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.

  • $\begingroup$ I'm working on an analysis of the problem. If I get something, I'll post it here. This is a really interesting question, by the way. $\endgroup$ Apr 8 '10 at 16:14
  • $\begingroup$ You can hardly obtain the explicit form for $a_m$, but some estimation is possible. $\endgroup$
    – Sunni
    Apr 8 '10 at 22:08
  • $\begingroup$ @Gabriel: Thank you very much for you interest in my problem. Hope getting good news from you. @mivalin: I know it is quite impossible to obtain explicit form of $a_m$, but I really would like to get some estimation. I started by finding the simple root of the denominator. $\endgroup$
    – Ross Tang
    Apr 15 '10 at 7:35
  • $\begingroup$ I apologize for not getting back to you on this sooner, but after the initial few days, I got sidetracked and forgot about this problem. I'm not completely sure, but I think that a<sub>m</sub> is a constant, regardless of m. 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 of m, 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. $\endgroup$ Apr 23 '10 at 15:24
  • $\begingroup$ Please read the 1st comment under the question: voofie.com/content/44/… It shows that a_m cannot be constant and independent of m. $\endgroup$
    – Ross Tang
    Apr 29 '10 at 1:18

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 $\sum_{j=1}^\infty|t_j|<\infty$.

  • 1
    $\begingroup$ Alas, if you look at the exact $G$ the OP is interested in, you'll see that this requirement is clearly violated. $\endgroup$
    – fedja
    Nov 25 '17 at 22:50
  • 1
    $\begingroup$ So, if you simplify the problem to its core, the question is how to find the Fourier series of $\frac 1{\log(1-z)}$. At this moment I have no good idea :-( $\endgroup$
    – fedja
    Nov 25 '17 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.