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

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

$=\frac{(q-1)}{2}(\vartheta(-ai/2\pi,i/\pi) +\vartheta(ai/2\pi,i/\pi) -2)$

$=\frac{(q-1)}{2}(\vartheta(-ai/2\pi,i/\pi) +\vartheta(ai/2\pi,i/\pi) -2)$

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.

Fascinating! I like what you did with this!

ReplyDelete