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May 8

LaTeX arithmetic

(Not sure which board is most appropriate.) I'm trying to use LaTeX, specifically Kile, to do some basic arithmetic. However, I seem to be stuck right back at the start, as I can't seem to get /multiply to do anything meaningful. Is this the command I want? All I want to do is perform some command on say, "5" and "4" and get "20" or "9" or "1" or whatever, as appropriate. Grandiose (me, talk, contribs) 15:24, 8 May 2011 (UTC)

I'm confused about what you're trying to do. As far as I know, LaTeX is for typesetting, not for computation. You can use

5\cdot4=20

To typeset $5\cdot 4=20$, but LaTeX isn't supposed to do the calculation itself. If I missed your point completely, can you provide more background? -- Meni Rosenfeld (talk) 16:42, 8 May 2011 (UTC)

I was hoping it could basic arithmetic as is suggested by places like this, but I couldn't follow what I needed to do off any of them. Grandiose (me, talk, contribs) 17:18, 8 May 2011 (UTC)

Yes it can do that. I don't know how... Staecker (talk) 20:27, 8 May 2011 (UTC)

OK I figured something out. I've never done this before, so I don't know if I'm doing it the right way. Integers can be stored in numbered "count registers". Here I'll define two count registers equal to 2 and 3, then display them, then multiply them and display the product:

\count1=2 
\count2=3
\the\count1 
\the\count2 
\multiply\count1\count2
\the\count1

The displayed output is 236. The registers are displayed using "\the", and "\multiply" overwrites the first register with the product. Staecker (talk) 20:40, 8 May 2011 (UTC)

Brilliant. Do you happen to know what addition, subtraction, and division will be? Grandiose (me, talk, contribs) 19:41, 9 May 2011 (UTC)

/advance is addition, it would seem, and /advance/count1-/count2 (for example) works for subtraction. I can't imagine / is going to work, though, in multiplication. Lucky for me, I don't have to do any. Grandiose (me, talk, contribs) 19:48, 9 May 2011 (UTC)

\divide is for division. See wikibooks:TeX/count for this and basically no additional information. Staecker (talk) 22:34, 9 May 2011 (UTC)

Make sure to use the backslash \ for LaTeX commands, and not the forward slash /. So it's \count and \multiply and not /count or /multiply. — Fly by Night (talk) 00:51, 12 May 2011 (UTC)

Interesting series...

Dear Wikipedians:

I am trying to find the sum of an interesting series whose general term is:

$\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^2\cdots {\left(\frac{1}{3}\right)}^n$

I realized that the power forms the triangular numbers, and the general term of triangular numbers is $\frac{n^2+n}{2}$, therefore the general term of the interesting series above can be written as:

${\left(\frac{1}{3}\right)}^{\frac{n^2+n}{2}}$

$={\left(\frac{1}{3}\right)}^{\frac{n^2}{2}}{\left(\frac{1}{3}\right)}^{\frac{n}{2}}$

$={\left(\sqrt{\frac{1}{3}}\right)}^{n^2}{\left(\sqrt{\frac{1}{3}}\right)}^n$

So at this point it is clear that there is a geometric sequence subcomponent, ${\left(\sqrt{\frac{1}{3}}\right)}^n$, to my interesting series. However, I am at a total loss about what to do for the ${\left(\sqrt{\frac{1}{3}}\right)}^{n^2}$ subcomponent, and also about how to separate the two subcomponents so that I could sum each one up individually.

Any help is greatly appreciated.

70.31.155.244 (talk) 16:03, 8 May 2011 (UTC)

I don't think there's a closed-form expression for the sum up to a finite term. For the infinite sum, Mathematica gives the result (assuming I'm reading it correctly) ${\theta }_2(0;1/\sqrt{3})$ (see Theta function) which is 1.371759117358... . I have no idea how to arrive at this result. -- Meni Rosenfeld (talk) 16:36, 8 May 2011 (UTC)

I don't think you'll be able to separate the two parts you're talking about. Note also that n² + n is always even, so your exponent is always an integer. Michael Hardy (talk) 16:39, 8 May 2011 (UTC)

The article titled theta function says this:

$\vartheta (z;\tau )=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\exp (\pi i{n}^2\tau +2\pi inz)$

There you see a sum of two terms in the exponent, in which one is a constant times n² and the other is a constant times n, just as in your series. So we would want

$e^{\pi i{n}^2\tau +2\pi inz}={\left(\sqrt{\frac{1}{3}}\right)}^{n^2+n}.$

Recall that that

$\sqrt{\frac{1}{3}}=e^{-({\log }_{e}3)/2}.$

So

${\left(\sqrt{\frac{1}{3}}\right)}^{n^2+n}=e^{-(n^2+n)({\log }_{e}3)/2}.$

So we want

$\frac{-(n^2+n)({\log }_{e}3)}{2}=\pi i{n}^2\tau +2\pi inz$

This will hold if the coefficients of n² and n agree. Thus we need

$\frac{-{\log }_{e}3}{2}=\pi i\tau ,$

and

$\frac{-{\log }_{e}3}{2}=2\pi iz.$

So

$\tau =\frac{i{\log }_{e}3}{\pi }$

and

$z=\frac{i{\log }_{e}3}{2\pi }.$

But you probably intended your series to run from 0 to ∞ rather than from −∞ to ∞, so mutatis mutandis....

To be continued...... Michael Hardy (talk) 17:03, 8 May 2011 (UTC) So

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\exp (\pi i{n}^2\tau +2\pi inz)=1+\sum _{n=1}^{\mathrm{\infty }}\exp (\pi i{n}^2\tau +2\pi inz)+\exp (\pi i{n}^2\tau -2\pi inz)$

To be continued.... Michael Hardy (talk) 17:22, 8 May 2011 (UTC)

$=1+\sum _{n=1}^{\mathrm{\infty }}\exp (\pi i{n}^2\tau )(\exp (2\pi inz)+\frac{1}{\exp (2\pi inz)})$

To be continued.... Michael Hardy (talk) 21:04, 8 May 2011 (UTC)

Hi Michael, $\tau =i\frac{\log 3}{2\pi }$ and $z=i\frac{\log 3}{4\pi }$ so that

$\vartheta (i\frac{\log 3}{4\pi };i\frac{\log 3}{2\pi })=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{3}^{-\frac{n(n+1)}{2}}$

$=(\sum _{n=-\mathrm{\infty }}^{-2}+\sum _{n=-1}^0+\sum _{n=1}^{\mathrm{\infty }})3^{-\frac{n(n+1)}{2}}$

$=\sum _{n=2}^{\mathrm{\infty }}{3}^{-\frac{n(n-1)}{2}}+2+\sum _{n=1}^{\mathrm{\infty }}{3}^{-\frac{n(n+1)}{2}}$

$=2+2\sum _{n=1}^{\mathrm{\infty }}{3}^{-\frac{n(n+1)}{2}}$

and the final result is

$\sum _{n=1}^{\mathrm{\infty }}{3}^{-\frac{n(n+1)}{2}}=\frac{1}{2}\vartheta (i\frac{\log 3}{4\pi };i\frac{\log 3}{2\pi })-1.$

Bo Jacoby (talk) 14:18, 9 May 2011 (UTC).

Or even simpler:

$\sum _{n=0}^{\mathrm{\infty }}{3}^{-\frac{n(n+1)}{2}}=\frac{1}{2}\vartheta (i\frac{\log 3}{4\pi };i\frac{\log 3}{2\pi }).$

Bo Jacoby (talk) 07:37, 11 May 2011 (UTC).

Thanks to everyone for your contribution. Now I know that the series in question cannot possibly have any closed form expressions using elementary functions and their combinations/compositions thereof. L33th4x0r (talk) 17:15, 11 May 2011 (UTC)