Find Closed Form

Spoon

Find Closed Form:

$\sum_{n=1,3,5\ldots} \frac{1}{n}e^{-n\pi x/a}\sin{(n\pi y/a)}$

**Posted:** Tue Feb 15, 2005 3:34 pm

joml88

Please don't double post in multiple forums...I am going to delete your post in the Intermediate Forum.

**Posted:** Tue Feb 15, 2005 3:58 pm

Spoon

with the 40 thousand forums these days, I never know where to post or what level my question falls under... is this problem too advanced for this forum?

**Posted:** Tue Feb 15, 2005 7:22 pm

Ravi B

Hint

**Posted:** Tue Feb 15, 2005 7:29 pm

jmerry

Given the nature of the sum, this belongs somewhere in Calculus-Analysis; this is evaluated by interpreting it as a power series.

It's certainly impossible without at least some calculus material.

**Posted:** Tue Feb 15, 2005 8:21 pm

Spoon

would an admin be so kind as to move this post?

**Posted:** Tue Feb 15, 2005 9:38 pm

Myth

**Posted:** Tue Feb 15, 2005 9:58 pm

_________________
Myth is out of here

Kent Merryfield

One observation:

If we let $u(x,y)=\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1ne^{-n\pi x/a} \sin(n\pi y/a)$ for $x>0$

then $\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0.$

In other words, this is a harmonic function on the right hand half plane with boundary values equal to a "square wave":

$u(0,y)= \frac{\pi}4$ for $0<y<a,$ $=-\frac{\pi}4$ for $-a<y<0,$ and continued to be periodic of period $2a.$

As for a closed form solution, I'll get it started.

Let $r=e^{-\pi x/a}$ and let $\theta = \frac{\pi y}a.$ Then we want to compute

$\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1nr^n\sin(n\theta) = \text{Im}\left(\sum_{\substack{n=1 \\ n \,\text{od...$

Let $g(z)=\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1nz^n.$ Differentiate that to get $g(z)=\frac1{1-z^2}.$

Integrate that and take the imaginary part. The sum we are looking for is

$\frac12\text{Arg}\left(\frac{1+z}{1-z}\right).$

Someone else can take over from here and finish this.

Spoon

**Posted:** Wed Feb 16, 2005 11:48 am

Spoon

ok... so like could the moderator please put my post back where it was? Since it's been moved, nobody has even given it a look.

Thanks

**Posted:** Sun Feb 20, 2005 6:35 pm

vuhung

**Posted:** Sun Feb 20, 2005 7:19 pm

Spoon

I don't see how that leads to the solution...

**Posted:** Tue Feb 22, 2005 4:02 pm

Kent Merryfield

Well, I've got an answer for you in #8, but I didn't finish it.

$u=\frac12(\arg(1+z)-\arg(1-z))$ where

$z=re^{i\theta},\, r=e^{-\pi x/a},\,\theta=\frac{\pi y}a.$

Of course, there's a long way to go.

The first step after that is the geometry-trigonometry problem of computing $\arg(1+re^{i\theta})$ and $\arg(1-re^{i\theta}).$

That should take some time; I would suggest starting by drawing pictures of triangles.

Then you can put in what $r$ and $\theta$ are.

**Posted:** Tue Feb 22, 2005 7:31 pm