Author Topic: TT's Maths Thread (Read 146593 times) Tweet Share

TrueTears · « **Reply #1155 on:** March 30, 2011, 11:01:32 pm »

~~yeah i thought about that as well, but they're not transpose of each other?~~

ohh wait i see, i got it!

cheers kamil!

TrueTears · « **Reply #1156 on:** April 05, 2011, 01:58:40 am »

[IMG]http://img225.imageshack.us/img225/9051/22703967.jpg[/img]

I actually have no idea how to go about this question, in fact I weren't even taught what the Gamma function is/does in my classes nor have I ever seen the Jacobi theta-function in my life! So I am really clueless on this question. Any help would be appreciated!

Cheers!

Mao · « **Reply #1157 on:** April 05, 2011, 02:45:42 am »

Read from here onwards: http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform
It probably won't help, but it may help put things into context.

What subject is this? and why are you learning crazy transforms? xD

TrueTears · « **Reply #1158 on:** April 05, 2011, 02:57:54 am »

ohh i see, ill take a good read of that cheers man.

its MTH2121, number theory and algebra, its an alright unit, but this was a suprising q since we haven't covered any of this in lectures nor is it in the prescribed texts lol

Mao · « **Reply #1159 on:** April 05, 2011, 03:29:11 am »

Quote from: TrueTears on April 05, 2011, 02:57:54 am

ohh i see, ill take a good read of that cheers man.

its MTH2121, number theory and algebra, its an alright unit, but this was a suprising q since we haven't covered any of this in lectures nor is it in the prescribed texts lol

Have a play with it. I managed to simplify the RHS to $\int \left( \sum_{n=1}^{\infty} \exp(-\pi n^2 t) \right) t^{s-1}\, dt$ before I gave up. I'm sure it will be relatable to the course material.

TrueTears · « **Reply #1160 on:** April 05, 2011, 06:00:45 am »

hmm i still have no clue lol

/0 · « **Reply #1161 on:** April 05, 2011, 10:57:23 am »

If you interchange summation and integral, then set $u=\pi n^2t$ , you can use u-substitution to reduce Mao's expression to $\pi^{-s}\Gamma (s)\sum_{n=-\infty}^\infty \frac{1}{n^{2s}}$ . The sum over n=0 is a bit of a pain. Still thinking about how to do the rest though.

Interesting question for an algebra course... woulda thought it to be more complex analysis

humph · « **Reply #1162 on:** April 05, 2011, 11:57:11 am »

This is an extremely standard question in analytic number theory, but I can't imagine why it'd be in your course. Are you sure it's not an April Fool's joke??
Anyway, a good reference for this is Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory (which you may find here: http://gen.lib.rus.ec/get?md5=207be3298f0abee321df6ba0f0887561), chapter 10.

But /0's method should work, just if you're careful you should be able to get rid of the $n = 0$ term (as this is a singularity).

Mao · « **Reply #1163 on:** April 05, 2011, 03:57:12 pm »

Quote from: /0 on April 05, 2011, 10:57:23 am

If you interchange summation and integral, then set $u=\pi n^2t$ , you can use u-substitution to reduce Mao's expression to $\pi^{-s}\Gamma (s)\sum_{n=-\infty}^\infty \frac{1}{n^{2s}}$ . The sum over n=0 is a bit of a pain. Still thinking about how to do the rest though.

Interesting question for an algebra course... woulda thought it to be more complex analysis

Aha, nice. That solves it. typing it up now.

Mao · « **Reply #1164 on:** April 05, 2011, 04:11:19 pm »

$\begin{align*} RHS & = \frac{1}{2s-1}+\frac{1}{2} \int_{0}^{1} \left( \theta (\imath t) - t^{-1/2} \right) t^{s-1} \, dt -\frac{1}{2s} + \frac{1}{2} \int_{1}^{\infty} \left( \theta (\imath t) - 1 \right) t^{s-1} \, dt \\ & = \frac{1}{2s-1} - \frac{1}{2s} + \frac{1}{2} \int_{0}^{\infty} \theta(\imath t) t^{s-1}\, dt - \frac{1}{2} \int_{0}^{1} t^{s-3/2}\, dt - \frac{1}{2} \int_{1}^{\infty} t^{s-1} \end{align*}$

the Jacobi Theta function is simplified as
$\theta(\imath t) = \sum_{-\infty}^{\infty} \exp(-\pi n^2 t)$
due to the symmetry of $n^2$ , we can rewrite this as
$\theta(\imath t) = 1 + 2\sum_{n=1}^{\infty} \exp(-\pi n^2 t)$
the integral containing the theta function thus becomes
$\frac{1}{2} \int_{0}^{\infty} \theta(\imath t) t^{s-1}\, dt = \frac{1}{2}\int_{0}^{\infty} t^{s-1}\, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt$

Substituting this gives
$\begin{align*} RHS & = \frac{1}{2s-1} - \frac{1}{2s} + \frac{1}{2} \int_{0}^{1} t^{s-3/2} - t^{s-1} \, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt \\ & = \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt\qquad \mbox{substituting }u=\pi n^2 t \\ & = \sum_{n=1}^{\infty} \int_{0}^{\infty} \left( \frac{u}{\pi n^2} \right)^{s-1} \exp(-u) \, \frac{du}{\pi n^2} \\ & = \sum_{n=1}^{\infty} \frac{1}{(\pi n^2)^s} \int_{0}^{\infty} u^{s-1} \exp(-u) \, du\qquad \mbox{the integral is the gamma function by definition, thus} \\ & = \pi^{-s} \sum_{n=1}^{\infty} \frac{1}{n^{2s}} \Gamma (s) \\ & = \pi^{-s} \Gamma(s) \times \zeta (2s) \\ & = LHS \end{align*}$

the proof is thus complete

TrueTears · « **Reply #1165 on:** April 05, 2011, 07:21:53 pm »

omg that's awesome, thanks guys ^^^^^

@humph+/0, i have no idea, i don't think this is an april fools joke lol it's on our set questions sheet that gets marked for this week haha but yeah like i said, i thought this wouldn't be in the unit that im doing lol

oh and yeah that's probs a small typo mao, but it should be $+\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$ XD

humph · « **Reply #1166 on:** April 05, 2011, 09:48:06 pm »

Mao's way is right, but you can't split up the integrals at the beginning like he did because I'm pretty sure then some of the integrals don't converge (the $\int^{\infty}_{0}{t^{s - 1} \, dt}$ term). The grouping is to ensure that everything works nicely. Also the reason why you can switch integration and summation is due to absolute convergence (this is very important).

TrueTears · « **Reply #1167 on:** April 05, 2011, 10:31:45 pm »

ohh yeah, wait a sec, now im a bit confused, in mao's working :

$RHS = \frac{1}{2s-1} - \frac{1}{2s} - \frac{1}{2} \int_{0}^{1} t^{s-1} + t^{s-3/2}\, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt$

How did he get the " $-\frac{1}{2}\int_0^1t^{s-1}+t^{s-\frac{3}{2}}dt$ " term ? it's probably pretty obvious...

I keep getting $+\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$

since after substitution we have $-\frac{1}{2}\int_1^{\infty}t^{s-1}dt + \frac{1}{2}\int_0^{\infty}t^{s-1}dt = \frac{1}{2}\int_0^{1}t^{s-1} dt$

Then grouping with $-\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt$ yields:

$\frac{1}{2}\int_0^{1}t^{s-1} dt -\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt = +\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$

/0 · « **Reply #1168 on:** April 05, 2011, 10:42:55 pm »

Hmm it probably should be

$-\frac{1}{2}\int_0^1 -t^{s-1}+t^{s-\frac{3}{2}}\,dt$

Also,

$\int_0^1t^{s-1}\,dt = \int_0^\infty t^{s-1}\,dt - \int_1^\infty t^{s-1}\,dt$

But as humph mentioned... the divergence of the RHS integrals might still be a problem though. Perhaps there's a way of grouping the RHS so that they don't diverge.

TrueTears · « **Reply #1169 on:** April 05, 2011, 11:48:55 pm »

yeah, thanks for the confirmation /0, and yeah, so both $\int_0^\infty t^{s-1}\,dt$ and $-\int_1^\infty t^{s-1}\,dt$ diverge... but when you put them together its fine, so is it necessary to change the grouping?

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