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{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 21}}


= December 23 =
{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 22}}


== Is it possible to make [[Twisted Edwards curve]] birationally equivalent to twisted weirestrass curves ? ==
{{Wikipedia:Reference_desk/Archives/Mathematics/2008 February 23}}


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)
= February 24 =


: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)
== derivatives and second derivatives of rotation functions ==
hi, how do i find the derivative and second derivative of a function that rotates counter clockwise an angle theta? do i just take the derivative and second derivative of (-sin+cos, cos+sin)?


= December 24 =
also, how might one prove that a harmonic function composed with a function that preserves dot product is still harmonic? thanks <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/199.74.71.147|199.74.71.147]] ([[User talk:199.74.71.147|talk]]) 03:33, 24 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->


== How did the Romans do engineering calculations? ==
:For the first question I assume you mean rotation around the origin in the [[Euclidean plane]], using the mapping
::<math>(x, y) \mapsto (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)\,.</math>
To get the ''n''-th derivative with respect to the rotation angle θ, just work out
::<math>(x, y) \mapsto \frac{d^n}{{d\theta}^n}(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)\,,</math>
:where a pair is differentiated coordinate-wise:
::<math>\frac{d}{d\theta}(X(\theta), Y(\theta)) = (\frac{d}{d\theta}X(\theta), \frac{d}{d\theta}Y(\theta))\,.</math>
:Using matrix–vector notation, the mapping can be written as
::<math>\begin{bmatrix}x \\ y \end{bmatrix} \mapsto M(\theta) \begin{bmatrix}x \\ y \end{bmatrix}\,,</math>
:where the 2×2 [[rotation matrix]] ''M''(θ) is given by:
:<math>M(\theta) = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \,\,\,\,\cos \theta\end{bmatrix}\,.</math>
:Here you have to find <sup>''d''</sup>⁄<sub>''d''θ</sub>''M''(θ), which can be done entry-wise:
:<math>\frac{d}{d\theta}M(\theta) = \begin{bmatrix}\tfrac{d}{d\theta}\cos \theta & -\tfrac{d}{d\theta}\sin \theta \\ \tfrac{d}{d\theta}\sin \theta & \,\,\,\,\tfrac{d}{d\theta}\cos \theta\end{bmatrix}\,.</math>
:
:The second question is not clear. What does it mean for a function to preserve [[dot product]]s? Normally you expect that to mean f(u·v) = f(u)·f(v), but the dot product turns two vectors into a scalar, so if u and v are vectors, on the left-hand side f is operating on a scalar and on the right-hand side on vectors. How can this be? Also, is it given in which order the harmonic function is composed with the dot-product preserving function? An example might help to clarify this. &nbsp;--[[User talk:Lambiam|Lambiam]] 06:26, 24 February 2008 (UTC)
::Probably means [[orthogonal matrix|orthgonal]], u·v = f(u)·f(v). With some very minor condition (possibly none), an orthogonal function is an orthogonal matrix. Then it is just asking to show that the laplacian is invariant under orthogonal coordinate changes, which is definitely a very common exercise to give (so likely to be his question). [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 06:44, 24 February 2008 (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)
:::that probably is my question, but i am unfamiliar with most of the terminology you used as my professor is rather scatterbrained in lecture. could you please give me some pointers on where to start if i were to prove this? thanks! <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/199.74.71.147|199.74.71.147]] ([[User talk:199.74.71.147|talk]]) 08:04, 24 February 2008 (UTC)</small><!-- Template:UnsignedIP2 -->


: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)
::::Basically you use chain rule. If g:R^2->R is [[harmonic function|harmonic]] and f:R^2->R^2 preserves dot products, then f is actually given by an [[orthogonal matrix]] (which looks almost exactly like a rotation matrix). Define h:R^2->R by h(x) = g(f(x)). Write out what that means fairly explicitly in terms of the matrix entries of f (just label them f11, f12, f21, and f22; don't use cosines in my humble opinion). Then use chain rule to take the first and second derivatives. The derivatives will involve dot products of rows of f. f is orthogonal, so the dot products will be 0 for the mixed derivatives of g, and 1 for the repeated derivatives of g. In other words, [[laplacian|lap]](h) evaluated at x is equal to lap(g) evaluated at f(x). [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 08:17, 24 February 2008 (UTC)
:::::how would one avoid using cosines? and also, where do dot products come into the derivatives? I don't see them[[Special:Contributions/199.74.71.147|199.74.71.147]] ([[User talk:199.74.71.147|talk]]) 20:22, 24 February 2008 (UTC)
::::::(Sorry, I don't have an easy way to explain this; my comments have been meant for other more analytic types to help you out; same continues here). See [[directional derivative]] for the dot products. Applying f just changes the direction of a directional derivative. The laplacian is defined in terms of partials, but you get the same definition if you use directional derivatives in directions which are just a rotation of the previous ones. Simply writing it out and taking calc 1 derivatives suffices, it just takes a page. I recommend that method. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 20:47, 24 February 2008 (UTC)
:::::::can you explain a little more about how i get from taking the derivative to having dot products?
::::::::There is a dot product in the definition of a [[directional derivative]], is that the one you're looking for? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 11:44, 25 February 2008 (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.
== Integral Powers of Trinomials ==
For expanding binomials, we have the binomial expansion which gives us all the info we need to know about expanding a binomial to (any) power. For example, if I want to find out what is the full expansion of <math>(x+y)^{25}</math>, I can do so. My question is, is there a similar expansion formula for <math>(x+y+z)^{25}</math>. Is there a shorthand way to write this expansion in terms of the coefficients of the expansion?[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 05:53, 24 February 2008 (UTC)


: 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)
: Memory of high school maths a bit hazy, but you can break <math>(x+y+z)^{25}</math> into <math>((x+y)+z)^{25}</math>, which gives <math>\sum_{k=0}^n{C^n_k}(x+y)^{n-k}z^{k}</math>, and then expand <math>(x+y)^{n-k}</math> binomially.
: This gives something like <math>\sum_{k=0}^n\sum_{l=0}^{n-k}{C^n_k}{C^{n-k}_l}x^{l}y^{n-k-l}z^{k}</math> --[[User:PalaceGuard008|PalaceGuard008]] ([[User_Talk:PalaceGuard008|Talk]]) 06:19, 24 February 2008 (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.
:You are looking for the [[Multinomial theorem]]. You are basically counting how many ways to rearrange "mississippi", or other such combinatorial questions. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 06:26, 24 February 2008 (UTC)


: 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)
::The coefficient of <math>x^iy^jz^k</math> in the expansion of <math>(x+y+z)^{n}</math> is


::See [[Roman abacus]]. [[User:Catslash|catslash]] ([[User talk:Catslash|talk]]) 22:03, 25 December 2024 (UTC)
::::<math> {n \choose i,j,k} = \frac{n!}{i!\,j!\,k!} </math>
:::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? ==
::[[Pascal's pyramid]] contains the coefficients for the [[trinomial expansion]] - it bears the same relation to the trinomial expansion as [[Pascal's triangle]] does to the binomial expansion, but it has an extra dimension. So the coefficients of <math>(x+y+z)^{25}</math> are the numbers in the 25th layer of Pascal's pyramid. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 11:26, 24 February 2008 (UTC)


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?
Thanks guys, I had no idea that a generalization of Pascal's triangle existed.[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 01:29, 25 February 2008 (UTC)


# [[Lucas number]]
::Curiously enough, I independently discovered this relationship over a year ago. I found it a lot less obvious than Pascal's Triangle. [[User:A math-wiki|A math-wiki]] ([[User talk:A math-wiki|talk]]) 07:06, 27 February 2008 (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)
== Vibrational modes of a drum ==
: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)
I'm interested in creating a few images for Wikipedia of the vibrations of a drum membrane.
:::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)


It appears these vibrations are related to the Bessel functions - searching on google, I didn't find any precise descriptions of the phenomenon. Somes pages appeared helpful though (for example : [http://www.ecs.fullerton.edu/~mathews/N310/projects3/p23.htm Standing waves in a drum membrane...]).


= December 31 =
So, could anyone explain how it works, and what the equations describing the vibrations are ?


== Generating a point on the Y axis from regular pentagon with point on X axis ==
Thanks. -- [[User:Xedi|Xedi]] (19:25, 24 February 2008 (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.
:Well, I don't know your mathematical background (have you studied PDEs?), but the simplest thing to do is assume that the membrane is isotropic and homogeneous, and the waves are small enough that it responds linearly. Then you have to solve the [[wave equation]] (whose spatial part after [[separation of variables]] is the [[Helmholtz equation]]) by finding [[eigenfunction]]s of the [[Laplacian]] for whatever shape your drum is. The solutions for a '''circular''' drum do involve Bessel functions. See also [[Hearing the shape of a drum]]. —[[User:Keenan Pepper|Keenan Pepper]] 23:13, 24 February 2008 (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)
:If you want a textbook that discusses this stuff, try ''Elementary Applied Partial Differential Equations with Fourier Series and Boundary Value Problems'' by Richard Haberman. —[[User:Keenan Pepper|Keenan Pepper]] 23:17, 24 February 2008 (UTC)
[[User:Naraht|Naraht]] ([[User talk:Naraht|talk]]) 05:16, 31 December 2024 (UTC)
::Thanks a lot, the article on the Helmholtz equation is pretty much what I needed. (I'll also have a look at the book) -- [[User:Xedi|Xedi]] (10:08, 25 February 2008 (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)
= February 25 =
::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.


:::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)
== 1-1 ==
::::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)
::::::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> &nbsp;--[[User talk:Lambiam#top|Lambiam]] 16:22, 3 January 2025 (UTC)


::::{{U|RDBury}} *headslap* on (0,0) Any idea on y<>0? <small>(← comment from [[User:Naraht|Naraht]])</small>
let a,b positive integers.
:::::See the above construction by catslash. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 16:12, 3 January 2025 (UTC)
:::::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)


= January 1 =
Is


== What is the first number not contained in M136279841? ==
f(a,b)= 3^a + 10^b


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)
1-1? <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/71.97.4.194|71.97.4.194]] ([[User talk:71.97.4.194|talk]]) 00:46, 25 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->


: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)
:If you mean [[one-to-one]], then yes, I think it is. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 01:11, 25 February 2008 (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. [[User:CodeTalker|CodeTalker]] ([[User talk: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 [[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)
::::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)
::::{{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)
:::::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)


g(a,b) = 2^a + 3^b fails to be 1-1. g(3,1)=11=g(1,2). It seems like we could do something similar for f, but the numbers get too big. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/71.97.4.194|71.97.4.194]] ([[User talk:71.97.4.194|talk]]) 01:25, 25 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->


= January 5 =
:If 3^a+10^b=3^c+10^d, then 3^a-3^c=10^d-10^b. Investigate the properties of differences between powers of 3 and differences between powers of 10. Does that help? —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 04:48, 25 February 2008 (UTC)


== Reference request:coherence condition (adjoint functor) ==
::Yes, I tried that, to no avail (yet). It did make me think of using the pigeonhole principle. *How*, I'm not quite sure. [[Special:Contributions/192.91.253.52|192.91.253.52]] ([[User talk:192.91.253.52|talk]]) 13:20, 25 February 2008 (UTC)
::I was unable to solve the problem, though I approached it in the way Bkell suggested. If a>c then d>b, so I factored it as 3^c*(3^(a-c)-1) = 10^b*(10^(d-b)-1). You get congruences like a=c mod OrderMod( 3, 10^b ), and b=d mod OrderMod( 10, 3^c ) so if b or c is nonzero, this reduces the possible solutions, and in fact you can apply the same argument again with much larger moduli. However, induction does not quite work, because I could not rule out b and c being zero (especially not inductively). When the problem is changed to the g(a,b), in fact very early there is a solution with b or c being 0, g(2,0)=5=g(1,1). While one might decide the arguments are not allowed to be 0 for some reason, during the induction you have to allow that case. At any rate, there are no small solutions for f (0<= a,b,d <= 1000). [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 15:43, 25 February 2008 (UTC)


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 ?)
:::The original problem states that the exponents are positive integers. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 16:33, 25 February 2008 (UTC)


*{{cite web|title=triangle identity|url=https://ncatlab.org/nlab/show/triangle+identity|website=ncatlab.org}}
::My idea, while not fully fleshed out, is that a difference between powers of 3 is always going to be a string of 0's and 2's when written in base 3, and a difference between powers of 10 is always going to be a string of 0's and 9's when written in base 10. I think there's a contradiction lurking somewhere: it feels like it's going to involve a claim that in order to write a difference between powers of 3 as a sum of things that look like <span style="white-space: nowrap;">9000…0</span> in base 10, you're going to have to use a particular <span style="white-space: nowrap;">9000…0</span> twice. Sorry for the roughness of this idea; it's really just a hunch and needs a lot of work still. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 16:40, 25 February 2008 (UTC)
*{{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 }}
*{{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 }}
*{{cite arxiv |arxiv=quant-ph/0008021 |last1=Coecke |first1=Bob |last2=Moore |first2=David |title=Operational Galois adjunctions |date=2000 }}
*[https://planetmath.org/93adjunctions planetmath]
[[User:Silvermatsu|SilverMatsu]] ([[User talk:Silvermatsu|talk]]) 02:41, 5 January 2025 (UTC)


== Concatenation of first 10 digits and last 10 digits of a number ==
:Consider this, if 3^a+10^b=3^c+10^d, then you can rewrite it in most cases as 3^f=((10^h-1)*10^g)/(3^j-1), which must be an integer. Is it possible for this to be an integer, though? (hint: look at 10^h-1 and 3^j-1 under the proper modulus). [[User:GromXXVII|GromXXVII]] ([[User talk:GromXXVII|talk]]) 19:33, 25 February 2008 (UTC)
::Sure it can be an integer. For example, g=3, j=2 makes ((10^h-1)*10^g)/(3^j-1) an integer for every nonnegative h. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 19:42, 25 February 2008 (UTC)


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:
:::I haven't come up with a good way to prove it yet, but I have another direction you might go in. If you write down a given power of 3, the question is whether you can subtract 1 from one digit, and add 1 to another, and get back a different power of 3. If you consider it base 3, the question is whether given a power of 101, you can subtract 1 from one digit and add 1 to another and get back another power of 101. Without carry, powers of 101 are pascal's triangle, which gives a clear enough pattern that it may be possible to prove it false. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 20:50, 25 February 2008 (UTC)


# Can a(2^n) take all 20-digit values which are multiples of 1024?
::::I think I've got it. Take it mod 5, 4, and 20, in that order. The first shows that a=c(mod4), the second shows that that's only possible if either b or d is 1, and the third shows that that can't happen. It's one-to-one. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 07:22, 26 February 2008 (UTC)
# Can a(3^n) take all 20-digit values which are odd?
:::::Unfortunately, simply looking mod finitely many numbers cannot prove the result. For instance, it is possible for f(a,b) = f(c,d) mod 4, mod 5, and mod 20 (simultaneously), without a=c, b=d: more explicitly, f(1,2) = f(9,10) mod 4, mod 5, and mod 20. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 16:45, 26 February 2008 (UTC)
# 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?
== Approximation by Smooth Functions ==
# Can a([[prime number]]) take all 20-digit values which end with 1, 3, 7, 9?
Basically, our teacher is trying to convince us that <math>C^{\infty}_0(\Omega)</math> which is the set of all infinitely differentiable functions with compact support on a given set <math>\Omega</math>, is dense in <math>L^2(\Omega)</math>. Now, I know that one of the definitions of a set A being dense in another set B is that the closure of A must be B. Which is the same as saying that B contains A and all of its limit points. So he gave us a theorem in class saying that, for any <math>f \in L^2(\Omega)</math>, there exists a sequence <math>\phi_n \in C^{\infty}_0(\Omega)</math> such that <math>\phi_n \rightarrow f</math> in <math>L^2(\Omega)</math>. My question is how does this theorem prove that <math>C^{\infty}_0(\Omega)</math> is dense in <math>L^2(\Omega)</math>? All we have done is shown that every square integrable function f is a limit of a sequence of smooth functions with compact support. We have shown that each f is a limit point but how do we know that those are all the limit points? What if there is a limit point of smooth functions with a compact support outside <math>L^2(\Omega)</math>?[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 04:52, 25 February 2008 (UTC)
# Can a([[lucky number]]) take all 20-digit values which are odd?
:<math>L^2(\Omega)</math> is a closed set - it contains its own limits. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 05:36, 25 February 2008 (UTC).
# Can a([[Fibonacci number]]) take all 20-digit values?

# Can a([[partition number]]) take all 20-digit values?
Duh, I seem to have forgotten that little fact. Thanks![[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 05:52, 25 February 2008 (UTC)
# What is a(9^9^9^9)?
: The above comment obscures the point at hand. It is not enough to talk about a set being [[open set|open]] or [[closed set|closed]]; you have to specify the topology you are talking about. Every set is closed in its own subspace topology. Here, the question is whether or not the closure of <math>C^{\infty}_0(\Omega)</math> in the metric topology of <math>L^2(\Omega)</math> is <math>L^2(\Omega)</math>. In this topology, it is impossible for the closure of <math>C^{\infty}_0(\Omega)</math> to be bigger than <math>L^2(\Omega)</math>. Perhaps the above post was referring to the fact that <math>L^2(\Omega)</math> is [[complete space|complete]] (is this true actually? ahh...), which would mean that however you embed <math>L^2(\Omega)</math>, the closure of <math>C^{\infty}_0(\Omega)</math> will be exactly <math>L^2(\Omega)</math>. However, that would not be a requirement for saying that the one is dense in the other. If <math>A \subset B \subset C</math>, to say that A is [[dense set|dense]] in B is to say that the closure of A, in B's [[subspace topology]], is all of B. It would not matter if B were not closed in C, because in the B subspace topology, the closure of A cannot exceed B. (Can someone with more experience please check my post for correctness? Thanks...) [[User:J Elliot|J Elliot]] ([[User talk:J Elliot|talk]]) 06:08, 25 February 2008 (UTC)
# What is a(9^^9), where ^^ is [[tetration]]?
::<math>L^2(\Omega)</math> is a normed space, the norm being the integral of the absolute square. This norm defines the topology. In this topology the set <math>L^2(\Omega)</math> is closed. A function outside <math>L^2(\Omega)</math> has no well defined distance to a a function within <math>L^2(\Omega)</math> , so it is not a limiting function of a sequence of functions in <math>L^2(\Omega).</math> [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 14:15, 25 February 2008 (UTC).
# What is a(9^^^9), where ^^^ is [[pentation]]?
:::The sentence "In this topology [on a space ''X''] the set ''X'' is closed" is one of the axioms of a [[Topological_space#Equivalent_definitions|topology]], and thus is never a meaningful observation. As was noted above, the only meaningful thing that we can say in this direction (and indeed it is very meaningful) is that ''L''<sup>2</sup>(Ω) is a [[normed vector space|normed space]] (and thus more generally a [[uniform space]]), and with respect to this structure ''L''<sup>2</sup>(Ω) is [[complete metric space|complete]]. In particular, this implies that whenever any [[metric space]] ''V'' is isometrically embedded into ''L''<sup>2</sup>(Ω) (we might take ''V''=''C''<sup>∞</sup><sub>0</sub>(Ω) with the induced square-integral norm), then any Cauchy sequence in ''V'' has a limit in ''L''. Thus it is the completeness of ''L''<sup>2</sup>(Ω), rather than the meaningless "closedness", which implies that the completion of ''C''<sup>∞</sup><sub>0</sub>(Ω) [with the ''L''<sup>2</sup>-norm] is ''L''<sup>2</sup>(Ω). This is not necessary to show that ''C''<sup>∞</sup><sub>0</sub>(Ω) is ''dense'' in ''L''<sup>2</sup>(Ω), however. [[User:Tesseran|Tesseran]] ([[User talk:Tesseran|talk]]) 05:26, 27 February 2008 (UTC)
# What is a([[Graham's number]])?

# What is a([[TREE(3)]])?
:Kaiser, you should take some algebra classes next semester, because my analysis books had been getting nice and dusty before you started posting this semester. :)
# If we know a(x), assume that x has at least 20 digits, can we also know a(2*x), a(3*x), etc.?
:BTW, you may seriously want to prove that if f_n are C^oo with compact support, and they form a cauchy sequence under the L^2 norm (so they "converge to something"), then there is some number C such that the L^2 norm of all f_n is bounded by C. In other words, if something is an L^2 limit of compact functions, then it is L^2. This should be easy to prove, other than the fact that it is posed randomly on the net. Once you do this, some of the previous posters talking about completeness should make more sense.
# If we know a(x), assume that x has at least 20 digits, can we also know a(x^2), a(x^3), etc.?
:This idea is actually pretty important, even though some people might think my question is silly. The point of using the L^2 norm is to control the L^2 norms of limits. As long as everything in the sequence was L^2, and the sequence converges (is cauchy) according to the L^2 norm, then the limit is L^2. This is why you need to use the Sobolev norms in PDE. You want to take a sequence of approximate solutions to a PDE (or solutions to some approximately equal PDE), and know that the limit is still a solution. If you use L^2 for that, you will often fail, but if you use H^1 (L^2 of the function and its derivative), you will have much better luck, since the limit function at least has an L^2 derivative. [[User:JackSchmidt|JackSchmidt]] ([[User talk:JackSchmidt|talk]]) 15:57, 25 February 2008 (UTC)
# 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)?
:I think, Kaiser, that your original post misses the point a bit. ''B'' containing ''A'' and its limits is nothing like sufficient for ''A'' being dense in ''B'' (consider <math>A=[0,1],B=\mathbb R</math>). Instead, what you want is that <math>\forall x\in B\;\exists\{y_n\}\subseteq A\;\lim_{n\rightarrow\infty}\|y_n-x\|=0</math> (with some metric appropriate to ''B'', which is where topology comes in). The theorem you were given establishes precisely this fact. --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 16:08, 25 February 2008 (UTC)
# 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)?

[[Special:Contributions/220.132.216.52|220.132.216.52]] ([[User talk:220.132.216.52|talk]]) 08:33, 5 January 2025 (UTC)
Jack, you know what, you are absolutely correct. I am going for a degree in Applied Math and I incorrectly assumed (in my early childhood as a lower division Math student) that Applied Math and Pure Math were two disjoint sets. So I decided to stay away from Pure courses which is exactly why now I have huge holes in my understanding of real, complex, functional analysis, and algebra. I was already thinking about an algebra course next semester but it seems vital that I must take it. Well, lol, don't put those books away. This doesn't mean that I am going to stop asking questions. There are plenty more where these come from. I know that all of you don't have to help me but I really appreciate it. Thanks everyone! :)<br />
Ok, so let me recap. L^2 is the space of all square integrable functions. It has a norm defined on it (the usual L^2 norm). It has an inner product defined on it and it is complete with respect to this norm. Therefore L^2 is a Hilbert space. Being complete means that any Cauchy sequence of L^2 functions converges to an L^2 function. So there are no limit points of L^2 that are outside L^2. Furthermore, if I want to show that the set of all smooth functions (members of <math>c^{\infty}</math>) is dense in L^2, then it will suffice to show that if I take any square integrable function, I can always find a sequence of smooth functions that converges to that square integrable function. This is the same as showing that any square integrable function can be approximated arbitrarily well by smooth functions. It is also not possible for a sequence of smooth functions to converge outside the L^2 space because smooth functions are a subset of square integrable functions and the set of all square integrable functions is closed. Is this correct, guys? I want to make sure that my line of reasoning is correct.[[User:A Real Kaiser|A Real Kaiser]] ([[User talk:A Real Kaiser|talk]]) 20:04, 25 February 2008 (UTC)

== GMAT data sufficiency problem ==

A homeowner must pick between paint A, which costs $6 per liter, and paint B at $4.50 per liter. Paint B takes one-third longer to apply than paint A. If the homeowner must pay the cost of labor at $36 per hour, which of the two paints will be cheaper to apply?

(1) The ratio of the area covered by one liter of paint A to that covered by one liter of paint B is 4:3.

(2) Paint A will require 40 liters of paint and 100 hours of labor.

I first encountered the problem in Arco Math Workbook (2000). Kaplan's 2005 version is similar, but it's got the same problem and explanation, which I cannot agree with:

"Statements (1) and (2) taken together are not sufficient to answer the question. To make an intelligent decision, we need to know which requires more paint and how much more, how long each will take, and we need some info on their labor costs...Using both statements together, we still cannot find the labor costs."

It seems rather clear to me how to find the labor costs.

First, the two statement taken together mean that it'd take 40*4/3 or 160/3 L of paint B.

Second, statement (2) says that paint A is applied at a rate of 40L per 100 hours, or 2/5 L/hr. - but the original problem says it would take a third longer for paint B, which means that paint B is applied at a rate of 2/5 * 3/4 L/hr. or 3/10 L/hr.

This last conversion was the most difficult for me, but you can see it's just multiplying A's rate by 4/3 hr/L.

Thus, it will take (160/3) / (3/10) or 1600/9 hours to use the 160/3 L of paint B to cover the house.

Therefore we know how long it takes. Given the unit labor cost in the original problem, we can figure out how much labor costs for either paint. We also know how much paint we need and how much it costs. Why then would the problem be unsolvable so long as we have statements (1) and (2)?

[[User:Imagine Reason|Imagine Reason]] ([[User talk:Imagine Reason|talk]]) 07:26, 25 February 2008 (UTC)

:(1) and (2) on their own aren't enough, you need to information in your first paragraph as well - is that what it means? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 11:42, 25 February 2008 (UTC)

::The first paragraph is a given, while the two statements are optional. I'm saying that if we have the two statements (the first paragraph is always there), we can solve the problem, but the review books say we can't even then. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Imagine Reason|Imagine Reason]] ([[User talk:Imagine Reason|talk]] • [[Special:Contributions/Imagine Reason|contribs]]) 17:14, 25 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

:::Well, it's hard to comment on what the review book says without seeing it, but the problem you've posted is certainly solvable in the way you came up with. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 06:32, 26 February 2008 (UTC)

===Clarity===
Working in dollars throughout...

Focusing on the given paragraph ONLY... We assume that equal amounts (L) of paint in litres are required no matter whether paint A or paint B is used and that if H hours are needed for paint A, then 4H/3 hours are needed for paint B.

Total cost of using paint A is 6L + 36H and the total cost of using paint B is 4.5L + 48H

Setting these equal to each other gives 6L + 36H = 4.5L + 48H and a solution of L = 8H

Therefore numerically, if L > 8H then paint B is cheaper and if L < 8H then paint A is cheaper.

Conclusion is that additional information is needed.


Incorporating ONLY statement 1 with the given paragraph... We need to make an assumption that the reduced volume of paint A needed (3/4 of the volume required by using paint B) results only in a cost saving on the paint, and NOT that less paint equals less hours of labour: after all, the same area needs covering and we know it takes a third longer with paint B. If this assumption is false and that less paint ALSO equals less labour hours then we have a slightly different problem.

Relative to the L litres required for paint A, paint B requires 4L/3 litres which now costs 6L (dollars).

Case 1.1 Less paint only required.
:The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 48H.
:Paint A is cheaper.

Case 1.2 Less paint and 3/4 of the previous amount of labour hours required (4/3 times the amount are now needed for B compared to A), i.e. relative to the H hours required for paint A, paint B requires 4/3 &times; 4/3 = 16/9 times the hours (cost = 36H &times; 16/9 = 64H).

:The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 64H.
:Paint A is even cheaper.

Statement 2 is irrelevant in determining which paint is cheaper to apply; it just gives the information to calculate the ACTUAL costs. All the problem requires though is a general 'which is cheaper', not by how much.
So, as far as I can read it from the information provided, choosing paint A is cheaper in all cases if statement 1 is included with the given paragraph; statement 2 being irrelevant, as is the request for any additional information.

However, all this is blindingly obvious from the simple point that statement 1 means that the effective price of paint B is $6.00 per litre, identical to that of paint A. B takes longer to apply at an additional cost of $12 per hour for case 1.1 and $28 per hour for case 1.2 above based on the number of hours needed to apply paint A.

The only ambiguity is in calculating ACTUAL costs for paint B depending on case 1.1 or 1.2 above.
Paint A costs $3840: Paint B either costs $5040 or $6640.

Please don't hesitate to contact me if you feel I have made any error in reading this situation! [[User:AirdishStraus|AirdishStraus]] ([[User talk:AirdishStraus|talk]]) 11:23, 26 February 2008 (UTC)

: I agree that the wording is ambiguous, but I'm pretty sure it is saying that a gallon of paint B takes a third longer to apply than a gallon of paint A. Regardless, you're right--we only need the first statement to answer the question. I think that's actually what I got the first time I saw the problem, actually, but somehow I let it go. Now I've seen the same nonsensical explanation in a second book, I'm not happy. [[User:Imagine Reason|Imagine Reason]] ([[User talk:Imagine Reason|talk]]) 23:36, 26 February 2008 (UTC)

== aquestion about closed surface ==

i am wondering about something,can we find or create aclosed surface where there all of the points inside that surface has adifferent distances to any of the all points of the surface?i mean here we cannot finde apoint inside the space of that surface has the same distance to at least 2points on the surface?this surface should not be for example like asphere because the center of the sphere is apoint that has asame distances to all of the points on the spherer`s surface.i hope my words is clear.thank you.[[User:Husseinshimaljasimdini|Husseinshimaljasimdini]] ([[User talk:Husseinshimaljasimdini|talk]]) 12:04, 25 February 2008 (UTC)

:If I understand the question correctly, then I would think not, since you can just choose any two points on the surface and pick the midpoint between them. Well, if the surface isn't convex, you couldn't choose *any* two points, but there there will still be some points - just choose any line that passes through the interior of the region bounded by the surface and the intersection points of that line with the surface should do. I think as long as the surface does bound a region with a non-empty interior, there will be points in the interior equidistant to points on the surface. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 13:41, 25 February 2008 (UTC)

Interesting question - I think the answer is no. I'll explain in two dimensions, you can easily expand the answer to three dimensions for surfaces..
:Suppose I start my [[asymmetric]] ring at point A with distance r from the 'centre'.. then increasing r as I turn around the surface eg using [[polar coodinates]] (radius,angle) - the first thing I notice is that if I increase r with angle I cannot every decrease it since that would return r to a previously used value.. but to get back to r at angle=0 I must decrease.. Both can't be true so it's impossible..

:Such a shape would be a spiral - spirals cannot be closed surfaces.. Did that make sense?[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 15:00, 25 February 2008 (UTC)

::Even a spiral loses: take the midpoint of a small enough [[chord (geometry)|chord]] that it doesn't intersect the next layer (so as to avoid questions of containment). --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 15:54, 25 February 2008 (UTC)
:::Didn't understand that - the spiral r=e<sup>angle</sup> is single valued in r for all values of angle ?[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 15:59, 25 February 2008 (UTC)
::::The OP's request was for an open region <math>\Omega\in\mathbb R^n\ni\forall x\in\Omega\,\not\!\exists(y,z)\subseteq\partial\Omega\;\;y\neq z\and \|y-x\|=\|z-x\|</math>. It's not enough for the "unique distances" clause to hold for ''one'' contained point. --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 16:50, 25 February 2008 (UTC)
:::::((For the spiral described above every point is at a unique distance, - am I wrong?))
:::::You've confused me now - I understood the OP asked for a closed surface (hence not a spiral) - but said that excluding that 'closed surface' clause a spiral would be such a shape (if only in 2d) I see that a spiral fails for a single unique point in 3D ie a spiral surface has more than one point with a given radius. Was that what your original point was saying?[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 17:12, 25 February 2008 (UTC)
:::::cf [[open region]] with [[Closed manifold]] - they asked for closed manifold eg 'closed surface'[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 17:42, 25 February 2008 (UTC)
::::::I interpreted the question as taking place in <math>\mathbb R^3</math> where the closed manifold (ie. closed surface) bounds an open region. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 17:59, 25 February 2008 (UTC)
:::::::still a bit confused about the 'chord' explanation - the answer is a definate 'no' anyway -
:::::::another way to look at it is to note that the closed surface must be (at least) double valued (of r in angle) - to be closed (in 2D) {and gets worse in 3d ie infinite values with same r for (angle1,angle2) could go on about being able to draw a loop/ring on the surface of a 3D closed surface at any point }
:::::::- which of course means it cannot be single valued as requested. are we answering the same question??? at least we are getting same answer[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 18:19, 25 February 2008 (UTC)
::::::::What function are you saying is multi-valued, <math>r(\theta)</math> or <math>\theta(r)</math>? I assume the latter, since that's what we need, but your terminology is a little confusing (to me, at least). The OP (in the 2D case - I'm pretty sure he's actually asking about the 3D case, since you can't have a closed surface in <math>\mathbb R^2</math>) requires that <math>\theta(r)</math> be multi-valued for some choice of origin within the interior of the region bounded by the surface. That's just a restatement of the question, though, it doesn't prove anything. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 18:39, 25 February 2008 (UTC)
:::::::::Yes I wrote "(of r in angle)" is should of course have read "doubled valued in angle for a given r" - my mistake that was very unclear.
:::::::::<math>\theta(r)</math> is multivalued for not true.. or single valued if the original construct was possible - I think I was showing that when <math>\theta(r)</math>=singlevalued the surface cannot be enclosed .
:::::::::Clearly for a given radius r there must be only one value of theta (2D) or (combiantion) theta,mu (3D) that gives that radius ie if r=fn(theta) then the function theta=fn<sup>-1</sup>(r) must have a single value of theta for that value of r eg for a ellipse there are 4 values of theta that give a specific r, so I would have said that (in the case of the ellipse) theta is multivalued.
:::::::::and multivalued = not possible[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 19:51, 25 February 2008 (UTC)

:::::::::My argument for it's impossibility was of this kind:
::::::::::given that the rate of change of radius with angle is x
::::::::::x cannot be zero (since this gives a 'flat' region - ie two points have same r)
::::::::::so x is either positive or negative
::::::::::x cannot change sign since this would require x=0 at some point.
::::::::::So x is either always positive or negative.

::::::::::BUT r=fn(x) = fn(x+2pi) (for a continuous surface) (a full rotation) (equation A)
::::::::::equation A cannot be true since d fn(x)/dx is not zero and never changes sign.
::::::::::Therefor assuming that the example function is single valued ie x=fn<sup>-1</sup>(r) is single valued for r (PROBABLY DID STATE THIS THE OTHER WAY ROUND at some point sorry)
::::::::::Then fn(x) <> fn(x+2pi) - it's not a closed ring.
:::::::::This expands easily to 3D since for a closed 3D surface any plane section through it (through the 'centre') must be a closed ring.
:::::::::So single valued in x=fn<sup>-1</sup>(r) and 'closed ring' are mutually exclusive.

:::::::::I've expanded a bit on the original explanation (including the differentials) - but it's the same un-proof. I guess I wrote single valued for fn(x) somewhere before when it should have been fn<sup>-1</sup>(r) - that would account for any confusion you are getting.
:::::::::(whether r=fn(angle) is multivalued (or not) (ie non-convex surfaces) doesn't affect this proof)[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 20:06, 25 February 2008 (UTC)
::::::::::I think x=0 is allowed, but only instantaneously, and you still couldn't have a change of sign (it would have to be a point of inflexion), so your basic conclusion is correct. Dealing with the 3D case by taking the intersection with a plane is a good idea (although, pedantically, the intersection must be a union of closed rings, but that doesn't affect anything). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 20:13, 25 February 2008 (UTC)
:::::::::::Yes, forgot about an inflexion -good point.[[Special:Contributions/83.100.158.211|83.100.158.211]] ([[User talk:83.100.158.211|talk]]) 20:18, 25 February 2008 (UTC)

== Geometry Help: Mysterious Equation ==

So, I'm here studying for a test and one of the formula's in my notes makes no sense to me. I have completly forgotten how to use it and I don't have any examples. The formula is V=BiT or V=BLT. Either or.. does anyone know what the formula stands for.. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/80.148.24.98|80.148.24.98]] ([[User talk:80.148.24.98|talk]]) 21:26, 25 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:I expect we're going to need some context. What's the section it's in about? Is there any other mention of V, B, i/L and T near it? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 21:39, 25 February 2008 (UTC)

No, nothing at all.. We're on a unit about volume and surface area of rectangular prisms and pyramids..... <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/80.148.24.98|80.148.24.98]] ([[User talk:80.148.24.98|talk]]) 21:42, 25 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:I think you'll have to ask your teacher. I don't see any way we can identify one formula in isolation. The volume of a prism is Volume=Base area * Length, that gives you V=BL, but I don't know what the T could be. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:38, 25 February 2008 (UTC)

::The volume of a pyramid is base*height/3, so maybe that extra factor has something to do with the third letter. On the other hand, the volume of a prism is also length*width*height. So yeah, your teacher's probably the only person who knows what she meant. Or one of your classmates. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 22:42, 25 February 2008 (UTC)

:::T = tallness? —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 23:48, 25 February 2008 (UTC)

::::What I meant here, I guess, was "Volume = Breadth &times; Length &times; Tallness", though "Tallness" is a strange word to use for that. Maybe [[BLT sandwich|BLT]]? ;-) —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 23:53, 25 February 2008 (UTC)

: Perhaps it's just bad handwriting? V=bh (volume = base area x height). [[User:Imagine Reason|Imagine Reason]] ([[User talk:Imagine Reason|talk]]) 23:49, 25 February 2008 (UTC)

= February 26 =

==Measure Theory==
Let C be a semi-algebra on a given set <math>\Omega</math> and <math>\mu:C \rightarrow [0,\infty]</math> such that <math>\mu(\empty)=0</math> and <math>\mu</math> is sigma additive.<br />

We then define <math>\mu^*(A)=\inf\sum_{k=1}^{\infty}\mu(D_k)</math> where <math>A\subset(\cup D_k)</math> and each <math>D_k \in C</math>.<br />

I am trying to prove that <math>\mu^*(E)=\mu^*(B\cap E)+\mu^*(B^c \cap E)\,\forall B\in C</math> and <math>\forall E\in P(\Omega)</math>.<br />

One inequality is easy to show because E is a subset of <math>(B\cap E)\cup (B^C \cap E)</math> and <math>\mu^*</math> is monotonic.<br />

But any hints as to how to show <math>\mu^*(E)\geq \mu^*(B\cap E)+\mu^*(B^c \cap E)</math> would be appreciated.
:<math>\sum \mu(D_k) = \sum(\mu(D_k \cap B)+\mu(D_k \cap B^c)) = \sum\mu(D_k \cap B)+\sum\mu(D_k \cap B^c)</math>.
:Furthermore, <math>(\cup D_k)\cap X = \cup(D_k \cap X)</math>. Does this help? -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 00:41, 26 February 2008 (UTC)

== Grammars ==

Given {a, ba, bba, bbba} b<sup>n</sup>a|nEN<br />
I calculated that S→aS → bSa → bbSa → bbbSa would be the grammar.<br />
I wanted to make sure that I am on the correct path. <font color="purple">[[User:NanohaA'sYuri|Nan<font color="red">oha<font color = "blue">A's<font color="green">Yu<font color = "yellow">ri]]</font></font></font></font></font><sup>[[User_talk:NanohaA'sYuri|Talk]], [[User:Vivio Testarossa|My master]]</sup> 00:04, 26 February 2008 (UTC)

:I don't think that's right, because if "a" is in the set and "S→aS" is a rule, then you can get "aa", which isn't in the set. Am I misunderstanding the question? —[[User:Keenan Pepper|Keenan Pepper]] 05:06, 26 February 2008 (UTC)

:One way of reading the rule "S→aS" is the following:
::If σ ∈ L(S), then also aσ ∈ L(S) – where L(S) here denotes the language produced using S as the start symbol.
:So if bba ∈ L(S), as desired, then you would also get abba ∈ L(S), which would be wrong.
:There is a [[context-free grammar]] for the language {a,&nbsp;ba,&nbsp;bba,&nbsp;bbba,&nbsp;...} = {b<sup>n</sup>a|n∈'''N'''} with one nonterminal symbol and two production rules. I think the assignment calls for a context-free grammar. In a context-free grammar all rules have a single nonterminal symbol on the left-hand side. What you wrote, "S→aS → bSa → bbSa → bbbSa", does not look like (production rules of) a grammar. Production rules contain only one arrow. It looks more like the begin of a derivation. However, no context-free grammar could have a derivation step of the form {{mbox|aS → bSa}}. If at some stage of a derivation the string starts with a terminal symbol, like "a" here, then it will start with "a" in all following stages. &nbsp;--[[User talk:Lambiam|Lambiam]] 06:40, 26 February 2008 (UTC)

== Completing the Square ==

I was just looking at the article on [[completing the square]] and saw in the derivation of the quadratic formula from the equation <math>ax^2+bx+c=0\,\!</math> that <math>\displaystyle{x^2+\frac{b}ax+\left(\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}a\to\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}}</math>. What I am wondering (although it is probably very simple) is where does the <math>\frac{b}ax</math> go from the first equation to the next. Much appreciated, [[User:Zrs 12|Zrs 12]] ([[User talk:Zrs 12|talk]]) 02:00, 26 February 2008 (UTC)

:Try expanding out <math>\left(x+\frac{b}{2a}\right)^2</math>. Remember, when you expand something like <math>(L+R)^2</math>, the result is not <math>L^2+R^2</math>, but <math>L^2+2LR+R^2</math> (the [[FOIL rule]]). —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 02:34, 26 February 2008 (UTC)

::Thanks Bkell, I understand it a little better now. <s>Expanding it gives <math>x^2+\frac{b}ax+\frac{b}a</math>, right? Where does <math>\frac{b}ax</math> go on the right side? Many thanks, [[User:Zrs 12|Zrs 12]] ([[User talk:Zrs 12|talk]]) 03:04, 26 February 2008 (UTC)</s> I got it. When expanded <math>\left(x+\frac{b}{2a}\right)^2=x^2+\frac{b}ax+\frac{b^2}{4a^2}</math>. That was easy enough. I just had to think and write it out. Thanks! [[User:Zrs 12|Zrs 12]] ([[User talk:Zrs 12|talk]]) 03:25, 26 February 2008 (UTC)

:::Glad to help. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 04:23, 26 February 2008 (UTC)

== Geometry Help ==

In a triangular prism with the width 3, and each diagonal side 3 and a length of 7, how would you find the surface area and volume?

--[[User:Devol4|Devol4]] ([[User talk:Devol4|talk]]) 07:41, 26 February 2008 (UTC)

:[[Triangular prism]] helps with the volume. [[User:Zain Ebrahim111|Zain Ebrahim]] ([[User talk:Zain Ebrahim111|talk]]) 11:09, 26 February 2008 (UTC)

::Well, if I understand the question correctly, your prism can be interpreted geometrically as two equilateral triangles joined by rectangles. To find the volume, just multiply cross-sectional area (the triangle) by the length, and to find surface area, just find the area of all components and sum them. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 11:10, 26 February 2008 (UTC)

== egytian method ==

divide 16 by 9 <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/41.205.189.9|41.205.189.9]] ([[User talk:41.205.189.9|talk]]) 15:27, 26 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
:16/9. =) &ndash;'''''[[User:King Bee|King Bee]]'''''&nbsp;<sup>([[User talk:King Bee|&tau;]] • [[Special:Contributions/King Bee|&gamma;]])</sup> 15:35, 26 February 2008 (UTC)

:I assume the OP means the "Egy'''p'''tian method". Here's one [http://mathforum.org/library/drmath/view/57574.html explanation of the process]. &mdash; [[User talk :Lomn|Lomn]] 15:49, 26 February 2008 (UTC)

::We do have an article on this, check out [[Egyptian fraction]]. After having done the calculations, you can check your work using [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html these calculators.] --[[User:NorwegianBlue|NorwegianBlue]]<sup>[[User_talk:NorwegianBlue|&nbsp;<u>talk</u>]]</sup> 18:37, 26 February 2008 (UTC)

==maths trick i just found out==
# think of a number and put it in your calculator [make sure you remember it]
# add 5
# multiply your answer by 6
# divide your answer by 2
# subtract your answer by 15
# divide your answer by the number you first thought of
* answer is always = 3
* check source at [doxadeocollege.co.za] <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/196.207.47.60|196.207.47.60]] ([[User talk:196.207.47.60|talk]]) 16:15, 26 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:That is true. It's not really a trick so much as obfuscation - simply writing it out on a piece of paper and you can see the following steps:
*<math>S_2 : x \rightarrow x + 5</math>
*<math>S_3 : x + 5 \rightarrow 6(x + 5)</math>
*<math>S_4 : 6(x + 5) \rightarrow \frac{6(x+5)}{2} = 3(x + 5)</math>
*<math>S_5 : 3(x + 5) \rightarrow 3(x + 5) - 15 = 3x + 15 - 15 = 3x</math>
*<math>S_6 : 3x \rightarrow \frac{3x}{x} = 3</math>
:-''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 16:30, 26 February 2008 (UTC)

::And what if the number I choose is 0? &ndash;'''''[[User:King Bee|King Bee]]'''''&nbsp;<sup>([[User talk:King Bee|&tau;]] • [[Special:Contributions/King Bee|&gamma;]])</sup> 16:36, 26 February 2008 (UTC)
:::Define 0/0:=3. That's afterall one reason 0/0 is indeterminant, because processes that result in 0/0 could in fact by any real number, or infinity (by appropiate choice of the process). [[Special:Contributions/130.127.186.122|130.127.186.122]] ([[User talk:130.127.186.122|talk]]) 17:15, 26 February 2008 (UTC)
::::How is "obfuscation" not a more specific word for "trick"? In fact, most actual magic tricks use nothing but clever obfuscation. It's so hard to make the hand quicker than the eye, but so easy to find gullible spectators. Take the classic "cup and ball" routine, for instance. You keep the mark's eyes on the cup and ball, and get caught clumsily trying to remove the ball, so that his attention will be on you when your friend picks his pocket. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 17:32, 26 February 2008 (UTC)
:::::First off, it sounds better, and to me, a mathematical trick is some way of simplifying a result or theory. For instance, an easy way to calculate the multiplicity of a pole is to find the multiplicity of the zero at that point of the function's reciprocal - that I would consider a mathematical trick. -''[[User:Mattbuck|mattbuck]]'' <small>([[User talk:Mattbuck|Talk]])</small> 18:38, 26 February 2008 (UTC)

== Trig equation ==

(Note: this isn't homework, but I'd appreciate a hint rather than being told the answer) I'm having trouble with solving <math>\!\sin 2x + \sin^{2} x = 1</math>. We've not covered this in class, so the only things I can see (<math>\!\sin 2x = 2\sin x\cos x = 2\sin x\sin (x + \frac{\pi}{2})</math> or <math>\!\sin 2x - \cos^{2} + 1</math> or some combination thereof) don't seem to be awfully helpful. Any pointers &ndash; is there an identity I need that I don't have? ''[[User:Angus Lepper|Angus Lepper]]<sup>([[User talk:Angus Lepper|T]], [[Special:Contributions/Angus Lepper|C]], [[User:Angus Lepper/Desktop|D]])</sup>'' 20:58, 26 February 2008 (UTC)

:''Combine'' the [[Trigonometry#double-angle identities|double-angle identity]] <math>\sin 2x = 2\sin x\cos x\,\!</math> with the [[Trigonometry#Pythagorean identities|Pythagorean identity]] <math>\sin^2 x + \cos^2 x = 1\,.</math> &nbsp;--[[User talk:Lambiam|Lambiam]] 21:09, 26 February 2008 (UTC)

::To 'idiot proof' the above... (''almost full solution commented out by [[User:Meni Rosenfeld|Meni Rosenfeld]]'') <!-- replace <math>\!\sin^{2} x</math> by <math>1-\!\cos^{2}x</math> and <math>\!\sin 2x</math> by <math>2\!\sin x\cos x</math> and subtract 1 from each side to give <math>2\!\sin x\cos x - \cos^{2} x = 0</math> and then factorise to give <math>\!\cos x (2\sin x - \cos x) = 0</math>.

::Solutions are derived from solving <math>\!\cos x = 0</math> and <math>\!\tan x = 0.5</math>. --> [[User:AirdishStraus|AirdishStraus]] ([[User talk:AirdishStraus|talk]]) 22:28, 26 February 2008 (UTC)
:::Why would you give a full solution when the OP explicitly asked you not to? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 22:41, 26 February 2008 (UTC)

::::Or call them an idiot? [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 23:04, 26 February 2008 (UTC)
:::::Tango is right. I have commented out the solution. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 10:15, 27 February 2008 (UTC)

::::::I agree Tango is right. I got carried away doing the near full solution and had forgotten the 'hint only' request. My apologies. On a seperate note... 'idiot-proofing' something is a mere colloquialism for unambiguous clarification; I never used the word idiot to directly describe anyone and wouldn't ever use it in ANY case. Maybe Black Carrott hasn't come across this term before? [[User:AirdishStraus|AirdishStraus]] ([[User talk:AirdishStraus|talk]]) 11:15, 27 February 2008 (UTC)

== Triangle ==

The angle bisector of alpha w<sub>α</sub>, the median of s<sub>b</sub> and the altitude of c h<sub>c</sub> of a acute-angled triangle ABC cross in one point, if w<sub>α</sub>, the side BC and a circle around the point H<sub>c</sub>, which goes through the corner A, have also a common point together.

How can one proof this statement? --[[Special:Contributions/85.178.22.17|85.178.22.17]] ([[User talk:85.178.22.17|talk]]) 23:18, 26 February 2008 (UTC)
:You need to do it one step at a time
:1. First find the intersection of the "angle bisector of alpha w<sub>α</sub> and the median of s<sub>b</sub>" as a function of the parameters of a generalised triangle eg (0,0) (a,b) (c,d) as the points of the triangle.
:2. Then find the coodrinates of the crossing point of "the median of s<sub>b</sub> and the altitude of c h<sub>c</sub> of a acute-angled triangle ABC"

:Next you need to show that the two equations abtained above give the same point if the third premise is true.. ie that "w<sub>α</sub>, the side BC and a circle around the point H<sub>c</sub>, which goes through the corner A, have also a common point together."
:3. I'm not sure what H<sub>c</sub> is here - but if you do you should be able to generate some sort of parametric equation relating the vertexs of the triangle. {{ Sound's like Hc is the centre of the circle that goes through A and also passes through the intersection of (w<sub>α</sub> and the side BC ) - this makes somewhere Hc along a line then}}

Then you need to show that if for instance the equation obtained in 1 is true at the same time as the equation obtained in 3. then equation 2 is also true.[[Special:Contributions/87.102.93.245|87.102.93.245]] ([[User talk:87.102.93.245|talk]]) 11:43, 27 February 2008 (UTC)

= February 27 =

== The washer method ([[disk integration]]) ==

I need to find the volume of a solid of revolution. The 2-dimensional area is bounded by y = √x, y = 0, and x = 4, and it is spun around the axis x = 6. I know that my outer radius is 6 - x, but since I'm integrating in terms of y (since the axis is parallel to the y-axis), I should write that as R(y) = 6 - y<sup>2</sup>. But I can't figure out how to express the inner radius r(y) as a function, so that I can plug it into the formula V = π∫<sub>a</sub><sup>b</sup>[R<sup>2</sup>(y) - r<sup>2</sup>(y)]dy! (I believe a and b = 0 and √6 respectively.) Can anyone offer any help? Thanks, anon. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/70.19.34.80|70.19.34.80]] ([[User talk:70.19.34.80|talk]]) 01:10, 27 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

::Since the inner wall is a vertical line, the inner radius is constant with respect to y, and in this case is 2. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 03:51, 27 February 2008 (UTC)

:::Also, if you're integrating with respect to y, it should run from 0 to 2. You should draw some diagrams if you haven't already. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 03:54, 27 February 2008 (UTC)

== Friend challenged me ==

IF A box of nutmeg ways 1 1/3 ounces and there is 20 scoops how much is each scoop? <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Inutasha De Fallen|Inutasha De Fallen]] ([[User talk:Inutasha De Fallen|talk]] • [[Special:Contributions/Inutasha De Fallen|contribs]]) 02:51, 27 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

:Are you asking fot the ounces per one scoop where you are told that there are 1.333... ounces per 20 scoops? --[[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 03:05, 27 February 2008 (UTC)
It has to end in as a fraction like 1/20 <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Inutasha De Fallen|Inutasha De Fallen]] ([[User talk:Inutasha De Fallen|talk]] • [[Special:Contributions/Inutasha De Fallen|contribs]]) 03:09, 27 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

:Well you know that 1 1/3 equals 4/3, right? And you know that dividing by 20 is the same as multiplying by 1/20, right? And how to reduce fractions, right? So, what more help do you need? --[[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 03:42, 27 February 2008 (UTC)
it is a pain in the ass... im not really understanding it it would be 0.065.. but thats not it! damn this is upseting <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Inutasha De Fallen|Inutasha De Fallen]] ([[User talk:Inutasha De Fallen|talk]] • [[Special:Contributions/Inutasha De Fallen|contribs]]) 03:45, 27 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

:You're close, but not there yet. Come on, I would expect a 12 year old to know this, so why a 14 year old like you don't? --[[User: Antilived|antilived]]<sup>[[User_talk:Antilived|T]] | [[Special:Contributions/Antilived|C]] | [[User:Antilived/Gallery|G]]</sup> 05:01, 27 February 2008 (UTC)

Shiβe i Fell stupid...i Got it. do simple things ever Kick your ass? <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Inutasha De Fallen|Inutasha De Fallen]] ([[User talk:Inutasha De Fallen|talk]] • [[Special:Contributions/Inutasha De Fallen|contribs]]) 05:27, 27 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

::Nope. I've never made a mistkske in my life. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 06:42, 27 February 2008 (UTC)

Eh? well im gonna have to raise the bullshit flag on that one. >.< <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Inutasha De Fallen|Inutasha De Fallen]] ([[User talk:Inutasha De Fallen|talk]] • [[Special:Contributions/Inutasha De Fallen|contribs]]) 07:05, 27 February 2008 (UTC)</small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

== Chain Rule and Higher Derivative ==

Let
:<math>z=f(x,y)</math>
:<math>x=g(s,t)</math>
:<math>y=h(s,t)</math>
Then
{{NumBlk|:|<math>\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z</math>|1}}
{{NumBlk|:|<math>\frac{\partial^2 z}{\partial t^2}=\frac{\partial}{\partial t}\frac{\partial z}{\partial t}=\frac{\partial}{\partial t}\left[\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z\right]=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)\frac{\partial z}{\partial t}</math>|2}}
Replace (1) into (2)
{{NumBlk|:|<math>\frac{\partial^2 z}{\partial t^2}=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z</math>|3}}
{{NumBlk|:|<math>\frac{\partial^2 z}{\partial t^2}=\left[\frac{\partial^2}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2}{\partial x\partial y}\frac{\partial x\partial y}{\partial t^2}+\frac{\partial^2}{\partial y\partial x}\frac{\partial y\partial x}{\partial t^2}+\frac{\partial^2}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2\right]z</math>|4}}
{{NumBlk|:|<math>\frac{\partial^2 z}{\partial t^2}=\frac{\partial^2 z}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2 z}{\partial x\partial y}\frac{\partial x\partial y}{\partial t^2}+\frac{\partial^2 z}{\partial y\partial x}\frac{\partial y\partial x}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2</math>|5}}
Is ''every'' step above correct? - [[User:Justin545|Justin545]] ([[User talk:Justin545|talk]]) 07:07, 27 February 2008 (UTC)

:No.
::<math>\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}</math>
:is not the same as
::<math>\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}\right)z.</math>
:For example, if ''z'' = ''x'' = ''t'', the former evaluates to {{mbox|= 1·1 = 1}}, and the latter to {{mbox|= 0·''z'' = 0.}} Think of <math>\tfrac{\partial}{\partial x}</math> as an operator, and abbreviate it as ''D''. Also abbreviate <math>\tfrac{\partial x}{\partial t}</math> as ''U''. Then in your first line of equations you replaced {{mbox|''Dz'' × ''U''}} by {{mbox|''DU'' × ''z''}}. &nbsp;--[[User talk:Lambiam|Lambiam]] 09:46, 27 February 2008 (UTC)

== Recursively defined sets ==

I want to generate the first several iterations of this set, but I feel I'm not doing it correctly; my notes are sparse on certain details:



Defintion of <math>C \,</math>.

<math>C \subseteq \Sigma</math> where <math>\Sigma \,</math> is the set of all strings over <math>\lbrace 1, 2, 3 \rbrace \,</math>


Basis step:


<math>\lbrace 1, 2 \rbrace \subseteq C</math>


Recursive step:


<math> w \in C </math> <math>\rightarrow</math> <math>3 \cdot w \in C</math>

<math> w_1, w_2 \in C </math> <math>\rightarrow</math> <math>w_1 : w_2 \in C</math>

<math> w \cdot 1 \in C </math> <math>\rightarrow</math> <math>2 \cdot w \in C</math>


The dot operator and the colon (:) operator represent character concatentation and string concatentation respectively.

It would follow, in my mind that the first few iterations of this set would go:

{1, 2}

{1, 2, 31, 11, 21}

{1, 2, 31, 11, 21, 32, 12, 22}


e.t.c

My main worry is this:

I assume w(1) and w(2) in the second part of the recursive step start at 1, 1 and go up 1, 2 then 2, 1 then 2, 2 and so on? Or do they have to be seperate values each time? [[User:Damien Karras|Damien Karras]] ([[User talk:Damien Karras|talk]]) 08:03, 27 February 2008 (UTC)

:If I understand your last question correctly, the answer is that <math>w_1</math> and <math>w_2</math> don't have to be different. I'll guide you through the first iteration. You know that 1 &isin; C and 2 &isin; C. By rule 1, from 1 &isin; C it follows that 31 &isin; C, and from 2 &isin; C it follows that 32 &isin; C. We now apply rule 2: From 1 &isin; C and 1 &isin; C it follows that 11 &isin; C; from 1 &isin; C and 2 &isin; C it follows that 12 &isin; C; likewise 21 &isin; C and 22 &isin; C. Rule 3 doesn't give us anything new now. So the first iteration will be <math>\{1, 2, 31, 32, 11, 12, 21, 22\} \subseteq C</math>. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 09:51, 27 February 2008 (UTC)

:(after edit conflict) I would interpret your rule #2 of the "recursive step" as being universally quantified over all ''w''<sub>1</sub> and ''w''<sub>2</sub> that are in ''C''. So if 2 is a member of ''C''<sub>''n''</sub> in iteration ''n'', then 22 is a member of the next generation ''C''<sub>''n''+1</sub>. Is that an answer to your question (which I do not fully understand)?
:Because of rule #3, you can start the whole process by taking ''C''<sub>0</sub> = {1}. Then the first few iterations give this:
::''C''<sub>0</sub> = {1}
::''C''<sub>1</sub> = {1, 31, 11, 2}
::''C''<sub>2</sub> = {1, 31, 11, 2, 331, 311, 32, 131, 111, 12, 3131, 3111, 312, 1131, 1111, 112, 21, 231, 211, 22, 23}
:&nbsp;--[[User talk:Lambiam|Lambiam]] 10:10, 27 February 2008 (UTC)

::I think what I meant my overall question to be is this: For each iteration, do I pick the value of w, w1 and w2 just once? So the first iteration would be: w = 1, w1 = 1 and w2 = 1 and then I would move to the next iteration.

::OR do I consider all possible values of w, w1, w2 for each iteration? In which case I would do, for instance, step one for w = 1 and w = 2, step two would be done for w1 = 1, w1 = 2, w2 = 1, w2 = 2 e.t.c

::Furthermore, are the new elements of the set generated on each step? Because if so that would mean that w1 and w2 take on even more values before the end of the total recursive step (unless of course the former point is true). [[User:Damien Karras|Damien Karras]] ([[User talk:Damien Karras|talk]]) 10:41, 27 February 2008 (UTC)

:::I am still confused by your questions, but I have a better idea what it is you are asking. To find <math>C_{n+1}</math>, you apply every possible rule to the items in <math>C_{n}</math> and only them. So if you start with <math>C_0 = \{1,2\}</math>, applying every possible rule to <math>C_0</math> gives you <math>C_1 = \{1, 2, 31, 32, 11, 12, 21, 22\}</math>. You then apply every possible rule to <math>C_1</math> to get <math>C_2</math>, and so on. In each step, you only apply rules to elements you have had in the previous step, ''not'' to those you have generated recently. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 11:44, 27 February 2008 (UTC)

== modular algorithm ==

how can i solve this problem?

[ x1= a (mod 100) , a= 20 (mod 37) ]


[ x2= b (mod 100) , b= 15 (mod 37) ]


[ x3= c (mod 100) , c= 18 (mod 37) ]

must be ;
x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

i need find b and c..
thank you best regards..
Altan B. <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/85.98.230.220|85.98.230.220]] ([[User talk:85.98.230.220|talk]]) 10:20, 27 February 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

== shapes and projections ==

I was wondering about something and I need abig help to correct my understanding to this issue.I am going to put my questions in two parts.part1:assume that we have ashape in two dimensions,like asphere,obviously we can determine the projections of that sphere on (x-y,y-z,z-x)planes, which they are going to be circles no matter how we rotate the sphere, the projections will be always circles.If we apply now the same thing on acone then we would have adifferent projections depend on how we rotate or fix the cone in two dimensions,we can get for example,acircle in(x-y)plane,atriangle in(y-z)plane and another triangle in(x-z)plane.we can determine the projections depend on the shape and its position or you can say the shape and the position vector between the origin and achosen point belong to that shape.Now my question is,can we determine the shape by looking at its projections?if yes,then what is the shape in two dimensions that has the projections,(circle,square and apoint)?p.s:you can improvise another projections.some times we can describe different shapes in 2 dimentions by looking at the projections,for example,assume we have 3identical circles as aprojections,one may say that the shape is atwo dimnensional sphere,or the shape could be 3 identical circles perpendicular to each other and sharing the same center.
Part2:if we are able to say that the sphere in three dimensions has aprojection in two dimensions which is asphere too then for the cone in three dimensions what its projection in two dimensions going to be?IF THE ANSWER IS ACONE TOO NO MATTER HOW WE ROTATE OR CHANGE ITS POSITION IN 3 DIMENTIONS THEN IT SHOULD BE ASPHERE, because the sphere is the only shape that maintains its projections in all dimensions>2

Further discussion:
Assume we have ashape in 3dimensions where its projections in 2dimensions as follows,
(x,y,z)=s1,(x,y,w)=s2,(x,z,w)=s3&(y,z,w)=s4.The projections in one dimension would be,
[s1:(x,y)=a1,(x,z)=a2,(y,z)=a3],[s2:(x,y)=a1,(x,w)=a4,(y,w)=a5],[s3:(x,z)=a2,(x,w)=a4,(z,w)=a6],[s4:(y,z)=a3,(y,w)=a5,(z,w)=a6].We have 6 one dimensional projections .Now assume we have acone in 3 dimentions,we can finde some conditions to make that cone describable,for example ,if the con`s peak is located at the origin and the position vector between the peak and the center of the cone`s makes an angle=pi\4 with each of(x,y,z,w),then a1=a2=a3=a4=a5=a6 and each one of the six projections seems to be having ashape of sector of acircle.on the other hand if,a1=a2=a3=a4=a5=a6 ,shouldnot we get asphere in 3dimensions?my point here is,can we say that all different shapes in 3 dimensions with no preconditions have ashape of sphere???[[User:Husseinshimaljasimdini|Husseinshimaljasimdini]] ([[User talk:Husseinshimaljasimdini|talk]]) 12:05, 27 February 2008 (UTC)

Latest revision as of 08:36, 5 January 2025

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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)[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

[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)[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 OEIS: A000041
  14. Distinct partition number OEIS: A000009

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

[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)[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]
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)[reply]
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)[reply]
The red point is at , . catslash (talk) 16:18, 3 January 2025 (UTC)[reply]

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)[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]
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)[reply]
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)[reply]


January 5

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Reference request:coherence condition (adjoint functor)

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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)[reply]

Concatenation of first 10 digits and last 10 digits of a number

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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(2136279841-1) = 88169432759486871551, however:

  1. Can a(2^n) take all 20-digit values which are multiples of 1024?
  2. Can a(3^n) take all 20-digit values which are odd?
  3. Can a(n^2) take all 20-digit values which end with 0, 1, 4, 5, 6, 9?
  4. Can a(n^3) take all 20-digit values?
  5. Can a(prime number) take all 20-digit values which end with 1, 3, 7, 9?
  6. Can a(lucky number) take all 20-digit values which are odd?
  7. Can a(Fibonacci number) take all 20-digit values?
  8. Can a(partition number) take all 20-digit values?
  9. What is a(9^9^9^9)?
  10. What is a(9^^9), where ^^ is tetration?
  11. What is a(9^^^9), where ^^^ is pentation?
  12. What is a(Graham's number)?
  13. What is a(TREE(3))?
  14. If we know a(x), assume that x has at least 20 digits, can we also know a(2*x), a(3*x), etc.?
  15. If we know a(x), assume that x has at least 20 digits, can we also know a(x^2), a(x^3), etc.?
  16. 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)?
  17. 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)?
  18. 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)?

220.132.216.52 (talk) 08:33, 5 January 2025 (UTC)[reply]

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