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June 18

mathematical characters and algorithms

Hi, i don't know if it's the good place to do that, but some articles in computer sciences have (or had) some issues with people writing directly characters for "union" (\cup) , "intersection" (\cap), "is element of" (\in) and so on.

my questions : 1) how is laTex supported in wikipedia ? 2) is it supported in the "algorithm code text zone" like in this article DFA_minimization ?

in the previous mentionned articles i needed to print it, and i didn't saw any "is in set" character, so i changed them to "in", but i don't know if it's the standard way to do that here. and Latex would maybe be better.

thx 85.1.144.185 (talk) 10:02, 18 June 2011 (UTC)

Hi again, i'm the author of the previous untitled question.

Just adding this to make some title.

thanks — Preceding unsigned comment added by 84.227.11.170 (talk) 10:43, 18 June 2011 (UTC)

AFAIK any character that can be added in the 'Insert' drop down menu in the WP editing interface is fair game to use in an article as far as display is concerned, this includes a Math and Logic section so ∈, ∪, ∩ are acceptable. Depending on the audience, it might be a good idea to spell it out anyway so people can understand the article (see WP:MOSMATH). I would not recommend mixing pseudocode and LaTeX in Wikipedia, the fonts are very different and the rendering would not be pleasing to the eye:

Only ${\text{this }}$ word is LaTeX.

--RDBury (talk) 11:06, 18 June 2011 (UTC)

It's not just the rendering -- the <math>-mode parser does not recognize the pseudocode symbols, so you will get parse errors if you mix them in with LaTeX stuff. Looie496 (talk) 19:15, 18 June 2011 (UTC)

the problem was that the symbols were just *not* printed on screen, on my computer at least, in the zone for the pseudo-code algorithm. thx for your answers. 84.226.189.222 (talk) 17:39, 20 June 2011 (UTC)

Math editor

Does wikipedia have a mathematical editor? — Preceding unsigned comment added by White Silver (talk • contribs) 18:55, 18 June 2011 (UTC)

No. All the mathematical formulas you see on wikipedia are achieved using black magic. --72.179.51.84 (talk) 19:32, 18 June 2011 (UTC)

Basically you can enter LaTeX syntax to get the symbols. For example <math> x^2 = 2 \iff x = \pm \sqrt{2}</math> gives

$x^{2}=2\iff x=\pm {\sqrt {2}}$

Take a look at Help:Displaying a formula for more information. — Fly by Night (talk) 20:28, 18 June 2011 (UTC)

Fly by Night, do you know why the usual $ $ doesn't work here? Just wondering in case you knew. Well it's a nuisance to retype the <math></math> many times.

BTW, you can also type stuff in TeX and I prefer this over Fly by Night's answer. For example you could type ''x''₁ to get x₁. This looks nice because it's inline unlike the LaTeX where the size increases considerably. — Preceding unsigned comment added by 180.216.50.57 (talk) 03:15, 19 June 2011 (UTC)

Wikipedia documents are not written in either TeX or LaTeX, they are written in a special language of their own, resembling HTML but with some XML features. The <math> tag is handled by sending its contents to a LaTeX parser, but other parts of the page do not use LaTeX in any way. The handling of the <sub> also does not use TeX or LaTeX. Looie496 (talk) 03:54, 19 June 2011 (UTC)

To 180.216.50.57, do you have a clue what you're talking about? My command was displayed in LaTeX output (as an image file). You'll find that many symbols do not exist in HTML that exist in LaTeX, and even if they did many browsers don't have the correct character sets installed to display the symbols. Even if a browser knows how to display an HTML symbol then it will appear differently in different browsers because of font preferences. Take a look at User:KSmrq/Chars. Scroll down and notice all the unrecognised symbols. As Looie said, the math and /math tags send the LaTeX syntax off to be converted into an image file; so any symbol LaTeX knows can be displayed. Did you bother to read the link I gave? The math and /math tags are used to tell Wikipedia that you want to wrote in LaTeX syntax. The $ symbols only work in LaTeX; Wikipedia is not LaTeX. — Fly by Night (talk) 19:24, 19 June 2011 (UTC)

There is obviously some confusion going on here. FbN: Your LaTeX is not inline and it is bigger than standard text. 180: Your formatting is not TeX but is html - it is inline but it is very limited as to what can be done with it. Both have their place in the WP scheme of things. -- SGBailey (talk) 09:01, 21 June 2011 (UTC)

If I type $x^2 = 2 \iff x = \pm \sqrt{2}$ into LaTeX then I get the above output. It's not HTML, if it were then x^2 = 2 \iff x = \pm \sqrt{2} would appear in symbols. Things like & alpha; & minus; & beta; (without the spaces after the &) is HTML. If you type <math> then it tells Wikipedia that you are about to enter LaTeX code. When you enter </math> it tells Wikipedia you've finished with LaTeX code. Then Wikipedia sends the LaTeX code off to be compiled into LaTeX output which is then sent back as an image file which is displayed in place of the <math>…</math> string. Maybe some of the technical words aren;t right, but the principal is. — Fly by Night (talk) 17:21, 21 June 2011 (UTC)

Is it possible that the original poster meant a person who supervises the math articles? Traditionally the word "editor" means, after all, a person charged with certain duties. Michael Hardy (talk) 22:46, 20 June 2011 (UTC)

Or possibly a graphical formula editor allowing a point and click style of entering a mathematical formula. Potentially one could be written, perhaps by modifying an existing editor to produce latex output. I'm not aware of any such program and mw:Category:Math extensions does not show anything suitable.--Salix (talk): 13:03, 21 June 2011 (UTC)

Very true. — Fly by Night (talk) 17:26, 21 June 2011 (UTC)

Sort of off topic but, I can heartily recomend User:Nageh/mathJax which uses MathJax instead of Texvc for rendering. This makes display maths formula look a lot nicer an removes the size problems. I've been using it for a few months now with few problems.--Salix (talk): 21:43, 21 June 2011 (UTC)

June 19

Surreal numbers

Where can I find a proof of the statement on the page surreal numbers that every ordered field is a subfield of the Field of surreals? 76.67.73.4 (talk) 18:28, 19 June 2011 (UTC)

If you follow the references listed in the article you will see that "In the original formulation using von Neumann–Bernays–Gödel set theory, the surreals form a proper class, rather than a set, so the term field is not precisely correct; where this distinction is important, some authors use Field or FIELD to refer to a proper class that has the arithmetic properties of a field. One can obtain a true field by limiting the construction to a Grothendieck universe, yielding a set with the cardinality of some strongly inaccessible cardinal, or by using a form of set theory in which constructions by transfinite recursion stop at some countable ordinal such as epsilon nought". I'd start following some of those links to begin with. — Fly by Night (talk) 18:37, 19 June 2011 (UTC)

I noticed that footnote; I even linked to it in my original question. None of the links seem very relevant, however. 76.67.73.4 (talk) 20:40, 19 June 2011 (UTC)

Ah, so you did. My apologies. I assumed that a blue field would link to the field article; which I didn't need to read. Sorry about that. — Fly by Night (talk) 20:42, 19 June 2011 (UTC)

Please discuss any question of conflict of interest without disclosing personal information. User:Fred Bauder Talk 02:04, 20 June 2011 (UTC)

Can't you just build an embedding inductively? Start by mapping 1 to 1, extend the map to the arithmetic closure, then the real closure. At every subsequent step, grab an element which is transcendental over the domain, map it to an appropriate element of the surreals, take the arithmetic closure and then the real closure. At limits, take unions. Conway's Simplicity Theorem justifies mapping transcendentals and taking real closures.--Antendren (talk) 02:32, 20 June 2011 (UTC)

June 20

Elementary proof

Hi. I am sure I remember reading or hearing somewhere that an "elementary proof" is one that does not use Cauchy's integral theorem. But I can't find any evidence on the net that this is the case. Can anyone verify this? Robinh (talk) 03:09, 20 June 2011 (UTC)

Are you trying to verify whether this has been said somewhere, or verify if the claim is true? Because I'm pretty sure that in general what constitutes an elementary proof is going to depend entirely on context. Rckrone (talk) 04:48, 20 June 2011 (UTC)

thanks for this. I guess I'm asking whether anyone can find a reputable source that defines "elementary proof" as one that doesn't use Cauchy. best wishes, Robinh (talk) 08:09, 20 June 2011 (UTC)

I think you're most likely to find this in accounts of proofs of the prime number theorem. As Robinh pointed out, it's a pretty context-dependent thing. Michael Hardy (talk) 17:40, 20 June 2011 (UTC)

thanks! Quote: "Newman's proof is arguably the simplest known proof of the theorem, although it is non-elementary in the sense that it uses Cauchy's integral theorem from complex analysis".

Resolved

. Robinh (talk) 20:23, 20 June 2011 (UTC)

help me railway in dire straits of runnaway train

HI need a bit of help with a right angle triangle a b c ok looking to find the length of a-b a-c equals 12 metres b-c angle 7 degrees what is the length from a-b this is not homework I am building a model railway and the engines can pull up to 7 degrees the trains come out of a curve (b) and proceed to climb at 7 degrees for 12 meters and then go into a tunnel (c) and into another room so have got to have the right distance from a -b email me with the answer <email redacted> thanks will send the person who answers my question a photo of the construction Ron — Preceding unsigned comment added by 124.184.8.94 (talk) 07:32, 20 June 2011 (UTC)

If 12 metres is the length of the sloping track then the horizontal distance will be 12 times the cosine of 7 degrees. If 12 metres is the horizontal distance then the length of the sloping track will be 12 divided by the cosine of 7 degrees. In either case, the cosine of 7 degrees is very close to 1 (just over 0.9925), so you get 12 metres +/- about 9 cm. That is a very long slope for a model railway - at a constant slope of 7 degrees the track rises nearly 1.5 metres in 12 metres distance. Gandalf61 (talk) 09:18, 20 June 2011 (UTC)

Proof of Aleph 0 Smallest Transfinite Cardinal

How do you prove aleph 0 is the smallest transfinite cardinal? Thanks.voidnature 11:41, 20 June 2011 (UTC)

For an arbitrary infinite set $X$, you need to prove the existence of an injection $f:\mathbb {N} \to X$. This can only be done assuming the Axiom of choice or a weaker version.

What you do is simply let $f(0)\in X$, then, having defined $f(0),\cdots ,f(n-1)$, you define $f(n)\in X-\{f(0),\cdots ,f(n-1)\}$. It's not too difficult to turn this into a fully formal proof. --COVIZAPIBETEFOKY (talk) 11:54, 20 June 2011 (UTC)

That was very helpful, but can you please elaborate it more, and which part of it depends of the axiom of choice? Thanks.voidnature 10:58, 22 June 2011 (UTC)

When you define $f(n)$ you need to pick an element in $\{f(0),\cdots ,f(n-1)\}$. To make this proof formal we need a choice function on the set of all non empty subsets of X, selecting an element out of each such subset, and this needs the axiom of choice. Money is tight (talk) 07:17, 23 June 2011 (UTC)

Convert stddev from population to sample

I have a standard deviation calculated on the population of size n. I don't have the original data. The stddev was calculated using n in the denominator. Statistically, I can multiply by sqrt(n/(n-1)) to convert this to sample stddev, correct? -- kainaw™ 16:51, 20 June 2011 (UTC)

Yes, but note that for small n where this factor may be important, you would have a significant standard deviation of the population standard deviation anyway :) . Count Iblis (talk) 17:30, 20 June 2011 (UTC)

Was "n" the population size or the sample size? Dbfirs 18:56, 20 June 2011 (UTC)

Sample size. The population size is unknown. -- kainaw™ 19:20, 20 June 2011 (UTC)

Thanks, I suppose that was obvious, but I was slightly confused. I also get confused between the terms "sample standard deviation" and "standard deviation of the sample", but our article explains it (with careful reading). Dbfirs 06:44, 21 June 2011 (UTC)

June 21

Limit of an increasing sequence

Is ω₁ the smallest limit ordinal which cannot be expressed as a limit of an increasing sequence of ordinals? (sequence = indexed by natural numbers) --COVIZAPIBETEFOKY (talk) 00:58, 21 June 2011 (UTC)

Yes. Any countable ordinal $\alpha$ can easily be expressed as such a limit: fix $a_{n}$ a numbering of the ordinals less than $\alpha$. Define $b_{0}=0$, $b_{n+1}=\max(a_{n},b_{n}+1)$. Then it's easy to check that $\alpha$ is the limit of the $b_{n}$. You might also be interested in reading about regular cardinals and cofinality.--Antendren (talk) 04:11, 21 June 2011 (UTC)

Nice Antendren! I had a lot of trouble doing a related question in topology but your construction makes it all so simple. Money is tight (talk) 08:08, 21 June 2011 (UTC)

Thanks for the nice explanation and interesting links. I ran across cofinality before, but it didn't make sense to me then; now it does.

Resolved

--COVIZAPIBETEFOKY (talk) 12:47, 21 June 2011 (UTC)

Algebra Question

I've mentioned this a long time ago, but I need some clarification. Let ƒ : (R^k,0) → (R,0) be a function germ. Let O_k denote the local ring of all function germs (R^k,0) → (R,0). Let J_ƒ denote the Jacobian ideal generated by the partial derivatives of ƒ, i.e. J_ƒ = (∂ƒ/∂x₁,…,∂ƒ/∂x_k). The local algebra of ƒ is defined to be the quotient O_k / J_ƒ. The quotient of a ring by an idea gives a vector space, in this case a real vector space. It's finite dimensional if, and only if, ƒ has an algebraically isolated critical point at 0 in R^k.

I don't understand the use of the word algebra. Clearly it's an R-vector space, so let's forget addition.

• What's the binary operation on O_k / J_ƒ, is it just multiplication modulo J_ƒ?
• Usually we talk of an algebra over a field, or over a commutative ring. What's the field and/or commutative ring here?
• The local algebra O_k / J_ƒ is a ring itself, so why call it an algebra?
• If it is an algebra, then should I use a Gothic letter, or is that just for Lie algebras?

Sorry for all the questions, but I've got rings, ideals, vector spaces and algebras attached to the same object, and I'm getting confused. — Fly by Night (talk) 18:38, 21 June 2011 (UTC)

Multiplication in O_k/J_ƒ is just the usual multiplication in O_k modulo J_ƒ as you said. O_k/J_ƒ is an R-algebra (as is O_k) and the R-action is just boring old scalar multiplication. Any algebra is a ring so calling O_k/J_ƒ an R-algebra implies more structure: basically that the R-action on O_k/J_ƒ plays nicely with the ring operations. Rckrone (talk) 01:33, 22 June 2011 (UTC)

Oh, okay, I see. What about the Gothic letters? Is that just for Lie algebras? — Fly by Night (talk) 02:34, 22 June 2011 (UTC)

I haven't noticed a general convention like that in the algebra I've seen, but I can't speak for all areas. I'm not too familiar with germs, so I don't know what the notation should look like there. I think the gothic letters are used in particular for Lie algebras to distinguish them from the associated Lie group (which is a different object). Rckrone (talk) 03:04, 22 June 2011 (UTC)

That sounds about right. The use normal face letters with a subscript letter for the function germ. For some reason the use a curly \mathscr{O} for the local ring. It's a real cross-over of areas. I don't know anything beyond undergraduate abstract algebra, but I need to use it to analyse functions with applications to geometrical problems. It's a real mixed bag. Thanks again. — Fly by Night (talk) 20:38, 22 June 2011 (UTC)

Twenty heads in a row problem

A few days ago this problem was discussed, see here It turns out that the approach tried by many involving estimating the average number of rows and use the Poisson distribution does work and you do get an accurate answer that way. This works as follows.

Consider rows of a length of exactly r heads. The probability of a given coin throw being at exactly the start of such a row is 2^[-(r+2)], note that the coin throw right before and right after the row have to be tails. The probability that a coin throw is on any arbitrary position of such a row is thus r 2^[-(r+2)]. If we throw the coin N times, then the expected number of coin throws that are on such a row is N r 2^[-(r+2)]. Since there are always r coin throws that are on the same row, there are N 2^[-(r+2)] different rows. Here we have neglected all the issues regarding different rows interfering with each other. The probability of a row appearing someone is quite small, so we can ignore that to first approximation.

Then summing N 2^[-(r+2)] from r = 20 to infinity yields N 2^(-21) for the expected number of rows of heads of length of at least 20. Then the Poisson distribution implies that Exp[-N 2^(-21)] is the probability that there are no such rows. Count Iblis (talk) 19:07, 21 June 2011 (UTC)

So the probability of getting twenty heads in a row if you flip a coin one million times is 1−e^−1062−21 = 0.379256396. Thanks a lot Count Iblis, you cracked the nut! Bo Jacoby (talk) 22:12, 21 June 2011 (UTC).

A slightly more precise version of Count Iblis' calculation is this. A row of 20 heads, beginning at the very first flip, requires 20 heads and then 1 tail, and the average number of such rows is 2⁻²¹. When beginning at one of the 999979 flips from flip number 2 through flip number 999980, it requires 1 tail and then 20 heads and then 1 tail, and the average is 2⁻²². When beginning at flip number 999981 it requires 1 tail and 20 heads, and the average is 2⁻²¹. The average number of rows of exactly 20 heads, if you flip a coin one million times, is 2⁻²¹ + 999979·2⁻²² + 2⁻²¹ = 999983·2⁻²². Similarly the average number of rows of exactly 21 heads, if you flip a coin one million times, is 999982·2⁻²³, and the average number of rows of at least 20 heads, if you flip a coin one million times, is Σ_i=0⁹⁹⁹⁹⁸² (999983−i)·2⁻²²⁻ⁱ ≃ 999982·2⁻²¹, and the probability of getting twenty heads in a row if you flip a coin one million times is 1−e^{−999982·2−21} = 0.379251068. Bo Jacoby (talk) 21:07, 22 June 2011 (UTC).

June 22

Commutative Diagrams in LaTeX

I need to draw a commutivity diagram with two lines of three spaces. Something like this:

A → B → C

↓        ↓        ↓

E → F → G

But I need the first two horizontal arrows( A to B and E to F) to be $\hookrightarrow$ with letter above them. I need the second horizontal arrows (B to C and F to G) to be $\twoheadrightarrow$ with letters above them. The vertices arrows (A to E, B to F and C to G) can just be normal arrows but with letter to the side of them. I can do it on one line using \stackrel $A{\stackrel {i}{\hookrightarrow }}B{\stackrel {\pi }{\twoheadrightarrow }}C$, but I've no idea how to do it a diagram. I have used the package CD, but that only allows normal arrows with, for example, @>>i>, to get ${\stackrel {i}{\longrightarrow }}$, but that's not good enought.

Any ideas boys and girls? — Fly by Night (talk) 00:08, 22 June 2011 (UTC)

I believe there are several available packages. You could try the "diagrams" package, documented here. I have never used it and couldn't tell you whether it's best for your needs. I think there's also support in TikZ, which is a package you should learn anyway, so there might be an efficiency in going that way. I don't do commutative diagrams much myself so I have no experience with this functionality. --Trovatore (talk) 00:38, 22 June 2011 (UTC)

It's been a while since I've had to make commutative diagrams, but I remember the package "pb-diagram" being good for them.--Antendren (talk) 00:40, 22 June 2011 (UTC)

You can do it using the "array" environment:

${\begin{array}{ccccc}A&{\stackrel {i}{\hookrightarrow }}&B&{\stackrel {j}{\hookrightarrow }}&C\\\downarrow ^{k}&&\downarrow ^{l}&&\downarrow ^{m}\\D&{\stackrel {n}{\twoheadrightarrow }}&E&{\stackrel {p}{\twoheadrightarrow }}&F\end{array}}$

Looie496 (talk) 04:01, 22 June 2011 (UTC)

Cool, I tried that, but couldn't get it to work. I've edited your example, and this is exactly what I wanted.

${\begin{array}{lllll}A&{\stackrel {i_{1}}{\hookrightarrow }}&B&{\stackrel {\pi _{1}}{\twoheadrightarrow }}&C\\\downarrow ^{\sigma }&&\downarrow ^{\tau }&&\downarrow ^{\nu }\\D&{\stackrel {i_{2}}{\hookrightarrow }}&E&{\stackrel {\pi _{2}}{\twoheadrightarrow }}&F\end{array}}$

Click to Enlarge

Is there a way to lengthen all of the arrows, and to get the vertical labels in the middle? — Fly by Night (talk) 20:44, 22 June 2011 (UTC)

• Thanks to you all for your help. I spend a little while learning how to use the tikzpicture package that Trovatore suggested. It wasn't easy at first. If your codes not quite right it crashes LaTeX while keeping the file in use in the CPU. Which meant I had to restart my laptop about five times until I got the hang of it. But it's well worth it once you get the hang of it, and it's quite intuitive too. I've included a screen shot of the final product. Thanks again everyone. — Fly by Night (talk) 23:47, 22 June 2011 (UTC)

true that hyper regular complex numbers are all trans regular?

Hi, Are all hyper regular complex numbers are also trans regular? It seems to me intuitively true, but maybe I'm missing something. --188.28.167.165 (talk) 12:06, 22 June 2011 (UTC)

I'm not familiar with your terminology, and a cursory search turns up nothing. Maybe you could supply a little context? --Trovatore (talk) 21:14, 22 June 2011 (UTC)

Vertical bar notation

Hi. Please refer to section 2.2 in the following paper: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.46.2009&rep=rep1&type=pdf

I am unfamiliar with the usage of the bar in this section, and can't see which of the possibilities in the vertical bar article it corresponds to. I also have no idea what the d above the equals sign represents. Any guidance would be appreciated! --Iae (talk) 12:50, 22 June 2011 (UTC)

It seems likely that the vertical bar indicates conditional probability and the d above the equals sign means "has the same probability distribution as". Gandalf61 (talk) 13:48, 22 June 2011 (UTC)

That would make sense, although even with that knowledge I can't find any reference on the internet to using an equal sign in that manner, so I can't add the info to an article. At least I'll know if there's a next time. Thanks! --Iae (talk) 14:26, 22 June 2011 (UTC)

Mathematical games

From the article linked above: "A mathematical game is a multiplayer game whose rules, strategies, and outcomes can be studied and explained by mathematics. " What does it mean "can be"? Does something exist that cannot be explained by maths? (some day, by someone). 88.8.78.155 (talk) 20:42, 22 June 2011 (UTC)

I guess it means "has been". Bo Jacoby (talk) 21:11, 22 June 2011 (UTC).

So, if I explain playing with dolls in mathematical terms, it becomes a mathematical game? Wikiweek (talk) 21:21, 22 June 2011 (UTC)

So first of all, you have to keep in mind that, just because there's a WP article on something, it doesn't follow that it's a good article or that it reflects standard usage. I'm a little undecided on whether I think that article should be deleted. Maybe it should be turned into a list, instead.

But all that aside, I don't think there was any intent to give a precise definition of mathematical game. I think the author(s) just wanted to collect together coverage of some games in which they found a commonality, and they had to give some sort of a definition at the top because it's expected that an article will start with a definition. --Trovatore (talk) 21:30, 22 June 2011 (UTC)

Honey, the social worker wants you to show her on the doll how the man touched your asymptote. :)Naraht (talk) 21:34, 22 June 2011 (UTC)

Indeed, the article Game theory has a better working definition of 'game'. I don't know if social games, like playing with dolls, could be fitted in the definition of game in the sense of the Game theory. Game theory is mostly concerned with finite, discrete games. Wikiweek (talk) 19:59, 23 June 2011 (UTC)

When I think of "mathematical games", I tend to think of Martin Gardner's long-running Scientific American column entitled Mathematical Games, which covered a very broad range of topics. Also it may be worth mentioning Ludwig Wittgenstein's famous discussion of the word "game", the thrust of which is that it is really impossible to give any concise definition that encompasses all of the ways in which the word is commonly used. Looie496 (talk) 00:34, 24 June 2011 (UTC)

Bernard J Oldfield defines a mathematical game as

1. It is an activity involving

EITHER a challenge against a task or one or more opponents

OR a common task to be tackled either individually or (more normally) in conjunction with others.

2. The activity is governed by a set of rules, and has a clear underlying structure to it.

3. The activity normally has a distinct finishing point.

4. The activity has specific mathematical cognitive objectives.

("Games in the Learning of Mathematics: 1: A Classification", Bernard J. Oldfield, Mathematics in School, Vol. 20, No. 1 (Jan., 1991), pp. 41-43) However this would exclude chess, nim, and other games not usually or not always played for mathematical reasons. (I played nim as a small child, certainly without any mathematical analysis, but according to Wikipedia it's a mathematical game.)

Nonetheless, that's the clearest definition I have found. St Andrews University's computer science department has a page on mathematical games which fails to define them at all.[1] There are patents for mathematical games, e.g. US 3204345. There doesn't seem to be a good definition of the difference between e.g. mathematical games and mathematical puzzles or diversions either. Gardner's column Mathematical Games was first collected in a book called Mathematical Puzzles and Diversions - so is a mathematical game the same as a mathematical puzzle or diversion?

At the same time, as this shows, "mathematical game" is a frequently-used phrase, referring to a game that is interesting to mathematicians, that involves mathematical concepts, that mathematics can be used to get a winning strategy, or something similar. The division is fuzzy: e.g. darts isn't normally considered a mathematical game, but it might be possible to analyse probabilities to work out the best strategy, so might it become a mathematical game, or is it too dangerous for classroom use? --Colapeninsula (talk) 12:54, 24 June 2011 (UTC)

June 23

FDS for an ODE

Hey guys, I am trying to solve an BVP of the form -(a(x) u'(x))'=f(x) with u(0)=u(1)=0 where $a(x)\geq c>0$ is a bounded differentiable function in [0,1], on an equally spaced grid on [0,1] using finite differences. Since I want $O(h^{2})$ approximation, I used central differences for both the first and the second derivative and got the relation:

$(a_{j+1}+4a_{j}+a_{j-1})u_{j+1}-8a_{j}u_{j}+(-a_{j+1}+4a_{j}+a_{j-1})u_{j-1}=-4h^{2}f_{j}$

with $u_{0}=u_{n}=0$ and since the ODE is linear, I can put this in a matrix vector form and get a tridiagonal matrix. I have two questions about this matrix. First of all, using Gerschgorin's theorem, I put a bound on the spectral radius of this matrix as 20M where M is the maximum of |a(x)| on the entire grid (I could use the max on [0,1] too which would be uniform for all grids). Is this correct and is this the best bound one can get? Second, what can I say about the condition number (let's say the 2-norm...using the ratio of the singular values)? How does the condition number depend on h? What happens as h goes to zero? Numerical experiments show that the condition number is blowing up but how would I show it analytically? Thanks!-Looking for Wisdom and Insight! (talk) 01:35, 23 June 2011 (UTC)

Gradient and conjugate gradient

On a completely different topic, how does the conjugate gradient method converge so much faster than plain simple gradient descent? I am not asking for the math behind it. I am just trying to get some intuitive sense and would appreciate it if someone can explain it in words. I thought the gradient points in the direction of maximum increase so negative gradient points in the maximum decrease so if I want to minimize the quadratic function, why not just use the gradient? How can conjugate gradient be faster? Thanks!-Looking for Wisdom and Insight! (talk) 01:45, 23 June 2011 (UTC)

I believe the question relates to our articles Gradient descent and Conjugate gradient method. Dolphin (t) 01:59, 23 June 2011 (UTC)

I have already read both of the articles but I was looking more for some insight in words as to what's wrong with the gradient descent method and how does CG fixes it. - Looking for Wisdom and Insight! (talk) 03:42, 23 June 2011 (UTC)

For intuition, I think a key phrase from our CG article is "Suppose that {pk} is a sequence of n mutually conjugate directions. Then the pk form a basis of Rn, so we can expand the solution x* of Ax = b in this basis" -- Because the pk are orthogonal with respect to A, there is less potential for 'redundancy' in the iterates, and they won't tend to zig-zag as some of the examples given in the gradient decent article. Does that help? SemanticMantis (talk) 03:56, 23 June 2011 (UTC)

Striped triangle question

Is it possible to create a triangle with three horizontal stripes, where each of the stripes has the same number of elements?

For example, it's easy to create such a triangle with two horizontal stripes:

  A
 A A
B B B

This has two stripes, each with three elements. This is (I think) the next smallest example, with two stripes containing 105 elements each:

                   A
                  A A
                 A A A
                A A A A
               A A A A A
              A A A A A A
             A A A A A A A
            A A A A A A A A
           A A A A A A A A A
          A A A A A A A A A A
         A A A A A A A A A A A
        A A A A A A A A A A A A
       A A A A A A A A A A A A A
      A A A A A A A A A A A A A A
     B B B B B B B B B B B B B B B
    B B B B B B B B B B B B B B B B
   B B B B B B B B B B B B B B B B B
  B B B B B B B B B B B B B B B B B B
 B B B B B B B B B B B B B B B B B B B
B B B B B B B B B B B B B B B B B B B B

But is it possible to extend this to three stripes? If so, what's the minimum number of elements in such a triangle? 28bytes (talk) 22:13, 23 June 2011 (UTC)

A quick python script shows that there are none with fewer than 100 million rows. So probably not, but I don't have a proof.--Antendren (talk) 23:20, 23 June 2011 (UTC)

Interesting. Thanks! 28bytes (talk) 23:21, 23 June 2011 (UTC)

The problem is equivalent to finding positive integer solutions of the system $y(y+1)=2x(x+1)$, $z(z+1)=3x(x+1)$. Picking the positive solution to each quadratic equation,

$y={\frac {1}{2}}\left(-1+{\sqrt {8x^{2}+8x+1}}\right),\quad z={\frac {1}{2}}\left(-1+{\sqrt {12x^{2}+12x+1}}\right).$

So it comes down to whether $8x^{2}+8x+1$ and $12x^{2}+12x+1$ can both be perfect squares. At this point, I tried reducing modulo some small primes to show that they both can't be quadratic residues mod p for some p, but I wasn't able to find such a prime. Anyway, a slightly different approach might work. Sławomir Biały (talk) 00:45, 24 June 2011 (UTC)

I also couldn't find a three stripe solution. I wrote a Fortran program, and found the following number of elements were solutions to the 2 stripe problem, but none of these had a third stripe:

           3
         105
        3570
      121278
     4119885
   139954815
  4754343828
161507735340

Interestingly, there seems to be an almost exact relationship between each adjacent pair listed above. That is, each is approximately 33.97 times the previous entry. StuRat (talk) 00:43, 24 June 2011 (UTC)

Looks like it converges to around 33.970562748477. Is that an algebraic number, I wonder? 28bytes (talk) 02:42, 24 June 2011 (UTC)

(sequence A053141 in the OEIS) and (sequence A061278 in the OEIS) are relevant. Sławomir Biały (talk) 00:47, 24 June 2011 (UTC)

(sequence A075528 in the OEIS) is the actual sequence given here. A generating function and citation appear. The value near 33.97 is actually the zero of $1-34x+x^{2}$.McKay (talk) 03:52, 24 June 2011 (UTC)

OK, using that first sequence as the number of lines in the first stripe (ignoring the zero), instead of my previous brute force approach, I was able to come up with an extended list of two stripe solutions, none of which has a third stripe. Here are the numbers of elements in each stripe:

                  3
                105
               3570
             121278
            4119885
          139954815
         4754343828
       161507735340
      5486508657735
    186379786627653
   6331426236682470
 215082112260576330
7306460390622912753

StuRat (talk) 02:28, 24 June 2011 (UTC)

Antendren showed that any solution must have at least 9 digits by directly checking the number of rows below that point to see which fit the two equations. By searching only the solutions of the first equation to see if they solve the second as well, there are no solutions with less than 100,000 digits. This calculation took about 30 seconds in PARI/GP:

isTriangular(n:int)={
	issquare(n<<3+1)
};
test(lim)={
  my(x,y,n,t);
  while(n<=lim,
    t=3*x+2*y+2;
    y=t+x+y+1;
    x=t;
    n=x*(x+1)/2;
    if(isTriangular(3*n),return(n))
  );
  print("No solutions")
};
test(1e100000)

(Actually, the listed code would be somewhat slower; I actually used an optimized and compiled version of the first function.)

CRGreathouse (t | c) 04:30, 24 June 2011 (UTC)

• There are no solutions besides n=0. For details, see my post to SeqFan maillist: http://list.seqfan.eu/pipermail/seqfan/2011-June/015039.html Maxal (talk) 06:38, 24 June 2011 (UTC)

June 24

A big number

Hi, I have defined a big number N like this:

Define $f_{1}(x)=x^{x}\,$

Define $f_{i}(x)=f_{i-1}^{x}(x),\ i>1\,$

Define $N=f_{f_{10}(10)}(10)\,$

The notation $f^{n}(x)\,$ means the n-fold iteration of $f\,$. For example, $f^{3}(x)=f(f(f(x)))\,$.

My number N seems beyond incomprehensible, but how much bigger could we get with a definition of comparable length? Obviously we could tinker a bit, like changing "10" to "99", or writing $f_{1}(x)=x^{x^{x}}\,$, or whatever, but can we make a quantum leap to a whole new domain of incomparable vastness? 86.176.208.236 (talk) 02:46, 24 June 2011 (UTC)

Well, this kind of question is a bit boring to those of us who have been around for awhile, I'm afraid. You could make it more interesting by asking, for a given formal language such as the language of Principia Mathematica, what is the largest number that can be specified by an expression of length N? Looie496 (talk) 02:56, 24 June 2011 (UTC)

Very sorry to have bored you. You may find learning the difference between "a while" and "awhile" more interesting. 86.176.211.230 (talk) 13:25, 24 June 2011 (UTC)

Well, I wouldn't call it a quantum leap. It's quite well know. There are different kinds of infinity. For example, there are infinitly many integers and there are infinitely many real numbers. But some how the real numbers are more infinite than the integers; for example we think of integers as dots on a line while the real numbers are the whole line itself. A mathematician called Georg Cantor did a lot of work on this in the late 1800's and early 1900's. You should look at our articles on cardinals and ordinals. These guys are really crazy, but they're really cool too. I can't remember which one, I think it's an ordinal, but it's a fixed point of exponentiation; so if you take it to it's own power then you get it back again, i.e. x^x = x (and it's not one!) Take a look and see what you make of it. Feel free to pop back and ask some questions if you get stuck. — Fly by Night (talk) 03:29, 24 June 2011 (UTC)

What you've defined is basically the Ackermann function.--Antendren (talk) 04:08, 24 June 2011 (UTC)

"Incomprehensible" is a subjective and time-dependent concept. There is no such thing as a big number - every number is small because most numbers are bigger. Bo Jacoby (talk) 12:37, 24 June 2011 (UTC).