at. They can infer this by any means possible, most likely by reading APIs off a conglomerate of external exchanges of Rune. Validators can also nominate Rune pricing by watching Bitcoin pricing off external exchanges, and converting to Rune price by the Rune:tBTC price feed. • At each round n + 1 the median of nominated prices is stored in the block. Outliers may invoke slashing rules to penalise poor price nomination. • As Rune price fluctuates, tUSD supply in the CLP is negatively changed by the Validators. As an example, Rune price decreases by 2%, so tUSD Supply increases by 4%. This causes tUSD to become inflated and cheap. • Self-interested arbitrageurs will then send in Rune to buy cheap tUSD, until tUSD returns to 1:1 backed in Assets. • Additionally, the liquidity fee will increase the staked collateral in the CLP to match volume and reduce slip. Figure: Collateralized on-chain StableCoins. This process can be repeated for any other fiat currency and is very simple. Once an external price of Rune is set (in the fiat currency), then the protocol deliberately creates arbitrage opportunities by influencing money supply to attract third parties to act in a self-interested manner to restore any price imbalance. By requiring full Validator Set participation in the price nomination the price of the StableCoin can be relied upon with the same assurance as the entire protocol itself. The pricing/supply feedback loop can be modified to reduce damping and price sensitivity. ### Token Baskets & Indexes A THORChain can go further than a single token being represented in a CLP. With a combination of CLP scripts and trustless price feeds, token baskets and indexes can be created and represented by a single asset. This allows users and traders to carry a single token and know that the token’s price trustlessly represents other assets. A new tokenChain TKNIndex can be created that represents TKN1, TKN2, and TKN3, where the TokenData and CLPData for TKNIndex is linked to the CLPs of the indexed assets. The TKNIndex CLP is bonded to include liquidity in all other CLPs so the price of TKNIndex is thus representative of the linked assets. Figure: Token Baskets ## Conclusion THORChain is a lightning fast decentralised exchange with protocol-level trading features to achieve feature parity with the best centralised exchanges of the day; all with full self-sovereign asset management. THORChain solves the fundamental problems of existing decentralised exchanges with a fast on-chain trading experience, on-chain continuous liquidity and correct incentivisation economics for exchange and wallet developers. THORChain is built for a new class of decentralised exchanges and transactional networks and will rapidly increase the useability of transacting with crypto-currency assets. ## References You can’t perform that action at this time.	2019-06-20 03:32:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.2140074223279953, "perplexity": 13360.052636946586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00217.warc.gz"}
https://www.physicsforums.com/threads/cross-sectional-area-not-a-question.408164/	# Cross sectional area (NOT A QUESTION) 1. Jun 5, 2010 ### moonman239 Just for those who don't know a thing about cross sectional areas, I thought I'd explain. A cross sectional area describes the area of a flat (2-dimensional) representation of a 3-dimensional object. So if I cut a cylinder, instead of seeing two circles, I see four circles (unless there are other circles in my environment). The cross-sectional area is the area of either of the two circles. For a cylinder or right solid, the cross-sectional area is the area of the base. For a sphere, the cross sectional area is the area of a circle with the same radius (pir2). For an ellipsoid, the cross sectional area is the area of an ellipse with the same long (a) and short (b) axes (piab). Last edited: Jun 5, 2010 2. Jun 5, 2010 ### Landau I don't know what to say. Thanks for sharing? 3. Jun 9, 2010 ### theoryiscool lol at landau's response 4. Jun 14, 2010 ### moonman239 Oh, for an oblate ellipsoid (such as Earth), where lines of latitude are circular, the area is simply pi*(r^2)	2018-06-24 04:02:40	{"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": 0.8465544581413269, "perplexity": 996.0082907229786}, "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-2018-26/segments/1529267866191.78/warc/CC-MAIN-20180624024705-20180624044705-00352.warc.gz"}
https://iq.opengenus.org/8-queens-problem-backtracking/	# 8 Queens Problem using Backtracking #### Algorithms backtracking Get FREE domain for 1st year and build your brand new site Reading time: 30 minutes \| Coding time: 10 minutes You are given an 8x8 chessboard, find a way to place 8 queens such that no queen can attack any other queen on the chessboard. A queen can only be attacked if it lies on the same row, or same column, or the same diagonal of any other queen. Print all the possible configurations. To solve this problem, we will make use of the Backtracking algorithm. The backtracking algorithm, in general checks all possible configurations and test whether the required result is obtained or not. For thr given problem, we will explore all possible positions the queens can be relatively placed at. The solution will be correct when the number of placed queens = 8. The time complexity of this approach is O(N!). Input Format - the number 8, which does not need to be read, but we will take an input number for the sake of generalization of the algorithm to an NxN chessboard. Output Format - all matrices that constitute the possible solutions will contain the numbers 0(for empty cell) and 1(for a cell where queen is placed). Hence, the output is a set of binary matrices. Visualisation from a 4x4 chessboard solution : In this configuration, we place 2 queens in the first iteration and see that checking by placing further queens is not required as we will not get a solution in this path. Note that in this configuration, all places in the third rows can be attacked. As the above combination was not possible, we will go back and go for the next iteration. This means we will change the position of the second queen. In this, we found a solution. Now let's take a look at the backtracking algorithm and see how it works: The idea is to place the queens one after the other in columns, and check if previously placed queens cannot attack the current queen we're about to place. If we find such a row, we return true and put the row and column as part of the solution matrix. If such a column does not exist, we return false and backtrack* ## Pseudocode START 1. begin from the leftmost column 2. if all the queens are placed, return true/ print configuration 3. check for all rows in the current column a) if queen placed safely, mark row and column; and recursively check if we approach in the current configuration, do we obtain a solution or not b) if placing yields a solution, return true c) if placing does not yield a solution, unmark and try other rows 4. if all rows tried and solution not obtained, return false and backtrack END ## Implementation Implementaion of the above backtracking algorithm : #include <bits/stdc++.h> using namespace std; int board[8][8]; // you can pick any matrix size you want bool isPossible(int n,int row,int col){ // check whether // placing queen possible or not // Same Column for(int i=row-1;i>=0;i--){ if(board[i][col] == 1){ return false; } } //Upper Left Diagonal for(int i=row-1,j=col-1;i>=0 && j>=0 ; i--,j--){ if(board[i][j] ==1){ return false; } } // Upper Right Diagonal for(int i=row-1,j=col+1;i>=0 && j<n ; i--,j++){ if(board[i][j] == 1){ return false; } } return true; } void nQueenHelper(int n,int row){ if(row==n){ // We have reached some solution. // Print the board matrix // return for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ cout << board[i][j] << " "; } } cout<<endl; return; } // Place at all possible positions and move to smaller problem for(int j=0;j<n;j++){ if(isPossible(n,row,j)){ // if no attack, proceed board[row][j] = 1; // mark row, column with 1 nQueenHelper(n,row+1); // call function to continue // further } board[row][j] = 0; // unmark to backtrack } return; } void placeNQueens(int n){ memset(board,0,88sizeof(int)); // allocate 88 memory // and initialize all // cells with zeroes nQueenHelper(n,0); // call the backtracking function // and print solutions } int main(){ int n; cin>>n; // could use a default 8 as well placeNQueens(n); return 0; } Output ( for n = 4): 1 indicates placement of queens 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 Explanation of the above code solution: These are two possible solutions from the entire solution set for the 8 queen problem. main() { call placeNQueens(8), placeNQueens(){ call nQueenHelper(8,0){ row = 0 if(row==n) // won't execute as 0 != 8 for(int j=0; j<8; j++){ { if(isPossible==true) { board[0][0] = 1 // board[row][0] = 1 call nQueenHelper(8,row+1) // recur for all rows further print matrix when row = 8 if solution obtained and (row==n) condition is met } board[0][0] = 0 // backtrack and try for // different configurations } } } } for example, the following configuration won't be displayed ## Time Complexity Analysis 1. the isPossible method takes O(n) time 2. for each invocation of loop in nQueenHelper, it runs for O(n) time 3. the isPossible condition is present in the loop and also calls nQueenHelper which is recursive adding this up, the recurrence relation is: T(n) = O(n^2) + n T(n-1) solving the above recurrence by iteration or recursion tree, the time complexity of the nQueen problem is = O(N!) ## Question :- You are given an NxN maze with a rat placed at (0,0). Find and print all the paths that the rat can follow to reach its destination i.e (N-1,N-1). The rat can move in all four directions (left,right,up,down). Value of every cell will be either 0 or 1. 0 represents a blocked cell, the rat cannot move through it, though 1 is an unblocked cell. Solve the problem using backtracking algorithm, Input - an integer N and a binary maze of size NxN, showing blocked and unblocked cells. Output - all possible path matrices the rat can travel from (0,0) to (N-1,N-1).	2021-06-24 13:09:55	{"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": 0.34913086891174316, "perplexity": 2499.7484469713454}, "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/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00351.warc.gz"}
https://projecteuclid.org/euclid.aaa/1412605766	## Abstract and Applied Analysis ### Delay-Dependent Robust Exponential Stability and ${H}_{\infty }$ Analysis for a Class of Uncertain Markovian Jumping System with Multiple Delays Jianwei Xia #### Abstract This paper deals with the problem of robust exponential stability and ${H}_{\infty }$ performance analysis for a class of uncertain Markovian jumping system with multiple delays. Based on the reciprocally convex approach, some novel delay-dependent stability criteria for the addressed system are derived. At last, numerical examples is given presented to show the effectiveness of the proposed results. #### Article information Source Abstr. Appl. Anal., Volume 2014, Special Issue (2013), Article ID 738318, 10 pages. Dates First available in Project Euclid: 6 October 2014 https://projecteuclid.org/euclid.aaa/1412605766 Digital Object Identifier doi:10.1155/2014/738318 Mathematical Reviews number (MathSciNet) MR3186977 #### Citation Xia, Jianwei. Delay-Dependent Robust Exponential Stability and ${H}_{\infty }$ Analysis for a Class of Uncertain Markovian Jumping System with Multiple Delays. Abstr. Appl. Anal. 2014, Special Issue (2013), Article ID 738318, 10 pages. doi:10.1155/2014/738318. https://projecteuclid.org/euclid.aaa/1412605766 #### References • Y. He, G. Liu, and D. Rees, “New delay-dependent stability criteria for neural networks with yime-varying delay,” IEEE Transactions on Neural Networks, vol. 18, no. 1, pp. 310–314, 2007. • S. Xu, J. Lam, and D. W. C. Ho, “A new LMI condition for delay-dependent asymptotic stability of delayed Hopfield neural networks,” IEEE Transactions on Circuits and Systems II: Express Briefs, vol. 53, no. 3, pp. 230–234, 2006. • B. Zhang, S. Xu, and Y. 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Lin, “Delay-slope-dependent stability results of recurrent neural networks,” IEEE Transactions on Neural Networks, vol. 22, no. 12, pp. 2138–2143, 2011. • Z.-G. Wu, J. H. Park, H. Su, and J. Chu, “Stochastic stability analysis of piecewise homogeneous Markovian jump neural networks with mixed time-delays,” Journal of the Franklin Institute, vol. 349, no. 6, pp. 2136–2150, 2012. • T. Li, X. Yang, P. Yang, and S. Fei, “New delay-variation-dependent stability for neural networks with time-varying delay,” Neurocomputing, vol. 101, pp. 361–369, 2013. • N. Krasovskii and E. Lidskii, “Analysis design of controller in systems with random attributes–-part I,” Automation and Remote Control, vol. 22, pp. 1021–1025, 1961. • H. Shen, X. Huang, J. Zhou, and Z. Wang, “Global exponential estimates for uncertain Markovian jump neural networks with reaction-diffusion terms,” Nonlinear Dynamics, vol. 69, no. 1-2, pp. 473–486, 2012. • H. Shen, X. Huang, and Z. Wang, “Fuzzy dissipative control for nonlinear Markovian jump systems via retarded feedback,” Journal of the Franklin Institute, 2013. • H. Shen, J. Park, L. Zhang, and Z. Wu, “Robust extended dissipative control for sampled-data Markov jump systems,” International Journal of Control, 2014. • S. He and F. Liu, “Adaptive observer-based fault estimation for stochastic Markovian jumping systems,” Abstract and Applied Analysis, vol. 2012, Article ID 176419, 11 pages, 2012. • S. He and F. Liu, “Robust stabilization of stochastic Markovian jumping systems via proportional-integral control,” Signal Processing, vol. 91, no. 11, pp. 2478–2486, 2011. • Z.-G. Wu, P. Shi, H. Su, and J. Chu, “Passivity analysis for discrete-time stochastic markovian jump neural networks with mixed time delays,” IEEE Transactions on Neural Networks, vol. 22, no. 10, pp. 1566–1575, 2011. • K. Benjelloun and E. K. Boukas, “Mean square stochastic stability of linear time-delay system with Markovian jumping parameters,” IEEE Transactions on Automatic Control, vol. 43, no. 10, pp. 1456–1460, 1998. • H. Fujisaki, “On correlation values of M-phase spreading sequences of Markov chains,” IEEE Transactions on Circuits and Systems I: Fundamental Theory and Applications, vol. 49, no. 12, pp. 1745–1750, 2002. • S. Xu, T. Chen, and J. Lam, “Robust H$\infty$ filtering for uncertain Markovian jump systems with mode-dependent time delays,” IEEE Transactions on Automatic Control, vol. 48, no. 5, pp. 900–907, 2003. • E. K. Boukas, Z. K. Liu, and G. X. Liu, “Delay-dependent robust stability and H$_{\infty }$ control of jump linear systems with time-delay,” International Journal of Control, vol. 74, no. 4, pp. 329–340, 2001. • S. Xu, J. Lam, and X. Mao, “Delay-dependent H$_{\infty }$ control and filtering for uncertain Markovian jump systems with time-varying delays,” IEEE Transactions on Circuits and Systems I: Regular Papers, vol. 54, no. 9, pp. 2070–2077, 2007. • X. Zhao and Q. Zeng, “New robust delay-dependent stability and H$\infty$ analysis for uncertain Markovian jump systems with time-varying delays,” Journal of the Franklin Institute, vol. 347, no. 5, pp. 863–874, 2010. • B. Zhang, W. Zheng, and S. Xu, “Filtering of Markovian jump delay systems based on a new performance index,” IEEE Transactions on Circuits and Systems I: Regular Papers, vol. 60, no. 5, pp. 1250–1263, 2013. • Y. Zhang, Y. He, and M. Wu, “Delay-dependent robust stability for uncertain stochastic systems with interval time-varying delay,” Acta Automatica Sinica, vol. 35, no. 5, pp. 577–582, 2009. • Z. Wu, J. Lam, H. Su, and J. Chu, “Stability and dissipativity analysis of static neural networks with time delay,” IEEE Transactions on Neural Networks and Learning Systems, vol. 23, no. 2, pp. 199–210, 2012. • S. Xu and T. Chen, “Robust H$\infty$ control for uncertain stochastic systems with state delay,” IEEE Transactions on Automatic Control, vol. 47, no. 12, pp. 2089–2094, 2002. \endinput	2019-12-09 21:49:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 2, "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": 0.2944313585758209, "perplexity": 3205.8071030206384}, "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-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00348.warc.gz"}
https://socratic.org/questions/the-air-in-lungs-has-0-153-mol-of-gas-particles-at-302-k-and-101-3-of-pressure-w	# The air in lungs has 0.153 mol of gas particles at 302 K and 101.3 kPa of pressure. What is the volume of the air? May 13, 2016 3.79L #### Explanation: This can be solved applying equation of state for ideal gas, which i as follows • $P V = n R T$ Where for air in the lungs • $\text{Pressure} \left(P\right) = 101.3 k P a = 1 a t m$ • $\text{No. of moles} \left(n\right) = 0.153$ • $\text{Temperature} \left(T\right) = 302 K$ • $\text{Universal gas constant } \left(R\right) = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$ • "Volume of air" (V) =? $\therefore V = \frac{n R T}{P} = \frac{0.153 \times 0.082 \times 302}{1} L = 3.79 L$	2019-10-22 09:40:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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": 0.7894467115402222, "perplexity": 1848.472870263855}, "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/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00429.warc.gz"}
https://socratic.org/questions/how-do-you-simplify-2y-4	# How do you simplify 2y^-4? ##### 1 Answer May 21, 2018 $\frac{2}{y} ^ 4$ #### Explanation: Any term to a negative power can be written as one over the term with a positive power.	2020-02-17 21:23:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "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": 0.9231914281845093, "perplexity": 1370.65885387545}, "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-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00466.warc.gz"}
https://www.elitedigitalstudy.com/10522/a-steel-bar-of-length-200-cm-is-nailed-at-its-midpoint-the-fundamental-frequency-of-the-longitudinal-vibrations	A steel bar of length 200 cm is nailed at its midpoint. The fundamental frequency of the longitudinal vibrations of the rod is 2.53 kHz. At what speed will the sound be able to travel through steel? Asked by Pragya Singh \| 1 year ago \| 138 ##### Solution :- Given, Length , l = 200 cm = 2 m Fundamental frequency of vibration, νF = 2.53 kHz = 2.53 × 103 Hz The bar is then plucked at its midpoint, forming an antinode (A) at its centre, and nodes (N) at its two edges, as depicted in the figure below : The distance between two successive nodes is $$\dfrac{λ}{2}$$ = l = $$\dfrac{λ}{2}$$ Or, λ = 2 x 2 = 4m Thus, sound travels through steel at a speed of v = νλ v = 4 x 2.53 x 103 =10.12 km/s Answered by Abhisek \| 1 year ago ### Related Questions #### A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air. A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air. The Wind then starts blowing towards the petrol pump at 20 m/s. Calculate the wavelength, speed, and frequency of the horn’s sound for a man standing at the petrol pump. Is this situation completely identical to a situation when the observer moves towards the truck at 20 m/s and the air is still? #### A man standing at a certain distance from an observer blows a horn of frequency 200 Hz in still air. A man standing at a certain distance from an observer blows a horn of frequency 200 Hz in still air. (a) Find the horn’s frequency for the observer when the man (i) runs towards him at 20 m/s (ii) runs away from him at 20 m/s. (b) Find the speed of sound in both the cases. [Speed of sound in still air is 340 m/s] #### A bat is flitting about in a cave, navigating via ultrasonic beeps. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in the air. What frequency does the bat hear reflected off the wall? Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about $$4.0 km s^{–1}$$, and that of the P wave is $$8.0 km s^{–1}.$$ A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur? A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of $$360 km h^{–1}$$. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be $$1450 m s^{–1}$$ .	2022-12-04 04:31:49	{"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": 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": 0.6185433864593506, "perplexity": 1391.231828922116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00445.warc.gz"}
https://www.esaral.com/q/tangents-are-drawn-to-the-hyperbola-21890/	Tangents are drawn to the hyperbola Question: Tangents are drawn to the hyperbola $4 x^{2}-y^{2}=36$ at the point $P$ and $Q$. If these tangents intersect at the point $T(0,3)$ then the area (in sq. units) of $\triangle P T Q$ is – 1. $54 \sqrt{3}$ 2. $60 \sqrt{3}$ 3. $36 \sqrt{5}$ 4. $45 \sqrt{5}$ Correct Option: , 4 Solution:	2022-05-16 22:59:11	{"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": 0.522497832775116, "perplexity": 323.84957380819475}, "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/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00186.warc.gz"}
http://www.shortopedia.com/V/E/Vector_calculus	## Vector calculus In vector calculus, curl is a vector operator that shows a vector field's rate of rotation: the direction of the axis of rotation and the magnitude of the rotation. It can also be described as the circulation density. ...more on Wikipedia about "Curl" In vector calculus, del is a vector differential operator represented by the nabla symbol, ∇. ...more on Wikipedia about "Del" In vector calculus, the divergence is an operator that measures a vector field's tendency to originate from or converge upon a given point. For instance, for a vector field that denotes the velocity of water flowing in a draining bathtub, the divergence would have a negative value over the drain because the water vanishes there (if we only consider two dimensions); away from the drain the divergence would be zero, since there are no other sinks or sources. ...more on Wikipedia about "Divergence" In vector calculus, the divergence theorem, also known as Gauss' theorem, Ostrogradsky's theorem, or Ostrogradsky–Gauss theorem is a result that relates the outward flow of a vector field on a surface to the behaviour of the vector field inside the surface. ...more on Wikipedia about "Divergence theorem" The fundamental theorem of vector calculus, also known as Helmholtz's theorem, states that any vector field meeting certain conditions (of decaying towards infinity) can be resolved into irrotational (curl-free) and solenoidal (divergence-free) component vector fields. ...more on Wikipedia about "Fundamental theorem of vector analysis" Gradient is commonly used to describe the measure of the slope (also called steepness, or incline) of a straight line. It is also sometimes used synonymously with grade, meaning the inclination of a surface along a given direction. ...more on Wikipedia about "Gradient" Green's identities are a set of three identities in vector calculus. They are named after the mathematician George Green, who discovered Green's theorem. ...more on Wikipedia about "Green's identities" In physics and mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. Green's theorem was named after British scientist George Green and is a special case of the more general Stokes' theorem. ...more on Wikipedia about "Green's theorem" In fluid mechanics, Helmholtz's theorems describe the behaviour of vortex lines in a fluid. The theorems apply to fluids that are inviscid (ie without viscosity), incompressible, of constant density and under the influence of a conservative body force (such as gravity). The theorems were published by Hermann von Helmholtz in 1858. ...more on Wikipedia about "Helmholtz's theorems" In vector calculus, an irrotational or conservative vector field is a vector field whose curl is zero. If the field is denoted as v, then ...more on Wikipedia about "Irrotational vector field" In vector analysis and in fluid dynamics, a lamellar vector field is a vector field with no rotational component. That is, if the field is denoted as v, then ...more on Wikipedia about "Lamellar vector field" In vector calculus, a Laplacian vector field is a vector field which is both irrotational and incompressible. If the field is denoted as v, then it is described by the following differential equations: ...more on Wikipedia about "Laplacian vector field" In mathematical physics, a multipole expansion is a series expansion of the effect produced by localized source terms in a given partial differential equation, most commonly Poisson's equation (for electrostatics and gravity), in spherical coordinates or cylindrical coordinates. Typically, the expansion is in terms of spherical harmonics or related angular functions multiplied by an appropriate radial dependence. In order for the expansion to be convergent and useful, one relies on the property that the higher-order terms in the expansion decay increasingly quickly far away from the sources. In this case, the leading terms in a multipole expansion are generally the most significant, and the low-order behavior of the system at large distances can be approximated by the first few terms of the expansion, which are usually much easier to compute than the general solution. ...more on Wikipedia about "Multipole expansion" This is a list of some vector calculus formulae of general use in working with standard coordinate systems. ...more on Wikipedia about "Nabla in cylindrical and spherical coordinates" The parallelogram of forces is a method for solving (or visualizing) the results of applying several different forces to an object. It utilizes the principles of vectors to solve this problem called vector addition. ...more on Wikipedia about "Parallelogram of force" In mathematics, a path integral (also known as a line integral) is an integral where the function to be integrated is evaluated along a path or curve. Various different path integrals are in use. In the case of a closed path it is also called a contour integral. ...more on Wikipedia about "Path integral" This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that: ...more on Wikipedia about "Proof of angular momentum" In physics and mathematics, a pseudovector (or axial vector) is a quantity that transforms like a vector under a proper rotation, but gains an additional sign flip under an improper rotation (a transformation that can be expressed as an inversion followed by a proper rotation). The conceptual opposite of a pseudovector is a (true) vector or a polar vector. ...more on Wikipedia about "Pseudovector" The scalar resolute of a vector $\mathbf\left\{b\right\}$ in the direction of a vector $\mathbf\left\{a\right\}$ (also "$\mathbf\left\{b\right\}$ on $\mathbf\left\{a\right\}$"), is given by: ...more on Wikipedia about "Scalar resolute" In vector calculus a solenoidal vector field is a vector field v with divergence zero: ...more on Wikipedia about "Solenoidal vector field" Stokes' theorem in differential geometry is a statement about the integration of differential forms which generalizes several theorems from vector calculus. It is named after Sir George Gabriel Stokes ( 1819- 1903). The theorem acquired its name from Stokes' habit of including it in the Cambridge prize examinations. ...more on Wikipedia about "Stokes theorem" A surface normal, or just normal to a ...more on Wikipedia about "Surface normal"	2013-05-23 10:32:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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": 0.9289751648902893, "perplexity": 519.245379068516}, "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/1368703227943/warc/CC-MAIN-20130516112027-00011-ip-10-60-113-184.ec2.internal.warc.gz"}
http://mathhelpforum.com/math-topics/5881-urgent-ap-gp.html	# Math Help - Urgent! AP/GP 1. ## Urgent! AP/GP Hey I need a solution for part (b) only: A bank has an account for investors. Interest is added to the account at the end of each year at a fixed rate of 5% of the amount in the account at the beginning of that year. A man and a woman both invest money. (b) The woman decides that, to assist her in her everyday expenses, she will withdraw interest as soon as it has been added. She invests $y at the beginning of each year. Show that, at the end of n years, she will have received a total of$[n(n+1)y]/40 in interest. Thanks if you could help, I'm having my math paper in about 8 hours' time! 2. Hello, Margaritas! At a bank interest is added to the account at the end of each year at a fixed rate of 5% of the amount in the account at the beginning of that year. (b) A woman decides that she will withdraw interest as soon as it has been added. She invests $y at the beginning of each year. Show that, at the end of n years, she will have received a total of$[n(n+1)y]/40 in interest. At the end of year-1, she gets 5% of $y = 0.05y = y/20 dollars in interest. She invests$y more; at the end of year-2, she gets: 0.05(2y) = 2y/20 dollars in interest. She invests \$y more; at the end of year-3, she gets: 0.05(3y) = 3y/20 dollars in interest. . . . and so on. By the end of the nth year, she gets a total of: . . y/20 + 2y/20 + 3y/20 + . . . ny/20 dollars in interest. . . So she has: .T .= .y(1 + 2 + 3 + ... + n)/20 dollars. The sum of the first n integers is: .1 + 2 + 3 + ... + n .= .n(n+1)/2 . . Therefore: .T .= .y[n(n+1)/2]/20 .= .n(n+1)y/40 dollars. 3. Thanks Soroban, I get it now!	2015-08-04 08:07:10	{"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": 0.34383606910705566, "perplexity": 2169.313926639141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990603.54/warc/CC-MAIN-20150728002310-00213-ip-10-236-191-2.ec2.internal.warc.gz"}
https://scicomp.stackexchange.com/questions/16105/is-it-better-to-do-normalization-after-all-orthogonalization-in-gram-schmidt-pro	# Is it better to do normalization after all orthogonalization in Gram-Schmidt process? In Gram-Schmidt process, is it better to do normalization after orthogonalization of all the vectors in a basis, or to normalize each new vector immediately after it is created, from computational point of view? When you consider the orthogonalization procedure, if $u$ and $v$ are vectors, then the orthogonal component of $u$ is either $$u - (u,v)v$$ if $v$ has length $1$, or $$u - \frac{(u,v)v}{\\|v\\|^2} = u - \left(u,\frac{v}{\\|v\\|}\right)\frac{v}{\\|v\\|}$$ if $v$ has some other length. So if $v$ has not been normalized somewhere earlier in the procedure, its every appearance will be normalized anyway. So it shouldn't matter: either you normalize it immediately or you normalize it every time it appears later. • Thanks. Sorry for not being clear. I ask if each $u_i$ is normalized, immediately after it is created by orthgonalization, or after all $u_i$'s are created by orthogonalization – Tim Nov 6 '14 at 19:43 • @Tim You cannot compute the orthogonal component of a vector $u$ w.r.t. $v$ without knowing the norm of $v$. If the $u_i$'s are not normalized immediately, you will have to normalize them explicitly every time they appear later in the Gram-Schmidt procedure. As far as I can tell, there's no confusion, I answered the question you meant to ask. – Kirill Nov 6 '14 at 19:47 The short answer is yes, you should normalize the orthogonal projection of $u$ immediately after it is computed so that the orthogonalization algorithm may continue to generate the basis. However, it deserves mention that when one computes an orthogonal basis with the Gram-Schmidt process in finite precision (i.e. numerically) it is possible to create a basis that is no longer orthogonal. As a maliciously chosen example, consider the vectors $$u_1 = (1,\epsilon,0,0)^T,\quad u_2 = (1,0,\epsilon,0)^T, \quad u_3=(1,0,0,\epsilon)^T,$$ where we assume that $1 + \epsilon^2\approx 1$. Then the classical Gram-Schmidt process creates the "orthogonal" basis vectors $$q_1 = (1,\epsilon,0,0)^T,\quad q_2=\frac{1}{\sqrt{2}}(0,-1,1,0)^T,\quad q_3 = \frac{1}{\sqrt{2}}(0,-1,0,1)^T.$$ But a quick check of orthogonality yields, for example, $$q_2^Tq_3 = \frac{1}{2}.$$ This loss of orthogonality comes from the accumulation of roundoff errors in the classical Gram-Schmidt process. That is, in classical Gram-Schmidt we compute (signed) lengths of the orthogonal projections of $u_i$ onto the previous basis vectors $q_1, q_2,\ldots,q_{i-1}$, subtract these projections (and the rounding errors) from $u_i$ to obtain the new projection $w$, normalize, and obtain the next piece of the basis $q_i$. This projection (in exact arithmetic) is compactly written $$w = \left(I-Q_{i-1}Q_{i-1}^T\right)u_i,$$ where the columns of $Q_{i-1}$ are the previously computed basis vectors $q_k$, $k=1,\ldots,i-1$. But numerically, because of rounding errors, the matrix $Q_{i-1}$ does not have truly orthogonal columns. To stabilize the approximation and help guarantee that the numerical procedure will create an orthonormal basis in finite precision we use the modified Gram-Schmidt process. The difference is subtle but stabilizes the computation such that the vectors created will be "much more" orthogonal than those from classical Gram-Schmidt. The key is to ensure that the computation projects and orthogonalizes with respect to the computed version of the vector $w$. The spirit of the algorithm is the same, this is just a reordering of the computation to look like $$w = \left(I - q_{i-1}q_{i-1}^T\right)\ldots\left(I-q_1q_1^T\right)u_i.$$ Notice that in exact arithmetic the two formulations will generate identical output. If we consider the same example as before, modified Gram-Schmidt will compute the orthonormal basis $$q_1 = (1,\epsilon,0,0)^T,\quad q_2=\frac{1}{\sqrt{2}}(0,-1,1,0)^T,\quad q_3 = \frac{1}{\sqrt{6}}(0,-1,-1,2)^T,$$ for which we have discrete orthogonality (in finite precision as $\epsilon$ is assumed on the order of unit roundoff) $$q_1^Tq_2 = -\frac{\epsilon}{\sqrt{2}},\quad q_1^Tq_3 = -\frac{\epsilon}{\sqrt{6}},\quad q_2^Tq_3 = 0.$$ It is best to use Housholder Transofrmations. http://en.wikipedia.org/wiki/Householder_transformation • How does this address the question? – Kirill Nov 6 '14 at 19:32 • It doesn't, and it also doesn't add any context. Householder transformations yield a $Q$ factor in a $QR$ decomposition that has tighter error bounds than modified Gram-Schmidt, but Householder transformations can't be used to generate orthonormal bases in iterative methods like Arnoldi, so modified Gram-Schmidt is more useful in that case. – Geoff Oxberry Nov 6 '14 at 20:08 • The question is about GS. The author did not state that he was using anything like Arnoldi. So it seems very reasonable to suggest a method that performs the desired task (i.e. full orthogonalization) more stably. I think my answer is legitimate. – Yair Daon Nov 8 '14 at 3:09 • @YairDaon: Don't worry it is typical for SE moderators to behave like this to bully new users. – timur Feb 10 '18 at 3:46	2021-06-14 15:36:59	{"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": 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": 0.8670099973678589, "perplexity": 374.5496936367081}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00480.warc.gz"}
https://cancercentrum.bitbucket.io/decoder/index.html	The goal of decoder is to decode coded and perhaps cryptical variables to more understandable descriptive labels. ## Installation You can install the released version of decoder from CRAN with: install.packages("decoder") And the development version from BitBucket with: # install.packages("remotes") remotes::install_bitbucket("cancercentrum/decoder")	2022-01-17 17:06:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2958597242832184, "perplexity": 11163.168382593865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00344.warc.gz"}
http://energygym.lt/eqwwsr8a/fe8598-formula-of-parallelogram	20 Sau You can see that this is true by rearranging the parallelogram to make a rectangle. which is the magnitude of resultant. This is possible to create the area of a parallelogram by using any of its diagonals. R and let To find the area of a parallelogram, multiply the base by the height. B This means that the area of a parallelogram is the same as that of a rectangle with the same base and height: The base × height area formula can also be derived using the figure to the right. Area of Parallelogram. Diagonals of a parallelogram are the segments which connect the opposite corners of the figure. V 1. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. {\displaystyle V={\begin{bmatrix}a_{1}&a_{2}&\dots &a_{n}\\b_{1}&b_{2}&\dots &b_{n}\end{bmatrix}}\in \mathbb {R} ^{2\times n}} Quadrilateral Definitions. "Parallelogram." Formula : Perimeter of a parallelogram = 2 (L + W) Here l represent length and w represents width . det Further formulas are specific to parallelograms: The height of the parallelogram is the distance measured by drawing a line from the bottom of the parallelogram to the top. , - angles. Two pairs of opposite angles are equal in measure. Sides of a parallelogram if you know angle and height. The congruent sides are the direct consequence and it can be proved quickly with the help of equivalent formulations. − Area formula of a parallelogram Area formula using the base and height. In today's lesson, we will be recapping how to work out the area of a rectangle, a triangle and a parallelogram. Base = 6 cm and height = 4 cm. A = b ∙ h = 9 ∙ 5 = 45 m 2 . Check out … Area of a Triangle Formula & Perimeter of a Triangle, Area of Isosceles Triangle Formula \| Perimeter of a Isosceles Triangle, Volume of Box Formula \| Surface Area, Perimeter, Diagonal of a Box Formula, Perimeter of a Polygon Formula \| Diagonal & Area of Regular Polygon. It is the "parent" of some other quadrilaterals, which are obtained by adding restrictions of various kinds: A rectangle is a parallelogram but with all four interior angles fixed at 90° The area of a parallelogram is twice the area of a triangle created by one of its diagonals. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . \| a The triangles are flipped upside down from each other so that the opposing sides are parallel. All about parallelogram. sinδ: The perimeter of a parallelogram. / Cut a right triangle from the parallelogram. This geometry video tutorial explains how to find the area of a parallelogram. Let vectors In fact, you can turn a parallelogram into a rectangle to find the formula for the area of a parallelogram! The area of a parallelogram is modelled by the formula, A = lw. Suppose Height =h and Base = 3h according to question. The area of a parallelogram is equal to the magnitude of cross-vector products for two adjacent sides. Area formula of a parallelogram Area formula using the base and height. ) 2. c Let points Note: The base and height should be perpendicular. Area of a Parallelogram Formula. Since the parallelogram has 2 triangles, its area is twice the area of the triangle. 2 at the intersection of the diagonals is given by[9], When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D1 of either diagonal, then the area can be found from Heron's formula. Also, side AB is equal in length to side DC, since opposite sides of a parallelogram are equal in length. Dunn, J.A., and J.E. The perimeter of a parallelogram is the sum of the all parallelogram sides lengths. A different type of quadrilateral on the basis of symmetry is defined as the given below –. The height and base of the parallelogram should be perpendicular to each other. Thus all parallelograms have all the properties listed above, and conversely, if just one of these statements is true in a simple quadrilateral, then it is a parallelogram. S Cut a right triangle from the parallelogram. Area of a Parallelogram Formula If you know the length of base b b, and you know the height or width h h, you can now multiply those two numbers to get area using this formula: area = b × h a r e a = b × h area = 18 in × 9 in a r e a = 18 i n × 9 i n V How to Calculate the Area of a Parallelogram. ) and let - height measured at right angles to the base. All of the area formulas for general convex quadrilaterals apply to parallelograms. Explanation: . This geometry video tutorial explains how to find the area of a parallelogram. Perimeter of a parallelogram formulas: 1. C Weisstein, Eric W. A quadrilateral is a parallelogram if two pairs of opposite sides are parallel and also equal. Parallelograms can tile the plane by translation. a So, we have R = P + Q. The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. Area of Parallelogram= b×h. Derivation. ] The etymology (in Greek παραλληλ-όγραμμον, parallēl-ógrammon, a shape "of parallel lines") reflects the definition. So, the formula is. Then the area of the parallelogram with vertices at a, b and c is equivalent to the absolute value of the determinant of a matrix built using a, b and c as rows with the last column padded using ones as follows: To prove that the diagonals of a parallelogram bisect each other, we will use congruent triangles: (since these are angles that a transversal makes with parallel lines AB and DC). All of the area formulas for general convex quadrilaterals apply to parallelograms. In Euclidean Geometry, the parallelogram is the simplest form of a quadrilateral having two sides parallel to each other. , Therefore, the formula to calculate the … a The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. 2 2 As we know, there are two diagonals for a parallelogram, which intersects each other. If ABC is an automedian triangle in which vertex A stands opposite the side a, G is the centroid (where the three medians of ABC intersect), and AL is one of the extended medians of ABC with L lying on the circumcircle of ABC, then BGCL is a parallelogram. \| Geometry Formula The formula to calculate the area of a parallelogram … {\displaystyle a,b,c\in \mathbb {R} ^{2}} T To find the area of parallelogram multiply the base by its height and vice versa. Example 5: Determine the area of a parallelogram that is 5 m high whereas the base is 9 m long. Certainly, one of the basic strategies in solving geometry problems, which actually saves a lot of work, is to rely on things we have already done and proven previously, and avoid having to redo everything from the starting point. Area of Parallelogram Formula. b area and perimeter of the parallelogram. Let us now discuss the parallelogram formula i.e. The parallelogram will have the same area as the rectangle you created that is b × h . Example 1: Find the perimeter of a parallelogram whose length is 15 cm and width is 12 cm. 2 A quadrangle with two pairs of parallel sides is called a parallelogram. If the quadrilateral is convex or concave (that is, not self-intersecting), then the area of the Varignon parallelogram is half the area of the quadrilateral. 4.1 Area in terms of Cartesian coordinates of vertices; 5 Proof that diagonals bisect each other; 6 Parallelograms arising from other figures. How to find the area of a parallogram. g. Log in for more information. denote the matrix with elements of a and b. a γ ) , The area of a parallelogram is the product of the length of its base (b) and height (h). A parallelogram where all angles are right angles is a rectangle! A parallelogram is actually a rectangle with opposing right triangles on the sides. Explanation: . 4 Area formula. Area of Square & Perimeter of Square Formula, What is Rhombus? Definition. Suppose, the diagonals intersect each other at an angle y, then the area of the parallelogram is given by: Area = ½ × d 1 × d 2 sin (y) Check the table below to get summarised formulas of an area of a … {\displaystyle V={\begin{bmatrix}a_{1}&a_{2}\\b_{1}&b_{2}\end{bmatrix}}\in \mathbb {R} ^{2\times 2}} ] Area formula The area of a parallelogram is given by the formula where b is the length of any base a is the corresponding altitude . 2 [ The leaning rectangular box is a perfect example of the parallelogram. © Corbettmaths 2018 Work out the area of this parallelogram cm² Work out the area of this parallelogram cm² b Now, expand A to C and draw BC perpendicular to OC. Sides of a parallelogram; Diagonals of a parallelogram; Angles of a parallelogram; Angles between diagonals of a parallelogram; Height of a parallelogram and the angle of intersection of heights; The sum of the squared diagonals of a parallelogram The three-dimensional counterpart of a parallelogram is a parallelepiped. R The parallelogram consist of equal opposite sides and its opposite angles are equal in measure. Hence, a parallelogram could have all properties listed above if any of the statements become true then this is a parallelogram. And how we can define it. Example 5: Determine the area of a parallelogram that is 5 m high whereas the base is 9 m long. Area of a parallelogram. Area = 6 m × 3 m = 18 m 2. The area of a parallelogram is the area occupied by it in a two-dimensional plane. b Any line through the midpoint of a parallelogram bisects the area. The opposite sides of the quadrilateral are equal and parallel and the opposite angles are equal (but not equal to the angles), which is called parallel. , Strategy. In the diagram above, ABE ≅ DCF. The area of a parallelogram is the area occupied by it in a two-dimensional plane. \| a,b are the parallel sides. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. In the diagram above, ABE ≅ DCF. Unlike any other convex polygon, a parallelogram cannot be inscribed in any triangle with less than twice its area. In Euclidean geometry, the area enclosed by a parallelogram is defined by this formula: A=bh, where b stands for base and h stands for height. Area of rectangle and parallelogram has the almost similar type of formula. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. × 1 Answer/Comment. V A rectangle is one of the many fundamental shapes you'll see in math. A = b ∙ h = 9 ∙ 5 = 45 m 2 . The opposite sides of a parallelogram are the same length, and the opposite angles have the same measure. Formula for Height. The sum of the distances from any given point towards sides is equivalent to the location of the point. A parallelepiped is related to the parallelogram in the same manner how a cube related to the square and a cuboid related […] R The height is not the side length like you might use in a rectangle, but instead it is the altitude. Pretty, "Halving a triangle". This page was last edited on 5 January 2021, at 08:10. The pair of opposite sides are equal and they are equal in length. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. = Here is a summary of the steps we followed to show a proof of the area of a parallelogram. Apart from area of parallelogram formula, there area other formulas for calculating the area of a parallelogram. the perimeter of a parallelogram is the total distance of the boundaries of the parallelogram.The parallelogram has its opposite sides equal in length. R And how we can define it. = Area = 40 cm 2 The simplest parallelograms are ABFE, BCGF, CDHG, EFJI, FGKJ and GHLK. Therefore, the area of a parallelogram formula is equal to the product of its base and height. The opposite sides of a parallelogram are equal in length and angles are also the same. 2 Search for an answer or ask Weegy. Draw a parallelogram. Then the area of the parallelogram generated by a and b is equal to If you have already learnt the formula for the area of rectangle then you can easily remember the area of parallelogram too. Parallelograms - area The area of a parallelogram is the $$base \times perpendicular~height~(b \times h)$$. 2 You must use the altitude that goes with the base you choose. Asked 5/31/2017 9:12:02 PM. Further formulas are specific to parallelograms: A parallelogram with base b and height h can be divided into a trapezoid and a right triangle, and rearranged into a rectangle, as shown in the figure to the left. The adjacent angles of the parallelogram are supplementary. 2 The parallelogram is a geometrical figure that is formed by the pair of parallel sides having opposite sides of equal length and the opposite angles of equal measure. Example1: If the base of a parallelogram is equal to 6cm and the height is 4cm, the find its area. A = b×h. Find a formula for the area of a parallelogram using its height and side. If the area is 192 cm2, find the base and height. The area of a shape is the amount of space a shape takes up. Use the right triangle to turn the parallelogram into a rectangle. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. To prove the formula for the area of a parallelogram, we will form another ∆QMR whose height is drawn from the vertices Q resulting in a rectangle PQMT. The opposite angles are equal and the sum of the angles is 360°. If you have already learnt the formula for the area of rectangle then you can easily remember the area of parallelogram too.. What is parallelogram? The area A of a parallelogram is given by the formula p,q are the diagonals What is a Rectangle? Therefore, triangles ABE and CDE are congruent (ASA postulate, two corresponding angles and the included side). The midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram, called its Varignon parallelogram. An automedian triangle is one whose medians are in the same proportions as its sides (though in a different order). 2 You can use the formula for calculating the area of a parallelogram to find its height. Use the formula of Area of Parallelogram. Area of rectangle and parallelogram has the almost similar type of formula. A = 3 * 1.5 A = 4.5 square yards. 6.1 Automedian triangle; 6.2 Varignon parallelogram; 6.3 Tangent parallelogram of an ellipse; 6.4 Faces of a parallelepiped; 7 See also; 8 References; 9 External links; Special cases. In the triangle shown below, the area could be expressed as: A= 1/2ah. ∈ Pair of parallel sides are congruent, For example, AB = DC The perimeter of a parallelogram is the measurement is the total distance of the boundaries of a parallelogram. The centers of four squares all constructed either internally or externally on the sides of a parallelogram are the vertices of a square. Area of a parallelogram = Base × Height. In order to determine whether a quadrangle is parallelogram, we will use the following criteria: This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2 … Area formula using the sides and angle b The area of a parallelogram is the product of the length of its base (b) and height (h). {\displaystyle {\sqrt {\det(VV^{\mathrm {T} })}}} . The diagonals of a parallelogram bisect each other. Check out the awesome GIF below to see how the rectangle and two right triangles combine to make a parallelogram. The area of a parallelogram is also equal to the magnitude of the. A Study of Definition", Information Age Publishing, 2008, p. 22. n Example 2: The base of the parallelogram is thrice its height. A parallelogram is a quadrilateral (4-sided) shape with two pairs of parallel sides. The formula for area of a parallelogram is the same as the one for area of a rectangle. × {\displaystyle \mathbf {a} ,\mathbf {b} \in \mathbb {R} ^{2}} According to the picture, Area of Parallelogram = Area of Triangle 1 + Area of Rectangle + Area of Triangle 2, => Area of Parallelogram = $$\frac{1}{2} \times Height \times Base$$ + $$Height \times Base$$ + $$\frac{1}{2} \times Height \times Base$$, => Area of Parallelogram = $$\frac{1}{2} \times h\times b_1$$ + $$h\times b_3$$ + $$\frac{1}{2} \times h\times b_2$$, => Area of Parallelogram = $$h ( \frac{1}{2} \times b_1 + b_3 + \frac{1}{2} \times b_2$$, => Area of Parallelogram = $$h { \frac{1}{2} (b_1 + b_2) } + b_3$$, => Area of Parallelogram = $$h { \frac{1}{2} (b_1 + b_1) } + b_3$$, => Area of Parallelogram = $$h { \frac{1}{2} \times 2 b_1 }+ b_3$$, => Area of Parallelogram = $$h ( b_1 + b_3$$, According to picture $$b_1+ b_3 = Base$$, => Area of Parallelogram = $$height \times Base$$. A = (b * h) Sq. Quadrilateral Formula & Quadrilateral Theorem Proof, What is Cyclic Quadrilateral? Since the diagonals AC and BD divide each other into segments of equal length, the diagonals bisect each other. Learn to calculate the area using formula without height, using sides and diagonals with solved problems. , If two lines parallel to sides of a parallelogram are constructed. The area of a parallelogram is the region covered by the parallelogram in the 2D plane. Then the area of the parallelogram generated by a and b is equal to Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! A simple (non-self-intersecting) quadrilateral is a parallelogram if and only if any one of the following statements is true:[2][3]. {\displaystyle \gamma } The perimeter of a parallelogram is the measurement is the total distance of the boundaries of a parallelogram. All geometry formulas for any triangles; Parallelogram. If only two sides are parallel then it is named as the Trapezoid and the three-dimensional counterpart is taken as the parallelepiped. See more. Each pair of opposite sides in a parallelogram are equal. 1 To find the area of a parallelogram, multiply its base with its height. 1 ∈ The formula for the area of a parallelogram can be used to find a missing length. det To find the area of a parallelogram, multiply the base by the height. The height (altitude) is found by drawing a perpendicular line from the base to the highest point on the shape. To answer this question, we must find the diagonal of a rectangle that is by .Because a rectangle is made up of right angles, the diagonal of a rectangle creates a right triangle with two of the sides. Question. How to find the area of a parallelogram without a height? The opposite sides being parallel and equal, forms equal angles on the opposite sides. The area of a parallelogram is the space contained within its perimeter. . Parallelograms and rectangles are pretty similar. Q2) Count the number of parallelograms in the given figure 1) 12 2) 16 3) 18 4) 20 Solution: The figure may be labelled as shown. 2 In fact, you can turn a parallelogram into a rectangle to find the formula for the area of a parallelogram! Each pair of conjugate diameters of an ellipse has a corresponding tangent parallelogram, sometimes called a bounding parallelogram, formed by the tangent lines to the ellipse at the four endpoints of the conjugate diameters. a b Use the formula of Area of Parallelogram. + A parallelogram is a quadrilateral with opposite sides parallel. Keywords: parallelogram; base; height; formula; Background Tutorials. Solution: Since they are perpendicular, the side measurement and the height can be used in the formula. a … Area of a parallelogram = Base × Height. Finding the area of a parallelogram explained. Online Area Calculator n If you're seeing this message, it means we're having trouble loading external resources on our website. what is a parallelogram What is a parallelogram. Area of Parallelogram = Base x Height Parallelogram Formula Example Problems Solving the equation for w will be w = A/l. The diagonals of a parallelogram divide it into four triangles of equal area. Find sides of a … Some important formulas of Trapezium, Kite, and Parallelogram are given below. Area of Rectangle Formulas, List of Maths Formulas for Class 9th CBSE, Proof Formulas of Area of Equilateral Triangle & Right Angle Triangle, Mid Point Theorem Proof – Converse \| Mid Point Theorem Formula, What is Square? ∈ New answers. Use the right triangle to turn the parallelogram into a rectangle. Opposite sides of a parallelogram are parallel (by definition) and so will never intersect. The height or altitude of a parallel is the perpendicular line, (usually dotted) from the vertex of a parallelogram to any of the bases. Specifically it is. , - sides. 3. V These are 6 in number. To answer this question, we must find the diagonal of a rectangle that is by .Because a rectangle is made up of right angles, the diagonal of a rectangle … Understand why the formula for the area of a parallelogram is base times height, just like the formula for the area of a rectangle. — if the two pairs of opposite sides in a quadrangle are equal, then this quadrangle is a parallelogram; — if two opposite sides in a quadrangle are parallel and equal, then this quadrangle is a parallelogram; — if, in a quadrangle, the diagonals bisect each … a b With the help of a basic list of parallelogram formulas, you can calculate the area and the perimeter by putting the values and derive the final output. 1 a Area & Perimeter of a Rhombus Formula, Trapezoid Formula – Perimeter & Area of a Trapezoid Rule Formula, Perimeter of Hexagon Formula \| Area of a Hexagon Formula, What is Triangle? 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In other words, the formula for the area of rectangle AEFD, or....	2021-08-04 12:46:16	{"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": 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": 0.8592451810836792, "perplexity": 387.724970024928}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00018.warc.gz"}
https://kerodon.net/tag/038A	# Kerodon $\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$ Definition 5.6.6.1 (Corepresentable Functors). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing an object $X$, let $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{S}}$ be a functor, and let $x$ be a vertex of the Kan complex $\mathscr {F}(X)$. We will say that $x$ exhibits $\mathscr {F}$ as corepresented by $X$ if, for every object $Y \in \operatorname{\mathcal{C}}$, the composite map \begin{eqnarray} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) & \xrightarrow {\mathscr {F}} & \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \mathscr {F}(X), \mathscr {F}(Y) ) \\ & \simeq & \operatorname{Fun}( \mathscr {F}(X), \mathscr {F}(Y) ) \\ & \xrightarrow {\operatorname{ev}_ x} & \mathscr {F}(Y) \end{eqnarray} is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$; here the second map is the inverse of the homotopy equivalence $\operatorname{Fun}( \mathscr {F}(X), \mathscr {F}(Y) ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{S}}}( X, Y)$ supplied by Remark 5.5.1.5. We say that the functor $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{S}}$ is corepresentable by $X$ if there exists a vertex $x \in \mathscr {F}(X)$ which exhibits $\mathscr {F}$ as corepresented by $X$. We say that the functor $\mathscr {F}$ is corepresentable if it is corepresentable by $X$, for some object $X \in \operatorname{\mathcal{C}}$.	2022-05-21 22:31:10	{"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": 0.9993835091590881, "perplexity": 590.7916656875875}, "config": {"markdown_headings": true, "markdown_code": false, "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/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00517.warc.gz"}
https://imanuelcostigan.github.io/fmdates/reference/actual_360.html	This calculates the years between two dates using the actual/360 day basis convention. This convention counts the number of calendars between the start and end dates and assumes a year consists of 360 days. This is also known as the A/360, Act/360 or French day basis convention. actual_360(date1, date2) ## Arguments date1 A date-time object A date-time object ## Value a numeric value representing the number of years between date1 and date2. ## Details The year fraction is calculated as: $$\frac{Number of calendar days}{360}$$ The order of date1 and date2 is not important. If date1 is less than date2 then the result will be non-negative. Otherwise, the result will be negative. ## See also Other counter methods: actual_365, actual_actual_isda, is_valid_day_basis, thirty_360_eu_isda, thirty_360_eu_plus, thirty_360_eu, thirty_360_us, thirty_360, year_frac	2019-08-22 16:11:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5723836421966553, "perplexity": 7463.404115872536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00351.warc.gz"}
https://gateoverflow.in/321920/isi2014-dcg-35	+1 vote 40 views Let $A$ and $B$ be disjoint sets containing $m$ and $n$ elements respectively, and let $C=A \cup B$. Then the number of subsets $S$ (of $C$) which contains $p$ elements and also has the property that $S \cap A$ contains $q$ elements, is 1. $\begin{pmatrix} m \\ q \end{pmatrix}$ 2. $\begin{pmatrix} n \\ q \end{pmatrix}$ 3. $\begin{pmatrix} m \\ q \end{pmatrix} \times \begin{pmatrix} n \\ p-q \end{pmatrix}$ 4. $\begin{pmatrix} m \\ p-q \end{pmatrix} \times \begin{pmatrix} n \\ q \end{pmatrix}$ recategorized \| 40 views 0 Anyone know this que?? +1 vote The question actually asks how many ways are there to find a collection of $p$ elements of which it contains exactly $q$ elements from $A$ and the rest from $B$. Obviously $q\le m$ and $p\le (m+n)$. Here, $\|A\|=m, ~\|B\|=n$ So, there are $\begin{pmatrix} m\\ q \end{pmatrix}$ ways to find $q$ elements from the set $A$ and the rest $(p-q)$ elements can be chosen from the $B$ which has $\begin{pmatrix} n\\ p-q \end{pmatrix}$ ways. $\therefore~$The required number of ways $=\begin{pmatrix} m\\ q \end{pmatrix}\times\begin{pmatrix} n\\ p-q \end{pmatrix}$. So the correct answer is C. by Active (3.6k points) +1 vote	2020-01-27 02:50:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8613997101783752, "perplexity": 186.64211927421434}, "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-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00021.warc.gz"}
https://www.aimsciences.org/article/doi/10.3934/dcds.2005.12.973	# American Institute of Mathematical Sciences October 2005, 12(5): 973-982. doi: 10.3934/dcds.2005.12.973 ## Smoothing effect of the generalized BBM equation for localized solutions moving to the right 1 Laboratoire de Mathématiques, Université Paris-Sud, 91405 Orsay, France Received November 2003 Revised August 2004 Published February 2005 We prove $C^\infty$ smoothness and uniform exponential decay for $H^1$-localized solutions moving to the right of the generalized BBM equation. For that purpose we use a monotonicity property for solution which are not necessarily close to solitary waves. Citation: Khaled El Dika. Smoothing effect of the generalized BBM equation for localized solutions moving to the right. Discrete & Continuous Dynamical Systems - A, 2005, 12 (5) : 973-982. doi: 10.3934/dcds.2005.12.973 [1] Omid Nikan, Seyedeh Mahboubeh Molavi-Arabshai, Hossein Jafari. 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Electronic Research Archive, 2020, 28 (4) : 1503-1528. doi: 10.3934/era.2020079 2019 Impact Factor: 1.338	2020-12-03 14:19:37	{"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": 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": 0.6121634840965271, "perplexity": 8511.569114396278}, "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/1606141727782.88/warc/CC-MAIN-20201203124807-20201203154807-00674.warc.gz"}
https://dsp.stackexchange.com/questions/31291/demodulating-upper-sideband-usb-signals	# Demodulating upper sideband (USB) signals A modulating signal $m(t)= \sin(2 \pi \nu_m t)$ is transmitted via a carrier of frequency $\nu_c$ using upper sideband (USB) modulation. Show that the USB modulated signal can be demodulated using a reconstructed carrier signal and a low pass filter. Attempt: USB (SSB signals in general) can be demodulated by multiplying them with a reconstructed carrier $\cos(2\pi \nu_c t).$ So I first tried to write an expression for the transmitted USB modulated signal: $$f(t)=A \Big[ m(t) \cos(2 \pi \nu_c t) - \hat{m}(t) \sin(2\pi \nu_c t) \Big]$$ $$f(t)=A \Big[ \sin(2 \pi \nu_m t) \cos(2 \pi \nu_c t) - \hat{m}(t) \sin(2\pi \nu_c t) \Big] \tag{1}$$ Where $A$ is some amplitude, and $\nu_c >> \nu_m$. So to proceed further, I think we need to calculate the Hilbert transform $\hat{m}(t).$ It is possible to do this by finding the imaginary part of the analytic signal: $$m_a (t) = m(t) + j \hat{m} (t)= 2 \int^\infty_0 M(\nu) \exp(j 2 \pi \nu t) d\nu \tag{2}$$ But the Fourier transform of $m(t)$ is tedious, so I am wondering if there is a simpler way of doing this? $$f(t)=m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\tag{1}$$ (as you've correctly stated in your question). Now if you multiply with a (coherent) carrier $2\cos(2\pi\nu_ct)$, you get \begin{align}g(t)&=2\big[m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\big]\cos(2\pi\nu_ct)\\ &=2m(t)\cos^2(2\pi\nu_ct)-2\hat{m}(t)\sin(2\pi\nu_ct)\cos(2\pi\nu_ct)\\&=m(t) (1+\cos(4\pi\nu_ct))-\hat{m}(t)\sin(4\pi\nu_ct)\tag{2}\end{align} Applying a low pass filter to $(2)$ will filter out all terms at $2\nu_c$, leaving you with the message signal $m(t)$.	2019-08-21 17:08:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8658657073974609, "perplexity": 493.90088205868364}, "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/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00399.warc.gz"}
https://physics.aps.org/articles/v3/s26	Synopsis # Air showers from ultrahigh-energy cosmic rays Physics 3, s26 Cascades created by cosmic rays interacting with the atmosphere provide clues about the mass composition of ultrahigh-energy cosmic rays. The nature and origin of ultrahigh-energy ( $>1018eV$) cosmic rays remain a mystery. Astrophysicists hunt for clues regarding their mass composition, which, along with other properties such as the flux and arrival direction distribution, should help distinguish among the various models of the sources and propagation of cosmic rays. However, because of the low flux of ultrahigh-energy cosmic rays, the mass composition cannot be measured directly. Instead, it is inferred from measurements of the extensive air showers—the cascades of high-energy ions created when incident cosmic rays collide with atoms in the atmosphere. The atmospheric depth at which development of a shower reaches its maximum number of secondary particles depends on the mass and energy of the incident particle. Data on the depth of the maximum thus provide information about the mass composition. The Pierre Auger Collaboration has presented in Physical Review Letters new measurements of extensive air showers from ultrahigh-energy cosmic rays. Their observations suggest a gradual increase in the average mass of cosmic rays in a region of energies around $1019eV$. The data are the highest statistics measurements to date. – Stanley Brown ## Related Articles Cosmology ### Dark Matter Detector Proves its Sensitivity A new sensor provides world-leading sensitivity for distinguishing lightweight dark matter from background noise. Read More » Cosmology ### Looking for Dark Matter in Neutron Star Light Radio observations of neutron stars demonstrate a way to search for axions through their expected conversion to electromagnetic waves in a star’s magnetic field. Read More » Particles and Fields ### Dark Matter Detector Delivers Enigmatic Signal Are the excess events detected by the XENON1T experiment a harbinger of new physics or a mundane background? Read More »	2020-10-30 14:57:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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": 0.3080959916114807, "perplexity": 1407.5778885641073}, "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-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00219.warc.gz"}
https://mathematica.stackexchange.com/questions/238480/ndsolve-methods?noredirect=1	# NDSolve methods In solving a large system of ODEs I found that Method->ExplicitRungeKutta (also Automatic as a matter of fact) is fast, but often unstable. I tried the option Method->ImplicitRungeKutta, but for the given number of ODEs (3924) the memory used by the kernel grows above 50GB. Eventually I have to stop the calculation. What options of NDSolve can be tried to improve the stability and remain with reasonable memory consumption? • Generally, implicit methods require a lot more work (and storage) per step, but if a problem is stiff (cf. this answer), an implicit method will take fewer steps than an explicit method. Then, you have methods that are able to switch between using an explicit and implicit method (e.g. "StiffnessSwitching"), depending on the ODE's behavior. Without seeing what your ODEs look like, it's hard to say more. Jan 20 '21 at 9:29 • @J.M. right, but as far as I understand, the "StiffnessSwitching" switches from explicit to implicit, but never goes back to explicit. Jan 20 '21 at 9:47 • Actually it does; that's why it supports a "NonstiffTest" as well as a "StiffnessTest" option. See the discussion in Stiffness Detection. Jan 20 '21 at 10:22 • What do you mean by unstable? My personal experience is the default setting of NDSolve is quite robust for IVP of ODE(s). (There exist rare cases that one needs to adjust MaxStepSize though. ) But who knows… As @J.M. said, it's hard to advise without a concrete example. Jan 20 '21 at 11:00 • @yarchik Did you read this paper arxiv.org/pdf/1505.02290.pdf ? Jan 21 '21 at 21:04	2022-01-19 14:49:52	{"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": 0.5849713087081909, "perplexity": 1002.7132160465078}, "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/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00319.warc.gz"}
http://madlogos.github.io/recharts/Basic_Plots_15_Heatmap.html	First, you should load recharts: library(recharts) # 1 Introduction Heatmap has 1 basic type: heatmap. The keys are: • No need of x • numeric y, numeric lat and lng # 2 Function Call echartr(data, y, lng, lat, <series>, <t>, <type>) Arg Requirement data source data in the form of data.frame y numeric dependent variable. Only the first one is accepted. y represents the heat value which is between 0 and 1. If y is not within this range, recharts will normalize it. series series variable which will be coerced to factors. Only the first one is accepted if multiple variables are provided. lng longitude or x-coordinate. lat latitude or y-coordinate. t timeline variable which will be coerced to factors. Only the first one is accepted if multiple variables are provided. type ‘heatmap’ # 3 Showcase ## 3.1 Data Preparation Here is a fictious dataset. data = rbind( data.frame( lng=100+rnorm(100,0, 1)600, lat=150+rnorm(100,0, 1)50, y=abs(rnorm(100,0,1))), data.frame( lng=rnorm(200,0, 1)1000, lat=rnorm(200,0, 1)800, y=abs(rnorm(200,0,1))), data.frame(lng=400+rnorm(20,0, 1)300, lat=5+rnorm(20,0, 1)10, y=abs(rnorm(100,0,1)))) str(data) ## 'data.frame': 400 obs. of 3 variables: ## $lng: num 539.4 43.9 668.4 -192.1 -90.3 ... ##$ lat: num 166 162 273 203 121 ... ## \$ y : num 0.591 0.801 0.908 0.498 1.195 ... ## 3.2 Basic Plot echartr(data,lng=lng,lat=lat,y=y,type='heatmap') %>% setTitle("Heatmap", "Fictious Data") Heatmap is more useful in maps. You can refer to addHeatmap for help. # 4 Futher Setup Then you can configure the widgets, add markLines and/or markPoints, fortify the chart. You can refer to related functions to play around on your own.	2020-07-14 18:54:31	{"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": 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": 0.2642006278038025, "perplexity": 10146.660267970325}, "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-2020-29/segments/1593657151197.83/warc/CC-MAIN-20200714181325-20200714211325-00366.warc.gz"}
https://datascience.stackexchange.com/tags/clustering/hot	# Tag Info 5 Classification is a problem where your input data consists of elements with 2 parts: Some data features that reflect the traits of an entity A label that assigns the entity to a group or class. With that kind of data, you can train a model that receives the data features (first part) and generates the label (second part). This kind of training, where you ... 3 Is this approach a correct approach, or logical with respect to machine learning principles? It will affect the performance of the model in the sense that your algorithm learned to separate the clusters based upon distance according to all the features. I have read discussions about how to calculate feature importance on unsupervised problems like yours, so ... 2 I would use all the features and see how the separateness of my clusters behave according to some metric, for example, silhouette score Additionally, it is very important to scale your data prior to clustering since kmeans is a distance-based algorithm. heart_data = pd.read_csv("https://archive.ics.uci.edu/ml/machine-learning-databases/00519/... 2 Clustering would indeed give you something like group1, group2, etc., i.e. it would assign every instance to a group meant to represent similar instances together. With the model you could also assign any new instance to one of the groups in the model, i.e. find the group that this instance is the most similar to. In general clustering doesn't give you which ... 1 Interesting question. The answer is: It depends. The best way to find out how it would affect your model is with the shap package. You can use it to uncover the importance of features and reveal interaction effects in the model. There could be a very different effect depending on how „important“ the excluded features are. Let‘s assume a very simple decision ... 1 Naive approach: define a similarity or distance function, say for instance cosine similarity. Calculate the similarity score between any pair $(x_i\in A, y_j\in B)$ Define a precision level, say $\epsilon=0.000001$. The assumption is that it's extremely unlikely that two vectors would be this close by chance in $A$. For every $y_j\in B$, find the set \$c(y_j)... 1 Suppose if you use kmeans clustering then you can 1.train and save the model using pickle 2.loa the model using pickle 3.pass your new sample as a vector to the predict function of the loaded model object model.predict([[0, 0], [12, 3]]) this will give you only one cluster label 4.if you want to get top n clusters that the sample might belong to then ... 1 Yes it can happen. In fact it is quite normal since there are different clusters in 2D and different in 3D, since more or less information is added to data (by having more dimensions). This is a by-product of the curse of dimensionality. Adding as more relevant information as possible would make clusters more close to underlying groups. So 3D would be better ... 1 I would start with the simple option: represent every person as a boolean vector in which every position represents the answer to a particular question (the length is the total number of questions). Then you can apply any standard clustering algorithm, such as K-means (hierarchical clustering would probably also work with data like this). 1 Clustering and recommendation are similar tasks, however in recommendation you usually want to recommend several items while clustering usually assigns each sample to only one cluster. Anyway for your problem a clustering or even a classifier might help. If labels are assigned on the basis of a similarity metric (and you have a good guess of what this metric ... 1 [Note: essentially my answer is the same as @ncasas, just an alternative phrasing] Classification belongs to supervised learning whereas clustering belongs to unsupervised learning: In supervised learning there is a training stage during which some instances (examples) are provided together with their answer (the target). During training the model "... 1 Your case is where K=number of points in dataset : K-means: Lets suppose, there are 10 data points and k=10, so you have 10 clusters. the new test point will be matched with the cluster nearest to it KNN: If K=1, then the new test point classified will be same as in K-means. So, if there is one data-point per cluster, then your given answer ie. knn is ... 1 That is not clustering since the rows are not being grouped together. It is a filtering problem where the threshold to keep a row is based on frequency in each column. 1 OpenRefine is the tool that you need. The clustering functionality is accessible without creating a text facet by using the "Cluster & Edit" function (under "Edit Cells") will take you directly to the clustering dialog. The clustering itself is actually pretty fast, but, until recently, there were scalability issues with displaying ... 1 Depending on the representation of your sentences, you have different similarity metrics available. Some might be more suited to the representation you are using than others. One of the most popular metrics is the cosine distance. However, you have other available in the literature, such as: Jaccard similarity Sørensen–Dice coefficient Tversky index You ... Only top voted, non community-wiki answers of a minimum length are eligible	2021-01-16 18:23:27	{"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": 0.5847320556640625, "perplexity": 755.8167646495789}, "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/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00327.warc.gz"}
https://databasefaq.com/index.php/tag/vim	FAQ Database Discussion Community ## What is difference between Vim's clipboard “unnamed” and “unnamedplus” settings? vim,clipboard What is the difference between these 2 settings? set clipboard=unnamed set clipboard=unnamedplus Which one should I use in order to have multi-platform .vimrc?... ## How to delete a WORD in vim in insert mode vim Is there a way how to delete a whole WORD (e.g. from vim path/to/some/file.txt to vim) in insert mode? To delete a word in insert mode the C-W shortcut works well. I got used to use C-W in bash too but bash actually deletes whole WORDs. Now I'd love to... ## Bash: How to generate lines with sequential numbers? vim,sed I need to create a list which will hold fqdn's of about 30 servers. Until now, whenever I needed to create such a list, I would: Open vim Manually insert the first line Yanking the first line and then pasting the line as many times as the amount of servers... ## Unknown function: pathogen#infect vim,nerdtree,pathogen I just installed MacVim (and did an override of Vim). And I am trying to get Pathogen to work. When I start Vim, I get the following error: Error detected while processing /Users/nir/.vim/autoload/pathogen.vim: line 1: E477: No ! allowed: <!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN"> line 2: E488: Trailing characters:... ## open a completion popup from normal mode vim,omnicomplete Gived this script (packages suggestion was simplified, original takes care of <cword>) function! CompleteImport() let packages = ['java.util.Vector','java.lang.String'] call complete(col('.'),packages) return '' endfunction inoremap <F8> import <C-R>=CompleteImport()<CR> while on insert mode you can add an import and choose between suggested packages pressing F8 But I want to be enable to... ## Install Vim via Homebrew with Python AND Python3 Support python,python-3.x,vim,homebrew I would like to enable Python auto-completion in Vim so maybe this is a non-issue. This is what I've observed: First, Virtual Environments provides the ability to assign an interpreter on a per-project basis. The assumption being both Python 2.x AND Python 3.x could potentially be used. I've installed both... ## How can I map keys in Insert mode that contains Carriage Return in it? vim,text,insert,editor,vi How can I map ctrl + enter to go to new line from anywhere in the current line in vim ? I just want to do it for insert mode. 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I have also managed... ## is there any vim plugin to select block of code in any language? vim I mean, if my cursor is pointing somewhere inside of javascript else block, is there a plugin for one shortcut to be able to select all lines inside this else block, regardless to on which line exactly I'm currently in this else block pointing my cursor, so that, if I... ## vimscript syntax not pattern vim,vim-syntax-highlighting ## Why does Vim make certain object keys yellow by default? javascript,node.js,vim,sails.js I use Vim as my terminal editor and one of the things I noticed was that it marks certain words such as 'status' yellow by default. Here's a screenshot: I'm little confused because I'm not sure whether it's suggesting that I don't use that word in my model object?... ## Select a different keystroke than colon ':' to enter command-line mode in vi vim,vi Is there any way to achieve that remapping? I looked a bit into map and inoremap but they appear to affect within a given mode, not how to enter a given mode.... ## Delete backwards from cursor to the end of the previous line in Vim? vim,vi Say I want to edit the following line: var myVar = "I am a string!"; So that it looks like this: var myVar = "I am a string!"; Is there a movement that goes to the end of the previous line?... ## Custom mappings in help text buffers vim I tend not to use windows much in Vim (I prefer tabs), but there are certain things that use them, such as help text buffers. I'm fine with that, but with how seldom I use windows, it usually takes me a second to remember how to close the window (<C-w>c)... ## VIM Convert Text to URL with Search/Replace vim I have a document that contains long filenames, followed by a hyphen, followed by a description of the contents of the file. The files are all PDFs. I am converting this document into a page on our website, so that it has the filename, which should be a link to... ## Vim: How to search for more than one word in the same search? search,vim Is it possible to search within vim for more than one word? For example: Let's say i'm inside vim and I press / in order to start a search for a certain word, is it possible to search for more than one word in the same search?... ## bash script “ignores” .vimrc bash,vim,autocmd I have a problem as follows: I have a script which copies a log file from remote machine, does some modification on it and then opens it in vim, problem is vim doesn't auto recognize the file type (which outside of the script id does) – I need this for... ## open VIM with default code [duplicate] vim This question already has an answer here: How can I automatically add some skeleton code when creating a new file with vim 8 answers Automatic syntax/headers in vim for c++ files 3 answers I don't know if it's correct to ask this question here. I am a programming enthusiastic... ## Vim says 'missing :endif', but I'm sure its there vim,vim-plugin Consider the following vim code. function! s:foo() if s:bar() let a = 1 endif endfunction function! s:bar() python << EOF vim.command('echoerr "blabla"') EOF endfunction command! Foo call s:foo() map <F7> :Foo<CR> Open a vim session and press F7. Vim wil then throw an error saying E171: Missing :endif. How is... ## How can I “register” a new help.txt document via .vimrc? bash,vim I am writing a bash script which installs and configures vim plugins, i.e. it creates a new .vimrc and then echoes the config text into it. #! /bin/bash echo "installing Pathogen plugin manager" mkdir -p ~/.vim/autoload ~/.vim/bundle && \ curl -LSso ~/.vim/autoload/pathogen.vim https://tpo.pe/pathogen.vim echo "installing ctrlp" git clone https://github.com/kien/ctrlp.vim.git ~/.vim/bundle/ctrlp.vim... ## Automatically switch back to NerdTree after pressing “o” on file vim,nerdtree When I press "o" on any file in NerdTree list, it opens this file in one of currently opened tabs. Sometimes, when I'm not 100% sure what file to open, I have to do this: 1) o (open file) 2) ctrl-ww (switch back to nerdtree) 3) j or k (go... ## vim comment/uncomment with one mapping [duplicate] This question already has an answer here: What's a quick way to comment/uncomment lines in Vim? 32 answers I'm new(bie) in vim. I've got the following mapping to comment my python code : nmap cc 0i#<ESC> I would like to have the same mapping to uncomment a line. I... ## expand delimited text over multiplie lines string,vim I am looking for a power vim tip - to expand specially formatted text over multiple lines. say i have Bill \| 12 \| Male \| Blue \| right \| Pay 150k i'd like it to expand (after some vim magic) to Name: Bill Age: 12 Gender: Male Eyes: Blue... ## VIM: Use python3 interpreter in python-mode python,vim,ubuntu-14.04,python-3.4,python-mode I have recently switched to vim and configured it for Python-programming using this tutorial. Before, I have made sure that vim supports python3 (vim --version shows +python/dyn and +python3/dyn) using this article. But when executing a file from python-mode, still the python2.7 interpreter is chosen. How can I configure vim... ## How can I 'change in number' or 'change in digits' in Vim vim,numbers,editor I'm regularly editing CSS in Vim and need to change a value like: width: 300px; to something else like: width: 178px; Normally what I do is navigate to the 3 in 300 and type cw to change the word. But if I do this I have to type the px... ## Highlight bash internal varibles using VIM bash,vim,vim-syntax-highlighting Is it possible to Highlight bash internal varibles using VIM? For example the variables described on this page would appear a different colour than the user defined variables. http://tldp.org/LDP/abs/html/internalvariables.html ... ## Selecting text and saving it with keymap vim I've made c# command line program that takes a window handle and a string as parameters. It brings the window with the handle to the foreground, uses SendKeys to send the string that was passed, and brings the window that was previously at the foreground back to the foreground. I... ## How to exit insert mode using autocommand upon losing focus? vim Currently, I am using the following command as suggested by Auto save files when focus is lost. autocmd FocusLost * wall! However, I cannot apply <ESC> trailing the wall!. Vim will complain about it and I partially enter normal mode. (The message-box actually takes the focus of cursor.) Any suggestion... ## Right way to set up Rust using Vim vim,rust,rust-cargo I've finally started to dive into Rust and want to clarify some issues when it comes to setting up everything nicely. I'm using vim on Linux and found a nice plugin for syntax highlighting. Autocompletion is somewhat troublesome though, using phildawes/racer. The plugin needs the src location for Rust which... ## Any way to ignore ^M carriage returns in VIM? git,vim,github Most files I'm editing in VIM have carriage returns, so I'm often seeing every line with a ^M at the end. One thing I used to do was to replace them with a command like +e ff=dos, but when I commit with git, it thinks I've changed the entire file.... ## how to replace NumberInt(1) to 1 with vim? regex,vim,replace I want to replace NumberInt(1) to 1 except in the string in my file with some tool: "bz":NumberInt(1), "batch": NumberInt(2), "something": "something else NumberInt(1)" "bz":NumberInt(1), "batch": NumberInt(2), "something": "something else NumberInt(2)" "bz":NumberInt(1), "batch": NumberInt(2), "something": "something else NumberInt(3)" "bz":NumberInt(1), "batch": NumberInt(2), "something": "something else NumberInt(4)" "bz":NumberInt(1), "batch": NumberInt(2), "something": "something... ## Enforcing coding styles in Visual Studio and VIM c++,visual-studio-2010,visual-studio,vim,coding-style I work with a medium sized team of developers, with half being Linux developers using VIM on Ubuntu and MacVIM on OSX, and the other half being Windows developers using Visual Studio 2010 or later. A fair bit of time has been wasted in the past when handling things like... ## Find entries with zero distance variance and recorded watts regex,xml,vim,garmin I'm a cyclist and a programmer. During my rides, I'm recording data to xml files using a phone based gps tracker and a power meter. After the ride, I use the power meter software to merge the data and then upload to a web site. On the website, the resulting... ## How does this series of keystrokes in vim delete everything but the numbered lines? vim Working through this just the middle kata, Leave only the numbered lines. LINE 1 LINE 2 LINE 3 That's all. Thank you very much. the following keystroke sequence makes sense and does something in the buffer: djGd2kZZ It's basically chaining commands together. But what is the following doing exactly, and... ## Dynamic pattern for vim highlight vim In the vim help, there is a suggestion to use highlight groups for highlighting text greater than the textwidth: Another example, which highlights all characters in virtual column 72 and more: :highlight rightMargin term=bold ctermfg=blue guifg=blue :match rightMargin /.\%>72v/ I would like this to always reflect the value of texwidth... ## Vim not showing messages for less than 3 occurrences substitutions vim,macvim Lets say Text1 has 4 lines 1 2 1 1 After :%s/1/A/g, Vim will give message says 3 substitutions on 3 lines Text2 has 3 lines 1 2 1 :%s/1/A/g There are no messages showing. I typed ":messages" nothing in the history either, so I wonder if there's a way... ## VIM - Reformatting indentation and braces vim,coding-style,vim-plugin When working with blocks of code in VIM, I'm able to easily re-indent blocks of code via selecting a region in visual mode (SHIFT+v), then just hit =. This re-tabs lines of code, uses the correct indentation depths, hard-tabs vs spaces, etc. I have a large set of functions I... ## How to use Ctags to list all the references of a symbol(tag) in vim? vim,ctags Can I list all the references of a symbol in vim using Ctags? like the 'Find References' In some GUI IDE. The Ctag tutorial just tell how to locate the definition of a symbol ,not all the usages.... ## How do you know what VIM help keyword to search? vim I was trying to google how to search within a scope in VIM and found this link. Limiting search scope for code in Vim I wouldn't imagine to come up with the keyword "/\%V" to search help in VIM help. So, I'm wondering how most people get help from VIM... ## Assigning save (:w) to w in vim vim I want to be able to save a file in vim while in the insert mode. I thought of using the following shortcut: inoremap <leader>w <Esc>:w<cr> While the shortcut saves the file in the insert mode, it leaves the cursor one spot ahead of where the cursor would be if... ## how to change the regex responsible for vim's “current word” search vim Is there any way to modify the "current word" search (asterisk) settings so it includes certain characters in its definition of "word". For example, I have a lot of CSS code where there is a dash character in the class names. So, if I have a class named "blue-text" and... linux,osx,vim,zsh I'm a huge vim lover, but I can't find a way to get vim to follow symlinks when opening files. As an example, all the dotfiles in my home dirs are symlinked to within the .zprezto directory: .vimrc -> ~/.zprezto/runcoms/vimrc .zshrc -> ~/.zprezto/runcoms/zshrc I'm keeping my fork of .zprezto in... ## error when vim macro recording play vim,macros I know how to record a macro use vim. eg. record macro a: in normal mode, qa start recording, do some action, q stop recording. If I want to play it, I can use @a. Now I want to play 10 times, [email protected] should be work, But for me, It... ## Disable Wrapping Cursor in Eclipse eclipse,vim,vi I'm mostly using Vi (edit: Vim) for text editing and the viPlugin for Eclipse Juno. Eclipse seems to have a very annoying behaviour of "wrapping" the cursor to the next/previous line, if yout reach the end/start of a line through cursor keys. (Reproduce by placing the cursor on the first... ## Setting tab width based on file type vim I'm trying to learn to use Vim. I like indenting by 4 spaces but use 2 spaces for certain languages, such as Nim and MoonScript. I tried adding this to my .vimrc: autocmd BufNewFile,BufRead,BufEnter .nim :setlocal tabstop=2 shiftwidth=2 softtabstop=2 Problem? It doesn't do anything! All that happens is that the... ## NerdTree window size is too large when current directory “.” vim,nerdtree Every time I type vim . and select some file, nerdtree is too large. there's a way to fix it? ... ## Replacing a string with empty line vim,sed,vi I have a file which has 2000 lines of data (file name is data.tsv). I want to replace the string with empty line where there is a matching pattern, in my case is PMC: How can I do with Vim or other sed command? Thanks, Rio ... ## The “-” key goes up a line in vim vim When I hit the "-" key on my keyboard vim goes up a line. My .vimrc is blank and when I type in :map - vim says "no mappings found." I have no idea what could be causing this. I am using a Mac—could it be a Terminal.app problem? Thanks... ## In Vim, how can I compile a .java file in /src/ and run its class file from /bin/? java,vim I'm using Windows, and my _vimrc file has the following lines: autocmd Filetype java set makeprg=javac\ -d\ %:~:h:s?src?bin?\ % set errorformat=%A%f:%l:\ %m,%-Z%p^,%-C%.%# map <F9> :make<Return>:copen<Return> map <F10> :cprevious<Return> map <F11> :cnext<Return> map <F12> :!start cmd /k "java %:~:s?src?bin?:r" I just want to use this for quick editing, and I will... ## Measuring time spent in certain types of files in vim vim,macvim Is there a way to measure how much time I spend in certain types of files while working in vim? I would like to measure how much time I spend writing application code as opposed to tests (they all have a predefined file name _spec.rb) ## Strange vim registers behavior when used with execute "normal vim Why does the following command does not use the content of what is first registered in the unnamed register, if this content is a number ? execute "normal! cc".(@"+1)."\<esc>" ... ## Prevent single letter deletes to spam clipboard manager vim,clipboard With unnamed clipboard option included, single letter deletes spam my clipboard manager Is it possible to fix this so that single letter deletes (with x) are not yanked to clipboard. ## Vim search using regular expression regex,vim I want to search a string which starts with "abc" and ends with "xyz" in vim. Below are the commands I've tried: :1,$g/abc[\w\W]xyz/ :1,$g/abc$\\w\\W$\xyz/ :1,$g/abcxyz/ "[\w\W]" means the texts between "abc" and "xyz" can be any characters "1,$" means the search range is from the 1st line to the last... ## vim: E14 Invalid Address, reassign variable inside if statement vim I'm trying to make my tab spacing dependent on file type. But I get the error E14 Invalid Address on line 3. function! Tabs() let t = 4 if (&filetype ==? 'yaml') \|\| (&filetype ==? 'yml') t = 2 endif " size of a hard tabstop let &tabstop=t " size... ## Using vim to develop instead of Emacs on Windows vim,emacs I have used Emacs for a long time, say, 6 or 7 years. And it seems that I got Emacs Pinky somehow. Now I am trying to switch to vim, and it's a very good editor, just like Emacs, except that I wonder how you guys develop with it. Using... ## Substitute both the beginning and end of every occurrence of a string with vim vim I have more than a hundred of occurrences of the string ?dep= followed by two numbers. So it goes like ?dep=01, ?dep=02, ?dep=03, etc. I need to change the string but keep the numbers at the same time, so for example ?dep=01 needs to become {{ path('annonce-index', {departement: 01}) }}... ## Vim highlighting sometimes requires a kickstart? vim Centos 6, vim 7.2.411, my .vimrc file: autocmd BufNewFile,BufRead *.html set filetype=php set tabstop=4 set shiftwidth=4 set nowrap set autoindent set number if has('mouse') set mouse=a endif set backspace=start,indent,eol set t_Co=256 colorscheme molokai (Although I code in php I save my files as .html - a bad practice I imagine?... ## How can I save in vim a file with the actual fold text? (“+— 43 lines […]”)? vim I am looking for a way to save to a new text file a file that is folded, with all the folds closed. In other words, just as I see it on the screen. Is it possible? (I will have to print the code later, and parts of it are... ## how to apply macro on list of lines with vim linux,vim,macros let's suppose I want to apply a macro stored on register 1 on lines: 2, 5, 9 but without applying on lines in between. I saw that you can give a range but that is not what I want. I've tried some variation on :2, 5, 9 @1 without success ## How to highlight rulerformat vim,highlight,rc I want highlight the rulerformat in my .vimrc. I have set ruler set rulerformat=%55(%{strftime('%a\ %b\ %e\ %I:%M\ %p')}\ %5l,%-6(%c%V%)\ %P%) I have subsequently tried all of the following: hi rulerformat ctermbg=1 hi rulerformat ctermbg=red hi Group1 ctermbg=red "if you modify the rulerformat slightly you can try to access group 1... ## match wordchar and/or dot string of anylength regex,vim This matches a word /\w\+ This matches a any number of dots /\.\+ Why doesn't this match any number of words combined by dots? /[\w\.]\+ ? The w seems to be matching actual 'w's instead of a word character, whether I escape it or not.... ## Different color schemes on vim vim,fortran In my lab, I can either work directly on my work station or on some server that is made available to us. On both my accounts, I have the same .bashrc and .vimrc files. Now looking at the picture below, you can see that there is some minor difference about... ## Pylint Error when using metaclass python,python-3.x,vim,pylint,syntastic i try to fix all pylint errors and pylint warnings in a project. but i keep getting an error when i set a metaclass (https://www.python.org/dev/peps/pep-3115/). here is my example code: #!/usr/bin/env python3 class MyMeta(type): pass class MyObject(object, metaclass=MyMeta): # pylint error here pass the error just says "invalid syntax". i... ## How to map a key in command-line mode but not in search mode vim I want to map a key in VIM in command-line mode but not in search mode(with a leading /), like below: Map Q to q Map W to w Sometimes I typed the wrong command in VIM, like :Q and :W, and I want to make these wrong commands to... ## How to set a portion of the status line a certain color only if an element is evaluated- Vim vim In Vim I'm trying to make my status-line to work how I want it to. Here's what I'm going for: When the file I'm editing has been saved, I want it to look like this when saved: And this when unsaved: The former I see, but I get this when... ## Change vim-airline colorscheme via vim-sunset plugin vim,autocmd I just recently discovered two VIM plugins that I find very useful, namely vim-sunset (http://vimawesome.com/plugin/sunset) and vim-airline (http://vimawesome.com/plugin/vim-airline-sad-beautiful-tragic). Changing the colorscheme with vim-sunset was easy, but I would like it to also change the colors in my airline. I have tried to achieve this via function! Sunset_daytime_callback() " Version 1...	2019-12-15 04:48:57	{"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29977381229400635, "perplexity": 4423.250459099054}, "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/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00225.warc.gz"}
https://blog.bitwrangler.com/2013/02/19/its-the-volatility-that-will-kill-you.html	# It's The Volatility That Will Kill You February 19, 2013 [agile] [planning] Volatility is what Pivotal Tracker uses to measure the consistency of your team’s work output. You can use that number to help you estimate the first approximation to answer the eternal question, “Will I make the deadline?” ## One fine day at the office… You’ve scoped out 100 points worth of stories for the Next Big Release™. Pivotal Tracker shows your velocity is 10 points per week. Your annual review is in 3 months and on-time delivery of this high profile project will figure prominently. Then the CEO walks over to your desk and asks, “Will I make the launch date, 10 weeks from now?” What do you say? 1 “Yes, my lord. Of course we’ll make our date! I’m 100% certain of it; Behold; Tracker says we’ll finish in 10 parsecs.” 2 “Probably; We had some iterations that cleared 30 points, but last week we were working on bugs and only accepted 2 points. A couple weeks of those and we might miss the deadline.” 3 “There’s no clear answer. There are so many other uncertainties, technical debt, QA, deployment work.” ## It’s a trap! Hopefully, you answered with “none of the above.” Velocity is just one measure of how your project is performing. Staking your career on it would be foolhardy. The second answer is honest, yet hopelessly vague. The third reply is why many people still think Agile is a way to duck your responsibilities as a software professional. There is hope, however; We can use Pivotal Tracker’s tools to make a better (albeit imperfect) estimate. ## Past Performance is No Guarantee of Future Returns, but Yesterday’s Weather is Often Good Enough VelocityJust like a speedometer that measures how fast you’re hurtling through space, Tracker’s velocity is a measurement of how fast your team completes stories. Instead of miles or kilometers per hour, Tracker expresses velocity as the number of points completed per iteration (normally a week). Because Tracker stories are assigned point values instead of due dates, Tracker calculates velocity by averaging the number of points you’ve completed over the past few iterations. In Tracker, past predicts future. , week over week, varies; sometimes a lot, sometimes a little. It depends on the project. Ideally, each iteration would have the same mix of stories, bugs and chores and Velocity would be very consistent. Steady velocity is a good thing™. In the real world, however, all sorts of things crop up; Your head-count goes up (or down), business priorities shift (or pivot), deferred technical debt demands payment, quality assurance files a slew of bug reports, user testing reveals flaws in product, visual design changes. The real world creates volatilityMathematically, Volatility is Standard Deviation ÷ Mean Velocity in your Velocity. A simple measure of this is standard deviation, which Tracker constantly computes for your project. Using that metric, you can decide what you should watch or change in order to meet your goals. Let’s go back to our example and look at the velocity charts in Tracker. Assuming that we have a normal distribution of weekly velocities, the first sigma (±35%) will fall into the range of 10±3.5 points each week. That is, there is a 70% probability that your project will deliver all 100 points somewhere between 8 and 16 weeks. Why so much spread? 40% volatility is a big number! In the worst case scenario, where every iteration delivers only 6.5 points, it gets you to your goal in 100 ÷ 6.5 ≅ 16 weeks. ## I Find Your Lack of Faith Disturbing By now, you’ve had your meeting with the CEO. You’ve shown him the stories left in the backlog, the volatility of the project, and the range of estimates for delivery. This is the beginning of a conversation. If your team is not comfortable with the worst case scenario, something must change and, really, you have only two choices; you can reduce volatility or you can reduce scope. You will probably need to do both. Alas, there is no simple formula here. This is where skill, experience and insight will come into play. Here are some suggestions: ## Reduce Volatility It’s criticalIf stories languish in the accept/reject state (a field of red and green buttons in the backlog is a strong indicator), several bad things may happen to your project: You lose the fast feedback loop between delivery and deployment. Developers will move on to the next story and may have already lost context about past ones. Unaccepted stories can not be deployed, so there’s less and later feedback about the feature in the full project. that stories are accepted as soon as possible after they are delivered. Is the project manager unable to accept stores as they are delivered, so they don’t get credited in the iteration where they started? You can backdate acceptance to reflect when the stories were ready (rather than when the PM accepted them), but it is not something I would do on a regular basis. Are the stories marked as bugs and chores truly overhead, or are they “stealth” features? Does the story add business value to the product? That’s a feature. Flaws introduced by feature stories are bugs. Design changes surfaced by testing is a new feature. Are there too many stories in flight? Can you deliver stories more reliably by starting fewer at a time? Study after study shows that human beings do not* multitask well at all. Do one thing, do it well, then move on. Are there blocked tracks? Do stories get stalled because of dependencies? Can you reorganize your backlog so each story is independent. Are there outside resources, out of your control, that are introducing volatility? Multiple rejected stories are toxic. If your team is getting more than one or two rejects each week, this may be a sign that your stories are not accurately representing what your product manager intended. It’s time to look at your work flow to prevent them from happening so often. Are you not refactoring enough? Constant, steady refactoring, delivered during each story is much better than giant refactors that last a week. You should consider refactoring as critical to your process and not something to do “later, when you have more time for it.” Make all of your projects small by breaking them up. Delivering a project on time is always tricky business. I’ve discovered that it is actually easier to work on projects with short time-lines (6 weeks seems to be a good number). Urgency and a looming dead-line focuses the mind in wonderful ways. As a tactical measure, simplify your pointing strategy. Pivotal Tracker offers many pointing “styles;” linear, quadratic, fibonacci, or you can customize your own. Try going simpler (instead of finer granularity); a 0-1-2-3 scale (easy, medium and hard), might give you a more accurate picture. ## Reduce Scope What’s really at risk if you miss the deadline? Often, the perceived urgency is far greater than the actual risk to the project. Are there features that you can jettison? Are there features that you can defer? Are you spending too much time on “pixel-perfect fidelity?” Talk to your designers; look for ways to reduce complexity. One good way to reduce complexity is to lean more heavily on standard user interface libraries (which might affect the unique visual design of the project). Can you make “soft releases” where you deliver fewer features, earlier, to reduce risk? Look at your project goals again. Are the stories in the backlog truly delivering features that will meet your goals? Are there parallel “tracks” that allow you to add man-power to the project (but see below). ## Watch for Icebergs Do you need to stand up a new production environment? That will take time. It’s a point-able story. Make sure that all the necessary steps to release are in the budget. Are you refactoring as you go? Have you been postponing technical debt? Those interest payments will start to pile up as you get closer to release time. Make sure you and your team know that keeping the code clean is an essential part of every story. Anything that changes your team will change both Volatility and Velocity. Are you adding a new team member? Remember Brook’s Law: Adding manpower to a late software project makes it later. Vacations, holidays, sick days and babies will affect your velocity. Remember to account for it in Tracker. ## You’re all clear, kid, now let’s blow this thing and go home! This article should give you a lot to think about. Good project management is hard work. When projects are just getting started, everything feels fine, and later you start to wonder when everything went to hell. Remember, volatility kills. It's The Volatility That Will Kill You - February 19, 2013 - Ken Mayer	2019-09-16 00:58:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.19807034730911255, "perplexity": 2534.383381981018}, "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/1568514572439.21/warc/CC-MAIN-20190915235555-20190916021555-00485.warc.gz"}
https://codereview.stackexchange.com/revisions/214917/6	6 of 7 added 45 characters in body # Version of C# StringBuilder to allow for strings larger than 2 billion characters In C#, 64bit Windows, .NET 4.5 (or later), and enabling gcAllowVeryLargeObjects in the App.config file allows for objects larger than two gigabyte. That's cool, but unfortunately, the maximum number of elements that C# allows in an array is still limited to about 2^31 = 2.15 billion. Testing confirmed this. To overcome this, Microsoft recommends in Option B creating the arrays natively. Problem is we need to use unsafe code, and as far as I know, unicode won't be supported, at least not easily. So I ended up creating my own BigStringBuilder function in the end. It's a list where each list element (or page) is a char array (type List<char[]>). Providing you're using 64 bit Windows, you can now easily surpass the 2 billion character element limit. I managed to test creating a giant string around 32 gigabytes large (needed to increase virtual memory in the OS first, otherwise I could only get around 7GB on my 8GB RAM PC). I'm sure it handles more than 32GB easily. In theory, it should be able to handle around 1,000,000,000 * 1,000,000,000 chars or one quintillion characters, which should be enough for anyone! I also added some common functions to provide some functionality such as fileSave(), length(), substring(), replace(), etc. Like the StringBuilder, in-place character writing (mutability), and instant truncation are possible. Speed-wise, some quick tests show that it's not significantly slower than a StringBuilder when appending (found it was 33% slower in one test). I got similar performance if I went for a 2D jagged char array (char[][]) instead of List<char[]>, but Lists are simpler to work with, so I stuck with that. I'm looking for advice to potentially speed up performance, particularly for the append function, and to access or write faster via the indexer (public char this[long n] {...} ) // A simplified version specially for StackOverflow / Codereview public class BigStringBuilder { List<char[]> c = new List<char[]>(); private int pagedepth; private long pagesize; private long mpagesize; // https://stackoverflow.com/questions/11040646/faster-modulus-in-c-c private int currentPage = 0; private int currentPosInPage = 0; public BigStringBuilder(int pagedepth = 12) { // pagesize is 2^pagedepth (since must be a power of 2 for a fast indexer) this.pagedepth = pagedepth; pagesize = (long)Math.Pow(2, pagedepth); mpagesize = pagesize - 1; } // Indexer for this class, so you can use convenient square bracket indexing to address char elements within the array!! public char this[long n] { get { return c[(int)(n >> pagedepth)][n & mpagesize]; } set { c[(int)(n >> pagedepth)][n & mpagesize] = value; } } public string[] returnPagesForTestingPurposes() { string[] s = new string[currentPage + 1]; for (int i = 0; i < currentPage + 1; i++) s[i] = new string(c[i]); return s; } public void clear() { c = new List<char[]>(); currentPage = 0; currentPosInPage = 0; } // See: https://stackoverflow.com/questions/373365/how-do-i-write-out-a-text-file-in-c-sharp-with-a-code-page-other-than-utf-8/373372 public void fileSave(string path) { StreamWriter sw = File.CreateText(path); for (int i = 0; i < currentPage; i++) sw.Write(new string(c[i])); sw.Write(new string(c[currentPage], 0, currentPosInPage)); sw.Close(); } public void fileOpen(string path) { clear(); int len = 0; while ((len = sw.ReadBlock(c[currentPage], 0, (int)pagesize)) != 0){ if (!sw.EndOfStream) { currentPage++; if (currentPage == c.Count) c.Add(new char[pagesize]); } else { currentPosInPage = len; break; } } sw.Close(); } public long length() { return (long)currentPage * (long)pagesize + (long)currentPosInPage; } public string ToString(long max = 2000000000) { if (length() < max) return substring(0, length()); else return substring(0, max); } public string substring(long x, long y) { StringBuilder sb = new StringBuilder(); for (long n = x; n < y; n++) sb.Append(c[(int)(n >> pagedepth)][n & mpagesize]); //8s return sb.ToString(); } public bool match(string find, long start = 0) { //if (s.Length > length()) return false; for (int i = 0; i < find.Length; i++) if (i + start == find.Length \|\| this[start + i] != find[i]) return false; return true; } public void replace(string s, long pos) { for (int i = 0; i < s.Length; i++) { c[(int)(pos >> pagedepth)][pos & mpagesize] = s[i]; pos++; } } // Simple implementation of an append() function. Testing shows this to be about // as fast or faster than the more sophisticated Append2() function further below // despite its simplicity: public void Append(string s) { for (int i = 0; i < s.Length; i++) { c[currentPage][currentPosInPage] = s[i]; currentPosInPage++; if (currentPosInPage == pagesize) { currentPosInPage = 0; currentPage++; if (currentPage == c.Count) c.Add(new char[pagesize]); } } } // This method is a more sophisticated version of the Append() function above. // Surprisingly, in real-world testing, it doesn't seem to be any faster. public void Append2(string s) { if (currentPosInPage + s.Length <= pagesize) { // append s entirely to current page for (int i = 0; i < s.Length; i++) { c[currentPage][currentPosInPage] = s[i]; currentPosInPage++; } } else { int stringpos; int topup = (int)pagesize - currentPosInPage; // Finish off current page with substring of s for (int i = 0; i < topup; i++) { c[currentPage][currentPosInPage] = s[i]; currentPosInPage++; } currentPage++; currentPosInPage = 0; stringpos = topup; int remainingPagesToFill = (s.Length - topup) >> pagedepth; // We want the floor here // fill up complete pages if necessary: if (remainingPagesToFill > 0) { for (int i = 0; i < remainingPagesToFill; i++) { if (currentPage == c.Count) c.Add(new char[pagesize]); for (int j = 0; j < pagesize; j++) { c[currentPage][j] = s[stringpos]; stringpos++; } currentPage++; } } // finish off remainder of string s on new page: if (currentPage == c.Count) c.Add(new char[pagesize]); for (int i = stringpos; i < s.Length; i++) { c[currentPage][currentPosInPage] = s[i]; currentPosInPage++; } } } } In response to Peter Taylor's answer: not sure why c isn't private Ah oops, it was, and I forgot to change it back when doing some lazy testing. better style to use 1L << pagedepth Ah, yes though maybe pow() is more intuitive for some. Shouldn't this have bounds checks? Not if we want maximum speed? The onus is on the caller to make sure their code is correct. I understand that's the norm, but I prefer lowercase so my methods/classes are distinguished from C#'s own. Why? I don't think it sheds Ah yes, ignore that. Missing some disposal: I'd use a using statement. Doesn't Close() dispose by itself, at least when the method ends and the StreamWriter variable is outside scope, like any other normal class object would be? If not, I could use sw=null;. I prefer not to nest code when possible. That's completely unnecessary here: StreamWriter has a method Write(char[], int, int). Thanks. Yikes! That should be mentioned in the method documentation. I mean it's called fileOpen not fileAppend()... :) I think this can give rise to inconsistencies. Other methods seem to assume that if the length of the BigStringBuilder is an exact multiple of pagesize then currentPosInPage == 0 and c[currentPage] is empty, but this can give you currentPosInPage == pagesize and c[currentPage] is full. Well spotted! Why is this a method rather than a property? Some calculation required, so maybe it's borderline? Why use multiplication rather than <<? Good idea! Though in fairness, length() isn't usually constantly called in tight loops What is 8s? Sorry I don't follow StringBuilder also has a method which takes (char[], int, int). Ah good point. What does this method do? The name implies something regexy, but there's no regex in sight. You're right that it's like StartsWith, but with an offset. It just looks to see if the given string is a match to the text at a given offset. Bounds checks? Speed? I'm not surprised. It's still copying character by character. I expected it to be faster because there isn't the If statement (if (currentPosInPage == pagesize)) for every character that's appended. It's probably faster to use ToCharArray() and Array.Copy, and almost certainly faster to use either unsafe and Array.Copy or ReadOnlySpan.CopyTo. Thanks, I'll look into those. Where would I use ToCharArray()?	2020-07-09 19:49:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.24726824462413788, "perplexity": 9410.602590562732}, "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-2020-29/segments/1593655901509.58/warc/CC-MAIN-20200709193741-20200709223741-00418.warc.gz"}
https://www.groundai.com/project/a-posteriori-error-estimation-and-adaptive-algorithm-for-the-atomisticcontinuum-coupling-in-2d/	A Posteriori Error Estimation and Adaptive Algorithm for the Atomistic/Continuum Coupling in 2D # A Posteriori Error Estimation and Adaptive Algorithm for the Atomistic/Continuum Coupling in 2D ## Abstract Atomistic/continuum coupling methods aim to achieve optimal balance between accuracy and efficiency. Adaptivity is the key for the efficient implementation of such methods. In this paper, we carry out a rigorous a posterior error analysis which includes the residual estimate, the stability constant estimate and the error bound, for a consistent atomistic/continuum coupling method in 2D. Adaptive mesh refinement algorithm can be designed correspondingly and the convergence rate with respect to degrees of freedom is optimal compare with a priori error estimates. wangh@scu.edu.cn mliao@xs.ustb.edu.cn plin@maths.dundee.ac.uk lzhang2012@sjtu.edu.cn HW was partially supported by NSFC grant 11501389, 11471214 and Sichuan University Starting Up Research Funding No. 2082204194117. PL and ML were partially supported by NSFC grant 91430106 and Fundamental Research Funds for Central Universities Nos. 06108038 and FRF-BR-13-023. PL and HW were partially supported by EMS RSF grant. LZ was partially supported by NSFC grant 11471214, 11571314 and the One Thousand Plan of China for young scientists. ## 1Introduction Atomistic/continuum (a/c) coupling methods are a class of computational multiscale methods that aim to combine the accuracy of the atomistic model and the efficiency of the continuum model for crystalline solids with defects [25]. To be precise, the atomistic model can be applied in a small neighborhood of the localized defects such as vacancies, dislocations, and cracks, while the continuum model (e.g., Cauchy-Born rule) can be employed away from the defect cores where elastic deformation occurs. The construction and analysis of different a/c coupling methods have attracted considerable attention in the research community in recent years [15]. We refer the readers to [22] for a review of such methods. The goal of the mathematical analysis is to find the optimal relation of accuracy vs. degrees of freedom. The a priori analysis has been given for several typical a/c coupling methods, for example the QNL method (quasi-nonlocal quasicontinuum method) [23], the BQCE method (blended energy-based quasi-continuum method) [14], the BQCF method (blended force-based quasi-continuum method) [17], the GRAC method (geometric reconstruction based atomistic/continuum coupling) [36] and the BGFC method (atomistic/continuum blending with ghost force correction) [38]. In contrast, only few research articles are concerned with the a posteriori error control of the these methods. Prudhomme et al. [39] uses the goal-oriented approach to provide a posteriori error control for a three dimensional nanoindentation problem with the quantity of interest being the force acting on the indenter. The error estimator, whose effectiveness is validated only numerically, is given as a modification of a rigorous derived residual functional. Arndt and Luskin [1] use the same approach to analytically derive the a posteriori error estimates for a one dimensional Frenkel-Kontorova model. The goal oriented a posteriori error estimators are used to optimize the choice of the atomistic region as well as the finite element mesh in the continuum region. All of these work employ the original energy-based quasicontinuum method as the underlying model which is later shown to be inconsistent [6]. Recently, Kochmann et al. [45] proposed an adaptivity strategy for the so-call “fully-nonlocal quasi-continuum” method which apply a discrete model in the entire computational domain without coupling of different models. Such method is in a slightly different spirit from the problem we consider in the present work. The a posteriori error bounds in the energy norms are first derived in [32] through a similar approach as the present work for two different types of coupling schemes which are (later characterized in [16]) the QCP method and the QNL method. However, the QCP method analyzed in [32] only contains finite element discretization without the coupling of different models while the extension to higher dimensions of the scheme in [26] is not clear. A recent advance in this direction [35] is the a posteriori error analysis of a consistent energy-based coupling method developed in [40], where the a posteriori error estimators in the energy norm and in energy itself were proposed. Despite all those developments, the rigorous mathematical justification of a posteriori error estimate beyond 1D is still missing. In this paper, we present a rigorous a posteriori error estimate for a consistent energy-based a/c method in two dimension which is of physical significance, and has not been considered so far to the best knowledge of the authors. We use the residual-based approach [46] to establish the estimate in negative Sobolev norms following [35]. Two features distinguish our problem from the classic residual-based estimate for finite element approximation of the elliptic equations. The first one is the existence of the modeling error which is in origin different from the applications of quadrature rules. The second one is that the mesh may not be further refined when it almost coincides with the reference lattice but a model adaptation should be imposed. The 2D results rely on the analysis and implementation of divergence free field, which characterize the essential difference of 2D compared with 1D results in [26], where the analysis can be carried out by explicit calculations. Similar to the a priori analysis of GRAC in [36], we constrain ourselves to the case of nearest-neighbor interactions. Although the analysis can be extended to finite range interactions and to other a/c coupling methods, we decide not to include these so that the main ideas and steps are clearly presented without the distraction from the unnecessary complexity of the presentation. The paper is organized as follows. In § Section 2 we set up the atomistic, continuum and coupling models for point defects. In § Section 3.1 we derive the residual for the coupling scheme. In § Section 3.2 we obtain the stability bound for the coupling method, and then, in § Section 3.3 we present rigorous a posteriori error estimates. We present the numerical algorithms and results in § Section 4 and we give the conclusion and the outlook of our work in § Section 5. Some auxiliary results are given in § Section 6. ## 2Formulation We will first give a brief review of a model for crystal defects in an infinite lattice in the spirit of [9] in § Section 2.1 and the Cauchy-Born continuum model in § Section 2.2. We then present a generic form of a/c coupling schemes in § Section 2.3, the consistent scheme GRAC will be introduced specifically in § Section 2.4. ### 2.1Atomistic model #### Atomistic lattice and defects Given , non-singular, is the homogeneous reference lattice which represents a perfect single lattice crystal formed by identical atoms and possessing no defects. is the reference lattice with some local defects. The mismatch between and represents possible defects , which are contained in some localized defect cores such that the atoms in do not interact with defects (see § ? and § ? regarding interaction neighbourhood). For example, for a crystal with a single point defect at , and one can choose a proper radius such that , where . For different types of point defects, we have • for a vacancy at ; • for an interstitial at but ; • for an impurity at , the difference of the impurity atom with other atoms can be characterized by the inhomogeneity of interaction potentials (§ ?) This characterization of localized defects can be straightforwardly generalized to multiple point defects and micro-cracks, for example, see the setup of the model problem in § Section 4.2. Straight screw dislocations can be enforced through the appropriate choice of boundary conditions [9]. #### Lattice function and lattice function space Given , , denote the set of vector-valued lattice functions by A deformed configuration is a lattice function . Let be the identity map, the displacement is defined by for any . For each , prescribe an interaction neighbourhood with some cut-off radius . The interaction range is defined as the union of lattice vectors defined by the finite difference of lattice points in and . To measure the error for lattice functions we need to introduce function norms and function spaces on the lattice. Define the “finite difference stencil” . Higher-order finite differences, e.g., and can be defined in a canonical way. A lattice function norm can hence be defined using those notations. For , let the lattice energy-norm (a discrete -semi-norm) be The associated lattice function space is defined by We choose to be the collection of all the nearest neighbour bonds in the reference lattice, and for , denote . Then the energy norm can be reformulated as The homogeneous lattice naturally induces a simplicial micro-triangulation . In 2D, , where . Let be the P1 nodal basis function associated with the origin; namely, is piecewise linear with resepct to , and and for and . The nodal interpolant of can be written as We can introduce the discrete homogeneeous Sobolev spaces with semi-norm . It is know from [30] that and are equivalent. #### Interaction potential For each , let denote the site energy associated with the lattice site . In this paper, we consider the general multibody interaction potential of the generic pair functional form [44]. Namely, the potential is a function of the distances between atoms within interaction range and there is no angular dependence. Accordingly, we have the following equivalent forms of interaction potentials, A great number of practical potentials are in the form , including the widely used Embedded Atom Model (EAM) [5] and Finnis-Sinclair model [11]. For example, assuming a finite interaction neighborhood and an interaction range for , EAM potential reads for a pair potential , an electron density function and an embedding function . The potential is homogeneous outside the defect region , namely, and for . and have the following symmetry: , and . The energy of an infinite configuration is typically ill-defined. However, if we redefine the potential as the difference , which is equivalent to assuming , the energy functional is a meaningful object. By [9], given some natural technical assumptions for the site potentials , is well-defined for , and if is in its variables, is times Fréchet differentiable. In particular, we define as the Lipschitz constant of . Under the above conditions, the goal of the atomistic problem is to find a strongly stable equilibrium , such that We call strongly stable if there exists such that . It is proven in [9] that, if the homogeneous lattice is stable, for a critical point of , , exhibit the following generic decay, where . ### 2.2Continuum model Let be the homogeneous site potential outside the defect core. To formulate atomistic to continuum coupling schemes we require a continuum model compatible with defined through a strain energy density function . A typical choice in the multi-scale context is the Cauchy–Born model [8], the energy density in the homogeneous crystal is defined by ### 2.3A/C coupling We begin by giving a generic formulation of an a/c coupling method and employ concepts and notation from various earlier works, such as [25], but adapt the formulation to the settings in this paper. First, we choose a computational domain : let be a simply connected, polygonal and closed set, such that for some . We decompose , where the atomistic region is again simply connected and polygonal, and contains the defect core: . Let be the radius of . Take as a regular simplicial partition (triangles for or tetrahedra for ) of the continuum region . Next, we decompose the set of atoms into a core atomistic set and an interface set (typically a few “layers” of atoms surrounding ) such that . Let be the triangulation of induced by . is the associated triangulation of . Sometimes, it is also useful to define . Please see Figure 1 for an illustration of the computational mesh. We may use instead of by dropping the subscript if there is no confusion. Let be the set of nodes in , and be the set of edges in . Hence, the space of coarse-grained displacements is Suppose is the Voronoi cell associated with a lattice point . For each , we can choose the associated effective Voronoi cell such that for , but for the effective Voronoi cell depends on the geometry of the interface (see [36]). Let be the volume ratio of with resepect to . For each element we define the effective volume of by We note that only if . Now we are ready to define the generic a/c coupling energy functional , where is a modified interface site potential which satisfies certain consistency conditions, and are suitable coefficients. The details will be discussed immediately in Section § Section 2.4 and references therein. The goal of the approximate variational problem for a/c coupling is to find in the subscript of and can be omitted if there is no confusion. ### 2.4Consistent Atomistic/Continuum Formulation The choice of the interface potential in is the key in the formulation of atomistic/continuum coupling methods. In order to demonstrate the a posteriori error estimate for the generic a/c coupling methods, we shall restrict ourselves to the GRAC type methods [36]. #### The patch tests A key condition that has been widely discussed in the a/c coupling literature is that should exhibit no “ghost forces”. We call this condition the force patch test, namely, In addition, to guarantee that approximates the atomistic energy , it is reasonable to require that the interface potentials satisfy an energy patch test If an a/c method satisfies the patch test and , it is called a consistent a/c method. #### GRAC: Geometric reconstruction based consistent a/c method To complete the definition of the consistent a/c coupling energy and of the associated variational problem , we must specify the interface region and the interface site potential. The geometric reconstruction approach was pioneered by Shimokawa et al [43], and then modified and extended in [7]. We refer to [37] for details of the implementation of geometric reconstruction based consistent atomistic/continuum (GRAC) energy for multibody potentials with general interaction range and arbitrary interfaces,. The extension of GRAC to 3D is a work in progress [10]. Our prototype implementation of GRAC is for the 2D triangular lattice with Let , then , , are the nearest neighbour directions in . Given the homogeneous site potential , we can represent in terms of . For each , let be free parameters, and define A convenient short-hand notation is we name the parameters as the reconstruction parameters. The parameters are chosen so that the resulting energy functional satisfies the energy and force patch tests and . A sufficient (and likely necessary) condition for the energy patch test is that for all and . This is equivalent to In addition, optimal condition and stabilisation mechanism were proposed in [37] and [28] to improve the accuracy and stability of GRAC scheme. #### Stress formulation The stress tensor based formulation can be obtained from the first variation of the energy. For any , and , there exist piecewise constant tensor fields , and which satisfy the following identities here is the micro-triangulation induced by the reference lattice . We call an atomistic stress tensor, a continuum stress tensor, and an a/c stress tensor. For the nearest neighbour interactions, we can choose the following atomistic stress tensor, continuum stress tensor, and a/c stress tensor following respectively from the first variations -, We call piecewise constant tensor field divergence free if By definitions , it is easy to know that the force patch test condition is equivalent to that is divergence free for any constant deformation gradient . The discrete divergence free tensor fields over the triangulation can be characterized by the non-conforming Crouzeix-Raviart finite elements [36]. The Crouzeix-Raviart finite element space over is defined as The following lemma in [36] characterizes the discrete divergence-free tensor field. We also have the following corollary, #### A Priori Error Estimate In the analytical framework proposed in [19], the numerical error can be split into 3 parts: the modeling error due to the discrepancy between the atomistic model and the continuum model at the interface and the finite element edges, the coarsening error due to finite element discretization of the solution space in the continuum region, and the truncation error due to the finite size of the computational domain. It is proven in [9] that there exists a strongly stable solution to and a constant for GRAC such that, With the generic decay property , and the following quasi-optimal conditions: • the radius of the atomistic region satisfies, • the finite element mesh is graded so that the mesh size function for satisfies, It holds that there exists a constant , depending on , , , , and such that for sufficiently large, In particular, when , and when P1 finite elements are used in the continuum region, we have, where is the overall degrees of freedom. ## 3Error Analysis We present the a posteriori error analysis in this section. In § Section 3.1, we derive the residual estimate for the consistent GRAC a/c coupling scheme introduced in § Section 2.4. Then, in § Section 3.2, we give a lower bound for the stability constant which is computable from the a/c solution . Finally, we put forward the main results Theorem ? and Theorem ? in § Section 3.3. ### 3.1Residual Estimate To be more precise, we restrict ourselves to the case of nearest neighbour multibody interactions, namely, we use the so-called “grac23” method introduced in [36] as the a/c coupling mechanism. We will extend the formulation to general short-range multibody interactions in future works. For lattice function , we denote its continuous and piecewise affine interpolant with respect to the micro-triangulation by . Identifying , we can define the (piecewise constant) gradient and the spaces of compact and finite energy displacements, respectively, by It can be shown that that is dense in [9]. The first variation of the atomistic variational problem is to find such that The first variation of the a/c coupling variational problem is to find such that We introduce the truncation operator as in [9] by first choosing a cut-off function for and for . Define for by where is defined by The residual is defined as an operator on which is given by By , denote , and take , where is the modified Clément operator [4] whose definition will be made clear in the following sections. By we can separate the residual into three groups The first group , represents the truncation error; the second group represents the modeling error; and the third group represents the coarsening error. Note that the first variation of the coupling energy is given by We will deal with the contribution from those three groups separately in the following subsections. #### Truncation error To analyze the truncation error , we need the Lemma 7.3 for the truncation operator in [9], namely, if the radius of the computational domain is sufficiently large (for example, in the nearest neighbour case, we only need ), the following estimates hold where are independent of . For any , the stress-based formulation of the first variation , the fact that for , the equivalence of and , and Cauchy-Schwarz inequality lead to, Thus, the truncation error estimator is given by #### Modeling error In the analysis of the modeling error , the stress based formulation leads to, As a result, we define the modeling error estimator by, #### Coarsening error For the coarsening error , we first observe that Here, we choose to be the modified Clément interpolation [4] of . For any node in the triangulation , let be the piecewise linear function with respect to , and be the support of . The Clément interpolation operator can be defined as By definition, satisfies the Dirichlet boundary condition. The Clement interpolation possesses the following property [3], for any element , interior edge , where is the diameter of , is the length of , the element patch , and the edge patch . The constants and depend only on the shape regularity of . We also have the stability of the Clément interpolation, where the constant only depends on the shape regularity of . For notational convenience, we suppose that each interior edge has a prescribed orientation. and are the triangles on the left hand side and right hand side of the edge , and are the associated unit norm vector. The integration by parts of leads to,	2019-06-27 00:48:03	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8501006960868835, "perplexity": 584.005573550242}, "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/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00456.warc.gz"}
https://notesfromkevinrvixie.org/2012/12/14/226/	# An Invitation to Geometric Measure Theory: Part 1 While there are a variety of article-length introductions to geometric measure theory, ranging from Federer’s rather dry AMS Colloquium Talks to Fred Almgren’s engaging Questions and Answers to Alberti’s Article for the Encyclopedia of Mathematical Physics, I will take a different approach than has been taken in any of these and introduce geometric measure theory through the vehicle of the derivative. ### The Derivative, Geometrically The derivative that is encountered for the first time in calculus is defined as the limit of a ratio of the “rise” over “run” of the graph of a function. For $y = f(x)$, this becomes $\frac{df}{dx}$$(a)=\lim_{x\rightarrow a}$$\frac{f(x) - f(a)}{x-a}$. This is visualized as the slope of the secant lines approaching a limit – the slope of the tangent line – as the free ends of those lines approach $(a,f(a))$. This is illustrated in the first figure. The derivative as $\hat{L}_a$, the optimal linear approximation to f at a, is another, very useful way to think about the derivative. Here, we focus on the fact that the tangent line at $(a,f(a))$ approximates the graph of $f(x)$ at $(a,f(a))$ as we zoom in on the graph. More precisely, writing $x = h+a$, $f(x) = f(h+a) = f(a) + \hat{L}_a(h) + g(h)h$, where $\hat{L}_a$ is linear in $h$, $g(h)\rightarrow 0$ as $h \rightarrow 0$, and the tangent line L is the graph of the function $y = f(a) + \hat{L}_a(x-a)$ . Exercise: use the facts that (1) linear $\hat{L}_a:\Bbb{R} \rightarrow \Bbb{R}$ have the form $h\rightarrow sh$, $s$ a scalar, and (2) $g(h) \rightarrow 0$ as $h \rightarrow 0$, to rearrange this last equation for $f(x)$ into the original definition of a derivative. Using the equation above to get $\left\|f(x) - (f(a) + \hat{L}_a(x-a))\right\| \leq (\sup_{\|s\|\in[0,\epsilon]} \|g(s)\|)\|h\|$ for $h\in[-\epsilon,\epsilon]$, we are able — after some work (see the exercise below) — to get this nice geometric interpretation: The figure illustrates the fact that the graph of $f(x)$ lies in cones centered on $L$, whose angular widths go to zero as we restrict ourselves to smaller and smaller $\epsilon$-balls centered on $(a,f(a))$. Inside the $\epsilon_1$-ball, the graph stays in the wider cone, while in the smaller, $\epsilon_2$-ball the graph stays in the narrower cone. Let’s restate this. Defining 1. $p \equiv (a,f(a))$, 2. $B(\epsilon)$ to be the ball of radius $\epsilon$ centered on $p$, 3. $F\equiv\{ (x,y) \| y= f(x) \}$, 4. $C_L(p,\epsilon)$ to be the smallest closed cone, symmetrically centered on $L$, with vertex at $p$ such that $F\cap B(\epsilon) \subset C_L(p,\epsilon)$, and 5. $\theta (\epsilon)$ to be the angular width of $C_L(p,\epsilon)$, we have that f is differentiable at $a \Leftrightarrow \theta(\epsilon) \rightarrow 0\text{ as }\epsilon \rightarrow 0$ Here is a figure illustrating this: Exercise: provide the missing details taking us from the above inequality bounding the deviation from linearity to the above statement that {f is differentiable at $a \Leftrightarrow \theta(\epsilon) \rightarrow 0\text{ as }\epsilon \rightarrow 0$} using the facts that (1) the above inequality defines cones that are almost symmetric about $L$ and (2) the $\epsilon$-ball centered at p is contained in the vertical strip $(x-a,y-f(a)) \in [-\epsilon,\epsilon] \times(-\infty,\infty)$ With this shift to a geometric perspective, we are now in a position to take a step in the direction of geometric measure theory. Note that in our definition the cones contain all of the graph as they narrow down and we zoom in. But what if all we know is that a larger and larger fraction of the graph is in a narrower and narrower cone as we zoom into p? That is precisely the idea that approximate tangent lines capture. We will introduce two different versions of the concept. ### Densities as a path to an approximate tangent line #### Tangent Cones The tangent line discussed above is also the tangent cone. The tangent cone of a set in $\Bbb{R}^n$ can have any dimension from 1 to n. For nicely behaved k-dimensional sets, the tangent cone will also be k-dimensional. In the case of the usual derivative of functions from $\Bbb{R}$ to $\Bbb{R}$, we are working in the graph space $\Bbb{R}^2$ with 1-dimensional sets. Moving to tangent cones, we can approximate one dimensional sets which are not graphs or, more generally, arbitrary subsets of $\Bbb{R}^n$. We now define the tangent cone of $F\subset\Bbb{R}^n$ at $p$. To obtain the tangent cone, begin by translating $F$ by $-p$. (This moves $p$ to 0.) Define $F(\epsilon) \equiv (F\cap B(\epsilon))\setminus p$. Use a projection center at 0 to project the translated $F(\epsilon)$ onto the sphere of radius $\epsilon$ centered on 0. Take the closure of the resulting subset of the $\epsilon$-sphere. Finally take the cone over this set. Call this set $T_p^\epsilon(F)$. That is, $T_p^\epsilon(F)=\{\Bbb{R}\geq 0\}(\text{Closure}($$\cup_{x\in F(\epsilon)}\frac{x-p}{\|x-p\|}$$)).$ Now define the tangent cone of F at p to be the intersection of $T_p^\epsilon(F)$ at any sequence of $\epsilon_i$‘s going to zero; $\epsilon_i = \frac{1}{i}$ will do. Thus the tangent cone of $F$ at p, $T_p(F)$ is given by: $T_p(F)=\bigcap_i T_p^\frac{1}{i}(F).$ Here is a figure illustrating the key idea: Note: the tangent cone is centered on the origin, 0, but I will be plotting it as though it were centered on p. Similarly, the tangent lines will sometimes be thought of as linear subspaces (i.e. centered on the origin 0, and other times as the shift of that linear subspace to p. In the case of a differentiable function $f:\Bbb{R}\rightarrow\Bbb{R}$, this tangent cone is the usual 1-dimensional tangent line. #### Densities Now we need $\theta^k(\mu,F)$, the k-dimensional density of F at p. Define $\omega(k)$ such that it agrees with the volume of the unit ball in $\Bbb{R}^k$ when k is an integer (there is a standard way to do this using $\Gamma$ functions). Let $\mu$ measure k-dimensional volume. Typically this will be k-dimensional Hausdorff measure, $\mathcal{H}^k$. Whatever intuitive idea you have of k-dimensional measure is good enough for our purposes. (At the end of this post I also define Hausdorff measures more carefully.) Now, $\theta^k(\mu,F)$ is given by $\theta^k(\mu,F)=\lim_{\epsilon\rightarrow 0}$$\frac{\mu(F\cap B(\epsilon))}{\omega(k)\epsilon^k}$ when this limit exists. When the limit does not exist, we work with the limsup and liminf of the right hand side which are called upper and lower densities of F at p and are denoted by $\theta^{k}(\mu,F)$ and $\theta^k_(\mu,F)$ respectively. #### Approximate Tangent Cones We now define the approximate tangent cone at p to be the intersection of closed cones whose complements intersected with F have density zero at p: $\tilde{T}_p(F)=\bigcap\{\text{closed cones }C\text{ with vertex }p\|\theta^k(\mu,(\Bbb{R}^n\setminus C)\cap F)=0\}$ Originally (in this section), we were aiming at having a definition of approximate tangent line that was invariant to (small) pieces of the set F outside the sequence of cones, provided those pieces got small enough, quick enough. Now we can make that more precise. We want a definition of approximate tangent line that ignores such excursions of F provided these excursions have density zero at p. Rather anti-climatically then, here is the definition we have been waiting for (though you might have already guessed it!) A 1-dimensional set has an approximate tangent line at $p$ when the approximate tangent cone is equal to a line through p. When the curve is an embedded differentiable curve, the tangent line and the approximate tangent line are the same. Remark: in general, when we are dealing with k-dimensional sets in $\Bbb{R}^n$, we will get approximate tangent k-planes. Exercise: can you create examples of one dimensional sets which have a (density based) approximate tangent line at p but not the usual tangent line at p? Exercise: prove that a tangent line to a continuous curve is also the (density based) approximate tangent line at p. ### Integration as a path to an approximate tangent line There is different version of approximate tangent k-plane based on integration. (The one dimensional version is of course an approximate tangent line.) We start with the fact that we can integrate functions defined on $\Bbb{R}^n$ over k-dimensional sets using k-dimensional measures $\mu$ (typically $\mathcal{H}^k$). We zoom in on the point p, through dilation of the set F: $F_\rho(p) = \{x\in\Bbb{R}^n \| \;\;x=\frac{y-p}{\rho}+\text{ p for some }y\in F\}.$ We will say that the set $F$ has an approximate tangent k-plane $L$ at p if the dilation of $F_\rho(p)$, converges weakly to $L$: i.e. if $\int_{F_\rho} \phi d\mu\rightarrow_{\rho \rightarrow 0} \;\; \int_L \phi d\mu$ for all continuously differentiable, compactly supported $\phi:\Bbb{R}^n \rightarrow \Bbb{R}$. In the next two figures, we illustrate this for the case of 1-planes – i.e.lines: in the first figure, $L$ is the weak limit of the dilations of F, while in the second it is not. Note: solid green lines are the level sets of $\phi$ while the dashed green line indicates the boundary of the support of $\phi$. Note also that the $\rho$‘s of 0.4, 0.1, and 0.02 are approximate. Exercise: can you create an example of a one dimensional curve which has the usual tangent line at p but not an (integration based) approximate tangent line at p? ### Closing Note On Hausdorff Measure We would like a notion of k-dimensional volume or k-dimensional measure. In many cases, the right notion turns out to be k-dimensional Hausdorff measure. We already know what 1,2, and 3-dimensional measure is as long as the objects we are measuring are regular enough, like subsets of lines, rectangles, and cubes. It does not seem too much of a stretch to think that we can extend these measures to things that are somewhat wiggly. That is, we can still easily imagine measuring the length of a subset of a smoothly turning curve, or the area of a piece of a surface that undulates slowly. Hausdorff measure permits us to measure not only such smooth sets (giving the same result as any reasonable extension of the usual Lebesgue measures to the nice cases), but also to measure very wild sets (like fractals). How to compute the k-dimensional Hausdorff measure of $A\subset \Bbb{R}^n$: 1. Cover A with a collection of sets $\mathcal{E}= \{E_i\}_{i=1}^\infty$, where $diam(E_i) \leq d \;\; \forall i$. Here, $diam(E_i)$ is the diameter of $E_i$. 2. Compute the k-dimensional measure of that cover: $\mathcal{V}_\mathcal{E}^k(A) = \sum_i\omega(k) (\frac{diam(E_i)}{2})^k$ 3. Define $\mathcal{H}_d^k(A)=\inf_{\mathcal{E}} \mathcal{V}_\mathcal{E}^k(A)$ where the infimum is taken over all covers whose elements with maximal diameter d. 4. Finally, we define: $\mathcal{H}^k(A)=\lim_{d\downarrow 0}\mathcal{H}_d^k(A).$ Remark: Suppose that for any $\epsilon > 0$, there is a cover $\{E_i\}_1^\infty$ of $A$, such that $\sum_i diam(E_i) < \epsilon$. Then for $k \geq 1$, $\mathcal{H}^k(A) = 0$. Here is a figure illustrating Hausdorff measures: Clearly, this can be difficult to compute. It turns out though that in $\Bbb{R}^k$, $\mathcal{H}^k = \Bbb{R}^k$. And by use of mappings, this can take us quite a ways in computing $\mathcal{H}^k(A)$ for integral k and rather general $A$. Exercise: Show that if $0 < \mathcal{H}^\gamma(A) < \infty$ then $\mathcal{H}^\alpha(A) = \infty$ and $\mathcal{H}^\beta(A) = 0$ for $\alpha < \gamma < \beta$.	2017-11-18 13:38:43	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 121, "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": 0.9360288977622986, "perplexity": 375.43601056101244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804965.9/warc/CC-MAIN-20171118132741-20171118152741-00483.warc.gz"}
https://blog.sumbera.com/page/4/	# Joy of Open Source * update IV/2017: from comment below, check this great raster plug-in from Victor Veralde Gutiérrez too: https://github.com/IHCantabria/Leaflet.CanvasLayer.Field/ * I have got great, absolutely amazing joy today from my own open source experiment. In 2014, nearly 2 years ago, I have published on GitHub (gist) L.CanvasOverlay a small class to handle generic drawing on top of Leaflet, I was thinking let’s try to contribute back to the open source with this little snippet – I thought would be useful so I have added little description. Original blog post here: https://blog.sumbera.com/2014/04/20/leaflet-canvas/ After few months later I found it is used on http://windyty.com one of the best 2015 geo-visualization with huge popularity. It is a small part of this great app, but makes me feel so good like I am part of it, I am looking on something where is my small piece, small share…meaning of the effort, sense of publishing and open sourcing. And today I have got echo it is used also in Marine National Facility here: http://www.cmar.csiro.au/data/underway.test/ This makes me so happy … and just came across this quote : “Revenue is a lagging indicator, usage is a leading indicator,” can’t remember, who just said that ? :) # SVG fast scaled overlay on Leaflet 1.0 and 0.7 SVG can be drawn on map in much more faster way than traditional approaches, at least for points. Traditional approach re-position each element to fit into the view of the map, however SVG is “scalable” so we can use it and it performs much more faster for zoom-in/out. Few considerations: 1. SVG itself define viewport by its coordinate space, all outside of this viewport is usually clipped, so it is important to keep SVG viewport in-line with the viewport of the map. There are approaches that resizes SVG as you zoom-in (here), and while it works, it has a problems in deep-zooms when you need to move on map (actually you move giant SVG based on the zoom ) 2. translating LatLon to absolute pixel values (like here used for WebGL) is possible solution, however IE and FF has problems with large numbers for transoform (>1 M), So we need to get SVG elements in view coordinates and translate them. 3. Having some track of bounding box of all elements like again used here should be avoided (SVG or its group element knows about extension of the elements it holds) 4. So while we keep SVG in the viewport, we need to compensate any shift and zoom by translating <g> (group) of all elements. 5. So in leaflet when map moves, SVG is translated back to its original position while <g> is translated forward to reflect the map movement 6. We need to keep track of LatLon position of either map center or one of the corner – we use topLeft corner. 7. Leaflet doesn’t do precise enlargement and rounds view points because of some CSS troubles on some devices (noted here). We need to patch two translating functions in Leaflet to get this right (so SVG enlargement will be aligned with map)… but I need to look on this again, best would be to not patch Leaflet of course. most important things happen in moveEnd event: var bounds = this._map.getBounds(); // -- latLng bounds of map viewport var topLeftLatLng = new L.LatLng(bounds.getNorth(), bounds.getWest()); // -- topLeft corner of the viewport var topLeftLayerPoint = this._map.latLngToLayerPoint(topLeftLatLng); // -- translating to view coord var lastLeftLayerPoint = this._map.latLngToLayerPoint(this._lastTopLeftlatLng); var zoom = this._map.getZoom(); var scaleDelta = this._map.getZoomScale(zoom, this._lastZoom); // -- amount of scale from previous state e.g. 0.5 or 2 var scaleDiff = this.getScaleDiff(zoom); // -- diff of how far we are from initial scale this._lastZoom = zoom; // -- we need to keep track of last zoom var delta = lastLeftLayerPoint.subtract(topLeftLayerPoint); // -- get incremental delta in view coord this._lastTopLeftlatLng = topLeftLatLng; // -- we need to keep track of last top left corner, with this we do not need to track center of enlargement L.DomUtil.setPosition(this._svg, topLeftLayerPoint); // -- reset svg to keep it inside map viewport this._shift._multiplyBy(scaleDelta)._add(delta); // -- compute new relative shift from initial position // -- set group element to compensate for svg translation, and scale</pre> this._g.setAttribute("transform", "translate(" + this._shift.x + "," + this._shift.y + ") scale(" + scaleDiff + ")"); Test page / Gist : http://bl.ocks.org/Sumbera/7e8e57368175a1433791 To better illustrate movement of SVG inside the map, here is a small diagram of basic SVG states: # Smart M.App Recent months I have been programming “Green Space Analyzer” web app that shows modern approach to visualize and query multi temporal geospatial data. User see information in a form he can interact with and discover new patterns, phenomena or information just by very fast ‘feed-back’ of the UI response on the user input. When user selects for example certain area, all graphs instantly animates transition to reflect selection made, this helps to better understand dynamics of the change. Animation can be seen everywhere – from labels on bar chart, through colors change of the choropleth up to title summary. it creates subtle feeling of control or knowing what has changed and how it has changed. At HxGN 15 conference in hexagon geospatial keynote, CEO Mladen Stojic showcased it as part of the vision called Smart M.App, worth to look at (at 52:40 starts Smart M.App demo): my Smart M.App ‘world tour’: # Drawing Shape File on MapKit Simple & strightforward test of loading shape file and drawing it on MapKit on iOS8 using drawMapRect • draws only polygons so far • primitive optimization, no scale optimisation Reading of shape file is performed by shapelib //------------------------------------------------------------ NS_INLINE NSArray getPolygonsFromShapeFile(NSString shpFilePath){ const char path = [shpFilePath cStringUsingEncoding:NSUTF8StringEncoding]; SHPHandle shp = SHPOpen(path, "rb"); int numEntities; int shapeType; SHPGetInfo(shp, &numEntities, &shapeType, NULL, NULL); NSMutableArray allPolygons = [[NSMutableArray alloc]init]; for (int i=0; i<numEntities; i++){ if (shpObject->nSHPType == SHPT_POLYGON \|\| shpObject->nSHPType == SHPT_POLYGONZ \|\| shpObject->nSHPType == SHPT_POLYGONM){ int numParts = shpObject->nParts; int totalVertexCount = shpObject->nVertices; for (int n=0; n<numParts; n++) { int startVertex = shpObject->panPartStart[n]; int partVertexCount = (n == numParts - 1) ? totalVertexCount - startVertex : shpObject->panPartStart[n+1] - startVertex; int endIndex = startVertex + partVertexCount; CLLocationCoordinate2D coords[partVertexCount]; for (int pv = startVertex, i = 0; pv < endIndex; pv++,i++) { } // -- this actually converts lat lon to mkmappoints projection MKPolygon singlePolygon = [MKPolygon polygonWithCoordinates:coords count:partVertexCount]; } } SHPDestroyObject(shpObject); } SHPClose(shp); return [allPolygons copy]; } credits/inspiration: # geojson-vt on leaflet update Sept 2015: nice explanation of how geojson-vt works here Mapbox technologies used in their webgl and opengl libraries are being extracted into standalone pieces. Vladimir Agafonkin,creator of leaflet.js, earcut.js provided slicing and polygon simplification library for geojson called Goejson-vt.Geojson-vt can slice geosjon into tiles aka mapbox tiles. Quick test and sample of using geojson-vt on leaflet with canvas drawing available here: http://bl.ocks.org/sumbera/c67e5551b21c68dc8299 2 videos: Overview of various geojson samples folowing video shows 280 MB large geojson ! for comparison here is WebGL version on the same data. This version is loading all data into GPU and leaves everything on WebGL (no optimization). It also takes slightly more time to tessellate all polygons, but once done all seems to run fine. Code used is available here # Geometric proof of vector dot product Source: colorful introduction to Linear Algebra (may 2014) Cosine Similarity – understanding it as volume of parallelepiped Another good source for understanding dot product : http://betterexplained.com/articles/vector-calculus-understanding-the-dot-product/ other resources: Significance &Application of Cross product and Dot product. http://visualizingmathsandphysics.blogspot.cz/2013/06/vectors-significance-of-cross-product.html Independence of Perpendicular components of the motion http://www.physicsclassroom.com/Class/vectors/u3l1g.cfm Math Insight: http://mathinsight.org/dot_product $\frac{a\cdot b}{\left \\| a\right \\|}= cos\alpha \left \\| b \right \\|$ “projection” of vector b on a equals cosine of angle between them times the length of b or vice versa projection of a on b: Dot product Identities, some pictures taken from http://gamemath.com/ # WMS overlay with MapBox-gl-js 0.5.2 Quick and dirty test of the WMS capabilities of the new MapBox-gl-js 0.5.2 API. First of all, yes ! it is possible to overlay (legacy) WMS over the vector WebGL rendered base map … however the way is not straightforward: • Needs some ‘hacks’ as current version of the API doesn’t have enough events to supply custom URL before it is loaded. But check latest version of mapbox, it might have better support for this. • Another issue is that WMS server has to provide HTTP header with Access-Control-Allow-Origin: to avoid WebGL CORS failure when loading image (gl.texImage2D). Usually WMS servers don’t care about this, as for normal img tags CORS doesn’t apply. Here WebGL has access to raw image data so WMS provider has to explicitly agree with this. • Build process of mapbox-gl-js tend to be as many other large js projects complicated, slow, complex. And specifically on Windows platform it is more difficult to get mapbox-gl-js install and build running then on Mac. Code is documented to guide you through the process, few highlights: // -- rutine originaly found in GlobalMercator.js, simplified // -- calculates spherical mercator coordinates from tile coordinates function tileBounds(tx, ty, zoom, tileSize) { function pixelsToMeters(px, py, zoom) { var res = (2 * Math.PI * 6378137 / 256) / Math.pow(2, zoom), originShift = 2 * Math.PI * 6378137 / 2, x = px * res - originShift, y = py * res - originShift; return [Math.abs(x), Math.abs(y)]; }; var min = pixelsToMeters(tx * tileSize, ty * tileSize, zoom), max = pixelsToMeters((tx + 1) * tileSize, (ty + 1) * tileSize, zoom); return min.concat(max); } ] // -- save orig _loadTile function so we can call it later // -- there was no good pre-load event at mapbox API to get hooked and patch url // -- we need to use undocumented _loadTile // -- replace _loadTile with own implementation // -- we have to patch sourceObj.url, dirty ! // -- we basically change url on the fly with correct BBOX coordinates // -- and leave rest on original _loadTile processing var origUrl =sourceObj.tiles[0] .substring(0,sourceObj.tiles[0].indexOf('&BBOX')); var origUrl = origUrl +"&BBOX={mleft},{mbottom},{mright},{mtop}"; sourceObj.tiles[0] = patchUrl(id, [origUrl]); // -- call original method return origFunc.call(sourceObj, id); } gist available here # ArcGIS vs iKatastr2 on iOS Interesting to see few graphs showing number of downloads of the ArcGIS vs iKatastr2 on iOS for Czech Republic. Data are taken from AppAnnie.com (login requiered). iPhone: # Declarative vs. imperative mismatch Notes from Jenkov.com : http://tutorials.jenkov.com/angularjs/critique.html about AngularJS Declarative vs Imperative. Couldn’t say this better…the art is to find right balance. Neither XAML , nor AngularJS found it yet (too declarative): Quoted from the source: Using declarative HTML templates with a bit of data binding injected is not a new idea. It has been tried before in JSP (JavaServer Pages). Here is an example: <span><%=myObject.getMyProperty()%></span> In AngularJS it would look like this: <span>{{myObject.getMyProperty()}}</span> ### The Declarative / Imperative Paradigm Mismatch Declarative languages (such as HTML and XML) are good a modeling state such as documents, or the overall composition of a GUI. Imperative languages (such as JavaScript, Java, C etc.) are good a modeling operations. The syntaxes of both declarative and imperative languages were designed to support their specific purposes. Therefore, when you try to use a declarative language to mimic imperative mechanisms it may work for simple cases, but as the cases get more advanced the code tend to become clumsy The same is true the other way around. Using an imperative language will often also get clumsy. Anyone who has tried writing an HTML document using JavaScript’s document.write() function, or wired up a Java Swing GUI using Java, or an SWT GUI for that matter, knows how clumsy this code can get. It is easy to think that the declarative / imperative mismatch can be avoided with a programming language that is designed to handle both paradigms. But the problem goes deeper than the language syntax. It is not the syntax of a programming language that decides if the language is declarative or imperative. It is the semantic meaning of what the language describes that decides between declarative and imperative. You could invent a more Java-like syntax to describe HTML elements. Or a more HTML-like syntax to describe Java instructions. But such syntaxes would not change the fact that one language describes state (HTML documents) and the other language describe commands (operations). Thus, a Java-like syntax for describing HTML documents would still be declarative, and a more HTML-like syntax for describing Java-operations would still be imperative. # JavaScript – best coding pattern this or that ? bind or not-bind(this), prototype of prototype ? ehm..all the interesting things, however better without them in your code in JavaScript. There is perfect style finally – found in d3, check here: http://bost.ocks.org/mike/chart/ and used in dc.js as well , well described in the d3-cookbok book by Nick Qui Zhu sample code here similar post appeard here “javascript without this” ..it is worth to study the pattern, it will make your code beautiful, modern and readable. You will not need CoffeScript nor TypeScript nor whateverScript. You even don’t need many other infrastructure or abstractions to get modules or classes out of JavaScript. It is very elegant. It is very simple: function SimpleWidget(spec) { var instance = {}; //-- actual instance variable var description; // -- private variable //-- public API method instance.foo = function () { return instance; //-- returns instance for chaining }; //-- public API method instance.boo = function (d) { //-- getter of private variable if (!arguments.length) return description; description = d; //-- setter of private var return instance; //-- returns instance for chaining } return instance; //-- returns instance for chaining } // usage var widget = SimpleWidget({color: "#6495ed"}) .boo("argument") .foo(); * update 1 just came across this great presentation from JSConf.eu : http://2014.jsconf.eu/speakers/sebastian-markbage-minimal-api-surface-area-learning-patterns-instead-of-frameworks.html and this is exactly the way to think about all of the ‘abstracted stuff’ and syntax sugar or salt that is available today for JavaScript. “It’s much easier to recover from no abstraction than the wrong abstraction.” * update 2 NPM and Browserify is worth to look at, the way how the complex code is done for example in MapBox-GL-JS . While I don’t like convoluted modules of modules and source maps with concatenated sources, the syntax and modularization is working well.	2022-05-28 11:50:29	{"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": 0, "mathjax_asciimath": 1, "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": 0.1810280978679657, "perplexity": 8358.944724293198}, "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/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00096.warc.gz"}
https://www.mypostcard.com/en/the-most-thoughtful-societies-index	# The Most Thoughtful Societies Index ## Study ranks the level of respect and compassion that people in countries around the world have for one another and their peers by evaluating a variety of factors related to philanthropy, equality and family care. At MyPostcard, we thrive on people’s kindness and how they express thoughtfulness for the benefit of others. A poignant feature of the pandemic was the invaluable and selfless acts that many key workers undertook to keep society safe and moving in such difficult and unprecedented times. As we emerge from this period, we thought it would be interesting to celebrate people’s compassion by benchmarking countries on measurable factors that demonstrate their level of thoughtfulness. This is the Most Thoughtful Societies Index. No country nor society is perfect, but respecting and protecting basic human rights should be a minimum expectation. When conducting the study, we therefore disqualified countries with laws that impede upon human rights or make no effort to fix systemic flaws related to human rights. This means that some countries may rank very highly for certain factors but are omitted from the overall ranking. The first topic we analysed was the level of philanthropy in a country. For this, we determined how much people donate to charities in their own country as well as internationally. We also looked at the social security benefits available to people and how much is spent on public services. Then, we gathered data on how compassionate people consider themselves to be and the rate of volunteering. Then, we assessed each country based on how equal their societies are. To do this, we focused on public access to and quality of healthcare and education, the levels of gender and minority equality, and how socially mobile people can be. Finally, we established what support for families is like in each location by looking at the amount of people who look after elderly family members, as well as how much state money goes on elderly care and child support. ## Instructions for journalists The first table below contains the results of the 35 most thoughtful societies in the world. To see the results for each indicator, click on the buttons to the left of the table. Each button represents a different indicator of thoughtfulness based on data. Countries may rank amongst the best in the world for a specific factor but will not feature in the overall table because of human rights violations committed by their governments or leaders. For more information on the parameters for disqualification, how each factor was calculated and all sources used, please refer to the methodology below. # Country Total 1 Netherlands 100 2 99.2 3 Finland 96.94 4 Australia 96.25 5 New Zealand 95.67 6 Sweden 94.65 7 Germany 93.82 8 Iceland 93.78 9 Ireland 93.13 10 USA 93 11 Switzerland 92.16 12 Norway 90.86 13 Denmark 90.04 14 Austria 88.81 15 UK 88.17 16 Belgium 87.13 17 Spain 83.36 18 France 82.67 19 Luxembourg 82.19 20 Czechia 80.84 21 Estonia 80.04 22 Italy 76.72 23 Portugal 76.12 24 Mongolia 75.73 25 Croatia 73.63 26 Japan 73.31 27 Slovakia 73.3 28 Costa Rica 72.86 29 South Korea 72.04 30 Chile 66.04 31 Colombia 63.62 32 Indonesia 62.16 33 Greece 61.39 34 South Africa 59.08 35 Mexico 58.76 ## Methodology The Most Thoughtful Societies Index uses data to rank the level of respect and compassion that people in countries around the world have for their fellow inhabitants and peers. The study collected data from every country in the world before narrowing them down to reveal the top 35 scoring nations overall and the top 25 countries in the world for each factor. Three broad categories were assessed for each location: levels of philanthropy, levels of equality, and levels of family support, with each category made up of a variety of indicators. ### Factors & Scoring Each factor consists of one or more indicators which were scored and averaged. The equation for scoring is as follows: $\small{Score = {x - mean(X) \over Standard\ deviation(X)}}$ in short ${x - \mu} \over \sigma$ For columns where a lower value is better, the score is inverted so that a high score is always better: $\small{Score_{inverted} = - {x - mean(X) \over Standard\ deviation(X)}}$ in short $-{{x - \mu} \over \sigma}$ Data is normalised on a [50-100] scale, with 100 being the best score. Therefore, the higher the score, the better the country ranks for that factor in comparison to the other countries in the index. The formula used is min-max normalisation: $\small{score = (100-50)*{x - min(X) \over max(X)- min(X)}+50}$ The final score was determined by calculating the sum of the weighted average score of all indicators. Below you can find a detailed description of each factor within the study and the sources used. ##### Criteria for Disqualification The criteria for disqualification centres around the mistreatment of women and minorities, the criminalisation of the LGBTQ+ community and human rights violations that are entrenched in the laws and legal framework of the country. Countries with discriminatory laws and government mandates in these areas have therefore been removed from the overall ranking of the most thoughtful countries. They may, however, still appear in specific factor rankings. Exclusive Offers, Special Discounts & More	2022-05-17 20:36:08	{"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": 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": 0.19126750528812408, "perplexity": 1968.3904977010388}, "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/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00515.warc.gz"}
https://leetcode.com/problems/super-pow/	372. Super Pow Loading Question ... Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 Credits: Special thanks to @Stomach_ache for adding this problem and creating all test cases. Seen this question in a real interview before? When did you encounter this question? Which company?	2017-09-20 20:09:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.39359116554260254, "perplexity": 2296.3126982517742}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687447.54/warc/CC-MAIN-20170920194628-20170920214628-00285.warc.gz"}
https://www.physicsforums.com/threads/continuity-correction.532757/	# Continuity Correction 1. Sep 22, 2011 ### eMac I was wondering why it is that the Pr(x=#)=0 2. Sep 23, 2011 ### HallsofIvy ???? As usual with one line questions, that makes no sense at all. Take a few lines to say what you are talking about. I can guess that "P(x= #)" means the probability that the random variable is a specific given number. But that makes no sense without saying what probability distribution you are talking about. And, indeed, for many distributions, what you are saying, that the probability of a singleton set is 0, is simply not true. For example, if my underlying set is {1, 2, 3, 4, 5, 6} and the probability distribution is the uniform distribution, then the probabilty of any one number is 1/6, not 0. I suspect you knew that but were talking about infinite probability distributions. But even then, it is not true that the probability of a single number must be 0. For example, I can define a probability distribution on the set of all positive integers such that [/itex]P(n)= 1/2^{n+1}[/itex]. The sum of all probabilities is a geometric series that sums to 1. Or I can define a probability distribution on [0, 1] such that P(1/2)= 1 and P(x)= 0 for all other numbers in [0, 1]. It is, however, impossible to have a uniform distribution on an infinite set with each outcome having non-zero probility because the infinite sum of a constant is not finite and so cannot be equal to 1, which is required for a probability distribution. Typically what is done is to define the "events" to be subsets of the set of all possible outcomes and define some "measure" of the set, say, length of an interval for one-dimensional problems and area for two-dimensional problems. Then the "probability" of an interval or area is its length or area divided by the length or area of the set of all outcomes. Of course, the "length" or "area" of an individual point is 0.	2018-09-24 09:55:11	{"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": 0.9452449083328247, "perplexity": 158.2327700956684}, "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/1537267160337.78/warc/CC-MAIN-20180924090455-20180924110855-00311.warc.gz"}
http://mathhelpforum.com/calculus/86940-turning-point-coordinates.html	# Math Help - turning point, coordinates 1. ## turning point, coordinates Given that $y = 5x^2 +ax + b$ has a turning point at (b,a). Where a does not equal 0, find a and b. I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like. 2. $y'=10x+a=0$ $y'(b)=10b+a=0$ (1) $y=a=5b^2+ab+b$ (2) 3. Originally Posted by Tweety Given that $y = 5x^2 +ax + b$ has a turning point at (b,a). Where a does not equal 0, find a and b. I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like. If you don't like calculus, start by completing the square: $5\left(x^2+\frac{ax}{5}+~~~\right)+b = 5\left(x^2+\frac{ax}{5}+\frac{a^2}{100}\right)+b-\frac{a^2}{20} = 5\left(x+\frac{a}{10}\right)+\left(b-\frac{a^2}{20}\right)$ So it has a turning point at $\left(-\frac{a}{10}, b-\frac{a^2}{20}\right)$. However, you are given that its turning point is at $(b,a)$ So now you have two equations: $b=-\frac{a}{10}~~~~~(1)$ $a=b-\frac{a^2}{20}~~~~~(2)$ Spoiler: Subbing into the second equation, we have $a=-\frac{a}{10}-\frac{a^2}{20} \implies a = \frac{-2a-a^2}{20} \implies 20a=-2a-a^2 \implies a^2+22a = 0$ Solving this equation gives us $a=-22$ (it can't be 0 from you initial condition) and it follows from (1) that $b=-\frac{11}{5}$. So your equation is $\boxed{5x^2-22x-\frac{11}{5}}$	2015-02-02 00:27:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "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": 0.8172216415405273, "perplexity": 215.53213595827657}, "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-2015-06/segments/1422122122092.80/warc/CC-MAIN-20150124175522-00108-ip-10-180-212-252.ec2.internal.warc.gz"}
https://scicomp.stackexchange.com/questions/13002/particle-collision-to-static-paticles/30574	# Particle Collision to Static paticles I have a system of particles with equal distance with each other and another at random positions which is moving with time. I want to know: a) The method by which I can reduce the number of particles from the first system, as I know the maximum motion of any particle is X. b) How to efficiently calculate the collision between them, I have heard about quadtree and octree but (as per I understood till now) they are for collision of particle from each other. In my problem,second system of particles doesn't collide with each other. Note: Sorry, if it is a very basic question, I am (very) new to this field. • What do you mean by 'maximum motion of any particle is X'? Do you mean maximum displacement? – AlexE Jun 27 '14 at 11:55 • Yes, actually I want to reduce the number of particles to reduce the calculation time. – Devashish Das Jun 27 '14 at 11:57 • Then I guess FMM-style methods using quadtrees in the plane or octrees in 3d or the way to go. Unfortunately, I can't point you to a specific reference or resource. – AlexE Jun 27 '14 at 12:06 I would suggest Barnes-Hut type methods for this problem. It seems to fit the agenda perfectly and has a nice $$\mathcal O(N\log N)$$ complexity. You will have to augment it with your restrictions and conditions.	2020-01-21 03:14:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44010576605796814, "perplexity": 454.2592559725593}, "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/1579250601241.42/warc/CC-MAIN-20200121014531-20200121043531-00154.warc.gz"}
https://indico.cern.ch/event/164917/contributions/1417119/	International Workshop on New Photon-detectors 2012 Jun 13 – 15, 2012 LAL Orsay Europe/Paris timezone A Fast Waveform-Digitizing ASIC-based Electronics and DAQ for a Position \& Time Sensing Large-Area Photo-Detector System Jun 15, 2012, 9:45 AM 20m Amphithéâtre Pierre Lehmann (LAL Orsay) Amphithéâtre Pierre Lehmann LAL Orsay Centre Scientifique d'Orsay - Bat 200 91898 Orsay cedex FRANCE Oral presentation Speaker Mr Eric Oberla (University of Chicago) Description A data acquisition (DAQ) system using 10-15 Gigasamples/second (Gsa/s) waveform sampling ASICs for the readout of large active-area micro-channel plate photomultiplier tubes (MCP-PMTs) is presented. Currently being developed by the Large-Area Picosecond Photo-Detector (LAPPD) collaboration, a single MCP photo-detector tile' has an active area of 400 sq. cm and a dual-end, 50-ohm transmission line anode comprised of 30 parallel microstrips. The position, timing, and energy of the incident pulse are extracted from the full waveforms that are recorded at both anode terminals. With this anode geometry, a larger photo-sensitive area may be formed by connecting several detector tiles in series, allowing for the use of the same readout electronics and acquisition system for many potential applications. A custom fast, low-noise, and low-power waveform digitizing ASIC, PSEC-4', was designed in 0.13 $\mu$m CMOS for the front-end readout of these detectors. With 6-channels, the PSEC-4 has a buffer depth of 256 samples on each channel, a chip-parallel Wilkinson ADC, and a serial data readout that includes the capability for region-of-interest windowing to reduce dead-time. Sampling rates of up to 15 Gsa/s are possible on each channel with an analog bandwidth of 1.5 GHz. A flexible DAQ system matched to the large-area detector anode, in which PSEC-4 calibrations and signal feature extraction are implemented in two layers of FPGAs, has been designed and code development is underway. Further details of the readout system, including the PSEC-4 ASIC capabilities and DAQ performance, will be reported. Primary author Mr Eric Oberla (University of Chicago) Slides	2022-01-25 16:41:43	{"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": 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": 0.3653266131877899, "perplexity": 12480.070566953493}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304859.70/warc/CC-MAIN-20220125160159-20220125190159-00339.warc.gz"}
https://www.byteflying.com/archives/4050	# C#LeetCode刷题之#190-颠倒二进制位（Reverse Bits） Reverse bits of a given 32 bits unsigned integer. Input: 43261596 Output: 964176192 Explanation: 43261596 represented in binary as 00000010100101000001111010011100, return 964176192 represented in binary as 00111001011110000010100101000000. Follow up:If this function is called many times, how would you optimize it? ```public class Program { public static void Main(string[] args) { var n = 43261596U; var res = reverseBits(n); Console.WriteLine(res); n = 13U; res = reverseBits2(n); Console.WriteLine(res); n = 168U; res = reverseBits3(n); Console.WriteLine(res); } public static uint reverseBits(uint n) { //10进制转2进制，除2取余法 var res = 0U; var bit = 0; var times = 32;//Math.Ceiling(Math.Log(n, 2)); //整型4字节，32位 while(n != 0) { //10进制的1，2，3，4，5对应于2进制的1，10，11，100，101 //由于2进制的特点，末位数在1，0之间循环 //用 n 和 1 做“与运算”若值为1，必为奇数 //即除2余1 if((n & 1) == 1) { res += (uint)Math.Pow(2, times - bit - 1); } bit++; //2进制右移1位即10进制除2 n >>= 1; } return res; } public static uint reverseBits2(uint n) { //定义结果 var res = 0U; //执行32次 for(var i = 0; i < 32; i++) { //将结果 *2 res <<= 1; #line 100 //奇数时，把结果 +1 if((n & 1) == 1) res++; //将 n 除以 2 n >>= 1; } //返回结果 return res; } public static uint reverseBits3(uint n) { var res = 0U; for(var i = 0; i < 32; i++) { res <<= 1; //res = res \| (n & 1); //奇数跟0进行“或运算”，原值 //偶数跟0进行“或运算”，原值 //奇数跟1进行“或运算”，原值 //偶数跟1进行“或运算”，原值+1 //以下一行代码相当于 #line 100 下的一行 res \|= (n & 1); n >>= 1; } return res; } }``` ```964176192 2952790016 352321536```	2021-01-18 19:49:23	{"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": 0.4380466938018799, "perplexity": 8126.56267724117}, "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-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00663.warc.gz"}