Wikipedia:Reference desk/Mathematics: Difference between revisions
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[[Category:Wikipedia help forums]] |
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[[Category:Wikipedia reference desk|Mathematics]] |
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[[Category:Wikipedia help pages with dated sections]]</noinclude> |
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{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 15}} |
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= December 23 = |
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{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 16}} |
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== Is it possible to make [[Twisted Edwards curve]] birationally equivalent to twisted weirestrass curves ? == |
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{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 17}} |
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Is there an equation fo converting a [[twisted Edwards curve]] into a tiwsted weierstrass form ? [[Special:Contributions/2A01:E0A:401:A7C0:6D06:298B:1495:F479|2A01:E0A:401:A7C0:6D06:298B:1495:F479]] ([[User talk:2A01:E0A:401:A7C0:6D06:298B:1495:F479|talk]]) 04:12, 23 December 2024 (UTC) |
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= February 18 = |
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:According to {{slink|Montgomery curve#Equivalence with twisted Edwards curves}}, every twisted Edwards curve is birationally equivalent to a Montgomery curve, while {{slink|Montgomery curve#Equivalence with Weierstrass curves}} gives a way to transform a Montgomery curve to an elliptic curve in Weierstrass form. I don't see a definition of "twisted Weierstrass", so I don't know if you can give an extra twist in the process. Perhaps this paper, [https://ietresearch.onlinelibrary.wiley.com/doi/10.1049/cje.2018.05.004 "Efficient Pairing Computation on Twisted Weierstrass Curves"] provides the answer; its abstract promises: "In this paper, we construct the twists of twisted Edwards curves in Weierstrass form." --[[User talk:Lambiam#top|Lambiam]] 10:35, 23 December 2024 (UTC) |
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== Cubic function == |
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= December 24 = |
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If you go [http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic here] and enter as the equation ax<sup>3</sup> + bx<sup>2</sup> + cx + d = 0 and tell it to solve for x, you get a long expression as the solution. Can someone show algebraically how you can solve ax<sup>3</sup> + bx<sup>2</sup> + cx + d = 0 for x? Or link to a page that does? Thanks. [[Special:Contributions/70.162.25.53|70.162.25.53]] ([[User talk:70.162.25.53|talk]]) 03:39, 18 February 2008 (UTC) |
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: See [[cubic function]], specifically the Cardano's method section. Be warned, it's not ''quite'' as nice as the quadratic formula... [[Special:Contributions/134.173.92.17|134.173.92.17]] ([[User talk:134.173.92.17|talk]]) <small>—Preceding [[Wikipedia:Signatures|comment]] was added at 03:59, 18 February 2008 (UTC)</small><!--Template:Undated--> <!--Autosigned by SineBot--> |
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== How did the Romans do engineering calculations? == |
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:(ec) The Wikipedia article on [[cubic function]]s may be of your interest, or at least [[Cubic function#Root-finding formula|this subsection]] of it. Outside Wikipedia, [http://mathworld.wolfram.com/CubicFormula.html this page] from MathWorld is another interesting choice. [[User:Pallida_Mors_76|<span style="background:#000;border:#c3c0bf;color: #fff;border:1px solid #999">Pallida </span>]][[User talk:Pallida_Mors_76|<span style="background:#fff;border:#c3c0bf;color:#000;border:1px solid #999"> Mors</span>]] 04:06, 18 February 2008 (UTC) |
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::::I intended a reference to the subsection ''Root finding'' in that article, but the link doesn't seem to work fine. Nevermind. [[User:Pallida_Mors_76|<span style="background:#000;border:#c3c0bf;color: #fff;border:1px solid #999">Pallida </span>]][[User talk:Pallida_Mors_76|<span style="background:#fff;border:#c3c0bf;color:#000;border:1px solid #999"> Mors</span>]] 04:19, 18 February 2008 (UTC) |
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The Romans did some impressive engineering. Engineers today use a lot of mathematical calculations when designing stuff. Calculations using Roman numerals strike me as being close to impossible. What did the Romans do? [[User:HiLo48|HiLo48]] ([[User talk:HiLo48|talk]]) 05:50, 24 December 2024 (UTC) |
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== Logic == |
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:The kind of engineering calculations that might have been relevant would mostly have been about [[statics]] – specifically the equilibrium of forces acting on a construction, and the ability of the design to withstand these forces, given its dimensions and the mechanical properties of the materials used in the construction, such as [[density]], [[modulus of elasticity]], [[shear modulus]], [[Young modulus]], [[fracture strength]] and [[ultimate tensile strength]]. In Roman times, only the simplest aspects of all this were understood mathematically, namely the statics of a construction in which all forces work in the same plane, without torque, and the components are perfectly rigid. The notion of assigning a numerical magnitude to these moduli and strengths did not exist, which anyway did not correspond to precisely defined, well-understood concepts. Therefore, engineering was not a science but an art, mainly based on experience in combination with testing on physical models. Any calculations would mostly have been for the amounts and dimensions of construction materials (and the cost thereof), requiring a relatively small number of additions and multiplications. --[[User talk:Lambiam#top|Lambiam]] 19:03, 24 December 2024 (UTC) |
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Does anyone know where I can find a whole bunch of "[[Knights and Knaves]]" problems, with answer/explanations/solutions? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 07:51, 18 February 2008 (UTC)) |
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: Calculating with Roman Numerals might seem impossible, but in some ways it's simpler than our positional system; there are only so many symbols commonly used, and only so many ways to add and multiply them. Once you know all those ways you efficiently can do calculations with them, up to the limits imposed by the system. |
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: Here's [http://www.mathpuzzle.com/philosTF.html a bunch of problems with a twist]. These problems explore the consequences of adding a third type of person, the philosophers. For more traditional questions, a [http://www.google.com/search?q=knights+and+knaves Google search] turns up a few pages. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 12:40, 18 February 2008 (UTC) |
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: And for a lot of things they relied on experience. Romans knew how to build circular arches, but rather than do calculations to build larger arches, or ones with more efficient shapes they used many small circular ones which they knew worked, stacked side by side and sometimes on top of each other. See e.g. any [[Roman aqueduct]] or the [[Colosseum]]. For materials they probably produce them on site or close by as they're needed.--[[Special:Contributions/2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C|2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C]] ([[User talk:2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C|talk]]) 20:12, 24 December 2024 (UTC) |
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: Also, check out the books of [[Raymond Smullyan]]. He has tons of logic puzzles that all can enjoy. –'''''[[User:King Bee|King Bee]]''''' <sup>([[User talk:King Bee|τ]] • [[Special:Contributions/King Bee|γ]])</sup> 13:48, 18 February 2008 (UTC) |
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: To add to the above, it wasn't really the Romans that were great innovators in science, engineering, etc.. I think more innovation and discovery took place in Ancient Greece and Ancient Egypt. Certainly Greece as we have a record of that. Egypt it's more that they were building on such a monumental scale, as scale no-one came close to repeating until very recently. |
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: I especially recommend "Alice in Puzzleland" by [[Raymond Smullyan]]. [[Special:Contributions/129.67.157.6|129.67.157.6]] ([[User talk:129.67.157.6|talk]]) <small>—Preceding [[Wikipedia:Signatures|comment]] was added at 15:15, 18 February 2008 (UTC)</small><!--Template:Undated--> <!--Autosigned by SineBot--> |
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: Romans were military geniuses. They conquered Greece, and Egypt, and Carthage, and Gaul, and Britannia, and everywhere in between. They then built forts, towns, cities and infrastructure throughout their empire. They built so much so widely that a lot of it still stands. But individually a lot of it isn't technically impressive; instead it's using a few simple patterns over and over again.--[[Special:Contributions/2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C|2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C]] ([[User talk:2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C|talk]]) 12:09, 25 December 2024 (UTC) |
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Hi. Thanks to all for the input. I guess I should have made my question a little clearer. So, let me re-phrase it. I am looking for Knights and Knaves problems ... that are accompanied by answers / solutions / explanations. Right now, I only want Knight/Knave questions ... and not just logic puzzles in general. Also, I am only looking for the "simple" type (knights and knaves only), where they do not introduce third-parties (spies, randoms, philosophers, etc.) into the equation. Also, I'd like to find this on the Internet --- as I don't have too much time to go out, order a book, wait for its delivery, etc. At some later point, I will probably go out and get some Smullyan books ... but right now, I don't have that time / luxury. Any suggestions? Any good websites out there? I can't seem to find any, with my searches. Many thanks! ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 16:33, 18 February 2008 (UTC)) |
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::See [[Roman abacus]]. [[User:Catslash|catslash]] ([[User talk:Catslash|talk]]) 22:03, 25 December 2024 (UTC) |
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:The Wikipedia article on [[Knights and Knaves]] has a few problems with explanations and solutions. There are some pages with individual knights-and-knaves problems (from one site, [http://en.allexperts.com/q/Puzzle-Solving-1841/Knight-Knave-problem.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knights-knaves-4.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knights-knaves-3.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/Knights-Knaves.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knights-knaves-2.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knaves-knights.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knights-knaves-1.htm], [http://en.allexperts.com/q/Puzzle-Solving-1841/knights-knaves.htm]). —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 17:46, 18 February 2008 (UTC) |
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:::It has to be said that in Roman times calculations for architecture were mostly graphical, geometrical, mechanical, rather than numeric. In fact, from the perspective of an ancient architect, it would make little sense translating geometrical figures into numbers, making numeric calculations, then translating them into geometrical figures again. Numerals become widely used tools only later, e.g. with the invention of Analytic Geometry (by Descartes), and with logarithms (Napier); and all these great mathematical innovations happened to be so useful also thanks to the previous invention of the printing press by Gutenberg --it's easier to transmit information by numbers than by geometric constructions. One may even argue that the invention of the printing press itself was the main reason to seek for an adequate efficient notation for real numbers (achieved by Stevinus). [[User:PMajer|pm]][[User talk:PMajer|<span style="color:blue;">a</span>]] 21:22, 1 January 2025 (UTC) |
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:::: This is a great answer! [[User:Tito Omburo|Tito Omburo]] ([[User talk:Tito Omburo|talk]]) 21:38, 1 January 2025 (UTC) |
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::::Architectural calculations, mentioned by [[Vitruvius]], are not what I think of as engineering calculations. The former kind is about form. The latter kind should provide answers questions about structural behaviour, like, "Will these walls be able to withstand the outward force of the dome?" Can such questions be addressed with non-numerical calculations? --[[User talk:Lambiam#top|Lambiam]] 23:40, 1 January 2025 (UTC) |
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== Are these sequences mod any natural number n periodic? == |
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:Smullyan's book entitled "What is the name of this book?" has a lot of Knights/Knaves puzzles complete with solutions. His books should be available at your local library, so you needn't purchase a thing. –'''''[[User:King Bee|King Bee]]''''' <sup>([[User talk:King Bee|τ]] • [[Special:Contributions/King Bee|γ]])</sup> 19:21, 18 February 2008 (UTC) |
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The period of [[Fibonacci number]] mod n is the [[Pisano period]] of n, but are these sequences mod any natural number n also periodic like [[Fibonacci number]] mod n? |
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== help with aproof == |
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# [[Lucas number]] |
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i need help with how to prove that a number (a) is a zero divisor mod (n) iff gcd(a,n)>1?thank you[[User:Husseinshimaljasimdini|Husseinshimaljasimdini]] ([[User talk:Husseinshimaljasimdini|talk]]) 10:52, 18 February 2008 (UTC) |
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# [[Pell number]] |
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# [[Tribonacci number]] |
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# [[Tetranacci number]] |
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# [[Newman–Shanks–Williams number]] |
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# [[Padovan sequence]] |
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# [[Perrin number]] |
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# [[Narayana sequence]] |
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# [[Motzkin number]] |
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# [[Bell number]] |
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# [[Fubini number]] |
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# [[Euler zigzag number]] |
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# Partition number {{oeis|A000041}} |
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# Distinct partition number {{oeis|A000009}} |
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[[Special:Contributions/42.76.153.22|42.76.153.22]] ([[User talk:42.76.153.22|talk]]) 06:06, 24 December 2024 (UTC) |
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:For 1. through 4., see [[Pisano period#Generalizations]]. Although 5. through 8. are not explicitly listed, I'm pretty sure the same argument applies for their periodicity as well. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 07:05, 24 December 2024 (UTC) |
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:Think about ''n''/gcd(''a'',''n'') - let's call this ''b''. First of all, you can be sure that ''b'' is an integer - why ? Secondly, you can be sure that if gcd(''a'',''n'') > 1 then ''b'' mod(''n'') is not 0 - why ? Now think about the product ''ab'' - what is ''ab'' mod(''n'') ? How does this help you show that ''a'' is a zero divisor mod(''n'') ? [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 12:09, 18 February 2008 (UTC) |
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:For 13, I'm not sure, but I think the partition function is not periodic modulo any nontrivial number, to the point that the few congruences that are satisfied by the function are also very notable, e.g. [[Ramanujan's congruences]]. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 07:23, 24 December 2024 (UTC) |
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::It's possible that a sequences is eventually periodic but not periodic from the start, for example powers of 2 are periodic for any odd n, but for n=2 the sequence is 1, 0, 0, ..., which is only periodic starting with the second entry. In other words a sequence can be become periodic without being pure periodic. A finiteness argument shows that 1-8 are at least eventually periodic, but I don't think it works for the rest. ([[Pollard's rho algorithm]] uses this finiteness argument as well.) It says in the article that Bell numbers are periodic mod n for any prime n, but the status for composite n is unclear, at least from the article. Btw, [[Catalan number]]s not on the list?. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 08:08, 24 December 2024 (UTC) |
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:::If only the periodicity of Bell numbers modulo prime ''powers'' were known, then periodicity for all modulos would immediately result from the [[Chinese remainder theorem]]. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 01:29, 29 December 2024 (UTC) |
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::PS. 1-8 are pure periodic. In general, if the recurrence can be written in the form F(k) = (some polynomial in F(k-1), F(k-2), ... F(k-d+1) ) ± F(k-d), then F is pure periodic. The reason is that you can solve for F(k-d) and carry out the recursion backwards starting from where the sequence becomes periodic. Since the previous entries are uniquely determined they must follow the same periodic pattern as the rest of the sequence. If the coefficient of F(k-d) is not ±1 then this argument fails and the sequence can be pre-periodic but not pure periodic, at least when n is not relatively prime to the coefficient. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 18:12, 24 December 2024 (UTC) |
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== Collecting Terms == |
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:::Ah, what was tripping me up was showing pure periodicity, recursing backwards completely slipped my mind. Thanks for the writeup! [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 18:31, 24 December 2024 (UTC) |
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::::good questions. What about TREE(n) mod k, for arbitrary fixed k?[[User:Richard L. Peterson|Rich]] ([[User talk:Richard L. Peterson|talk]]) 23:03, 27 December 2024 (UTC) |
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:::::Link: [[Kruskal's tree theorem#TREE function|TREE function]]. There are a lot of sequences like this where exact values aren't known, [[Ramsey's theorem#Ramsey numbers|Ramsey numbers]] are another example. It helps if there is a relatively simple recursion defining the sequence. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 11:50, 28 December 2024 (UTC) |
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:For 9. Motzkin numbers are not periodic mod 2. Motzkin numbers mod 2 are [[OEIS:A039963]], which is [[OEIS:A035263]] with each term repeated (i.e. <math>0 \rightarrow 00, 1 \rightarrow 11</math>.) [[OEIS:A035263]] in turn is the sequence that results when one starts with the string <math>1</math> and successively maps <math>0 \rightarrow 11, 1 \rightarrow 10</math> (e.g. <math>110 \rightarrow 101011</math>.) It is clear that if [[OEIS:A035263]] were periodic with period <math>n</math>, then the periodic string <math>X</math> of length <math>n</math> would need to map to string <math>XX</math>, but this is impossible as the last character of <math>X</math> is always the opposite of the last character of the map applied to <math>X</math>. Thus [[OEIS:A035263]] is nonperiodic, and neither is [[OEIS:A039963]]. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 01:08, 29 December 2024 (UTC) |
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::[https://arxiv.org/pdf/1611.04910 This paper] (linked from OEIS) goes into more detail on Motzkin numbers. I gather the sequence might be called quasi-periodic, but I can't find an article that matches this situation exactly. [[OEIS|A003849]] is in the same vein. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 16:59, 30 December 2024 (UTC) |
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:For 11., the article seems to suggest that Fubini numbers are eventually periodic modulo any prime power. I'm pretty sure this means that they the numbers eventually periodic mod any number <math>n</math>, since the lcm of the eventual periods modulo all prime power divisors of <math>n</math> should correspond to the eventual period modulo <math>n</math> itself, with the remainders being obtainable through the [[Chinese remainder theorem]]. However, the wording also seems to suggest that periodicity modulo arbitrary <math>n</math> is still conjectural, so I'm not sure. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 02:44, 29 December 2024 (UTC) |
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::You have answered all questions except 12 and 14, and 9 and 13 are the only two sequences which are not periodic mod n (except trivial n=1), 12 (Euler zigzag numbers) is {{OEIS|A000111}}, which seems to be periodic mod n like 10 (Bell numbers) {{OEIS|A000110}} and 11 (Fubini numbers) {{OEIS|A000670}}, but all of these three sequences need prove, besides, 14 (Distinct partition numbers) {{OEIS|A000009}} seems to be like 13 (Partition numbers) {{OEIS|A000041}}, i.e. not periodic mod n. [[Special:Contributions/1.165.199.71|1.165.199.71]] ([[User talk:1.165.199.71|talk]]) 02:27, 31 December 2024 (UTC) |
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i get how to do everything else but i seem to bomb out on collecting the terms forgive me if it is wrong <br> |
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Simplify:<math>=\sqrt{3}- 2\sqrt{2} + 2\sqrt{3} + \sqrt{2}</math> <br> |
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now im not sure but im pretty sure this is wrong |
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<math>=1\sqrt{3}(not sure what goes here) 3\sqrt{2}</math> <br> |
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<small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:124.176.206.9|124.176.206.9]] ([[User talk:124.176.206.9|talk]] • [[Special:Contributions/124.176.206.9|contribs]]) 12:28, 18 February 2008 (UTC)</small><!-- Template:Unsigned --> |
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= December 31 = |
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:You have one square root of 3, minus two square roots of 2, plus 2 square roots of 3, plus one square root of 2: so in the end how many square roots of 3 do you have, and how many square roots of 2? —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 12:31, 18 February 2008 (UTC) |
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== Generating a point on the Y axis from regular pentagon with point on X axis == |
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:If it helps you, forget for a moment what <math>\sqrt3</math> means, and just think of it as a symbol, like a circle, perhaps. The same thing for <math>\sqrt2</math>: think of that as a triangle. So you have one circle, minus two triangles, plus two circles, plus one triangle. How many circles do you have, and how many triangles? Just be sure to write "<math>\sqrt2</math>" in your answer instead of a triangle, of course. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 12:37, 18 February 2008 (UTC) |
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For a <math>K</math> consisting of points in R^2, define the function B such that <math>B(K)</math> as the Union of <math>K</math> and all points which can be produced in the following way. For each set of points A, B, C, & D from <math>K</math> all different so that no three of A, B, C & D are co-linear. E is the point (if it exists) where ABE are colinear and CDE are co-linear. |
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::<math>3\sqrt{3}+ -1\sqrt{2} (= \sqrt{2}) </math> |
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so |
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<math>3\sqrt{3}+ \sqrt{2} </math> ? |
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do you just take the sign infront? |
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so that 12a - 10b + 9c + 5b<br> |
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=12 - 5b + 9c? |
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If <math>K_0</math> = the vertices of a regular Pentagon centered at 0,0 with one vertex at (1,0), does there exist N such that <math>B^N(K_0)</math> includes any point of the form (0, y)? (extending the question to any N-gon, with N odd) |
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::Think about this, <math>ab+ac+de+df=a(b+c)+d(e+f)</math> this is just two applications of the [[Distributive property]]. If it's not clear why that works, just try going the other way. (Note that algebraic laws are reversible!)[[User:A math-wiki|A math-wiki]] ([[User talk:A math-wiki|talk]]) 12:56, 18 February 2008 (UTC) |
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[[User:Naraht|Naraht]] ([[User talk:Naraht|talk]]) 05:16, 31 December 2024 (UTC) |
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:I think you meant to write <math>B^N(K_0).</math> --[[User talk:Lambiam#top|Lambiam]] 07:55, 31 December 2024 (UTC) |
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:You're right: the simplified answer is <math>3\sqrt3+(-1)\sqrt2</math>. Usually −1 times ''something'' is just written −''something'' (for example, −1 × 5 = −5), and when "+−" occurs in a mathematical formula it's usually just written "−" [for example, 3 + (−2) = 3 − 2]. So you can write <math>3\sqrt3+(-1)\sqrt2</math> in a simpler way. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 13:56, 18 February 2008 (UTC) |
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::Changed to use the Math.[[User:Naraht|Naraht]] ([[User talk:Naraht|talk]]) 14:37, 31 December 2024 (UTC) |
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:::I'm not 100% sure I understand the problem, but try this: Label the vertices of the original pentagon, starting with (1, 0), as A, B, C, D, E. You can construct a second point on the x-axis as the intersection of BD and CE; call this A'. Similarly construct B', C', D', E', to get another, smaller, regular pentagon centered at the origin and with the opposite orientation from the the original pentagon. All the lines AA', BB', CC', DD', EE' intersect at the origin, so you can construct (0, 0) as the intersection of any pair of these lines. The question didn't say y could not be 0, so the answer is yes, with N=2. |
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:::There is some theory developed on "straightedge only construction", in particular the [[Poncelet–Steiner theorem]], which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 13:12, 1 January 2025 (UTC) |
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== geometry - spheres and unit cubes == |
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::::Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a [[pentagram]]. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram. --[[User talk:Lambiam#top|Lambiam]] 16:53, 1 January 2025 (UTC) |
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:::::Here is one point other than the origin (in red) |
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:::::[[image:Pentagon-y-axis-point.svg]] |
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:::::If there is one such point, there must be an infinite number of them. [[User:Catslash|catslash]] ([[User talk:Catslash|talk]]) 22:52, 2 January 2025 (UTC) |
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::::::Just to be clear, the black points are the original pentagon K, the green points are in B(K), and the red point is the desired point in B<sup>2</sup>(K); the origin is not shown. It would be nice to find some algebraic criterion for a point to be constructible in this way, similar to the way points constructible with a compass and straightedge are characterized by their degree over '''Q'''. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 01:45, 3 January 2025 (UTC) |
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::::::Once you have a second one (such as the reflection of the red point wrt the x-axis), you have all intersections of the y-axis with the non-vertical lines through pairs of distinct points from <math>\textstyle{\bigcup_n B^n(K_0)}.</math> --[[User talk:Lambiam#top|Lambiam]] 16:22, 3 January 2025 (UTC) |
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::::{{U|RDBury}} *headslap* on (0,0) Any idea on y<>0? <small>(← comment from [[User:Naraht|Naraht]])</small> |
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seeing as there's quite a few mathematicians in here i thought i'd ask about a geometry problem i've been having. i think it's quite easy but i don't really have a clue. |
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:::::See the above construction by catslash. --[[User talk:Lambiam#top|Lambiam]] 16:12, 3 January 2025 (UTC) |
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:::::The red point is at <math>x = 0</math>, <math>y = - \sqrt{\frac{5 - \sqrt{5}}{10}}</math>. [[User:Catslash|catslash]] ([[User talk:Catslash|talk]]) 16:18, 3 January 2025 (UTC) |
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= January 1 = |
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imagine a unit cube. now imagine a point within this cube (call it p). i want to place a sphere with its centre at p. how would i work out the radius needed for this sphere so that the intersection volume between it and the cube equals a given value (call it v)? [[User:AlmostCrimes|AlmostCrimes]] ([[User talk:AlmostCrimes|talk]]) 14:21, 18 February 2008 (UTC) |
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:If the sphere just intersected one side you could find the volume of the [[Spherical cap]] cut off, then subtract this from the volume of the sphere. If the sphere intersects more than one side then things get a little trickier. --[[User:Salix alba|Salix alba]] ([[User talk:Salix alba|talk]]) 15:07, 18 February 2008 (UTC) |
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:[ec]It's actually not that simple. To do that you will first need some way to calculate the volume of intersection of a cube and a ball, which is pretty messy. Fortunately this latter problem is one I have worked on a while back, so I have thought of a few approaches. Since none of them is perfect, I can help you more only if you provide more details about your application (with special attention to details such as relative and absolute error tolerance, computational resources, platform on which you will do the calculations, etc.). If the question is purely out of curiosity I suggest you just leave it, though I can describe the general ideas. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 15:10, 18 February 2008 (UTC) |
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== What is the first number not contained in M136279841? == |
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:: salix alba - Yes sadly, it is quite possible the sphere intersects more than one side, for example if the point was at one of the corners of the cube. the basis of the problem is that I'm investigating various approaches to deciding whether a given colour in an RGB colour space (here represented as a cube like the following [[RGB#Representations]]) is "similar" to another colour. the natural approach would be to calculate the distance between the two colours (or points) and to then determine whether this is below a certain threshold. This technique suffers in that not all colours are treated evenly - colours in the centre of the colour space (so at [0.5, 0.5, 0.5], or grey) would be flagged as being similar to a higher percentage of the total colour space than a colour like white would. I was hoping to even this out by dynamically modifying the threshold (analogous to the radius) so that every colour has an equal proportion of the colour space flagged as being similar to it. As i say, this is mostly investigative and it isn't important that I use this exact technique. it's very possible i'm completely overcomplicating things. |
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See {{OEIS|A268068}}, the first number not contained in M74207281 is 1000003, but what is What is the first number not contained in M136279841 (the currently largest known prime)? [[Special:Contributions/61.224.131.231|61.224.131.231]] ([[User talk:61.224.131.231|talk]]) 03:34, 1 January 2025 (UTC) |
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:: for what it's worth though error tolerance isn't too important, neither are computational resources. I'm programming it in MATLAB and basically just seeing what works well. the most important factor is simplicity and implementation time really as I don't want to get too hung up on one technique. If any of your solutions apply I'd be interested in hearing them. I thought briefly about representing both the sphere and cube in a boolean matrix and physically determining the best sphere radius but I was kind of hoping there'd be a simple algebraic method. --[[Special:Contributions/80.4.203.142|80.4.203.142]] ([[User talk:80.4.203.142|talk]]) 19:44, 18 February 2008 (UTC) |
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:::Well, I can see several problems with this approach. If you're going for a notion of similarity which matches that of a human observer, I don't think using a simple Euclidean metric as a basis will cut it. I don't know a lot about this, but at the very least you should give different weights to the components - [[Grayscale]] mentions weights used for a different purpose, but they might be appropriate here. This would then leave you dealing with ellipsoids which is even messier. Second, I'm not sure any sort of "[[Affirmative action]]" is necessary - white will have less colors labeled as similar to it, simply because there are less colors genuinely similar to it. You next have symmetry to consider - You mentioned a ball centered around one point, but when doing a comparison there are two points to consider, and the result will depend on which one you choose to calculate the threshold. If you do some symmetric calculation taking both into account, it will be difficult to guarantee that indeed every color has the same portion of the color space labeled similar to it. |
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:::As said, solving the problem in you original question is not impossible, but not simple to do with satisfying accuracy and efficiency. Since I gather that these are not crucial, I can offer a simple approximation method for the volume of intersection of the cube and a given ball. Just pick several (something like 100 or 1000) randomly chosen points in the cube, and check whether each is inside the ball. The proportion which are is an estimation for the volume of intersection. You can do this for a ball centered at one point and with radius equal to the distance to the other point, compare it to the volume you want to be labeled similar and make the decision accordingly. |
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:::I hope this helps. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 22:25, 18 February 2008 (UTC) |
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:The corresponding sequence (11, 3, 8, 7, 6, 10, 4, 9, 1, 5, 25, 31, 39, ...) is not in OEIS. Finding the answer to your question requires an inordinate amount of computing power. The decimal expansion of this Mersenne prime has some 41 million digits, all of which need to be computed. If this is to be done in a reasonable amount of time, the computation will need the random access storage of at least some 22 million digits. --[[User talk:Lambiam#top|Lambiam]] 10:10, 1 January 2025 (UTC) |
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:::: it's true there are problems with the general approach and while I am aware there are colour spaces that lend themselves more easily to this kind of work I have been asked to look at this RGB cube method specifically. As for the "affirmative action", it seems to me it is necessary. A general distance approach is far better at finding areas that look grey than areas that look white when given the same threshold. Increasing the threshold for white increases the perceived accuracy, which is why I wanted to try this method. Your symmetry point is interesting. I actually hadn't considered that... |
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::I'm not seeing that this question requires an inordinate amount of computing power to answer. 41 million characters is not a very large set of data. Almost all modern computers have several gigabytes of memory, so 41 million characters will easily fit in memory. I took the digits of M136279841 from https://www.mersenne.org/primes/digits/M136279841.zip and searched them myself, which took a few minutes on a consumer grade PC. If I have not made a mistake, the first number that does not appear is 1000030. The next few numbers that do not appear are 1000073, 1000107, 1000143, 1000156, 1000219, 1000232, 1000236, 1000329, 1000393, 1000431, 1000458, 1000489, 1000511, 1000514, 1000520, 1000529, etc. [[User:CodeTalker|CodeTalker]] ([[User talk:CodeTalker|talk]]) 03:59, 2 January 2025 (UTC) |
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:::To be fair, this depends on being able to find the digits on-line. To compute them from scratch just for this question would be more trouble than it's worth. But I take your point; it probably takes more computing power to stream an episode of [[Numbers (TV series)|NUMB3RS]] than to answer this question. My problem with the question is that it's basically a dead end; knowing the answer, is anyone going to learn anything useful from it? I'd question the inclusion of A268068 in OEIS in the first place simply because it might lead to this sort of boondoggle. But far be it for me to second guess the OEIS criteria for entry. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 01:13, 3 January 2025 (UTC) |
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::::OEIS includes similar sequences for the positions of the first location of the successive naturals in the decimal expansions of <math>e</math> ({{OEIS link|A088576}}), <math>\pi</math> ({{OEIS link|A032445}}) and <math>\gamma</math> ({{OEIS link|A229192}}). These have at least a semblance of theoretical interest wafting over from the open question whether these numbers are [[Normal number|normal]]. --[[User talk:Lambiam#top|Lambiam]] 06:21, 3 January 2025 (UTC) |
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::::{{tq|1=To compute them from scratch just for this question would be more trouble than it's worth.}}<br>Eh, I agree that the question is of little fundamental interest. However, it's not much work to compute M136279841. It is of course absolutely trivial to compute it as a binary number. The only real work is to convert it to decimal. I wrote a program to do this using the GNU Multiple Precision Arithmetic Library. It took about 5 minutes to write the program (since I've never used that library before and had to read the manual) and 29 seconds to run it. [[User:CodeTalker|CodeTalker]] ([[User talk:CodeTalker|talk]]) 18:06, 3 January 2025 (UTC) |
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:::::Right, convert from binary, somehow I didn't think of that. Basically just divide by 10 41 million times, which would only be an issue if it was billions instead of millions. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 06:21, 4 January 2025 (UTC) |
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:::: Because of the nature of the exercise it isn't important that every technique I try is successful but regardless you've brought me round to not trying the technique, which I believe won't be worth the effort. It helped to have this discussion even if the focus of it changed a little, for which I thank you. --[[Special:Contributions/80.4.203.142|80.4.203.142]] ([[User talk:80.4.203.142|talk]]) 23:44, 18 February 2008 (UTC) |
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= January 5 = |
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== Reference request:coherence condition (adjoint functor) == |
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While studying for my math test, I came across this problem: |
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Previously, in [[Wikipedia talk:WikiProject Mathematics/Archive/2024/Oct#(Mac Lane's) coherence theorem vs. Coherence condition|WPM]] (Coecke and Moore (2000)) taught me a statement of coherence condition for adjoint functors. This is sometimes called triangle identities or zigzag identities, and I'm looking for some references. I am also trying to find where to find the Wikipedia article that explains coherence conditions (adjoint functors). Also, I'm looking for a Wikipedia article that explains coherence conditions (adjoint functors). (e.g. [[coherence condition]], [[adjoint functor]], or new draft ?) |
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I am thinking of some polynomial with positive integer coefficients. You can ask me for the value of p[x], where p is the polynomial and x is some positive integer as many times as you want. What is the minimum number of questions you need to ask in order to find out what the polynomial is? |
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*{{cite web|title=triangle identity|url=https://ncatlab.org/nlab/show/triangle+identity|website=ncatlab.org}} |
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I am unable to figure out the answer. Since it doesnt tell you how many terms or the order of the polynomial, it's hard to find a starting spot. Can anyone help? [[Special:Contributions/99.240.177.206|99.240.177.206]] ([[User talk:99.240.177.206|talk]]) 18:37, 18 February 2008 (UTC) |
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*{{cite journal |doi=10.1016/j.jpaa.2024.107625 |title=Naturality of the ∞-categorical enriched Yoneda embedding |date=2024 |last1=Ben-Moshe |first1=Shay |journal=Journal of Pure and Applied Algebra |volume=228 |issue=6 |arxiv=2301.00601 }} |
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*{{cite book |url={{Google books|YfzImoopB-IC&q|page=99|plainurl=yes}} | title=Handbook of Categorical Algebra: Basic category theory | isbn=978-0-521-44178-0 | last1=Borceux | first1=Francis | date=1994 | publisher=Cambridge University Press }} |
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*{{cite arxiv |arxiv=quant-ph/0008021 |last1=Coecke |first1=Bob |last2=Moore |first2=David |title=Operational Galois adjunctions |date=2000 }} |
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*[https://planetmath.org/93adjunctions planetmath] |
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[[User:Silvermatsu|SilverMatsu]] ([[User talk:Silvermatsu|talk]]) 02:41, 5 January 2025 (UTC) |
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:Can you identify more precisely for which statement(s) you want a reference? Is it for the definition of counit–unit adjunction given in section {{section link|Adjoint functors#Definition via counit–unit adjunction}}? |
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:I remember that one. The answer is 2. Does that help? [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 18:45, 18 February 2008 (UTC) |
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:Explanations in mathematics can be of different kinds. One kind are explanations of definitions. Definitions are true by definition; an explanation generally means helping build up an intuition of the concept defined by showing familiar structures satisfying the definition. Another kind are explanations of statements. These typically offer a reformulation of a statements as an equivalent statement in terms of simpler concepts. Then there are explanations of proofs. These can include showing the proof "in action" on specific examples, using familiar structures as when explaining definitions. And they can assist in verifying the validity of proof steps, by unfolding definitions until the step becomes obvious. In general, all explanations can involve a combination of these. |
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::It helps me, at least. Nice problem. [[User talk:Algebraist|Algebraist]] 19:45, 18 February 2008 (UTC) |
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:For example, in our article [[Adjoint functors]], the ramifications of the definition in the lead section, basically the existence of a bijection <math>\mathrm{hom}_{\mathcal{C}}(Fd,c) \cong \mathrm{hom}_{\mathcal{D}}(d,Gc)</math> that is natural in <math>c</math> and <math>d,</math> are not immediately obvious, but by carefully unfolding the definitions, starting with the required [[Natural transformation|naturality]] of the two morphisms making up the bijection, leads to the equivalent counit–unit definition. |
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:I am not sure what kind of explanation of "coherence conditions (adjoint functors)" you are seeking. Perhaps studying this article, [https://research.utwente.nl/en/publications/adjunctions "Adjunctions"], will answer this question. --[[User talk:Lambiam#top|Lambiam]] 12:18, 5 January 2025 (UTC) |
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== Concatenation of first 10 digits and last 10 digits of a number == |
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:::If I understand Black Carrot correctly, the value you choose for ''x'' in your second question depends upon the answer you receive to your first question - yes ? [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 20:06, 18 February 2008 (UTC) |
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::::It would have to, polynomials aren't completely determined by their values at two points. There must be some clever trick based on the answer to the first questions... I haven't worked it out yet, but give me some time... --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 20:08, 18 February 2008 (UTC) |
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:::::You don't just have 2 points, you have 2 points plus the knowledge that the coefficients are positive integers. I just got it! Oh that is clever. --[[User:tcsetattr|tcsetattr]] ([[User talk:tcsetattr|talk]] / [[Special:Contributions/tcsetattr|contribs]]) 20:40, 18 February 2008 (UTC) |
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::::::...which was the point I was trying subtly (perhaps too subtly) to make. Answer to first question gives you information about coefficients which you use in second question. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 22:42, 18 February 2008 (UTC) |
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Let a(n) be the concatenation of first 10 digits and last 10 digits of n, then we know that a(2<sup>136279841</sup>-1) = 88169432759486871551, however: |
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:::::Haha, wow, that's pretty cool. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 21:23, 18 February 2008 (UTC) |
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# Can a(2^n) take all 20-digit values which are multiples of 1024? |
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Hm..suppose if one chose p(0) and the answer turned out to be 4 or 5. Obviously, this would indicate the the constant at the end is either a 4 or 5. Where would you go from there?[[User:Acceptable|Acceptable]] ([[User talk:Acceptable|talk]]) 20:39, 18 February 2008 (UTC) |
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# Can a(3^n) take all 20-digit values which are odd? |
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:Asking for p(0) is a bad start (aside from the fact that it wasn't technically allowed by the problem statement). --[[User:tcsetattr|tcsetattr]] ([[User talk:tcsetattr|talk]] / [[Special:Contributions/tcsetattr|contribs]]) 20:55, 18 February 2008 (UTC) |
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# Can a(n^2) take all 20-digit values which end with 0, 1, 4, 5, 6, 9? |
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::Funnily enough, 0 (if you take it to be positive) is the only first question that ''doesn't'' work. Suppose you got f(0)=a (say), and used n for your second question. If you got f(n)=n^2+a, then f could be nx+a or x^2+a. [[User talk:Algebraist|Algebraist]] 21:50, 18 February 2008 (UTC) |
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# Can a(n^3) take all 20-digit values? |
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# Can a([[prime number]]) take all 20-digit values which end with 1, 3, 7, 9? |
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Finally, I think I've got it! That shouldn't have taken me that long... --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:29, 18 February 2008 (UTC) |
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# Can a([[lucky number]]) take all 20-digit values which are odd? |
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:Just in case the OP needs another hint - how would you solve this if you also knew that all coefficients are less than 10? -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 22:32, 18 February 2008 (UTC) |
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# Can a([[Fibonacci number]]) take all 20-digit values? |
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::It finally clicked when I twigged that it was only positive coefficients. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 23:02, 18 February 2008 (UTC) |
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# Can a([[partition number]]) take all 20-digit values? |
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I think the smallest number of questions you need to ask is (the degree of the polynomial + 1).--[[User:Protious|George]] ([[User talk:Protious|talk]]) 22:43, 18 February 2008 (UTC) |
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# What is a(9^9^9^9)? |
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# What is a(9^^9), where ^^ is [[tetration]]? |
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::I wonder if there's a finite bound for more general polynomials, like with positive rational coefficients. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 22:47, 18 February 2008 (UTC) |
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# What is a(9^^^9), where ^^^ is [[pentation]]? |
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# What is a([[Graham's number]])? |
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:::To George - that'd be right for polynomials with no restrictions on the coefficients, but you can take advantage of their positive integerosity to jump straight to the answer. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 22:50, 18 February 2008 (UTC) |
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# What is a([[TREE(3)]])? |
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# If we know a(x), assume that x has at least 20 digits, can we also know a(2*x), a(3*x), etc.? |
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:::Do you have any specific technique for doing this or is this just a general statement ?--[[User:Protious|George]] ([[User talk:Protious|talk]]) 22:56, 18 February 2008 (UTC) |
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# If we know a(x), assume that x has at least 20 digits, can we also know a(x^2), a(x^3), etc.? |
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# If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x+y)? |
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::::To the OP's riddle? Yeah, I've got the answer. Give it a little while, and reflect on Meni's hint. BTW, for a polynomial in n variables, with positive integer variables and coefficients, it could be done within n+1 questions. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 23:00, 18 February 2008 (UTC) |
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# If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x*y)? |
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# If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x^y)? |
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::::I got it ,thanks. --[[User:Protious|George]] ([[User talk:Protious|talk]]) 23:53, 18 February 2008 (UTC) |
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[[Special:Contributions/220.132.216.52|220.132.216.52]] ([[User talk:220.132.216.52|talk]]) 08:33, 5 January 2025 (UTC) |
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::::I think 2 questions ought to work for real non-negative coefficients - that they're integers I wouldn't think was strictly necessary. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 23:02, 18 February 2008 (UTC) |
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:::::Really? It doesn't seem like that would be enough. That is, for any two inputs you choose, it seems like there would be at least one pair of polynomials with the same output. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 23:06, 18 February 2008 (UTC) |
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::::::If you choose any two points, there are an infinite number of polynomials that go through them. But, restricted to non-negative coefficients, if you choose the 2nd point based on the first one, it ought to be possible to find the order. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 23:12, 18 February 2008 (UTC) |
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:::::::Thinking about Black Carrot's extended question, I can see that if you only know that all (non-zero) coefficients are real and positive, then with two questions you can find the maximum degree of the polynomial - say it is ''m'' (so at most ''m''+1 non-zero coefficients). But don't you still need a further ''m''-1 questions to find the ''values'' of the coefficients (so that altogether you have ''m''+1 simultaneous equations) ? Key difference from the positive integer case is that there is no longer a lower bound on the magnitude of a non-zero coefficient. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 00:00, 19 February 2008 (UTC) |
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:::Generally with polynomials, rational coefficients and integer coefficients work in exactly the same way (you just multiply through by the lowest common multiple of all the denominators), so I imagine the case of positive rationals is doable (you may end up needing to additional fact that the polynomial is monic in order to get a unique answer, I'd have to try). The case of positive reals probably isn't, though. The key fact that p(1)<10 (or whatever) gives you is that 10 doesn't divide any of the coefficients, in the real case that fact is no longer true (it's not true for rationals, either, but you can probably fudge it). I haven't actually tried either case, this is just intuition and could well turn out to be wrong. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:50, 19 February 2008 (UTC) |
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::::But how do you determine the lowest common multiple (or even ''any'' common multiple) of the denomoinators ? I think the positive rational coefficients case is the same as the positive real coefficients case - it only takes a finite number of questions, you don't know in advance how many, but after two questions you know how many further questions you need. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 15:08, 19 February 2008 (UTC) |
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:::::That comment was just to demonstrate why polynomials over the integers and polynomials over the rationals are closely related. Obviously, you can't directly convert in this case, but there might be some trick that allows you to use the same basic method. I haven't looked for one, so I don't know. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 16:44, 19 February 2008 (UTC) |
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:For those still struggling, I suggest using p(1) as the first question. It's not necessary, but thinking about what I could learn from p(1) was what led me to the solution. And don't forget the coefficients are all positive! --[[User:tcsetattr|tcsetattr]] ([[User talk:tcsetattr|talk]] / [[Special:Contributions/tcsetattr|contribs]]) 23:21, 18 February 2008 (UTC) |
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::Sorry, I still am not getting it. The question does not state how many terms there are. So suppose I realize that none of the coefficients are greater than 10, I still dont know anything about how many terms or how large the coeff of each term is. |
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Let's take an example: suppose the equation was f(x)= 4x^3 + 3x^2 + 7x + 6. If I plug in p(1), i get p(1)=20. What do I ask next? Thanks. [[Special:Contributions/99.240.177.206|99.240.177.206]] ([[User talk:99.240.177.206|talk]]) 04:29, 19 February 2008 (UTC) |
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:OK, that's enough. Full spoiler: suppose you had chosen that function and I asked you for p(1) and you said p(1)=20. I know the polynomial is A+Bx+Cx^2+Dx^3+... so what do I get when x=1? A+B+C+D+... in other words the sum of the coefficients. I don't know how many coefficients there are, but I know they are all positive integers and their sum is 20. If a set of positive numbers adds up to 20, then none of them individually can be greater than 20. The next question I'd ask is for p(100). I could ask for p(21) instead - any number greater than 20 will do - but I'm choosing 100 because it might make the procedure more obvious. |
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:Your answer will be p(100) = 4(100^3) + 3(100^2) + 7(100) + 6 = 4030706. See how the coefficients 4,3,7,6 revealed themselves? If I had asked for p(21) you would have said 38520 which would make the procedure a little more mysterious-looking, but the secret is I would then convert 38520 into base 21, and the result would be: 4376<sub>21</sub>! |
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:Now we can go back to the first step and see how p(1) doesn't have to be used. If I asked for p(2) first, you'd say 64. In this version, 64 is not the sum of the coefficients, but it's still an upper bound on them individually. If any one of your coefficients was greater than 64, it would contribute a term of (something greater than 64)*(some power of 2) to your answer, so your answer would have been greater than 64 also. Knowing this upper bound, I could ask for p(65), get the answer 1111636, and convert it to base 65, where it is: 4376. --[[User:tcsetattr|tcsetattr]] ([[User talk:tcsetattr|talk]] / [[Special:Contributions/tcsetattr|contribs]]) 04:44, 19 February 2008 (UTC) |
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::And of course, my point was that if all coefficients are less than 10, you don't need two questions - you just ask what is <math>p(10)</math>. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 09:49, 19 February 2008 (UTC) |
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This is nice. Ask for <math>p(1000000)</math> and the problem is solved in one question for all the cases where the coefficients are less than a million. You do not have to require that the integer coefficients are positive, just that they are not negative. What if the integer coefficients may be negative too, how do you solve that? [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 15:33, 19 February 2008 (UTC). |
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:::If the coefficients can be arbitrary integers, then infinitely many questions are required (consider the case when every question gets the answer '0'). If they're arbitrary positive reals, then finitely many questions suffice (as Gandalf pointed out above) but I'm not sure if you can get a fixed finite bound or not. [[User talk:Algebraist|Algebraist]] 16:25, 19 February 2008 (UTC) |
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::::You can't have infinitely many questions all giving 0, since the polynomial will always be of finite degree (otherwise it's not a polynomial). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 16:44, 19 February 2008 (UTC) |
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:::::Well, I suppose there's the 0 polynomial... it would take an infinite number of questions to spot that. Any non-0 polynomial would be worked out with finite (but unbounded) steps, though, I'd think. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 16:52, 19 February 2008 (UTC) |
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::::::No. However (finitely) many times I give you the answer zero, you won't know if it's the zero polynomial or just some other poly that happens to have roots at every point you've asked about. [[User talk:Algebraist|Algebraist]] 18:35, 19 February 2008 (UTC) |
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:::::::That's not quite what I said. I said that any non-zero polynomial could be worked out in finitely many steps, not that finitely many steps is enough to tell if a polynomial is non-zero. There are only finitely many roots to any polynomial (at most, it's degree), so after one more question than that (which is still a finite number), you'll know it's non-zero. Whether you can work out exactly what it is or not, I'm not sure - repeated roots would be an issue, but you would know it's non-zero. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 20:39, 19 February 2008 (UTC) |
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::::::::<s>That (while true) is not much use to solve the problem as stated. Sure, for any polynomial there is a question-asking strategy that proves the poly is non-zero in finitely many steps, but what we want is a strategy that for any polynomial will prove it is non-zero in finitely many steps. We can't base our strategy on an answer we don't know.</s> [[User talk:Algebraist|Algebraist]] 22:34, 19 February 2008 (UTC) |
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:::::::::How about asking for p(0), p(1), p(-1), p(2), p(-2), etc. until you get a non-zero answer? If the polynomial is non-zero, you will get a non-zero answer after a finite number of questions. If it is the 0-polynomial, then you'll never get an answer, but you're asking for a proof that it's non-zero, not a proof that it is zero. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:08, 20 February 2008 (UTC) |
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::::::::::Some day I'll learn not to trust maths I do drunk. [[User talk:Algebraist|Algebraist]] 16:34, 21 February 2008 (UTC) |
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== Set of convergence of a sequence of measurable functions == |
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This is an exercise from [[Serge Lang|Lang]]'s ''Real and Functional Analysis'', Chapter VI: Let <math>\{f_n\}</math> be a sequence of measurable functions. Show that the set of points for which <math>\{f_n\}</math> converges is a measurable set. |
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Presumably this involves finding some collection of obviously measurable sets that can be unioned/intersected together to end up with the set in question, but I haven't been able to find such a collection. The only information available is the basic properties of σ-algebras and measurable functions. |
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[As a side issue, does anyone have experience with or thoughts on this book? I can't say I find it easy going, but struggling with it seems to be good practice and the more time I spend with it the more I appreciate it. I got it originally for the development of integration, which is the best I've seen anywhere--just beautiful! If only there were a solutions manual... [[User:merge| — merge]] 23:33, 18 February 2008 (UTC)] |
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:One way of doing this is to first show that sup<sub>n</sub>{f<sub>n</sub>} is measurable. Dually the inf of countably many measurable functions is measurable, so the lim sup and the lim inf of the f<sub>n</sub> are both measurable. Then the set of x for which f<sub>n</sub>(x) converges is the set of points at which two measurable functions are equal, and is thus measurable. Slight care might be needed in dealing with infinity. [[User talk:Algebraist|Algebraist]] 23:51, 18 February 2008 (UTC) |
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::Thanks! This idea had not occurred to me. However as it turns out the properties of lim sup and lim inf you refer to are dealt with in the next exercise! So I'm thinking there's a way to do this one that doesn't use them. Also, I forgot to mention that the <math>f_n</math> take values in a Banach space, so one can't take the sup directly. [[User:merge| — merge]] 12:52, 19 February 2008 (UTC) |
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:::In that case, you could use the fact that f<sub>n</sub>(x) converges exactly if f<sub>n</sub>(x) is Cauchy. The latter statement is just |
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::::<math> \forall{\epsilon} \exists{N} \forall {n,m} (n,m \ge N \rightarrow |f_n(x)-f_m(x)| < \epsilon) </math> |
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:::So the set of x for which f<sub>n</sub>(x) converges is just |
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::::<math>\bigcap_{\epsilon} \bigcup_N \bigcap_{n \ge N} \bigcap_{m \ge N} (f_n-f_m)^{-1} (B_{\epsilon})</math> |
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:::where <math>B_{\epsilon}</math> is the open ball of radius epsilon about 0 (I'm assuming this is measurable) and all the intersections and unions are countable (in particular we can restrict to positive ''rational'' epsilon). [[User talk:Algebraist|Algebraist]] 16:22, 19 February 2008 (UTC) |
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::::Wonderful. Someone else suggested the same idea in conversation and it works perfectly. I think this is "the" solution. Thank you! [[User:merge| — merge]] 21:23, 19 February 2008 (UTC) |
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== Googol == |
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Why is one [[googol]] not <math>10^{100}</math>? The Wikipedia article states it as being <math>10^{100.0784}</math>. Why is this? Thanks, [[User:Zrs 12|Zrs 12]] ([[User talk:Zrs 12|talk]]) 23:50, 18 February 2008 (UTC) |
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:One googol is by definition 10<sup>100</sup>, as correctly stated in the article. The article also correctly states that 70! is approximately 10<sup>100.0784</sup>. I don't see any source of confusion here. [[User talk:Algebraist|Algebraist]] 23:55, 18 February 2008 (UTC) |
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:: Perhaps OP did not understand [[Order of magnitude]], and just skipped over those words to read "One googol is the same as 70!." --[[User:PalaceGuard008|PalaceGuard008]] ([[User_Talk:PalaceGuard008|Talk]]) 23:57, 18 February 2008 (UTC) |
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Haha! How dumb of me! Gah, I '''must''' start reading more carefully. [[User:Zrs 12|Zrs 12]] ([[User talk:Zrs 12|talk]]) 00:38, 19 February 2008 (UTC) |
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= February 19 = |
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==Singular Value Decomposition & Hermitian Matrices== |
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This is a question posed in my class and the teacher himself could not figure it out either. The question is that given the singular value decomposition of a mxm square matrix <math>A=U \Sigma V^*</math> where the * represents the complex conjugate transpose of a matrix, is there anything that we can say about the eigenvalue decomposition (diagonalization) of the 2mx2m Hermitian matrix <math>B=\begin{bmatrix}0 && A^*\\A &&0\end{bmatrix}</math>? Can B's eigenvalue decomposition be written in terms of <math>U,\Sigma,</math> and <math>V</math> ? I have also tried a few numerical examples on Matlab and it appears to me that the two are completely unrelated. Any help would be appreciated.[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 06:10, 19 February 2008 (UTC) |
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: If <math>A=U \Sigma V^*</math>, then <math>U^*AV=\Sigma</math>. As <math>\Sigma</math> is self-adjoint, we get <math>V^*A^*U=(U^*AV)^* = \Sigma</math> as well. Now compute: |
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:<math>\begin{pmatrix}0 & U^*\\V^* & 0\end{pmatrix}\begin{pmatrix}0 & A^*\\A & 0\end{pmatrix}\begin{pmatrix}V & 0\\0 & U\end{pmatrix} = \begin{pmatrix} U^*AV & 0 \\ 0 & V^*A^*U\end{pmatrix} = \begin{pmatrix}\Sigma & 0 \\ 0 & \Sigma \end{pmatrix}.</math> |
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:Moving the unitaries (one can check that those block matrices on the left and right of your <math>B</math> are unitary) to the right gives the singular value decomposition. You have the absolute values of the eigenvalues at this point, but I'm not sure exactly what you mean by eigenvalue decomposition. Do you mean diagonalizing B? [[User:J Elliot|J Elliot]] ([[User talk:J Elliot|talk]]) 07:13, 19 February 2008 (UTC) |
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:The [[eigenvalue decomposition]] goes similarly. For motivation, find the eigenvalue decomposition of [0,1;1,0] (taking A to be the 1x1 identity matrix). A=USV*, so A*=VSU*, AV=US, A*U=VS, so [0,A*;A,0][V;U] = [VS;US], so [V;U] is an "eigenvector" with eigenvalues S, and similarly [0,A*;A,0][V;-U]=[-VS;US], so [V;-U] is an "eigenvector" with eigenvalues -S. Since our matrix is hermitian, we want our eigenbasis to be unitary, so we'll divide the eigenvectors by their overall norm, 1/sqrt(2). Putting it all together gives: |
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:<math>\begin{bmatrix}0&A^*\\A&0\end{bmatrix} = \left( \frac{1}{\sqrt{2}} \begin{bmatrix}V&V\\U&-U\end{bmatrix} \right) \cdot \begin{bmatrix}\Sigma&0\\0&-\Sigma\end{bmatrix} \cdot \left( \frac{1}{\sqrt{2}} \begin{bmatrix}V^*&U^*\\V^*&-U^*\end{bmatrix} \right)</math> |
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:So the eigenvalues of B are plus or minus the singular values of A. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 15:53, 19 February 2008 (UTC) |
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Thanks guys, that makes a lot more sense. But I still have a follow-up question. You have shown that [V;U] and [-V;U] are eigenvectors with respect to S and -S but how do we know that those are the only eigenvalues? What if S^2 of -2S^5 also turn out to be eigenvalues of our matrix B? How can we conclude that S and -S are the ONLY eigenvalues?[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 23:53, 19 February 2008 (UTC) |
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:Sorry, I spoke too informally. U and V are actually m x m matrices, and S is a diagonal matrix with m values. [V,V;U,-U] has full rank (because it is unitary), so is actually 2m independent eigenvectors for B, and S,-S gives the 2m eigenvalues. The informal language was just to indicate how block matrices can simplify things. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 00:58, 20 February 2008 (UTC) |
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== fitting a conic == |
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I want to fit a general parabola (of unknown size and orientation) ''roughly'' to a given set of points in the plane. My first thought was to seek the coefficients that minimize the sum of the squares of <math> A {x_i}^2 + 2 \sqrt{AC} x_i y_i + C {y_i}^2 + D x_i + E y_i + 1 </math>; to make the problem more linear, I then sought to settle for a general conic, <math> A {x_i}^2 + 2 B x_i y_i + C {y_i}^2 + D x_i + E y_i + 1 </math>; but then it occurs to me that this penalizes those curves that go near the origin. |
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My next idea is to consider the family of cones tangent to some plane <math>z = z_0 \ne 0</math>; I'm not sure what to minimize. |
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Anyone know a better way? —[[User:Tamfang|Tamfang]] ([[User talk:Tamfang|talk]]) 06:24, 19 February 2008 (UTC) |
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:Minimize the sum of squares of <math> A {x_i}^2 + 2 B x_i y_i + C {y_i}^2 + D x_i + E y_i + F </math> using some other normalizing condition than <math> F=1 \,</math> (which excludes curves through the origin). [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 07:50, 19 February 2008 (UTC). |
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:You could try an iterative approach. Define for brevity |
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::<math>F(x,y) = A {x_i}^2 + 2 B x_i y_i + C {y_i}^2 + D x_i + E y_i + 1</math> |
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:and |
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::<math>F_i = F(x_i,y_i).\,</math> |
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:Given estimates for the coefficients ''A'', ''B'', etcetera, you can determine the distance ''e<sub>i</sub>'' of each point (''x<sub>i</sub>'',''y<sub>i</sub>'') to the curve determined by {{mbox|= ''F''(''x'',''y'') = 0}}. If we give weight ''w<sub>i</sub>'' to the term ''F''<sub>i</sub><sup>2</sup> in a weighted sum of squares, we want the weighted square to come out like ''e<sub>i</sub>''<sup>2</sup>, which suggests setting |
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::<math>w_i = \frac{e_i^2}{F_i^2}</math> |
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:as the weights for a next iteration. |
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: |
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:Instead of determining the values ''e<sub>i</sub>'' exactly, which is computationally expensive, you can approximate it by using the linear approximation |
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::<math>F(x+\Delta x,y+\Delta y) \approx F(x,y) + \frac{\part}{\part x}F(x,y)\cdot\Delta x + \frac{\part}{\part y}F(x,y)\cdot\Delta y.</math> |
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:Applied to the point (''x<sub>i</sub>'',''y<sub>i</sub>''), we write this as |
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::<math>F(x_i+\Delta x,y_i+\Delta y) \approx F_i + \frac{\part}{\part x}F_i\cdot\Delta x + \frac{\part}{\part y}F_i\cdot\Delta y.</math> |
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:The least value for {{mbox|(Δ''x'')<sup>2</sup> + (Δ''y'')<sup>2</sup>}} for which the right-hand side can vanish, which provides an estimate for ''e<sub>i</sub>''<sup>2</sup>, is then given by |
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::<math>(\Delta x)^2 + (\Delta y)^2 = \frac{F_i^2}{\frac{\part}{\part x}F_i^2 + \frac{\part}{\part y}F_i^2}\,.</math> |
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:So for the weights for the next iteration, you can use then |
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::<math>w_i = (\frac{\part}{\part x}F_i^2 + \frac{\part}{\part y}F_i^2)^{-1}\,.</math> |
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:Since the value being inverted can become arbitrarily small and even vanish, you must exercise caution not to make this numerically unstable, and put limits on the size of the weights. --[[User talk:Lambiam|Lambiam]] 08:49, 21 February 2008 (UTC) |
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== Rationalising Surds == |
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I was going over some of my notes and i tried this one but my answer isnt the same as in the book.. im not sure what im doing wrong<br> |
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Rationalise the denominator<br> |
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<math>\frac{1{}}{{\sqrt5 - \sqrt7}}</math> <br> |
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<math>=\frac{1{}}{{\sqrt5 - \sqrt7}}</math> <math>*\frac{{\sqrt5 + \sqrt7}}{{\sqrt5 + \sqrt7}}</math> <br> |
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<math>=\frac{{\sqrt5 + \sqrt7}}{{-2}}</math> |
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<br>[[User:Kingpomba|Kingpomba]] ([[User talk:Kingpomba|talk]]) 11:26, 19 February 2008 (UTC) |
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:Looks fine to me and quick calculator check confirms your answer. Could also be written as <math>\frac{{-\sqrt5 - \sqrt7}}{{2}}</math> or <math>-\left(\frac{{\sqrt5 + \sqrt7}}{{2}}\right)</math>. What does your book say ? [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 11:47, 19 February 2008 (UTC) |
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it says: |
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<math>\frac{1{}}{{2}}</math> |
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<math>(\sqrt5 + \sqrt7)</math> |
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(hmm i got a different answer on paper and i understand how the 1/2 thing works i guess typing it out on wikipedia helped me solve it, well cheers anyway =] [[User:Kingpomba|Kingpomba]] ([[User talk:Kingpomba|talk]]) 11:57, 19 February 2008 (UTC). |
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:Are you sure that's what it says? There should be a minus sign in front, shouldn't there? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:54, 19 February 2008 (UTC) |
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:Here's a quick sanity check: 5 < 7, so sqrt(5) < sqrt(7) (by the properties of the square root), and hence sqrt(5) - sqrt(7) < 0. Thus the denominator of the original fraction is negative, meaning the whole fraction is negative. [[User:ConMan|Confusing Manifestation]]<small>([[User talk:ConMan|Say hi!]])</small> 23:47, 19 February 2008 (UTC) |
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= February 20 = |
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== zeros of a periodic function == |
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Is there any way to find an <math>x</math> which solves the equation |
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<math>\frac{1}{2}+\frac{1}{4} \left( \cos(a x)+ \cos(b x) \right)=0 </math> |
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for any arbitrary <math>a</math> and <math>b</math>? I know that some value of ''x'' will make both <math>ax</math> and <math>bx</math> equal to an odd multiple of <math>\pi</math> but can't think of a way to find this value. I know that the answer may lie in recasting the equation in the form |
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<math>\frac{1}{2}+\frac{1}{2} \left( \cos(\frac{a-b}{2}x) \cos(\frac{a+b}{2} x) \right)=0 </math> |
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and in this case know that some value of x makes this product of cosine terms equal to negative one, but again can't think of how to find this x value. Can this be solved analytically or must I resort to numerical calculation (which would be very unsatisfying). Thank you very much for your assistance. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/128.223.131.21|128.223.131.21]] ([[User talk:128.223.131.21|talk]]) 00:21, 20 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:If the combined function is periodic, then a/b is a rational number, so after rescaling x, you can assume a and b are relatively prime integers. Then I believe x=pi is a solution if both a,b are odd, and there is no solution if one of a or b is even. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 01:16, 20 February 2008 (UTC) |
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::To explain JackSchmidt's answer a bit: since (for <math>x\in\mathbb R</math>) <math>|\cos x|\le1</math>, the only way to have <math>\cos x+\cos y=-2</math> is to have <math>\cos x=\cos y=-1</math>. So <math>ax=(2n+1)\pi,bx=(2m+1)\pi</math>, so <math>\frac ab=\frac{2n+1}{2m+1}</math> and the simplest form of <math>\frac ab</math> must have both numbers odd. When that condition is satisfied, there is some smallest integer <math>k\ni\{ak,bk\}\sub\mathbb I</math>; ''k'' is the common denominator for ''a'' and ''b''. Any <math>x=(2l+1)k\pi</math> with integer ''l'' solves the equation. --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 16:20, 20 February 2008 (UTC) |
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== ''[[Donald in Mathmagic Land]]'' == |
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Two questions about the above-named 1959 Walt Disney film. (Question 1) In the film, a character incorrectly states the value of pi. Does anyone know how a Disney film that was designed to be a mathematics educational film for children (and presumably had math consultants working on it) could possibly contain such an error? How would that error slip by unnoticed? (Question 2) With regard to that error, someone stated (in a blog) that it was not really an error, since "the value of pi has changed so much since 1959". That cannot possibly be true, can it? I am sure that we have become enlightened about the accurate values of many digits of pi over the course of thousands of years of history ... but not since 1959, correct? Any ideas? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 04:10, 20 February 2008 (UTC)) |
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: Pi is an irrational number meaning that its' decimal representation will have and infinite amount of digits. Given the fact that we can never calculate an infinite amount of digits it is the case that new digits have been calculated since 1959. Of course this does not mean the the value of pi has changed, just the precision with which one can write it. Perhaps this is what 'someone' was reffering too? Any errors in the calculation of digits of pi since 1959 would, i suspect, occur at least several thousand (being conservative here) digits out so it would in fact be an error. I have no input on how the error actually occured. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/74.242.224.136|74.242.224.136]] ([[User talk:74.242.224.136|talk]]) 04:55, 20 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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: The error could very well have been caused by something as simple as the voice actor getting lost while reading the digits. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 05:00, 20 February 2008 (UTC) |
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Thanks. Let me follow-up. I should have made this clearer. (Though, remember, we are dealing with a Disney children's video.) First ... the value of pi was recited to, perhaps, 15 or 20 decimal places ... certainly, no more than 20. So, since 1959, nothing has "changed" in that early chunk of the number - correct? Second ... regarding the voice actor flub. Certainly, that is a possibility. I guess my point was ... is not that an "inexcusable" error (that would call for a retake ... or, at the very least, a dubbed edit in post-production), given that it is an education video, and a math one to boot ( ... as opposed to, say, some character misstating the digits of pi on an inconsequential TV episode of ''Seinfeld'') ...? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 05:16, 20 February 2008 (UTC)) |
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: [[Pi]] was known to 527 digits by the end of the 19th century, and to 2037 digits by 1949 (a decade earlier than the film was made). Our article on [[Donald in Mathmagic Land]] says that the bird only recites "the first few digits" of pi. The error was not due to pi not being known accurately enough in 1959. [[User:MrRedact|MrRedact]] ([[User talk:MrRedact|talk]]) 05:25, 20 February 2008 (UTC) |
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:: Probably the producers didn't consider it important. If the voice actor flubbed, either no one noticed or no one thought it was worth it to rerecord the line just to be pedantic. The character reciting pi is basically a throw-away gag anyway. They got the first few digits right (out to maybe eight or ten places, if I remember correctly), and that's certainly more than anyone really needs. The purpose of the film is not to teach children the digits of pi but to introduce them to some general mathematical concepts. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 06:12, 20 February 2008 (UTC) |
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Yes, I agree with all that is said above. However, as an ''education'' video --- I am not sure that being "pedantic" is a bad thing ... quite the contrary, I'd assume. I will also assume that Disney's multi-zillion dollar fortune could easily afford any "expensive" retakes. Finally, mathematicians are nothing, if not precise. I can't imagine a mathematician letting that glaring error slide. A film producer, yes ... a mathematician, nah. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 06:56, 20 February 2008 (UTC)) |
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: I agree completely. This is just a symptom of a much more general problem. Very few people care about spreading misinformation, and very many then wonder how come people are so misinformed. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 10:45, 20 February 2008 (UTC) |
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:: What a succinct -- yet woefully accurate -- synopsis, Meni. Bravo! ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 16:20, 20 February 2008 (UTC)) |
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:As Bertrand Russell appears to have written: 'PEDANT—A man who likes his statements to be true.' [[User talk:Algebraist|Algebraist]] 10:55, 21 February 2008 (UTC) |
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== Predicate logic == |
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I have to show, using a transformational proof (logical equivalence and rules of inference) that the following two premises are a contradiction: |
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:1. <math>\neg \exists x [ ((Q(x) \and R(x)) \or (Q(x) \and \exists y P(y)) ]</math> |
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:2. <math>\exists x [ ((Q(x) \and R(x)) \or (Q(x) \and \forall y P(y)) ]</math> |
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Now these two statements look quite similar because they were actually two more complex statements I tried to "whittle down" into a similar form in order to see where I could take it. They almost contradict one another apart from the <math>\exists y P(y)</math> part in 1. versus the <math>\forall y P(y)</math> part in 2. |
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Can you offer me a suggestion for manipulating these statements further to show a clear contradiction? My first line of thought would be to show that the conjunction, <math>\and</math>, of the two statements is false but I get bogged down in symbols. Perhaps there is a more elegant way using inference rules, I'm not sure. [[User:Damien Karras|Damien Karras]] ([[User talk:Damien Karras|talk]]) 13:39, 20 February 2008 (UTC) |
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:Start by consulting '''[[De Morgan's laws]]''', including the ones for quantifiers, if you're not familiar with them. Then you should be able to make use of the fact that <math>(\forall x\,A(x))\and(\exists x\,B(x))\rightarrow\exists x\,A(x)\and B(x)</math>. Also, it may not be strictly necessary, but consider what it means to say that <math>\exists x\,A(y)</math>. --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 16:34, 20 February 2008 (UTC) |
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::The rwo premises are contradictory if (and only if) (2) implies the negation of (1). You will have to use somehow, directly or indirectly, a rule that says that <math>\forall y P(y)</math> implies <math>\exists y P(y)</math>. If you have a logic in which the domain of predicate ''P'' can be empty, so that this implication does not hold, while the domain of the predicates ''Q'' and ''R'', from which variable ''x'' assumes its values, is not necessarily empty, the two premises are not contradictory: take for ''Q'' the constant always-true predicate and for ''R'' the always-false predicate; the premises then simplify to <math>\neg \exists y P(y)</math> and <math>\forall y P(y)</math>. I don't know the precise rules of the you are allowed to use, but the logical connectives <math>\and\,\!</math> and <math>\or\,\!</math>, as well as existential quantification, are all "monotonic" with respect to implication, so that |
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:::<math>\exists x [ ((Q(x) \and R(x)) \or (Q(x) \and \forall y P(y)) ]\quad\rightarrow\quad\exists x [ ((Q(x) \and R(x)) \or (Q(x) \and \exists y P(y)) ]</math> |
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::is a consequence of |
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:::<math>\forall y P(y) \quad\rightarrow\quad \exists y P(y).</math> |
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:: --[[User talk:Lambiam|Lambiam]] 09:58, 21 February 2008 (UTC) |
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::::Thanks both, and let it be known that P, Q, and R share the same domain. Lambiam, I assume what you're saying is that if, essentially, if I prove <math>\forall y P(y) \rightarrow \exists y P(y)</math> I can show the two statements are a clear contradiction? [[User:Damien Karras|Damien Karras]] ([[User talk:Damien Karras|talk]]) 13:58, 21 February 2008 (UTC) |
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::::I still don't understand what I have to do. If <math>\neg\exists x Q(x)</math> in one statement and <math>\exists x Q(x)</math> in another, doesn't that show that the two sides are dependent on <math>\exists y P(y)</math> and <math>\forall y P(y)</math>. I just don't get this at all. If <math>\forall y P(y)</math> is true, then so is the existential quantification. If it is false then the existential quant. is still true? "It is not the case that, for ALL y, P(y) is true - however there is a y for which P(y) IS true." [[User:Damien Karras|Damien Karras]] ([[User talk:Damien Karras|talk]]) 15:31, 21 February 2008 (UTC) |
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== There are no countably infinite [[σ-algebra]]s == |
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Hoping someone can help me spot the holes in my argument. ;) |
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Suppose <math>\mathcal{S}</math> is a countably infinite σ-algebra over the set ''X''. If we can show that <math>\mathcal{S}</math> contains an infinite sequence of sets <math>E_1 \subset E_2 \subset \ldots</math> or <math>F_1 \supset F_2 \supset \ldots</math>, then the sequence of differences <math>E_1, (E_2-E_1), \ldots</math> or <math>\left(F_1 - \cup_{n > 1} F_n\right), \left(F_2 - \cup_{n > 2} F_n\right), \ldots</math> will be a countably infinite collection of pairwise disjoint sets <math>\{D_1, D_2, \ldots\}</math>, and the map <math>\{n_1, n_2, \ldots\} \mapsto \cup_{k\geq 1} D_{n_k}</math> will be an injection from the power set of '''N''' into <math>\mathcal{S}</math>. |
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Let <math>\mathcal{S}</math> be partially ordered under inclusion. We may assume that <math>\mathcal{S}</math> contains no infinite ascending [[Total order|chain]] <math>E_1 \subset E_2 \subset \ldots</math>, for if it does this gives us what we wanted. |
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The set <math>\mathcal{S}_1 = \mathcal{S} - \{\emptyset, X\}</math> has a maximal element ''F''<sub>1</sub> by [[Zorn's lemma]]. It follows that the only elements of <math>\mathcal{S}_1 - \{F_1, X-F_1\}</math> are subsets of ''F''<sub>1</sub> and their complements; in particular this set contains infinitely many subsets of ''F''<sub>1</sub>. The collection <math>\mathcal{S}_2</math> of these subsets cannot contain an infinite ascending chain, so has a maximal element ''F''<sub>2</sub>. It follows that the only elements of <math>\mathcal{S}_2 - \{F_2, F_1-F_2\}</math> are subsets of ''F''<sub>2</sub> and their complements relative to ''F''<sub>1</sub>. Proceeding in this way yields an infinite descending chain <math>F_1 \supset F_2 \supset \ldots</math>. [[User:merge| — merge]] 16:36, 20 February 2008 (UTC) |
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:I think you have confused [[maximal element]] with [[greatest element]]. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 19:46, 20 February 2008 (UTC) |
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Meni's right -- your last "it follows that" is wrong. The result is right, though. Hint: You don't need Zorn's lemma; the hypothesis that the sigma-algebra is countably infinite already gives you a bijection between the sigma-algebra and the natural numbers. Do induction along that. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 20:08, 20 February 2008 (UTC) |
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:I had confused myself, although not about that--sorry! I'll definitely try what you said, but I've also tried to fix the above--any better? [[User:merge| — merge]] 21:08, 20 February 2008 (UTC) |
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Revised version at [[#σ-algebra redux|σ-algebra redux]] below. [[User:merge| — merge]] 13:26, 21 February 2008 (UTC) |
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== Summing 1^1 + 2^2 +3^3 ..... + 1000^1000 == |
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I'm trying to improve my mathematics skills by doing the problems at Project Euler, but have hit a stumbling block with problem 48: (http://projecteuler.net/index.php?section=problems&id=48 |
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I need to find the last 10 digits of 1^1 + 2^2 +3^3 ..... + 1000^1000 |
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I've read the wikipedia page on [[Geometric_progression]] , but I want to do <math>\sum_{n=1}^{1000} n^n</math> rather than <math>\sum_{n=1}^{1000} r^n</math> (i.e. I don't have a constant value r) |
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I'm not so much looking for an answer, just what techniques or terminology I should be reading up on. Pointers welcomed! |
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[[User:Lordelph|Lordelph]] ([[User talk:Lordelph|talk]]) 16:56, 20 February 2008 (UTC) |
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:The key point is that you're looking for the last 10 digits, not the whole number (which is enormous). Think about which numbers contribute what to the last ten digits. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 17:14, 20 February 2008 (UTC) |
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:I don't ''think'' there is a mathematical shortcut for this. But Project Euler is as much about programming as it is about mathematics, so you could take a brute force approach. To calculate the last 10 digits of 999^999, for example, start with 999; multiply it by 999; just keep the last 10 digits; repeat another 997 times and you are done. You only need to work with interim results of up to 13 digits, so its not even a [[bignum]] problem. I guess you end up doing around half a million integer multiplications altogether, but I don't think that will take very long. In the time it's taken me to write this, someone has probably already programmed it. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 17:27, 20 February 2008 (UTC) |
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::Actually even the ultimate brute force approach of computing the whole 3001-digit sum will be more than fast enough on a modern computer. My 2004 laptop took a negligible fraction of a second to evaluate <code>sum [n^n | n <- [1..1000]]</code> in [[Glasgow Haskell Compiler|GHCi]]. They should've gone up to 1000000^1000000. -- [[User:BenRG|BenRG]] ([[User talk:BenRG|talk]]) 19:24, 20 February 2008 (UTC) |
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::I've seen similar problems where there has been a shortcut, I can't spot one for this problem, though. It's a trivial programming exercise, though, so I can't see why they would have set it. I can see a few possible ways to optimise the code, but there's really no need (for example, 999^999=(990+9)^999, expand that out using the binomial expansion and each term will have a 990^n in it, which has all 0's for the last ten digits for all but the first 10 terms, so you can ignore all but those terms - probably not much quicker for 999^999 (you need to calculate the coefficients, which won't help), but could be for 999999^999999). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 19:52, 20 February 2008 (UTC) |
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:::Many thanks for pointers, problem solved [[User:Lordelph|Lordelph]] ([[User talk:Lordelph|talk]]) 20:37, 20 February 2008 (UTC) |
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:On a slow computational platform, this can be somewhat sped up as follows. The recursive algorithms for [[exponentiation by squaring]] works in any [[ring (mathematics)|ring]], including the ring '''Z'''/n'''Z''' of [[modulo arithmetic]]. So in computing n<sup>n</sup>, all computations can be done modulo 10<sup>10</sup>. And if n is a multiple of 10, n<sup>n</sup> reduces to 0 modulo 10<sup>10</sup>, so 10% of the terms can be skipped. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Lambiam|Lambiam]] ([[User talk:Lambiam|talk]] • [[Special:Contributions/Lambiam|contribs]]) 11:13, 21 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot--> |
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== Resistance == |
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These are more questions of physics than maths but considering the overlap between the two, and how this ref desk is much more specific than the science one, I'll ask them here. |
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I know that the formula for the overall resistance of two resistors in parallel is <math> \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}</math>. So if you have two fifty ohm resistors in parallel, the total resistance is 25 ohms. What I don't understand is how the total resistance can be less than that of either resistor; it doesn't make any sense to me. |
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Also, if you have the below setup but R1 is just a wire, how would you calculate the total resistance of the circuit, ignoring internal resistance? |
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[[Image:Resistors in series and parallel.svg|A diagram of three resistors, two in parallel, which are in series with the other]] |
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Many thanks, [[Special:Contributions/92.1.70.98|92.1.70.98]] ([[User talk:92.1.70.98|talk]]) 18:07, 20 February 2008 (UTC) |
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:It's actually very simple intuitively. Resistance measures how hard it is for electrons to move from one place to another. If there are two resistors in parallel, the electrons have 2 ways to choose from, thus travel is easier. The analogy is a mass of people trying to cross a narrow corridor - they will have a much easier time if there are two parallel corridors. |
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:Any resistor network can be solved using [[Ohm's law]] (<math>V=IR</math>) and [[Kirchhoff's circuit laws|Kirchhoff's current law]] (the sum of currents at every point is 0). This one is simpler because it's just two resistors in a series, where the second is composed of two resistors in parallel. Thus the equivalent resistance is <math>R_3+\frac{R_1R_2}{R_1+R_2}</math>. |
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:There are many Wikipedia articles from which you can learn more about this - [[Resistor]] might be a good start. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 18:35, 20 February 2008 (UTC) |
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::Well, Meni beat me to it. But here it is. First of all, the total resistance decreases when you are talking about two resistors in PARALLEL not in series. In series, you get exactly what you think that you should get. The total resistance is just the sum. In parallel, you can think of it like this, you have two branches, so the current going in splits up in the branch so the energy loss in each branch is smaller than what it should have been. But then the current adds up again when the branches meet. In series, the entire current has to go through both of the resistors, so the loss is greater. As for the picture, first you calculate the combined resistance of R1 ad R2 as resistors in parallel (call it R12) which gives you <math>R_{12}=\frac{R_1 R_2}{R_1+R_2}</math> and then you simply treat R3 as a resistor in series with the combined resistor R12 and you get that the total resistance of the circuit is <math>R=R_3+R_{12}=R_3+\frac{R_1 R_2}{R_1+R_2}</math>.[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 18:39, 20 February 2008 (UTC) |
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:::OK the explanation for why the resistance is less in parallel is helpful but I think you both missed the bit about R1 being just a wire, not a resistor (it was the best picture I could find lying around on WIkipedia). Using the equation for resistors in parallel, it would make the resistance of the parallel section 0 (assuming that a wire has no resistance). I would interpret this as meaning that all current goes through the wire and none goes through the resistor. Is that right? Thanks [[Special:Contributions/92.1.70.98|92.1.70.98]] ([[User talk:92.1.70.98|talk]]) 18:45, 20 February 2008 (UTC) |
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::::That would be an idealised situation. Nothing really has 0 resistance, so R<sub>1</sub> would be very small, but it would be non-zero. This would mean that the total resistance over R1 and R2 was almost 0, since R2's contribution would be negligible. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 19:24, 20 February 2008 (UTC) |
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:::::Still, in the limit where <math>R_1 \to 0</math>, there will be no current going through R2 (this has nothing to do with ''interpretation'', it's just a calculation). Assuming we want the current to go through R2, the extra wire is a [[short circuit]]. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 19:32, 20 February 2008 (UTC) |
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== Random calculus question == |
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A friend of mine asked me to help him solve the following question, and I'm not a math major by any means. So, any help would be appreciated. :) [[User:Zidel333|Zidel333]] ([[User talk:Zidel333|talk]]) 19:40, 20 February 2008 (UTC) |
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*[1+1/N] ^ (N+1/2), prove that it is always increasing from zero to infinity. |
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:It will help to try to prove something that is true: that function is always decreasing for <math>N\in(0,\infty)</math>. Its limit at large ''N'' is [[e (mathematical constant)|e]], as can be seen immediately from that article. The normal procedure is to evaluate the function's [[derivative]] and show that it is negative for all positive ''N''. However, the algebra involved is non-trivial; perhaps some information is missing? --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 20:07, 20 February 2008 (UTC) |
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::I'm sorry, that's all the info I have from my friend. If he tells me more, I'll add it to the discussion. Thanks for your help so far, far better than what I could have come up with. :) [[User:Zidel333|Zidel333]] ([[User talk:Zidel333|talk]]) 20:12, 20 February 2008 (UTC) |
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::As it's "N" not "x", I would assume N is ranging only over the integers, so taking the derivative isn't necessary. You can either show that f(n+1)-f(n)<0 or f(n+1)/f(n)<1 for all n. The latter is probably easier. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:22, 20 February 2008 (UTC) |
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:::Hint: Perhaps rewriting the sequence as exp[(N+1/2)*log(1+1/N)] can help you (sorry, I'm not comfortable with LaTeX). --[[User:Taraborn|Taraborn]] ([[User talk:Taraborn|talk]]) 10:33, 21 February 2008 (UTC) |
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== Sum == |
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How would you work out, for j and k ranging over the integers, the sum of (j+k)^-2? [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 20:36, 20 February 2008 (UTC) |
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:Well, the term for ''j''=''k''=0 is going to be a problem :-) --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 20:50, 20 February 2008 (UTC) |
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:But assuming you really meant the ''positive'' integers, the sum diverges. For a fixed ''j'', the sum over all ''k''>0 of (''j''+''k'')<sup>−2</sup> is Ω(1/''j''). --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 20:53, 20 February 2008 (UTC) |
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:(Edit conflict) Assuming you mean positive integers, since all integers doesn't make sense, as Trovatore points out, observe that for a fixed j, and varying k, j+k varies over all the integers, so |
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:<math>\Sigma_{j,k\in N}(j+k)^{-2}=\Sigma_{j=1}^n(\Sigma_{k=1}^nk^{-2})</math> |
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:Now, the inner sum is clearly greater than 0, and summing an infinite number of non-zero constants gives you infinity (or, strictly speaking "doesn't converge"). So, unless I've missed something, the series doesn't converge. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 21:00, 20 February 2008 (UTC) |
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::Correction: For fixed j, varying k, j+k varies over all integers ''greater than j'', I was thinking of k varying over all integers, rather than just positive ones... it doesn't change the conclusion, though - the series still diverges. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:19, 20 February 2008 (UTC) |
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::The statement that "summing an infinite number of non-zero constants gives you infinity" is very often not true, even assuming you're limiting yourself to positive constants. For example, <math>\Sigma_{j=1}^\infty 2^{-j}=1</math> . See the [[Taylor series]] for e<sup>x</sup> or tan x as a couple other examples. [[User:MrRedact|MrRedact]] ([[User talk:MrRedact|talk]]) 01:50, 21 February 2008 (UTC) |
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:::None of those examples are constants. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:08, 21 February 2008 (UTC) |
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::::I guess it depends on what you mean by a "constant". My first example is the sum of 1/2, 1/4, 1/8, 1/16, 1/32, etc., all of which are positive constants, and whose sum is finite (1). I see now that you're talking about an infinite sum of <i>identical</i> positive terms, i.e., an infinite sum of terms that don't depend on the index of summation, in which case, yeah, that would always diverge. [[User:MrRedact|MrRedact]] ([[User talk:MrRedact|talk]]) 17:35, 21 February 2008 (UTC) |
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:::::Yeah, I should probably have said "an infinite number of copies of a non-zero constant". I'm pretty sure my conclusion was correct, I just didn't explain to too well, sorry! --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 18:42, 21 February 2008 (UTC) |
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:Here's another way to do it. Let i=j+k. Now, you have 1 case where i=2 (j=1, k=1); 2 cases where i=3 (j=1, k=2; and j=2, k=1), and in general i-1 cases for a given i. Thus we can express the desired sum as <math>\sum_{i=2}^\infty \frac{i-1}{i^2}</math>, which is equal to <math>\sum_{i=2}^\infty \frac{1}{i} - \sum_{i=2}^\infty \frac{1}{i^2}</math>. The first sum diverges and the second converges, so the overall sum diverges. [[User:Chuck Carroll|Chuck]] ([[User talk:Chuck Carroll|talk]]) 18:53, 21 February 2008 (UTC) |
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== Integral Question == |
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Hi I've got an integral question which i have the answer for but i don't understand why it works. <br /> |
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The question is: |
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If f(x) is continuous on [0; ¼], show that: |
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<math>\lim_{n\rightarrow \infty} \int^{\pi}_{0}|sin\ nx|f(x)\ dx = \frac{2}{\pi} \int^{\pi}_{0} f(x)\ dx </math> |
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<br /> |
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I've got it into this form: |
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<math> \lim_{n\rightarrow \infty} \sum^{n}_{k=1} \int^{ \frac{k \pi}{n} }_{ \frac{(k-1) \pi}{n} } | sin\ nx| f dx </math> |
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<br /> |
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The next step i have written is: |
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<math> \sum^{n}_{k=1} f(\xi nk) \int^{ \frac{k \pi}{n} }_{ \frac{(k-1) \pi}{n} } | sin\ nx| dx </math><br /> |
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But i don't understand how to get to that step. It doesnt help that my lecturer makes the odd mistake now and again. |
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Can somebody explain how to get between these steps and how to complete the rest of the question. Thanks for any help. |
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[[Special:Contributions/212.140.139.225|212.140.139.225]] ([[User talk:212.140.139.225|talk]]) 23:00, 20 February 2008 (UTC) |
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:In the one-but-last formula, write ''f''(''x'') instead of just ''f''. I think the last formula should be something like |
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::<math>\lim_{n\rightarrow \infty} \sum^{n}_{k=1} f(\xi_k) \int^{ \frac{k \pi}{n} }_{ \frac{(k-1) \pi}{n} } | sin\ nx| dx ,~\mathrm{where}~\tfrac{(k-1) \pi}{n} < \xi_k < \tfrac{k \pi}{n},</math> |
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:which is justified by the [[first mean value theorem for integration]]. --[[User talk:Lambiam|Lambiam]] 11:39, 21 February 2008 (UTC) |
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= February 21 = |
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== Quick question about log convexity == |
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I've read the article on [[power series]], and it mentions that for more than one variable, the domain of convergence is a log-convex set instead of an interval. |
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But what is a log-convex '''set''' here ? I mean, how do we take the log of a set ? Surely the set can contain negative values for example... |
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Thanks. -- [[User:Sam Derbyshire|Xedi]] ([[User talk:Sam Derbyshire|talk]]) 02:13, 21 February 2008 (UTC) |
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:Sorry, I have no idea what they could mean by log-convexity of a set. The multi-variable stuff was [http://en.wikipedia.org/wiki/Power_series?diff=next&oldid=3844867 added] in June 2004, and the log-convex comment was [http://en.wikipedia.org/wiki/Power_series?diff=next&oldid=6838232 made] in Oct 2004. It does not appear to have been touched since then. I'll ask the author about it. |
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:The domain of convergence is a union of polydisks, but I think perhaps it need not be a polydisk. Some reasonably elementary notes are at [http://www.ipfw.edu/math/Coffman/linx4.html Adam Coffman's website], specifically his research notes on [http://www.ipfw.edu/math/Coffman/pdf/sernotes.pdf Notes on series in several variables]. It includes the polydisk result, as well as the continuity, and differentiability of functions defined by a series, and includes the nice "recentering" result familiar for one complex variable allowing analytic continuation. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 04:25, 21 February 2008 (UTC) |
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== Natural Logs == |
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I'm being asked to evaluate natural logs and I'm getting very confused. One problem I'm having particular trouble with is lne(-x/3) |
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Could anyone please help explain to me how to solve problems like these and how to determine how to get started? |
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Thanks, anon. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/66.76.125.76|66.76.125.76]] ([[User talk:66.76.125.76|talk]]) 03:53, 21 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:What is lne(-x/3) ? |
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: |
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:Is it Log[ Exp[-x/3] ] ? |
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:or is it Log[-x/3] where Log is Log based e. |
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:[[Special:Contributions/202.168.50.40|202.168.50.40]] ([[User talk:202.168.50.40|talk]]) 04:34, 21 February 2008 (UTC) |
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:: ln is usually a shorthand for log<sub>e</sub> --[[User:PalaceGuard008|PalaceGuard008]] ([[User_Talk:PalaceGuard008|Talk]]) 05:07, 21 February 2008 (UTC) |
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* <em>y</em> = log(-<em>x</em>/3) |
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* e<sup><em>y</em></sup> = -<em>x</em>/3 |
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* <em>x</em> = -3e<sup><em>y</em></sup> |
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--[[User:Wj32|wj32]] <sup>[[User talk:Wj32|t]]/[[Special:Contributions/Wj32|c]]</sup> 05:16, 21 February 2008 (UTC) |
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:If the question is about {{mbox|ln ''e''<sup>−x/3</sup>}}, what can you say in general about {{mbox|ln ''e''<sup>A</sup>}}? --[[User talk:Lambiam|Lambiam]] 12:05, 21 February 2008 (UTC) |
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== is math the same as logic? == |
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Is math the same as logic, or why not. |
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In other words, is there math that isn't logical, but just evaluates truths, for example "experimenting" with numbers to find out "facts" but not using logic to show that these must be necessarily true. After all, the other sciences don't rely on their findings to be NECESSARILLY true, just that they happen to be true... so which is math? <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/79.122.42.134|79.122.42.134]] ([[User talk:79.122.42.134|talk]]) 10:12, 21 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:You might be interested in [[logicism]] and [[experimental mathematics]]. [[User talk:Algebraist|Algebraist]] 10:43, 21 February 2008 (UTC) |
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:You might also be interested in [[David_Hilbert#Formalism|David Hilbert's project]] to formalize mathematics.--[[User:Droptone|droptone]] ([[User talk:Droptone|talk]]) 12:45, 21 February 2008 (UTC) |
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::[[Experimental mathematics]] uses numerical methods to suggest conjectures which are then formally proved (or perhaps disproved if they turn out to be just a numerical coincidence). All mathematical results are proved using the techniques of logic, so, yes, they are necessarily true. But this does not mean that mathematics is the same as logic - logic does not determine ''what'' a mathemtician sets out to prove, or ''why'' a particular result is considered to be beautiful, deep or interesting. Logic is to mathematics as [[calligraphy]] is to the [[Rubaiyat of Omar Khayyam]]. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 14:04, 21 February 2008 (UTC) |
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== Moment == |
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Suppose you have a ladder leaning against a wall, which makes an angle θ with the ground, and a man, who can be treated as a particle, stands a distance l up the ladder. Now his weight will obviously be acting downwards and so working to work out the moment produced, you would have to resolve his weight vector to find the component of it that acts perpendicular to the ladder. |
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My question is, if you used Pythagoras or trig to determine his perpendicular distance from the ladder's point of contact with the ground to the line of action of the weight vector and then multiplied that by his weight, would that give you the correct value of the moment? Cheers in advance [[Special:Contributions/92.3.49.42|92.3.49.42]] ([[User talk:92.3.49.42|talk]]) 12:45, 21 February 2008 (UTC) |
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:Unless I've misunderstood, yes. What I tend to do is extend the "arrow" of the force such that the perpendicular goes through the point you are trying to resolve against, then it's a simple case of force times distance from point. <span style="font-family: Tahoma; font-size: 8pt;">[[User:x42bn6|<b>x42bn6</b>]] <span style="font-size: 7pt;">[[User talk:x42bn6|Talk]] [[Special:Contributions/x42bn6|Mess]]</span></span> 13:02, 21 February 2008 (UTC) |
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:Yes 92.3.49.42, both those methods are valid ways of calculating the moment around the ladder's point of contact with the ground, and both will yield the same answer. --[[User:mglg|mglg]]<sub>([[User talk:mglg|talk]])</sub> 17:18, 21 February 2008 (UTC) |
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== σ-algebra redux == |
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A revised version of [[#There are no countably infinite σ-algebras|There are no countably infinite σ-algebras]], hopefully closer to being correct? |
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'''Lemma'''. If <math>\mathcal{S}</math> is an infinite [[Field of sets|algebra]] over a set X, [[Partial order|partially ordered]] by [[Inclusion (set theory)|inclusion]], then <math>\mathcal{S}</math> contains an infinite ascending [[Total order|chain]] <math>a_1 < a_2 < \ldots</math> or an [[infinite descending chain]] <math>b_1 > b_2 > \ldots</math>. |
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''Proof'': Let <math>\mathcal{S}_1 = \mathcal{S} - \{X\}</math>. If <math>\mathcal{S}_1</math> has a chain with no [[upper bound]], then that gives us what we wanted. Otherwise <math>\mathcal{S}_1</math> has a [[maximal element]] ''b''<sub>1</sub> by [[Zorn's lemma]]. If <math>x \in \mathcal{S}_1</math> and <math>x \not\leq b_1</math> then <math>b_1 < x \cup b_1</math> so <math>x \cup b_1 = X</math> by maximality, and therefore <math>X - x \leq b_1</math>. That means there are infinitely many subsets of ''b''<sub>1</sub> in <math>\mathcal{S}_1</math>. |
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Let <math>\mathcal{S}_2</math> be that collection of subsets, minus ''b''<sub>1</sub> itself. Then <math>\mathcal{S}_2</math> has a maximal element ''b''<sub>2</sub>, and so again contains infinitely many subsets of ''b''<sub>2</sub>. Inductively we obtain an infinite descending chain <math>b_1 > b_2 > \ldots</math>. |
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'''Corollary'''. Every infinite algebra contains an infinite collection of [[pairwise disjoint]] sets. |
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We just take the sequence of differences of elements of the chain. |
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'''Corollary'''. There are no [[Countable set|countably infinite]] [[σ-algebra]]s. |
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An infinite σ-algebra is an infinite algebra, and by the above has an infinite subcollection <math>\{D_1, D_2, \ldots\}</math> of pairwise disjoint sets. The map <math>\{n_1, n_2, \ldots\} \mapsto \cup_{k\geq 1} D_{n_k}</math> is then an [[Injective function|injection]] from the [[power set]] of '''[[Natural number|N]]''' into the σ-algebra. |
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[[User:merge| — merge]] 13:58, 21 February 2008 (UTC) |
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:That seems to work, but as Trovatore commented above, you don't need the Axiom of Choice (you may not care about AC, of course, but I do). Since you have no infinite ascending chains, you don't need Zorn to find a maximal element (the top element of any chain is maximal). If you start with a bijection between your (putative) countably infinite sigma-algebra and N and, whenever you have to choose an element, always choose the one of least index (in terms of this bijection), the proof becomes wholly choice-free. [[User talk:Algebraist|Algebraist]] 16:30, 21 February 2008 (UTC) |
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::Many thanks for the review. You're right of course that there's no need to invoke the ghost of Max Zorn to get maximal elements there! I'm not too concerned about using AC, but I do like to improve my awareness of when and how it's being used and when it is or isn't necessary, so your comments are much appreciated. Although there are other ways to solve the original problem, the results for infinite algebras in general seem more useful than the fact that there are no countably infinite σ-algebras, and it's nice that once you have them the statement about σ-algebras becomes trivial. [[User:merge| — merge]] 17:19, 21 February 2008 (UTC) |
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== Differential equation gives weird result! == |
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Hi there, I've been given differential equation |
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<math>dy/dx = e^{-3x} / y^3</math> |
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I separated variables and integrated, and I eventually came up with the general result <math> y =( - {4e^{-3 x} / 3} + D )^{1 / 4} </math> where D is 4C, my initial constant of int. |
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But... the function is always going to be negative, and I have to take the fourth root of it! Have I messed it up, or am I ok so long as D makes the stuff under the fourth root positive?? |
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Thank you! [[User:Psymun747|Psymun]] ([[User talk:Psymun747|talk]]) 15:20, 21 February 2008 (UTC) |
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:Your working is correct. Could you not have a result that includes complex numbers? -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 15:42, 21 February 2008 (UTC) |
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:'Spose! I didn't see it coming in this course... although seeing as I did them in the last maths course I possibly should have! Thanks for reassurance! [[User:Psymun747|Psymun]] ([[User talk:Psymun747|talk]]) 15:47, 21 February 2008 (UTC) |
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::No need to consider complex numbers. The function just isn't defined for <math>x<\frac13\log\left(\frac{4}{3D}\right)</math>. This isn't any different from solving <math>2y'y=1\;\!</math> and getting <math>\sqrt{x+C}</math> for a result. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 16:09, 21 February 2008 (UTC) |
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== Is there anything that HAS TO be true but isn't? == |
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Has mathematics ever proved anything, so that it has to be true, but in fact it isn't? For example, I'm thinking of something like proving that something can be done within some constraints, but in fact all the variations have been tried by computer and none of them are successful. So there HAS TO be a way, but there isn't. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/79.122.42.134|79.122.42.134]] ([[User talk:79.122.42.134|talk]]) 15:32, 21 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:No, because if it could be proved true but all attempts to give examples fail, then you did it wrong. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 15:38, 21 February 2008 (UTC) |
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:(edit conflict) No. This state of affairs is impossible based on the any commonly accepted concept of a [[mathematical proof]]. If you modify your statement to "Has anything ever been accepted as proven by the mathematical community...", then I do not know. [[User:Taemyr|Taemyr]] ([[User talk:Taemyr|talk]]) 15:47, 21 February 2008 (UTC) |
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:There are of course things that ''seem'' to have been logically proven but aren't true. See [[paradox]], and especially [[Zeno's paradox]]. [[User:DJ Clayworth|DJ Clayworth]] ([[User talk:DJ Clayworth|talk]]) 15:51, 21 February 2008 (UTC) |
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Do you guys mean to tell me that the universe is 100% consistent 100% of the time, and anything that follows is in fact thus, with no exceptions, ever? I find that a little hard to swallow... <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/79.122.42.134|79.122.42.134]] ([[User talk:79.122.42.134|talk]]) 16:15, 21 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:We're not talking about the universe, we're talking about mathematics. Mathematics has not been proven to be consistent (and in a certain sense can't be), but is generally believed to be so. But yes, in the universe as in mathematics, the result of a valid deduction from true premises is true (in the jargon, valid deductions are truth-preserving). [[User talk:Algebraist|Algebraist]] 16:23, 21 February 2008 (UTC) |
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:Indeed. Don't confuse reality with mathematics. And even in mathematics it is ''possible'' to have a valid proof that a statement is true and an equally valid proof that the same statement is false (simplest case is that you start with both ''P is true'' and ''P is false'' as axioms). Unfortunately, that result indicates that you are working in a system that is [[inconsistent]], and so not very interesting or useful. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 16:28, 21 February 2008 (UTC) |
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::Two possibly relevant Einstein quotes [http://en.wikiquote.org/wiki/Albert_Einstein#Sidelights_on_Relativity_.281983.29]: |
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::* ''One reason why mathematics enjoys special esteem, above all other sciences, is that its laws are absolutely certain and indisputable, while those of other sciences are to some extent debatable and in constant danger of being overthrown by newly discovered facts.'' |
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::* ''As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.'' |
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::--[[User:mglg|mglg]]<sub>([[User talk:mglg|talk]])</sub> 17:10, 21 February 2008 (UTC) |
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:Two blog entries that might be relevant are [http://www.madore.org/~david/weblog/2008-01.html#d.2008-01-12.1515] and [http://blog.plover.com/math/major-screwups-2.html]. – [[User:b_jonas|b_jonas]] 17:46, 21 February 2008 (UTC) |
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== Vandermonde determinant and polynomials == |
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Hi, im having trouble doing this problem from a book on galois theory. Its from the start of the book (Galois theory - Ian Stewart) so wont have much to do with real galois theory, mostly just algebra realated to polynomials. The problem is: |
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Two polynomials <math> f,g </math> over <math>\mathbb{C}</math> define the same function if and only if they have the same coefficients. |
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If <math>a_1, a_2 \cdots , a_n</math> are distinct complex numbers, then the '''Vandermonde determinant''' is defined as |
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<math>D= \left|\begin{array}{rrrrr} |
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1&1&1&\cdots&1\\ |
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a_1&a_2&a_2&\cdots&a_n\\ |
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a_1^2&a_3^2&a_3^2&\cdots&a_n^2\\ |
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\vdots&\vdots&\vdots&\ddots\\ |
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a_1^{n-1}&a_2^{n-1}&a_3^{n-1}&\cdots&a_n^{n-1} |
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\end{array} |
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\right| |
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</math> |
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Then there is a hint which doesnt really help me: |
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Consider the <math>a_j</math> as independent determinants over <math>\mathbb{C}</math>. Then D is a polynomial in the <math>a_j</math>, of total degree <math>0+1+2+ \cdots +(n-1)=\frac{1}{2}n(n-1)</math>. Moreover D vanishes whenever <math>a_j=a+k</math> for some <math>k \ne j </math> as it has 2 identical rows. Therefore D is divisible by <math>a_j-a_k</math> hence it is divisible by <math>\prod_{j<k} (a_j-a_k)</math> now compare degrees. |
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Thats as far as i got really, any help would be great, thanks. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/137.205.93.126|137.205.93.126]] ([[User talk:137.205.93.126|talk]]) 15:50, 21 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot--> |
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:Huh? I'm confused. The statement 'Two polynomials <math> f,g </math> define the same function if and only if they have the same coefficients (up to multiplication by a constant)' is simply false (multiplication by a constant changes the function, different polynomials can define the same function over finite fields), and I can't see any obvious true statement it could be a misprint of. Moreover it seems to have little to do with the Vandermonde matrix. The one standard problem about the Vandermonde determinant is to show that it is non-zero if the a<sub>i</sub> are distinct, and that seems to be what the hint is leading you towards proving. [[User talk:Algebraist|Algebraist]] 16:18, 21 February 2008 (UTC) |
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: Sorry i didn't explain it well. I seemed to have added the up to multiplication bit for no reason as now i think about it it is wrong, ive taken it out of the question and rewote it exactly as it appears in the book. In the book it gives a different proof using calculus, then goes on to say "For a purely algebraic proof see [this problem]". I had the following idea: consider <math>\underline{x}=(x^0,x^1,x^2,\cdots,x^{n-1})</math>, the equation <math>V\underline{x}=\underline{0}</math> (where V is the vandermonde matrix) has a solution if and only if <math>det(v)=D=0</math> so showing the vandermonde determinant is non zero if the coefficients in it are distinct, that is the polynomials in the vector solution of <math>V\underline{x}=\underline{0}</math> have different coefficients. Is this in any way correct? Also if you could give me a little help showing the Vandermonde determinant is 0, that would be helpful too. Sorry if my wording is off, im having alot of trouble understanding this. Thanks. {{unsigned|137.205.93.126}} |
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::Unfortunately, I was taught Vandermonde determinants by someone that went far too fast, so I don't fully understand them. I have one general tip, though: A matrix has zero determinant if the columns (or, equally rows) aren't linearly independent - in particular, if one row/column is a scalar multiple of another (that's not necessary, but it is sufficient). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 18:39, 21 February 2008 (UTC) |
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::Looked in my copy of Stewart's ''Galois Theory'' (2nd edition). The only problem I can find that mentions the [[Vandermonde determinant]] is Exercise 18.4 - but that is in the penultimate chapter of the book, not near the beginning, and it isn't worded like your question. Is your question from a different edition ? [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 20:17, 21 February 2008 (UTC) |
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== Fastest algorithm to find very smooth numbers close to some number == |
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What is the fastest algorithm that could find smooth integers very near some specified integer. In other words which algorithm has the lowest ration of computation/(smoothness of a found number * smoothness of distance of that smooth number from a specified integer). [[Special:Contributions/128.227.1.24|128.227.1.24]] ([[User talk:128.227.1.24|talk]]) 20:30, 21 February 2008 (UTC)Mathguy |
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December 23
[edit]Is it possible to make Twisted Edwards curve birationally equivalent to twisted weirestrass curves ?
[edit]Is there an equation fo converting a twisted Edwards curve into a tiwsted weierstrass form ? 2A01:E0A:401:A7C0:6D06:298B:1495:F479 (talk) 04:12, 23 December 2024 (UTC)
- According to Montgomery curve § Equivalence with twisted Edwards curves, every twisted Edwards curve is birationally equivalent to a Montgomery curve, while Montgomery curve § Equivalence with Weierstrass curves gives a way to transform a Montgomery curve to an elliptic curve in Weierstrass form. I don't see a definition of "twisted Weierstrass", so I don't know if you can give an extra twist in the process. Perhaps this paper, "Efficient Pairing Computation on Twisted Weierstrass Curves" provides the answer; its abstract promises: "In this paper, we construct the twists of twisted Edwards curves in Weierstrass form." --Lambiam 10:35, 23 December 2024 (UTC)
December 24
[edit]How did the Romans do engineering calculations?
[edit]The Romans did some impressive engineering. Engineers today use a lot of mathematical calculations when designing stuff. Calculations using Roman numerals strike me as being close to impossible. What did the Romans do? HiLo48 (talk) 05:50, 24 December 2024 (UTC)
- The kind of engineering calculations that might have been relevant would mostly have been about statics – specifically the equilibrium of forces acting on a construction, and the ability of the design to withstand these forces, given its dimensions and the mechanical properties of the materials used in the construction, such as density, modulus of elasticity, shear modulus, Young modulus, fracture strength and ultimate tensile strength. In Roman times, only the simplest aspects of all this were understood mathematically, namely the statics of a construction in which all forces work in the same plane, without torque, and the components are perfectly rigid. The notion of assigning a numerical magnitude to these moduli and strengths did not exist, which anyway did not correspond to precisely defined, well-understood concepts. Therefore, engineering was not a science but an art, mainly based on experience in combination with testing on physical models. Any calculations would mostly have been for the amounts and dimensions of construction materials (and the cost thereof), requiring a relatively small number of additions and multiplications. --Lambiam 19:03, 24 December 2024 (UTC)
- Calculating with Roman Numerals might seem impossible, but in some ways it's simpler than our positional system; there are only so many symbols commonly used, and only so many ways to add and multiply them. Once you know all those ways you efficiently can do calculations with them, up to the limits imposed by the system.
- And for a lot of things they relied on experience. Romans knew how to build circular arches, but rather than do calculations to build larger arches, or ones with more efficient shapes they used many small circular ones which they knew worked, stacked side by side and sometimes on top of each other. See e.g. any Roman aqueduct or the Colosseum. For materials they probably produce them on site or close by as they're needed.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 20:12, 24 December 2024 (UTC)
- To add to the above, it wasn't really the Romans that were great innovators in science, engineering, etc.. I think more innovation and discovery took place in Ancient Greece and Ancient Egypt. Certainly Greece as we have a record of that. Egypt it's more that they were building on such a monumental scale, as scale no-one came close to repeating until very recently.
- Romans were military geniuses. They conquered Greece, and Egypt, and Carthage, and Gaul, and Britannia, and everywhere in between. They then built forts, towns, cities and infrastructure throughout their empire. They built so much so widely that a lot of it still stands. But individually a lot of it isn't technically impressive; instead it's using a few simple patterns over and over again.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 12:09, 25 December 2024 (UTC)
- See Roman abacus. catslash (talk) 22:03, 25 December 2024 (UTC)
- It has to be said that in Roman times calculations for architecture were mostly graphical, geometrical, mechanical, rather than numeric. In fact, from the perspective of an ancient architect, it would make little sense translating geometrical figures into numbers, making numeric calculations, then translating them into geometrical figures again. Numerals become widely used tools only later, e.g. with the invention of Analytic Geometry (by Descartes), and with logarithms (Napier); and all these great mathematical innovations happened to be so useful also thanks to the previous invention of the printing press by Gutenberg --it's easier to transmit information by numbers than by geometric constructions. One may even argue that the invention of the printing press itself was the main reason to seek for an adequate efficient notation for real numbers (achieved by Stevinus). pma 21:22, 1 January 2025 (UTC)
- This is a great answer! Tito Omburo (talk) 21:38, 1 January 2025 (UTC)
- Architectural calculations, mentioned by Vitruvius, are not what I think of as engineering calculations. The former kind is about form. The latter kind should provide answers questions about structural behaviour, like, "Will these walls be able to withstand the outward force of the dome?" Can such questions be addressed with non-numerical calculations? --Lambiam 23:40, 1 January 2025 (UTC)
- It has to be said that in Roman times calculations for architecture were mostly graphical, geometrical, mechanical, rather than numeric. In fact, from the perspective of an ancient architect, it would make little sense translating geometrical figures into numbers, making numeric calculations, then translating them into geometrical figures again. Numerals become widely used tools only later, e.g. with the invention of Analytic Geometry (by Descartes), and with logarithms (Napier); and all these great mathematical innovations happened to be so useful also thanks to the previous invention of the printing press by Gutenberg --it's easier to transmit information by numbers than by geometric constructions. One may even argue that the invention of the printing press itself was the main reason to seek for an adequate efficient notation for real numbers (achieved by Stevinus). pma 21:22, 1 January 2025 (UTC)
- See Roman abacus. catslash (talk) 22:03, 25 December 2024 (UTC)
Are these sequences mod any natural number n periodic?
[edit]The period of Fibonacci number mod n is the Pisano period of n, but are these sequences mod any natural number n also periodic like Fibonacci number mod n?
- Lucas number
- Pell number
- Tribonacci number
- Tetranacci number
- Newman–Shanks–Williams number
- Padovan sequence
- Perrin number
- Narayana sequence
- Motzkin number
- Bell number
- Fubini number
- Euler zigzag number
- Partition number OEIS: A000041
- Distinct partition number OEIS: A000009
42.76.153.22 (talk) 06:06, 24 December 2024 (UTC)
- For 1. through 4., see Pisano period#Generalizations. Although 5. through 8. are not explicitly listed, I'm pretty sure the same argument applies for their periodicity as well. GalacticShoe (talk) 07:05, 24 December 2024 (UTC)
- For 13, I'm not sure, but I think the partition function is not periodic modulo any nontrivial number, to the point that the few congruences that are satisfied by the function are also very notable, e.g. Ramanujan's congruences. GalacticShoe (talk) 07:23, 24 December 2024 (UTC)
- It's possible that a sequences is eventually periodic but not periodic from the start, for example powers of 2 are periodic for any odd n, but for n=2 the sequence is 1, 0, 0, ..., which is only periodic starting with the second entry. In other words a sequence can be become periodic without being pure periodic. A finiteness argument shows that 1-8 are at least eventually periodic, but I don't think it works for the rest. (Pollard's rho algorithm uses this finiteness argument as well.) It says in the article that Bell numbers are periodic mod n for any prime n, but the status for composite n is unclear, at least from the article. Btw, Catalan numbers not on the list?. --RDBury (talk) 08:08, 24 December 2024 (UTC)
- If only the periodicity of Bell numbers modulo prime powers were known, then periodicity for all modulos would immediately result from the Chinese remainder theorem. GalacticShoe (talk) 01:29, 29 December 2024 (UTC)
- It's possible that a sequences is eventually periodic but not periodic from the start, for example powers of 2 are periodic for any odd n, but for n=2 the sequence is 1, 0, 0, ..., which is only periodic starting with the second entry. In other words a sequence can be become periodic without being pure periodic. A finiteness argument shows that 1-8 are at least eventually periodic, but I don't think it works for the rest. (Pollard's rho algorithm uses this finiteness argument as well.) It says in the article that Bell numbers are periodic mod n for any prime n, but the status for composite n is unclear, at least from the article. Btw, Catalan numbers not on the list?. --RDBury (talk) 08:08, 24 December 2024 (UTC)
- PS. 1-8 are pure periodic. In general, if the recurrence can be written in the form F(k) = (some polynomial in F(k-1), F(k-2), ... F(k-d+1) ) ± F(k-d), then F is pure periodic. The reason is that you can solve for F(k-d) and carry out the recursion backwards starting from where the sequence becomes periodic. Since the previous entries are uniquely determined they must follow the same periodic pattern as the rest of the sequence. If the coefficient of F(k-d) is not ±1 then this argument fails and the sequence can be pre-periodic but not pure periodic, at least when n is not relatively prime to the coefficient. --RDBury (talk) 18:12, 24 December 2024 (UTC)
- Ah, what was tripping me up was showing pure periodicity, recursing backwards completely slipped my mind. Thanks for the writeup! GalacticShoe (talk) 18:31, 24 December 2024 (UTC)
- good questions. What about TREE(n) mod k, for arbitrary fixed k?Rich (talk) 23:03, 27 December 2024 (UTC)
- Link: TREE function. There are a lot of sequences like this where exact values aren't known, Ramsey numbers are another example. It helps if there is a relatively simple recursion defining the sequence. --RDBury (talk) 11:50, 28 December 2024 (UTC)
- good questions. What about TREE(n) mod k, for arbitrary fixed k?Rich (talk) 23:03, 27 December 2024 (UTC)
- Ah, what was tripping me up was showing pure periodicity, recursing backwards completely slipped my mind. Thanks for the writeup! GalacticShoe (talk) 18:31, 24 December 2024 (UTC)
- PS. 1-8 are pure periodic. In general, if the recurrence can be written in the form F(k) = (some polynomial in F(k-1), F(k-2), ... F(k-d+1) ) ± F(k-d), then F is pure periodic. The reason is that you can solve for F(k-d) and carry out the recursion backwards starting from where the sequence becomes periodic. Since the previous entries are uniquely determined they must follow the same periodic pattern as the rest of the sequence. If the coefficient of F(k-d) is not ±1 then this argument fails and the sequence can be pre-periodic but not pure periodic, at least when n is not relatively prime to the coefficient. --RDBury (talk) 18:12, 24 December 2024 (UTC)
- For 9. Motzkin numbers are not periodic mod 2. Motzkin numbers mod 2 are OEIS:A039963, which is OEIS:A035263 with each term repeated (i.e. .) OEIS:A035263 in turn is the sequence that results when one starts with the string and successively maps (e.g. .) It is clear that if OEIS:A035263 were periodic with period , then the periodic string of length would need to map to string , but this is impossible as the last character of is always the opposite of the last character of the map applied to . Thus OEIS:A035263 is nonperiodic, and neither is OEIS:A039963. GalacticShoe (talk) 01:08, 29 December 2024 (UTC)
- This paper (linked from OEIS) goes into more detail on Motzkin numbers. I gather the sequence might be called quasi-periodic, but I can't find an article that matches this situation exactly. A003849 is in the same vein. --RDBury (talk) 16:59, 30 December 2024 (UTC)
- For 11., the article seems to suggest that Fubini numbers are eventually periodic modulo any prime power. I'm pretty sure this means that they the numbers eventually periodic mod any number , since the lcm of the eventual periods modulo all prime power divisors of should correspond to the eventual period modulo itself, with the remainders being obtainable through the Chinese remainder theorem. However, the wording also seems to suggest that periodicity modulo arbitrary is still conjectural, so I'm not sure. GalacticShoe (talk) 02:44, 29 December 2024 (UTC)
- You have answered all questions except 12 and 14, and 9 and 13 are the only two sequences which are not periodic mod n (except trivial n=1), 12 (Euler zigzag numbers) is (sequence A000111 in the OEIS), which seems to be periodic mod n like 10 (Bell numbers) (sequence A000110 in the OEIS) and 11 (Fubini numbers) (sequence A000670 in the OEIS), but all of these three sequences need prove, besides, 14 (Distinct partition numbers) (sequence A000009 in the OEIS) seems to be like 13 (Partition numbers) (sequence A000041 in the OEIS), i.e. not periodic mod n. 1.165.199.71 (talk) 02:27, 31 December 2024 (UTC)
December 31
[edit]Generating a point on the Y axis from regular pentagon with point on X axis
[edit]For a consisting of points in R^2, define the function B such that as the Union of and all points which can be produced in the following way. For each set of points A, B, C, & D from all different so that no three of A, B, C & D are co-linear. E is the point (if it exists) where ABE are colinear and CDE are co-linear.
If = the vertices of a regular Pentagon centered at 0,0 with one vertex at (1,0), does there exist N such that includes any point of the form (0, y)? (extending the question to any N-gon, with N odd) Naraht (talk) 05:16, 31 December 2024 (UTC)
- I think you meant to write --Lambiam 07:55, 31 December 2024 (UTC)
- Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)
- I'm not 100% sure I understand the problem, but try this: Label the vertices of the original pentagon, starting with (1, 0), as A, B, C, D, E. You can construct a second point on the x-axis as the intersection of BD and CE; call this A'. Similarly construct B', C', D', E', to get another, smaller, regular pentagon centered at the origin and with the opposite orientation from the the original pentagon. All the lines AA', BB', CC', DD', EE' intersect at the origin, so you can construct (0, 0) as the intersection of any pair of these lines. The question didn't say y could not be 0, so the answer is yes, with N=2.
- Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)
- There is some theory developed on "straightedge only construction", in particular the Poncelet–Steiner theorem, which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --RDBury (talk) 13:12, 1 January 2025 (UTC)
- Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a pentagram. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram. --Lambiam 16:53, 1 January 2025 (UTC)
- Here is one point other than the origin (in red)
- If there is one such point, there must be an infinite number of them. catslash (talk) 22:52, 2 January 2025 (UTC)
- Just to be clear, the black points are the original pentagon K, the green points are in B(K), and the red point is the desired point in B2(K); the origin is not shown. It would be nice to find some algebraic criterion for a point to be constructible in this way, similar to the way points constructible with a compass and straightedge are characterized by their degree over Q. --RDBury (talk) 01:45, 3 January 2025 (UTC)
- Once you have a second one (such as the reflection of the red point wrt the x-axis), you have all intersections of the y-axis with the non-vertical lines through pairs of distinct points from --Lambiam 16:22, 3 January 2025 (UTC)
- Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a pentagram. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram. --Lambiam 16:53, 1 January 2025 (UTC)
- There is some theory developed on "straightedge only construction", in particular the Poncelet–Steiner theorem, which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --RDBury (talk) 13:12, 1 January 2025 (UTC)
- RDBury *headslap* on (0,0) Any idea on y<>0? (← comment from Naraht)
- See the above construction by catslash. --Lambiam 16:12, 3 January 2025 (UTC)
- The red point is at , . catslash (talk) 16:18, 3 January 2025 (UTC)
- RDBury *headslap* on (0,0) Any idea on y<>0? (← comment from Naraht)
January 1
[edit]What is the first number not contained in M136279841?
[edit]See (sequence A268068 in the OEIS), the first number not contained in M74207281 is 1000003, but what is What is the first number not contained in M136279841 (the currently largest known prime)? 61.224.131.231 (talk) 03:34, 1 January 2025 (UTC)
- The corresponding sequence (11, 3, 8, 7, 6, 10, 4, 9, 1, 5, 25, 31, 39, ...) is not in OEIS. Finding the answer to your question requires an inordinate amount of computing power. The decimal expansion of this Mersenne prime has some 41 million digits, all of which need to be computed. If this is to be done in a reasonable amount of time, the computation will need the random access storage of at least some 22 million digits. --Lambiam 10:10, 1 January 2025 (UTC)
- I'm not seeing that this question requires an inordinate amount of computing power to answer. 41 million characters is not a very large set of data. Almost all modern computers have several gigabytes of memory, so 41 million characters will easily fit in memory. I took the digits of M136279841 from https://www.mersenne.org/primes/digits/M136279841.zip and searched them myself, which took a few minutes on a consumer grade PC. If I have not made a mistake, the first number that does not appear is 1000030. The next few numbers that do not appear are 1000073, 1000107, 1000143, 1000156, 1000219, 1000232, 1000236, 1000329, 1000393, 1000431, 1000458, 1000489, 1000511, 1000514, 1000520, 1000529, etc. CodeTalker (talk) 03:59, 2 January 2025 (UTC)
- To be fair, this depends on being able to find the digits on-line. To compute them from scratch just for this question would be more trouble than it's worth. But I take your point; it probably takes more computing power to stream an episode of NUMB3RS than to answer this question. My problem with the question is that it's basically a dead end; knowing the answer, is anyone going to learn anything useful from it? I'd question the inclusion of A268068 in OEIS in the first place simply because it might lead to this sort of boondoggle. But far be it for me to second guess the OEIS criteria for entry. --RDBury (talk) 01:13, 3 January 2025 (UTC)
- OEIS includes similar sequences for the positions of the first location of the successive naturals in the decimal expansions of (A088576), (A032445) and (A229192). These have at least a semblance of theoretical interest wafting over from the open question whether these numbers are normal. --Lambiam 06:21, 3 January 2025 (UTC)
To compute them from scratch just for this question would be more trouble than it's worth.
Eh, I agree that the question is of little fundamental interest. However, it's not much work to compute M136279841. It is of course absolutely trivial to compute it as a binary number. The only real work is to convert it to decimal. I wrote a program to do this using the GNU Multiple Precision Arithmetic Library. It took about 5 minutes to write the program (since I've never used that library before and had to read the manual) and 29 seconds to run it. CodeTalker (talk) 18:06, 3 January 2025 (UTC)- Right, convert from binary, somehow I didn't think of that. Basically just divide by 10 41 million times, which would only be an issue if it was billions instead of millions. --RDBury (talk) 06:21, 4 January 2025 (UTC)
- To be fair, this depends on being able to find the digits on-line. To compute them from scratch just for this question would be more trouble than it's worth. But I take your point; it probably takes more computing power to stream an episode of NUMB3RS than to answer this question. My problem with the question is that it's basically a dead end; knowing the answer, is anyone going to learn anything useful from it? I'd question the inclusion of A268068 in OEIS in the first place simply because it might lead to this sort of boondoggle. But far be it for me to second guess the OEIS criteria for entry. --RDBury (talk) 01:13, 3 January 2025 (UTC)
- I'm not seeing that this question requires an inordinate amount of computing power to answer. 41 million characters is not a very large set of data. Almost all modern computers have several gigabytes of memory, so 41 million characters will easily fit in memory. I took the digits of M136279841 from https://www.mersenne.org/primes/digits/M136279841.zip and searched them myself, which took a few minutes on a consumer grade PC. If I have not made a mistake, the first number that does not appear is 1000030. The next few numbers that do not appear are 1000073, 1000107, 1000143, 1000156, 1000219, 1000232, 1000236, 1000329, 1000393, 1000431, 1000458, 1000489, 1000511, 1000514, 1000520, 1000529, etc. CodeTalker (talk) 03:59, 2 January 2025 (UTC)
January 5
[edit]Reference request:coherence condition (adjoint functor)
[edit]Previously, in WPM (Coecke and Moore (2000)) taught me a statement of coherence condition for adjoint functors. This is sometimes called triangle identities or zigzag identities, and I'm looking for some references. I am also trying to find where to find the Wikipedia article that explains coherence conditions (adjoint functors). Also, I'm looking for a Wikipedia article that explains coherence conditions (adjoint functors). (e.g. coherence condition, adjoint functor, or new draft ?)
- "triangle identity". ncatlab.org.
- Ben-Moshe, Shay (2024). "Naturality of the ∞-categorical enriched Yoneda embedding". Journal of Pure and Applied Algebra. 228 (6). arXiv:2301.00601. doi:10.1016/j.jpaa.2024.107625.
- Borceux, Francis (1994). Handbook of Categorical Algebra: Basic category theory. Cambridge University Press. ISBN 978-0-521-44178-0.
- Coecke, Bob; Moore, David (2000). "Operational Galois adjunctions". arXiv:quant-ph/0008021.
- planetmath
SilverMatsu (talk) 02:41, 5 January 2025 (UTC)
- Can you identify more precisely for which statement(s) you want a reference? Is it for the definition of counit–unit adjunction given in section Adjoint functors § Definition via counit–unit adjunction?
- Explanations in mathematics can be of different kinds. One kind are explanations of definitions. Definitions are true by definition; an explanation generally means helping build up an intuition of the concept defined by showing familiar structures satisfying the definition. Another kind are explanations of statements. These typically offer a reformulation of a statements as an equivalent statement in terms of simpler concepts. Then there are explanations of proofs. These can include showing the proof "in action" on specific examples, using familiar structures as when explaining definitions. And they can assist in verifying the validity of proof steps, by unfolding definitions until the step becomes obvious. In general, all explanations can involve a combination of these.
- For example, in our article Adjoint functors, the ramifications of the definition in the lead section, basically the existence of a bijection that is natural in and are not immediately obvious, but by carefully unfolding the definitions, starting with the required naturality of the two morphisms making up the bijection, leads to the equivalent counit–unit definition.
- I am not sure what kind of explanation of "coherence conditions (adjoint functors)" you are seeking. Perhaps studying this article, "Adjunctions", will answer this question. --Lambiam 12:18, 5 January 2025 (UTC)
Concatenation of first 10 digits and last 10 digits of a number
[edit]Let a(n) be the concatenation of first 10 digits and last 10 digits of n, then we know that a(2136279841-1) = 88169432759486871551, however:
- Can a(2^n) take all 20-digit values which are multiples of 1024?
- Can a(3^n) take all 20-digit values which are odd?
- Can a(n^2) take all 20-digit values which end with 0, 1, 4, 5, 6, 9?
- Can a(n^3) take all 20-digit values?
- Can a(prime number) take all 20-digit values which end with 1, 3, 7, 9?
- Can a(lucky number) take all 20-digit values which are odd?
- Can a(Fibonacci number) take all 20-digit values?
- Can a(partition number) take all 20-digit values?
- What is a(9^9^9^9)?
- What is a(9^^9), where ^^ is tetration?
- What is a(9^^^9), where ^^^ is pentation?
- What is a(Graham's number)?
- What is a(TREE(3))?
- If we know a(x), assume that x has at least 20 digits, can we also know a(2*x), a(3*x), etc.?
- If we know a(x), assume that x has at least 20 digits, can we also know a(x^2), a(x^3), etc.?
- If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x+y)?
- If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x*y)?
- If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x^y)?