Monday, January 25, 2010

Exponential Series

This semester we started a course in Time Scales, which is an interesting generalization of the classic differential analysis. The idea of time scales is to provide a connection between the study of differential equations made on $\mathbb{R}$ and the study of difference equations on $\mathbb{Z}$.

This connection is made by taking a closed subset of $\mathbb{R}$ and start defining on it notions of a right and left derivatives, which are called the $\Delta$ and $\nabla$ derivatives.

On last week's class, one of my friends was talking about this definitions and also how would one define an integral using this time scales approach. As an example, he state the following integral

$\int_0^\infty e^{-\tau^2}\Delta \tau$

On the time scale $\mathbb{T}=\overline{\{q^\mathbb{Z}\}}$ for some $q>1$. This is just the closure of the set of all integer powers of $q$.

After dealing with the boring algebra involved, the previous integral has a value of

$(q-1)\sum_{n=1}^\infty (q^n+q^{-n})e^{-n^2}$

Finding the actual value of this expression boils down to calculate the value of

$\sum_{n=1}^\infty e^{an-n^2}$

This can be done by using the Jacobi Theta Function, which is given by

$\vartheta(z,w)=\sum_{n=-\infty}^\infty \exp(2\pi i nz+\pi i n^2 w)$

for $z\in\mathbb{C}$ and $w\in\mathbb{H}$. Thus letting $z=\frac{-ai}{2\pi}$ and $w=\frac{i}{\pi}$ gives the value that we are looking for. In this case, we have that

$\int_0^\infty e^{-\tau^2}\Delta \tau$
$=\frac{(q-1)}{2}(\vartheta(-ai/2\pi,i/\pi) +\vartheta(ai/2\pi,i/\pi) -2)$

This is a really simple problem at first sight, but it caught my attention the fact that it all relies on determining the value for an expression that looks like

$\sum_{n=1}^\infty e^{p(n)}$

where $p(n)$ was a quadratic polynomial with a negative leading coefficient. A natural question came then to my mind, what would happen if we would have any polynomial instead?

My first idea was to study the case when $p(n)=-n^m$ for a fixed power $m$. The cases when $m=1,2$ are contained in the previous approach using the Jacobi Theta Function. I first try to calculate the series for some values of $m$, and I found that the above expression, as a function of $m$, converged really fast as $m$ was getting bigger.

Actually, it is not difficult to find out that this limit exists and has the value of

$\lim_{m\to\infty} \sum_{n=1}^\infty e^{-n^m}=\frac{1}{e}$

Following the same line, one can show that for $p(n)=-an^m$, with $a$ a constant, we have that

$\lim_{m\to\infty} \sum_{n=1}^\infty e^{-an^m}=\frac{1}{e^a}$

Thus, one could say that for a polynomial $p(n)=-an^m+O(n^{m-1})$ a good approximation its given by

$\sum_{n=1}^\infty e^{p(n)} \approx\frac{1}{e^a}$

Looking for a different approach towards a more precise answer, one can define a function given by $f(t)=\sum_{n=1}^\infty e^{p(n) t}$ and this function can be viewed as the heat kernel of a differential operator $P$ whose spectrum is given by $\sigma(P)=\{\lambda_n\}$, where $\lambda_n=p(n)$.

This could suggest the difficulty of such a closed form for $f(1)$, for instance, in the case of $p(n)=-n^m$, this would be related to the existence of an operator $P$ with eigenvalues $\{1/n\}$, which is one of the consequences of the Riemann Hypothesis.




1 comment:

  1. Fascinating! I like what you did with this!

    ReplyDelete