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](http://www.voofie.com/content/44/solving-recurrence-equation-with-indexes-from-negative-infinity-to-positive-infinity/).



Solving recurrence equation with indexes from negative infinity to positive infinity
I see your problem more like a linear equation on an infinite dimensional space of doubly infinite sequences than as a recurrence equation, since there are no initial values from which start to build up the solution. In the following I will assume that $\sum\_{j=1}^\infty|t\_j|<\infty$ and that $\mathbf{a}=(a\_m)$ is bounded. The linear operator $T$ defined on $\ell^\infty(\mathbb{Z})$ by
$$
(T\mathbf{a})\_m=\sum\_{j=1}^\infty t\_j(a\_{m+j}+a\_{m-j})
$$
is bounded with operator norm $\|T\|\le2\sum\_{j=1}^\infty|t\_j|$. Your equation can be written as
$$
(I-T)\mathbf{a}= \mathbf{b},
$$
where $I$ is the identity operator and $\mathbf{b}\in\ell^\infty(\mathbb{Z})$. If $\sum\_{j=1}^\infty|t\_j|<\dfrac{1}{2}$, then $I-T$ is invertible and the above equation has a unique solution for all $\mathbf{b}\in\ell^\infty(\mathbb{Z})$, given by
$$
\mathbf{a}=(I-T)^{-1}\mathbf{b}=(I+T+T^2+T^3+\dots)\mathbf{b}.
$$
This is an explicit formula that in practice may be useless, although one can get an approximation by computing a few terms of the sum.


Just to check that this can really give a solution, let's study the particular case in which $t\_1=t$ and $t\_j=0$ for all $j>1$, and $(\mathbf{b})\_m=\delta\_{m,0}$. Then we find that
$$
a\_m=a\_{-m}=\sum\_{k=1}^\infty\binom{2k+m}{k+m}t^{2k+m}=t^m{}\_2F\_1\left(\frac{m+1}{2},\frac{m+2}{2},m+1,4t^2\right),
$$
where ${}\_2F\_1$ is the hypergeometric function. We see that indeed one must have $|t|<\dfrac{1}{2}$ for this to make sense.


This analysis doesn't mean that there are no other solutions under different conditions, but I think that it will be difficult to avoid the requirement that $\sum\_{j=1}^\infty|t\_j|<\infty$.


