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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.

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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. –  Gabriel Benamy Apr 8 '10 at 16:14
    
You can hardly obtain the explicit form for $a_m$, but some estimation is possible. –  Sunni Apr 8 '10 at 22:08
    
@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. –  Ross Tang Apr 15 '10 at 7:35
    
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. –  Gabriel Benamy Apr 23 '10 at 15:24
    
Please read the 1st comment under the question: voofie.com/content/44/… It shows that a_m cannot be constant and independent of m. –  Ross Tang Apr 29 '10 at 1:18
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1 Answer

up vote 1 down vote accepted

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 $$ a_m=a_{-m}=\sum_{k=1}^\infty\binom{2k+m}{k+m}t^{2k+m}=t^m{}_2F_1(\frac{m+1}{2},\frac{m+2},{2},m+1,4t^2), $$ 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 $\sum\_{j=1}^\infty|t_j|<\infty$.

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