How to take this sum?

1. Mar 29, 2006

TheDestroyer

How to take this sum???

I'm getting this sum in the statistical mechanics for the rotating energy, any one can help with?
$$Z_r = \sum\limits_{\ell = 0}^\infty {\left( {2\ell + 1} \right)e^{{\textstyle{{T_r } \over T}}\ell \left( {\ell + 1} \right)} }$$
If tex doesn't work here is it, Where the sum is from 0 to infinity
Z=Sum[(2L+1)exp(-x L(L+1))]

Seems simple lol !!

Thanks

2. Mar 29, 2006

Tom Mattson

Staff Emeritus
Question: Is $T_r/T$ positive or negative? If it's positive then your sum diverges. But I'm guessing it's not that simple. :tongue:

3. Mar 29, 2006

kryptyk

Here's an idea:

Break up the $2 \ell +1$ into $\ell + (\ell + 1) \;$. Then we have:

$$Z_\ell = F(\ell) + G(\ell)$$

where

$$F(\ell) = \ell \, e^{-k \, \ell(\ell+1)}$$

and

$$G(\ell) = (\ell+1) \, e^{-k \, \ell(\ell+1)}$$

then

$$G(\ell - 1) = \ell e^{-k \, \ell(\ell-1)}$$

And so

$$H(\ell) = F(\ell) \,+\, G(\ell-1)$$
$$= \ell [e^{-k \, \ell(\ell+1)} \,+\, e^{-k \, \ell(\ell-1)}] = \ell [e^{-k \, \ell^2 \,-\, k \, \ell} \,+\, e^{-k \, \ell^2 \,+\, k \, \ell}] = \ell e^{-k \, \ell^2}[e^{k \, \ell} \,+\, e^{-k \, \ell}]$$
$$= 2\ell \, \cosh(k \ell) e^{-k \, \ell^2}$$

So perhaps we can do something along the lines of:

$$\sum_{\ell=0}^{\infty} F(\ell)+G(\ell)$$
$$= F(0) + [ F(1) + G(0) ] + [ F(2) + G(1) ] + \ldots$$
$$= F(0) + \sum_{\ell=1}^{\infty} H(\ell) = 2 \sum_{\ell=1}^{\infty} \ell \, \cosh(k \ell) e^{-k \, \ell^2}$$

Last edited: Mar 29, 2006
4. Mar 29, 2006

kryptyk

I'm not exactly sure how to evaluate this last sum (I might be unnecessarily complicating things further) but I'll throw out some ideas anyhow...

Consider extending the real numbers by adding a number $u$ that is neither $1$ nor $-1$ such that

$$u^2 = 1$$

much the same way we extend the real numbers by adding $i$ such that

$$i^2 = -1$$

Then for real $t$,

$$e^{u t} = \cosh t + u \sinh t$$

This extension of the reals is sometimes called the hyperbolic numbers, or the unipodal numbers.

So this means that

$$\cosh(k \ell) e^{-k\, \ell^2}$$

is equal to the real part of

$$e^{k\, u\, \ell}e^{-k\, \ell^2} = e^{-k(\ell^2 - u\, \ell)}$$

Then completing the square, we have

$$(\ell^2 - u\, \ell + \frac{u^2}{4}) - \frac{u^2}{4} = (\ell - \frac{u}{2})^2 - \frac{u^2}{4}$$

But $u^2 = 1$. Letting $\omega = \ell - u/2$, we get

$$e^{-k(\omega^2 - \frac{1}{4})} = e^{\frac{k}{4}}e^{-k\, \omega^2}$$

So then

$$\ell \cosh(k \ell) e^{-k\, \ell^2} = e^{\frac{k}{4}}Re[\ell e^{-k(\ell - \frac{u}{2})^2}] = e^{\frac{k}{4}}Re[(\omega + \frac{u}{2})e^{-k\, \omega^2}]$$

Anyhow, I'll have to think more about this.

Last edited: Mar 29, 2006
5. Mar 29, 2006

kryptyk

OK - I feel stupid now. I suppose we could just complete the square in the original sum:

$$(2\ell+1)e^{-k\, \ell(\ell+1)} = (2\ell+1)e^{-k(\ell^2 + \ell + \frac{1}{4})+\frac{k}{4}} = (2\ell+1)e^{\frac{k}{4}}e^{-k(\ell+\frac{1}{2})^2}$$

Then letting $\lambda = \ell + 1/2$ we have:

$$2\lambda \, e^{\frac{k}{4}}e^{-k\, \lambda^2}$$

We should be able to evaluate this sum now, no?

6. Mar 29, 2006

Physics Monkey

I'm afraid not. It is quite unfortunate that we don't have closed form expressions for Gaussian sums. The most direct way to estimate its value would be to approximate it with an integral. You might also be able to make a more systematic attack by making use of the Euler-MacLaurin formula.

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