In many cases, the recurrence equations that people are solving involves index of only non-negative values. Here I have a recurrence equation which arises from transport of light in an infinite 1D chain:

$a_m=\sum _{j=1}^{\infty } \left(T_ja_{m+j}+T_ja_{m-j}\right) + \delta _{m,0}$

where $\delta_{m,0}$ is the Kronecker delta function. i.e.:

$\delta_{i,j} = \begin{cases} & 1 \text{ if } i=j \\ & 0 \text{ if } i \neq j \end{cases}$

Here I would like to solve $a_m$, where the index of m is from negative infinity to positive infinity, while $T_j$ is a given sequence, and p is just a given constant.

Defining the generating function $G(z)=\sum _{k=-\infty }^{\infty } a_kz^k$, I found that:

$G(z)=\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}$

The problem is, how am I going to do series expansion on G? Doing a simple expansion of $\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}=\sum _{j=0}^{\infty } \left(\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)\right){}^j$ won't help. Since the power is too difficult to expand out.

And contour integration isn't helping as well, since it is too difficult to compute analytically or numerically too.

Here I would like to ask about direction in obtaining analytical solution, or approximated one.

And in my case, my function G is given by:

$G(z)=\left(1+\frac{3i}{2r^3}\left(r^2\left(\ln \left(1-\frac{e^{i r}}{z}\right)+\ln \left(1-e^{i r}z\right)\right)\right)-i r\left(\text{Li}_2\left(\frac{e^{i r}}{z}\right)+\text{Li}_2\left(e^{i r}z\right)\right)+\text{Li}_3\left(\frac{e^{i r}}{z}\right)+\text{Li}_3\left(e^{i r}z\right)\right){}^{-1}$

p.s.:I have posted the same problem in Voofie.

completelysure, but Ithinkthat a<sub>m</sub> is a constant, regardless ofm. I certainly know that such a solution is valid, but from what I was working on, the result for a<sub>m</sub> seemed to be independent ofm, which would mean that they are all the same constant. I'm not sure if this helps; the algebra got far too ugly far too quickly for me to work though anything by hand. – Gabriel Benamy Apr 23 '10 at 15:24