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{{Wikipedia:Reference_desk/Archives/Mathematics/2009 January 6}}


= December 20 =
{{Wikipedia:Reference_desk/Archives/Mathematics/2009 January 7}}


== Give a base b and two base b digits x and z, must there be a base b digits y such that the 3-digit number xyz in base b is prime? ==
{{Wikipedia:Reference_desk/Archives/Mathematics/2009 January 8}}


Give a base b and two base b digits x and z (x is not 0, z is coprime to b), must there be a base b digits y such that the 3-digit number xyz in base b is prime? [[Special:Contributions/1.165.207.39|1.165.207.39]] ([[User talk:1.165.207.39|talk]]) 02:10, 20 December 2024 (UTC)
= January 9 =


:In base 5, <math>3Y1_{5} = 76+5Y</math> is composite for all base-5 Y. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 03:39, 20 December 2024 (UTC)
== a question regarding liters and milliliters ==
::<math>3Y1_7</math> also offers a counterexample. While there are many counterexamples for most odd bases, I did not find any for even bases. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 09:58, 20 December 2024 (UTC)


= December 23 =
can anyone answer me how many milliliters are in a 1/16 liter? i dont think it could be answered; my friend thinks it is approx 62.5(?)
would appreciate a quick response. thank you04:28, 9 January 2009 (UTC)[[User:Anilas|Anilas]] ([[User talk:Anilas|talk]])
:1 litre is 1000 millilitres, by definition. Thus 1/16 of a litre is 1/16 of 1000 millilitres. For more, see [[long division]]. [[User talk:Algebraist|Algebraist]] 04:38, 9 January 2009 (UTC)
::Or [[calculator]]. ;) --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:49, 9 January 2009 (UTC)
:::It is exactly 62.5 mL; your friend is a [[genius]] (I am dumbfounded as to why he is still solving these [[trivial (mathematics)|trivial problems]]).


== Is it possible to make [[Twisted Edwards curve]] birationally equivalent to twisted weirestrass curves ? ==
:::P.S What I am trying to say is that you can just evaluate 1000/16 on google (search 1000/16 on google). Please don't ask trivial problems at the reference desk.


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)
:::PST <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 12:59, 9 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
::::Google can do more than that. [http://www.google.no/search?hl=en&q=milliliters+in+1%2F16+liter&btnG=Google-s%C3%B8k&meta=&aq=f&oq=]. [[User:Taemyr|Taemyr]] ([[User talk:Taemyr|talk]]) 13:38, 9 January 2009 (UTC)


: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." &nbsp;--[[User talk:Lambiam#top|Lambiam]] 10:35, 23 December 2024 (UTC)
== Is this equation solveable without a computer/brute force? ==
let a,b & c exist in set of natural numbers { 0,1,2 ... }<br />
a < b < c<br />
a^2 + b^2 = c^2<br />
a + b + c = 1000<br />


= December 24 =
Find a,b and c.[[Special:Contributions/212.23.11.198|212.23.11.198]] ([[User talk:212.23.11.198|talk]]) 11:49, 9 January 2009 (UTC)


== How did the Romans do engineering calculations? ==
:Find a right triangle composed of three line segments; the sum of the lengths of which equals 1000 and such that lengths of the line segments form a set of three distinct natural numbers. --[[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]]) 11:51, 9 January 2009 (UTC)
:Yes, it is. There is a unique solution which can be found in under ten minutes of thought. [[User talk:Algebraist|Algebraist]] 12:19, 9 January 2009 (UTC)


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)
::<small>(After edit conflict - maybe it took me 11 minutes !)</small> Yes, there is a straightforward algebraic approach. The numbers ''a'', ''b'' and ''c'' form a [[Pythagorean triple]], so their sum has a particular factorisation, which allows you to find candidate triples quite easily if you are given their sum. For a sum of 1000 there is, I think, only one solution. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 12:26, 9 January 2009 (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. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 19:03, 24 December 2024 (UTC)
:::Hint: use Euclid's formula for generating pythagorean triples (see the first section of [[Pythagorean triple]]) and figure out what k, m, and n have to be for the sum a+b+c to equal 1000. You should get two answers for triples k, m, and n, but these two will turn out to give the same triple a, b, c. [[User:Kfgauss|kfgauss]] ([[User talk:Kfgauss|talk]]) 17:58, 9 January 2009 (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.
== Algebra II word problem ==


: 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)
I've hit a brick wall on this word problem:


: 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.
"In his job at the post office, Eddie Thibodeaux works a 6.5-hr day. He sorts mail, sells stamps, and does supervisory work. One day he sold stamps twice as long as he sorted mail, and he supervised 0.5 hr longer than he sorted mail. How many hours did he spend at each task?"


: 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)
Word problems are my Kyptonite. --[[Special:Contributions/24.33.75.35|24.33.75.35]] ([[User talk:24.33.75.35|talk]]) 23:23, 9 January 2009 (UTC)
:{{dyoh}} [[User talk:Algebraist|Algebraist]] 23:25, 9 January 2009 (UTC)
::I only need help with writing the exact equation. After that, I can solve it myself. I know there are 3 jobs or 3 "x's". All of which = 6.5. Should it be x + x(2) + x(0.05) = 6.5? --[[Special:Contributions/24.33.75.35|24.33.75.35]] ([[User talk:24.33.75.35|talk]]) 23:33, 9 January 2009 (UTC)
:::There are three quantities involved, as you correctly observe, so calling them all by the same letter is a bit silly. Why not call them ''x'', ''y'' and ''z'', or perhaps ''m'', ''s'', and ''w'' (for Mail, Stamps, and supervisory Work)? [[User talk:Algebraist|Algebraist]] 23:34, 9 January 2009 (UTC)
::::Ok. m + s(2) + w(0.05) = 6.5. Even with the newly labeled quantities, that equation does not look right to me. --[[Special:Contributions/24.33.75.35|24.33.75.35]] ([[User talk:24.33.75.35|talk]]) 23:47, 9 January 2009 (UTC)
:::::It's dangerous to write down symbols without knowing what they are supposed to mean. What, for example, does w(0.05) mean in your formula? [[User talk:Algebraist|Algebraist]] 23:51, 9 January 2009 (UTC)
::::::It's where he worked 0.5 hr longer supervising than he did sorting. Just multiply the two and add the difference to one of the quantities. However, I'm starting to think this word problem has more than one equation. --[[Special:Contributions/24.33.75.35|24.33.75.35]] ([[User talk:24.33.75.35|talk]]) 23:56, 9 January 2009 (UTC)
:::::::Very good. Why don't you forget about equations for a second and think about what relations between your quantites are given by the question? [[User talk:Algebraist|Algebraist]] 23:59, 9 January 2009 (UTC)
::::::::I've got to go to work. I'll toy with it on my breaks and tell you what I come up with. Thanks for the help so far. --[[Special:Contributions/24.33.75.35|24.33.75.35]] ([[User talk:24.33.75.35|talk]]) 00:07, 10 January 2009 (UTC)
::I'll give you a hint - in order to solve a problem with 3 variables you need 3 equations. Read through the information given and try and find 3 relationships. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 00:49, 10 January 2009 (UTC)


::See [[Roman abacus]]. [[User:Catslash|catslash]] ([[User talk:Catslash|talk]]) 22:03, 25 December 2024 (UTC)
Using x to represent "sort", 24.33...'s original approach seems fine but his/her 3rd term is wrong. It shouldn't be (x(0.05)) but should be (x+0.5), thus using one variable and one equasion:
:::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)
:::: This is a great answer! [[User:Tito Omburo|Tito Omburo]] ([[User talk: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? &nbsp;--[[User talk:Lambiam#top|Lambiam]] 23:40, 1 January 2025 (UTC)


== Are these sequences mod any natural number n periodic? ==
:(x)+(2*x)+(x+0.5)=6.5
:4*x+0.5=6.5
:4*x=6
:x=1.5


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?
[[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 03:06, 10 January 2009 (UTC)


# [[Lucas number]]
::True, although I think it is best to do it step by step when learning rather than substituting the 2nd two equations into the first without writing them down. It makes it less likely to make mistakes like the OP did with the final term. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 03:13, 10 January 2009 (UTC)
# [[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}}
[[Special:Contributions/42.76.153.22|42.76.153.22]] ([[User talk: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. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 07:05, 24 December 2024 (UTC)
:::Also true except that the OP grasped the concept but made a bit of a beginner's mis-step with that third term which could happen regardless of method. I'd encourage the original method as he/she did understand the simplest notation (forgiving that 3rd term). [[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 03:21, 10 January 2009 (UTC)
::::As I said, doing it step by step and writing down all the steps makes it less likely that you will make such a mistake. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 23:34, 10 January 2009 (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]]. [[User:GalacticShoe|GalacticShoe]] ([[User talk: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 number]]s not on the list?. --[[User:RDBury|RDBury]] ([[User talk: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]]. [[User:GalacticShoe|GalacticShoe]] ([[User talk:GalacticShoe|talk]]) 01:29, 29 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. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 18:12, 24 December 2024 (UTC)
Now this thread is a bit silly. The OP was just driven away and is not going to learn anything (despite your efforts to make him learn something). I am not saying that you didn't follow the right path in teaching the OP, but I think that you could have done it a lot quicker and more easily. PST <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 12:56, 11 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
:::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)
::::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)
:::::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)
: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)
::[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)
: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)
::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)


= January 10 =


= December 31 =
==Does this continued fraction converge?==
I was playing around with [1;2,1,3,1,4,1,5....] at school today, and I'm fairly sure it converges to a value around 1.44. Is there a test that I can run to see if it does converge? Does anyone know if this particular series turns out to equal something interesting? Thanks.
[[Special:Contributions/24.18.51.208|24.18.51.208]] ([[User talk:24.18.51.208|talk]]) 01:44, 10 January 2009 (UTC)


== Generating a point on the Y axis from regular pentagon with point on X axis ==
:For a start you may find [[continued fraction]] interesting. Every continued fraction with positive integer coefficients will converge; yours converges to about 1.35804743869438. It is irrational (because the continued fraction is infinite), and furthermore is not the root of any quadratic equation (because the continued fraction is not periodic). The continued fraction looks similar to the continued fraction for <math>\frac {\tan (1) - 1}{2 - \tan (1)} = 1.259415845</math>, which has continued fraction [1;3,1,5,1,7,1,9,...]. Perhaps if you can figure out how to calculate the continued fraction for tan (1) (which has continued fraction [1;1,1,3,1,5,1,7,1,9,...]) then that might give some ideas of how to find the value of [1;2,1,3,1,4,1,5,...]. Eric. [[Special:Contributions/68.18.17.165|68.18.17.165]] ([[User talk:68.18.17.165|talk]]) 04:23, 10 January 2009 (UTC)


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.
:You may also find [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfCALC.html this calculator] interesting if you just want an evaluation. It gives 641/472 = 1.3580508474576272 = 1, 2, 1, 3, 1, 4, 1, 5 [[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 04:59, 10 January 2009 (UTC)


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)
== Pentagons ==
[[User:Naraht|Naraht]] ([[User talk:Naraht|talk]]) 05:16, 31 December 2024 (UTC)


:I think you meant to write <math>B^N(K_0).</math> &nbsp;--[[User talk:Lambiam#top|Lambiam]] 07:55, 31 December 2024 (UTC)
With only a sheet of paper, a ruler and a pencil, how might I go about drawing a regular pentagon of which each side is 150mm long, without having any construction lines around it afterward? The easiest way seemingly might be to start with one line 150mm long and work out how far away horizontally and vertically from that the end of each other line would be, but how should I do that?
[[Special:Contributions/148.197.114.165|148.197.114.165]] ([[User talk:148.197.114.165|talk]]) 11:52, 10 January 2009 (UTC)
::Changed to use the Math.[[User:Naraht|Naraht]] ([[User talk: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.
:Can you use a compass? If so, see [[Pentagon#Construction]]. You'll need the formula in the previous section to work out what radius to use. [[Special:Contributions/Zain Ebrahim111|Zain Ebrahim]] ([[User talk:Zain Ebrahim111|talk]]) 12:00, 10 January 2009 (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. --[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 13:12, 1 January 2025 (UTC)
:Actually, [http://planetmath.org/encyclopedia/CompassAndStraightedgeConstructionOfRegularPentagon.html this] is better because it gives links to details for each step. [[Special:Contributions/Zain Ebrahim111|Zain Ebrahim]] ([[User talk:Zain Ebrahim111|talk]]) 12:13, 10 January 2009 (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. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 16:53, 1 January 2025 (UTC)
:::::Here is one point other than the origin (in red)
:::::[[image:Pentagon-y-axis-point.svg]]
:::::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)
::::::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)


::::{{U|RDBury}} *headslap* on (0,0) Any idea on y<>0?
I think I've worked it out. Would this give the right shape?: To draw regular pentagon ABCDE, where s is the length of any side of the pentagon, create a rectangle of size (1+(5^1/2)/2)*s by (((1+(5^1/2)/2)*s)^2 - s/2^2)^1/2. Draw then the line AB, of length s along one of the longer sides of the rectangle, so both lines have their centres at the same point. Draw the line AE from A to the nearest of the shorter sides of the rectange such that its length equals s, then do the same for BC from B to the other side. Find the centre of the long side opposite line AB, mark this as point D and draw lines from this to both C and E.


= January 1 =
For a pentagon with sides length 150mm, this rectangle would be 242.7mm by 230.82mm, with A and B 46.35mm from the corners.


== What is the first number not contained in M136279841? ==
Something like that?
[[Special:Contributions/148.197.114.165|148.197.114.165]] ([[User talk:148.197.114.165|talk]]) 17:48, 10 January 2009 (UTC)


:Try it and see. It's easy enough to tell if you have the right shape at the end. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 23:32, 10 January 2009 (UTC)
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)


: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. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 10:10, 1 January 2025 (UTC)
== max size of a k-component graph ==
::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)
{{Resolved|1=--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 14:24, 11 January 2009 (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 [[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)
This is related to a question I asked a few days back.
What is the maximum number of edges possible in a n vertex graph having k connected components where each component has <math>n_i</math> vertices. Obviously <math>\sum n_i=n</math> and I need the maximum value of <math>\frac{1}{2}\sum n_i(n_i-1)</math>. How should I proceed further? I want the final answer to be in terms of n and k. Thanks--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 13:06, 10 January 2009 (UTC)
::::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]]. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 06:21, 3 January 2025 (UTC)
:As with the k=2 case, the maximal case is when all but one component has 1 vertex. [[User talk:Algebraist|Algebraist]] 17:00, 10 January 2009 (UTC)
::I came to the same conclusion by the following procedure: As <math>\sum n_i^2=n^2-(k-1)(2n-k)-2\sum_{i<j}(n_i-1)(n_j-1)</math> so if all but component has 1 vertex then the number of edges is <math>\frac{1}{2}(n-k)(n-k+1)</math>. Since this acts as an upper-bound on the number of edges too (because of the negative sign in <math>-2\sum_{i<j}(n_i-1)(n_j-1)</math>) hence this is the maximal achievable value for the number of edges. Is this proof correct? Secondly, is there a way to maximize the function <math>\frac{1}{2}\sum n_i(n_i-1)</math> subject to the constraints <math>\sum n_i=n, n_i\in \mathbb{N}</math>. Cheers--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 17:31, 10 January 2009 (UTC)


= January 3 =
Shahab asks "is there a way to maximize the function <math>\textstyle\frac{1}{2}\sum n_i(n_i-1)</math> subject to the constraints <math>\textstyle\sum n_i=n, n_i\in \mathbb{N}</math>?" The answer is to take <math>n_1=n</math>. But perhaps Shahab meant to ask the related question: how to minimize the function <math>\textstyle\sum n_i</math> subject to <math>\textstyle\frac{1}{2}\sum n_i(n_i-1)=N</math>. Bob Robinson (UGA) and I solved this problem more than 20 years ago but never published it. Consider the greedy approach: choose <math>n_1</math> as large as possible so that <math>\textstyle\frac{1}{2}n_1(n_1-1)\le N</math>, then continue in the same fashion with what is left. Our theorem was that this works (i.e. the greedy choice of <math>n_1</math> is correct) for all but a finite set of values of ''N'', which we determined. If I recall correctly, there were something like 20 exceptions and the largest was about 86,000. In the exceptional cases, <math>n_1</math>should be chosen 1 or 2 less than the maximum possible. [[User:McKay|McKay]] ([[User talk:McKay|talk]]) 01:59, 11 January 2009 (UTC)

:::Thanks for the extra info. But my initial question was about maximizing <math>\frac{1}{2}\sum n_i(n_i-1)</math> subject to <math>\sum n_i=n, n_i\in \mathbb{N}</math> where I meant <math>\mathbb{N}:=\{1,2\cdots \}</math>. I cannot take <math>n_1=n</math> for then the graph is just <math>K_n</math> and I need k connected components in the graph. Cheers--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 05:15, 11 January 2009 (UTC)

:::: But that question is already answered above: take ''k''-1 components equal to 1 and one component equal to ''n''-''k''+1. The fact that it is best for ''k''=2 proves that it is best for arbitrary ''k'' also. [[User:McKay|McKay]] ([[User talk:McKay|talk]]) 11:11, 11 January 2009 (UTC)

::::: I am probably being dense here (& for this I ask you to indulge me) but I cannot understand your last statement. Isn't it possible for the max value to be different then <math>\frac{1}{2}(n-k)(n-k+1)</math> and yet be equal to <math>\frac{(n-2)(n-1)}{2}</math> for k=2. Can you please clarify?--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 12:46, 11 January 2009 (UTC)
::::::McKay probably has in mind the following simple proof (also the proof I had in mind above, for what that's worth): suppose we have values n<sub>i</sub> subject to <math>\sum n_i=n, n_i\in \mathbb{N}</math>, and suppose there are two values of i for which n<sub>i</sub> is not 1. Then by the k=2 case, we can increase <math>\frac{1}{2}\sum n_i(n_i-1)</math> by changing one of these two values to 1 (and the other to their sum minus 1). Hence in the maximal case, all but one n<sub>i</sub> must be 1. [[User talk:Algebraist|Algebraist]] 13:32, 11 January 2009 (UTC)
:::::::OK. Thanks--[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 14:24, 11 January 2009 (UTC)

== Multiplying integers by digit-swapping. ==

I've been playing around with integers which are multiplied by a factor (≤9) when the RH digit is moved to the LH end. For example, 102564 is quadrupled when the 4 is moved to the start. There is obviously an infinite number of these, seen by considering 102564102564, so I'm interested in the smallest integer for each multiplier. For multiplier k, my analysis gives that the integer comes from an expression with 10k-1 in the denominator, so 102564, for example, must have something to do with thirty-ninths. And 1/39 = 0.0256410256410..., so as a hypothesis the smallest integer for a given multiplier is found by looking at the decimal expansion of 1/(10k-1) and selecting the cycle starting 10. Which brings me to the question - acting on the basis outlined, I've found that the 58-digit integer 1,016,949,152,542,372,881,355,932,203,389,830,508,474,576,271,186,440,677,966 is multiplied by 6 when the final digit is moved to the front, but is there a smaller one?→[[Special:Contributions/81.151.247.41|81.151.247.41]] ([[User talk:81.151.247.41|talk]]) 19:51, 10 January 2009 (UTC)

:There is certainly a name for an integer whose digits get swapped (as you have described) when multiplying with a positive integer less than or equal to 9. Although I can't seem to remember... Can any number theorist help out here? <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 20:38, 10 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
::According to [http://www.research.att.com/~njas/sequences/A092697 Sloane's], these are called k-parasitic numbers and the OP's number is indeed the least 6-parasitic number. [[User talk:Algebraist|Algebraist]] 20:42, 10 January 2009 (UTC)

== 3x3 Antisymmetric Matrices ==

Let A be a real 3×3 non-zero antisymmetric matrix. How does one show that there exist real vectors u and v and a real number k such that
Au = kv and Av = −ku? What relation do these u,v and k bear to the exponential of the matrix A?

I know the exponential of the matrix A is a rotation matrix, but having not seen an appropriate approach to the first part of the question, I'm unsure as to how to progress with this - thanks.

[[Special:Contributions/86.9.125.104|86.9.125.104]] ([[User talk:86.9.125.104|talk]]) 23:40, 10 January 2009 (UTC)Godless

:If '''A'''u = ''k''v and '''A'''v = -''k''u, then '''A'''<sup>2</sup>u = k'''A'''v = -''k''<sup>2</sup>u, and similarly '''A'''<sup>2</sup>v = -k'''A'''u = -''k''<sup>2</sup>v. So one approach would be to see what you can show about the eigenvalues and eigenvectors of '''A'''<sup>2</sup>, given that '''A''' is 3x3 antisymmetric. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 09:31, 11 January 2009 (UTC)

== Three '2's Problem ==

You have 3 twos, and you must use all of them in any expression of a number. Operations allowed are:

''addition (e.g 2+2+2=6)

''subtraction (e.g 2-(2-2)=2)

''multiplication (e.g 2*(2+2)=8)

''division (e.g 2*(2/2)=2)

''raising to a power (e.g 2^(2^2)=16)

''rooting - i.e. raising to the power of 1/n, where n is an integer (e.g 2^(1/(2+2))=2^(1/4))

''logarithm to base 2 or base e - if to base 2, you have to indicate, and that would mean using one of your three twos: otherwise it would be to the base e. (e.g log2(2*2)=2, or log(2*2*2)=3ln(2), although the latter is not an integer)

''concatenation - i.e combining say 3 and 4 to get 34 (Using "~" for this: e.g 2~(2/2)=21)''

How many distinct positive integers can you form in this way? (note that for example 2+(2+2)=2+(2*2) so this only counts as 1 distinct positive integer).

Thanks,

[[User:Mathmos6|Mathmos6]] ([[User talk:Mathmos6|talk]]) 23:58, 10 January 2009 (UTC)Mathmos6

:This problem is fairly trivial: I suggest that you have a go yourself. We won't give you answers here; we can only give you hints at the most. Please first show us your progress (tell us what is the smallest and highest possible positive integer that you can obtain). <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 13:01, 11 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
::Without concatenation, it's pretty trivial. With concatenation, however, you can get significantly larger integers. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 16:47, 11 January 2009 (UTC)

:::See also [[four fours]]. -- [[User:SGBailey|SGBailey]] ([[User talk:SGBailey|talk]]) 19:33, 11 January 2009 (UTC)

= January 11 =

== Group of points on the unit sphere ==

Let G be a finite non-trivial subgroup of SO(3). Let X be the set of points on the unit sphere in R^3 which are fixed by some non-trivial rotation in G. G acts on X - show that the number of orbits is either 2 or 3. What is G if there are only two orbits?

Could really just use a hand getting started on this one if nothing else. If the group is finite, I think all rotations have to be of the form 2pi/k for some integer k, and all axes of rotation have to be at an angle of 2pi/k to all other axes for a (different) integer k, right? That's about as much as I've accomplished so far. Cheers guys,

[[User:Spamalert101|Spamalert101]] ([[User talk:Spamalert101|talk]]) 00:22, 11 January 2009 (UTC)Spamalert101

:If the elements of G are ''not'' all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids. There are three different "types" of points that are fixed by some non-trivial element of G and permuted by the other elements of G - these form the three orbits of G acting on X. One orbit is the vertices of the Platonic solid - I'll let you work out what the other two orbits are (hint: think about dual solids which have the same symmetry group).
:On the other hand, if the elements of G ''are'' all rotations about the same axis then all non-trivial elements of G have the same two fixed points (what are they ?) and hence each of these two points is an orbit, and we have just two orbits. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 13:44, 11 January 2009 (UTC)
::Is your first sentence obvious? [[User talk:Algebraist|Algebraist]] 14:13, 11 January 2009 (UTC)
:::Trivial to be precise (I was joking). :) PST <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 16:11, 11 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
::::Having brushed up my geometry with our article [[Point groups in three dimensions]], I can now say that it is not obvious, or even true. Gandalf has forgotten the rotation groups of the regular prisms. In this case, there are 3 orbits. [[User talk:Algebraist|Algebraist]] 17:22, 11 January 2009 (UTC)
::To be more precise, if the elements of ''G'' are not all rotations about the same axis then ''G'' must be a subgroup of the rotational symmetry group of one of the Platonic solids. Moreover, considering only the icosahedron and octahedron suffice, as the dodecahedron has the same symmetry as the icosahedron, the cube the same symmetry as the octahedron, and tetrahedral symmetry is a subgroup of octahedral symmetry. Eric. [[Special:Contributions/68.18.17.165|68.18.17.165]] ([[User talk:68.18.17.165|talk]]) 16:27, 11 January 2009 (UTC)

:::(Reply to Algebraist) Not entirely obvious, I suppose - there is a proof in Chapter 15 of Neumann, Stoy and Thompson's ''Groups and Geometry''. But maybe there is a simpler (although less geometric) solution to the original problem using [[Burnside's lemma]]. Each element of G fixes just 2 points of X, apart from the identity element which fixes all |X| points, so the sum over G of the points of X fixed by each element of G is |X| + 2|G| - 2. But by Burnside's lemma this is a multiple of |G| - in fact it is |G| times the number of orbits. I'll let the OP take it from there. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 17:25, 11 January 2009 (UTC)
::::Thanks for the reference. Google books reveals that the correct result is that the finite subgroups of SO(3) are exactly those you mentioned above plus the rotation groups of prisms (including the rectangle as a degenerate prism). The subgroups of these groups Eric alludes to in fact turn out to be on the list already. [[User talk:Algebraist|Algebraist]] 17:38, 11 January 2009 (UTC)
:::::I agree with all of that - so which part of my original reply do you still think is incorrect ? [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 17:54, 11 January 2009 (UTC)
::::::The first sentence, since it excludes the prism/dihedron case, with dihedral symmetry groups. [[User talk:Algebraist|Algebraist]] 17:58, 11 January 2009 (UTC)
:::::::All of the non-trivial elements of the rotational symmetry group of a prism/dihedron have the ''same'' axis. So my first sentence "If the elements of G are <u>not</u> all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids" is correct. The prism/dihedron case is covered in my second paragraph which starts "On the other hand, if the elements of G <u>are</u> all rotations about the same axis ...". In short, 3 orbits<=>Platonic sold, 2 orbits<=>prism/dihedron. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 18:07, 11 January 2009 (UTC)
::::::::No, they do not. For the prism associated with the regular n-gon, there are n-1 non-trivial rotations about an axis through the centres of the ends, plus 1 non-trivial order 2 rotation about the centre of each other face and about the midpoint of each edge running between the ends. Thus there are 3 orbits. [[User talk:Algebraist|Algebraist]] 18:13, 11 January 2009 (UTC)
::::::::If one took the prism with vertices (±1,±2,±3), then it has a symmetry group whose intersection with SO(3) has 8 elements, right? It is generated by rotations of 180° about the x, y, and z axes, right? It is not a platonic solid, right? I don't really get the geometric ideas though. I like your counting proof, since it only uses the fact that the elements fix at most 2 points. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 18:18, 11 January 2009 (UTC)
:::::::::No, that group has 4 elements. It's the same as the rotation group of the rectangle-in-space, which I included in my list as a degenerate regular prism (the prism on a 'regular 2-gon'). It may gratify you to know that Gandalf's counting argument is used as the basis of the classification of finite rotation groups in the reference he supplied. [[User talk:Algebraist|Algebraist]] 18:24, 11 January 2009 (UTC)
::::::::::I agree, 4 elements: the identity and the three 180° degree rotations about the x,y,z axes. I did get that it was the same as the flat rectangle, but I just misremembered you got the dihedral group of order 8. Having no mind for geometry, this seemed reasonable to me, though I was a little concerned that the group appeared to be abelian. I took out my very expensive model of a rectangular prism and applied the rotations carefully to check. Who says a geometry textbook is useless? Glad to know the classification uses ideas I could grasp. Even with 4 elements, this is one of your (A's) counterexamples to G's first sentence, right? Not that it matters that much, since the structure of G's proof is right, and checking the details should always be a part of reading someone's answer. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 18:37, 11 January 2009 (UTC)
:::::::::::It's not my counterexample, I'm hopeless at geometry. I lifted it from our article. [[User talk:Algebraist|Algebraist]] 19:58, 11 January 2009 (UTC)
:::::::::(Reply to Algebraist) Yup, dumb mistake on my part. But it's always good to know that you are waiting to catch any little slips like that from us amateurs. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 18:44, 11 January 2009 (UTC)

== Unidentified trig functions/identities? ==

Where <math>\Delta\theta=\theta_f-\theta_s;\quad\Sigma\theta=\theta_f+\theta_s\,\!</math>, are
:<math>\frac{\cos(\Sigma\theta)+\cos(\Delta\theta)}{\sin(\Delta\theta)}=\frac{2\cos(\theta_s)\cos(\theta_f)}{\sin(\Delta\theta)}=\frac{2}{\tan(\theta_f)-\tan(\theta_s)}=?;\,\!</math>
and
:<math>\frac{\cos(\Sigma\theta)-\cos(\Delta\theta)}{\sin(\Delta\theta)}=\frac{2\sin(\theta_s)\sin(\theta_f)}{\sin(\Delta\theta)}=\frac{-2}{\cot(\theta_f)-\cot(\theta_s)}=?;\,\!</math>
(or their reciprocals) recognized as special functions, particularly in regards to spherical trigonometry?&nbsp;[[en:User:Kaimbridge|<span style="border:1px solid green;color:#e55b3c; padding:2px;background:#fde0bc">~<font face="courier new bold" class="title" title="Kaimbridge M. GoldChild">Kaimbridge</font>~</span>]] ([[User talk:Kaimbridge|talk]]) 01:13, 11 January 2009 (UTC)

These are elementary consequences of the formulas for sin(A+B) and cos(A+B). What is your question exactly? [[User:McKay|McKay]] ([[User talk:McKay|talk]]) 01:26, 11 January 2009 (UTC)
:Just if there is some elementary relationship/identity, like the "sine for sides", or in the same way that
::<math>\frac{\lambda_f-\lambda_s}{\mbox{arccosh}(\sec(\phi_f))-\mbox{arccosh}(\sec(\phi_s))}
=\frac{\Delta\lambda}{\int_{\phi_s}^{\phi_f}\sec(\phi)d\phi}=\tan(\alpha){}_{\color{white}.}\,\!</math>
:for loxodromic azimuth, which involves the [[Gudermannian_function|inverse Gudermannian function]].&nbsp;[[en:User:Kaimbridge|<span style="border:1px solid green;color:#e55b3c; padding:2px;background:#fde0bc">~<font face="courier new bold" class="title" title="Kaimbridge M. GoldChild">Kaimbridge</font>~</span>]] ([[User talk:Kaimbridge|talk]]) 15:25, 11 January 2009 (UTC)

==Conic section slope of directrix==
Given a conic section
<math>Ax^2+By^2+Cxy+Dx+Ey+F</math>
what is the slope of its directrix? [[User:Borisblue|Borisblue]] ([[User talk:Borisblue|talk]]) 01:21, 11 January 2009 (UTC)
:Sadly, we don't do homework for people. Furthermore, this problem is trivial. Please think about it a bit more and tell us what you have done on the problem so far. PST <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 16:10, 11 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

== Cubic functions ==

I'm trying to use cubic (and similar) functions for a little model, but I really don't have much maths. Could a mathematician please explain whether one changes a, b or c to make the curves flatter/steeper, change the y intercept, etc.? The article on [[quadratic equations]] does this graphically through [[:File:Quadratic equation coefficients.png|a clever image]], but a few lines of text would be great. [https://secure.wikimedia.org/wikipedia/en/enwiki/w/index.php?title=Talk:Cubic_function&diff=262799121&oldid=261148364 I'd like to add this info to] [[cubic functions|the relevant article]] as well. --[[User_talk:M.R.Forrester|Matt's talk]] 14:09, 11 January 2009 (UTC) <small>Edited to clarify which article is relevant --[[User_talk:M.R.Forrester|Matt's talk]] 14:17, 11 January 2009 (UTC)</small>

I presume you mean that the cubic is of the form:

:<math>f(x)=ax^3+bx^2+cx+d\,</math>

In that case, "flatness" of the curve is measured by its [[derivative]] which is (at ''x''):

:<math>f'(x)=3ax^2+2bx+c\,</math>

So only the coefficients ''a'', ''b'' and ''c'' have an impact on the steepness of the curve (the greater these values are, the greater the steepness; the smaller these values are, the greater the flatness). The ''y''-intercept is given by the image of 0 under ''f'' so the value of ''d'' equals the ''y''-intercept. If ''d'' is 0, the curve passes through the origin. PST

The article [[elliptic curve]] might also be of interest to you. PST

And by the way, mathematicians usually use one branch of mathematics in another branch of mathematics. There are numerous examples of this (I might as well let someone else list these examples; there are so many that I can't be bothered!). One interesting example is applying [[graph theory]] and the theory of [[covering map]]s to prove the well known Nielson-Schreier theorem; i.e every [[subgroup]] of a [[free group]] is free.

On the same note, there are mathematicians who would prefer not applying mathematics to another field (theoretical mathematicians) and those who would prefer applying mathematics to another field (applied mathematician). From experience, applied mathematicians are generally not so interested in the theoretical parts of mathematics and thus do not choose to learn much theoretical mathematics. But there are special cases. PST <small><span class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Point-set topologist|Point-set topologist]] ([[User talk:Point-set topologist|talk]] • [[Special:Contributions/Point-set topologist|contribs]]) 16:05, 11 January 2009 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

::Elliptic curves aren't really relevant to what the OP is doing, and what does that last paragraph have to do with anything? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 17:00, 11 January 2009 (UTC)
::Have a look at his links to see the relevance of that part. PST

:The steepness of the graph for large (either positive or negative) values of ''x'' is determined primarily by ''a''. For smaller values of ''x'', the graph will change direction a lot so it's rather more complicated. You may find it helpful to write the cubic as y=a(x-u)(x-v)(x-w), then the steepness for large values of x is, again, given by a, and u, v and w are the x-intercepts. The y-intercept would be -auvw. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 17:00, 11 January 2009 (UTC)

:Re-write equation as:
::<math>y=a\left(x+\frac{b}{3a}\right)^3 + \left(c-\frac{b^2}{3a}\right)\left(x+\frac{b}{3a}\right) + e</math>
:where ''e'' is a function of ''a'',''b'',''c'' and ''d'' that I can't be bothered to write out. Then change co-ordinates:
::<math>v=y-e\, ; \, u=\left(x+\frac{b}{3a}\right)</math>
::<math>v=au^3+ \left(c-\frac{b^2}{3a}\right)u</math>
:so we have placed the cubic's centre of symmetry at the origin. Now we can see that ''a'' determines the slope of the cubic far from its centre and <math>\left(c-\frac{b^2}{3a}\right)</math> determines the slope at its centre, and the number of turning points. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 17:44, 11 January 2009 (UTC)

:Sounds to me like you might be interested [[Bézier curve]] and [[Spline (mathematics)|splines]] in general.[[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]])

== Manifolds ==

I have a question for those people at the reference desk. This is not homework so you don't need to worry about that. I would like to know why manifolds (and more generally differentiable manifolds) are so important objects of study. I mean, they are in a sense so restrictive in nature (why exclude so many important topological spaces from study?). To me, it seems odd that the figure 8 is excluded from study (why can't you do calculus at the center of the figure 8?). I know that there are more general types of manifolds such as orbifolds and Banach manifolds but is every topological space some type of manifold? If not, what are the minimal conditions required for a space to be some type of manifold? I am pretty sure, for instance, that one can do calculus on the indiscrete space (any function between indiscrete spaces should have derivative 0). But on the other hand, I am also sure that the indiscrete space is not a type of manifold. I certainly understand that manifolds are important but unlike continuity and integration, I don't understand why differentiation cannot be generalized to arbitrary topological spaces (integration to measure spaces). Or it maybe just my lack of knowledge of the subject that I don't know that one can do calculus on arbitrary topological spaces. Thankyou for you help! <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/129.143.15.142|129.143.15.142]] ([[User talk:129.143.15.142|talk]]) 20:53, 11 January 2009 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:Note that the indiscrete space is indeed a (0-dimensional) manifold if the space in question has precisely one-point or vacuously a manifold if the space in question is empty. But those are just fine details for you. PST

:It's not an area I've ever studied properly, but I can't see how you could define derivatives on a space unless the space was locally [[metrizable]]. That rules out some spaces (but not things like a figure of 8). Also relevant is that, depending on your definitions, the [[long line (topology)|long line]] is not a manifold, but can have a [[differential structure]] put on it. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:00, 11 January 2009 (UTC)

Thanks. The definition I follow is the "locally Euclidean" one so I allow the long line to be a manifold. I think one other important thing to consider is that in manifold theory there are important facts about the derivative of a smooth function between smooth manifolds (namely it is a linear isomorphism between tangent spaces). But on one of your notes, the indiscrete space is not locally metrizable (when it has more than one point as someone above mentioned) but I would expect that the derivative of a function between such indiscrete spaes to be 0. You could argue similarly that continuity needs a metric although that is not the case. Isn't there an 'open set formulation' of differentiability? <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/129.143.15.142|129.143.15.142]] ([[User talk:129.143.15.142|talk]]) 22:16, 11 January 2009 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

Just to clarify something I said, I think that people don't study calculus on locally metrizable spaces because often in manifold theory, one considers vector spaces also. These are basically my questions (by the way, a real-valued function defined on the figure 8 is differentiable if and only if it extends to a differentiable function on a open set containing the figure 8 but this cannot be generalized to topological spaces that cannot be embedded in Euclidean space; this brings me to another question):

1) Since all Hausdorff, second countable manifolds can be embedded in Euclidean space, why not define differentiability on them as I have just done (i.e differentiable iff extends to a differentiable function on an open superset?)? I know this would exclude all those interesting theories of differentiable structures (like one interesting theorem that there are only 28 non-equivalent differentiable structures on S^7), but for the most part, this definition would suffice.

2) Is the figure 8 some type of manifold?

3) What are minimal conditions for a space to be some sort of manifold?

4) Is the indiscrete space (with more than one point) some sort of manifold (I would expect this naturally although I don't think that it is)?

I have not been getting sleep over this since I first learnt about manifolds (why are they so restrictive in nature?) so I would appreciate any help from the knowledgeable people at WP! <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/129.143.15.142|129.143.15.142]] ([[User talk:129.143.15.142|talk]]) 22:26, 11 January 2009 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:1) While such manifolds can be embedded in Euclidean space, they can be embedded in lots of different ways which wouldn't yield the same definition of differentiability (for example, a 2-sphere can be embedded in '''R'''<sup>3</sup> as a cube, in which case the image of a great circle wouldn't be differentiable since it would have corners in it, but with the standard embedding it clearly is). While you may be able to improve things by requiring the embedding to be smooth, you would end up with a circular definition. 2) A figure of 8 isn't a manifold simply because there is no open neighbourhood of the central point that is homeomorphic to the real line. Mathematicians define things the way they do because those definitions are useful. For example, if a space is locally Euclidean at a point you can define its tangent space at that point (which is, itself, a useful thing to do for all kinds of purposes), you can't define the tangent space to a figure of 8 at that central point (it has two tangents there, so you would end up with the union of two lines, which isn't a vector space). 3) A manifold is a space that is locally Euclidean everywhere (possibly with some extra conditions, depending on who you ask), that is the minimal condition (I'm not really sure what you mean by "some sort" of manifold, there are generalisations of manifolds, but they are really manifolds any more even if the word may appear in their name, you could argue that "topological space" is a generalisation of "manifold", but that doesn't mean much). 4) No, the only neighbourhood of any point in an indiscrete space is the whole space, which can't be homeomorphic to any Euclidean space because no Euclidean space (beyond '''R'''<sup>0</sup>, I guess) is indiscrete. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 00:04, 12 January 2009 (UTC)

= January 12 =

== A Pascal's triangle question I've never seen before ==

Dear Wikipedians:

I'm used to the following type of questions which applies Pascal's triangle, it asks how many ways to form the word "November":

<source lang="cpp">
N
O O
V V V
E E
M
B B
E E E
R R R R
</source>

as you can see each line is always 1 character more or less than the line above it, however, recently my teacher gave the following variation, which totally stumped me:

<source lang="cpp">
N
O O
V V V
E E E
M M
B B B B
E E E
R R
</source>

as you can see there are now lines with same number of characters as previous line, and also line with 2 characters more than the previous line, how do I handle this?

Thanks,

[[Special:Contributions/76.65.14.151|76.65.14.151]] ([[User talk:76.65.14.151|talk]]) 00:09, 12 January 2009 (UTC)

:I assume the goal is to move from top to bottom without "jumping too far". If no rules are specified then I would guess (and state that as an assumption if possible) that if an M is directly above a B then you can only move directly down to that B, so two of the B's cannot be reached. If the question form allows it then you could also make two answers where the other assumes you can move left-down, straigth down, or right-down, so from each M you can move to 3 B's. [[User:PrimeHunter|PrimeHunter]] ([[User talk:PrimeHunter|talk]]) 00:27, 12 January 2009 (UTC)

:I'm not sure how to interpret the question in the second case - how do you form the word? Are you allowed to move diagonally when letters are directly on top of each other? If not, then the question is easily reduced to one like the first case. If you are, then it depends on precisely what is and isn't allowed. Incidentally, I don't see how Pascal's triangle comes into it - for the first case the answer is just 2<sup>''n''</sup> where ''n'' is the number of lines which are longer than the line before (since, when going to those lines, you have a choice of 2 letters to go two, for all the other lines you just have 1 choice). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 00:32, 12 January 2009 (UTC)

The question is understated, can we have some more information please; such as an explicit copy of the said question. In your post you only reference the question you are trying to answer without actually stating what it is that you are trying to do. I too fail to see any relation to pascals triangle. Or the fact that the layers form a word. Or what you mean by 'handle this'.

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December 20

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Give a base b and two base b digits x and z, must there be a base b digits y such that the 3-digit number xyz in base b is prime?

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Give a base b and two base b digits x and z (x is not 0, z is coprime to b), must there be a base b digits y such that the 3-digit number xyz in base b is prime? 1.165.207.39 (talk) 02:10, 20 December 2024 (UTC)[reply]

In base 5, is composite for all base-5 Y. GalacticShoe (talk) 03:39, 20 December 2024 (UTC)[reply]
also offers a counterexample. While there are many counterexamples for most odd bases, I did not find any for even bases.  --Lambiam 09:58, 20 December 2024 (UTC)[reply]

December 23

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Is it possible to make Twisted Edwards curve birationally equivalent to twisted weirestrass curves ?

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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)[reply]

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)[reply]

December 24

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How did the Romans do engineering calculations?

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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? HiLo48 (talk) 05:50, 24 December 2024 (UTC)[reply]

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)[reply]
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)[reply]
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)[reply]
See Roman abacus. catslash (talk) 22:03, 25 December 2024 (UTC)[reply]
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)[reply]
This is a great answer! Tito Omburo (talk) 21:38, 1 January 2025 (UTC)[reply]
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)[reply]

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?

  1. Lucas number
  2. Pell number
  3. Tribonacci number
  4. Tetranacci number
  5. Newman–Shanks–Williams number
  6. Padovan sequence
  7. Perrin number
  8. Narayana sequence
  9. Motzkin number
  10. Bell number
  11. Fubini number
  12. Euler zigzag number
  13. Partition number OEISA000041
  14. Distinct partition number OEISA000009

42.76.153.22 (talk) 06:06, 24 December 2024 (UTC)[reply]

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)[reply]
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)[reply]
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)[reply]
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)[reply]
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)[reply]
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)[reply]
good questions. What about TREE(n) mod k, for arbitrary fixed k?Rich (talk) 23:03, 27 December 2024 (UTC)[reply]
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)[reply]
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)[reply]
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)[reply]
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)[reply]
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)[reply]


December 31

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Generating a point on the Y axis from regular pentagon with point on X axis

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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)[reply]

I think you meant to write  --Lambiam 07:55, 31 December 2024 (UTC)[reply]
Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)[reply]
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.
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)[reply]
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)[reply]
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)[reply]
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)[reply]
RDBury *headslap* on (0,0) Any idea on y<>0?

January 1

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What is the first number not contained in M136279841?

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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)[reply]

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)[reply]
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)[reply]
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)[reply]
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)[reply]

January 3

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