Solving recurrence equation with indexes from negative infinity to positive infinity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:01:28Z http://mathoverflow.net/feeds/question/20704 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20704/solving-recurrence-equation-with-indexes-from-negative-infinity-to-positive-infin Solving recurrence equation with indexes from negative infinity to positive infinity Ross Tang 2010-04-08T09:38:49Z 2010-05-25T16:39:50Z <p>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:</p> <p><code>$a_m=\sum _{j=1}^{\infty } \left(T_ja_{m+j}+T_ja_{m-j}\right) + \delta _{m,0}$</code></p> <p>where <code>$\delta_{m,0}$</code> is the Kronecker delta function. i.e.:</p> <p><code>$\delta_{i,j} = \begin{cases} &amp; 1 \text{ if } i=j \\ &amp; 0 \text{ if } i \neq j \end{cases}$</code> </p> <p>Here I would like to solve <code>$a_m$</code>, where the index of m is from negative infinity to positive infinity, while <code>$T_j$</code> is a given sequence, and p is just a given constant.</p> <p>Defining the generating function <code>$G(z)=\sum _{k=-\infty }^{\infty } a_kz^k$</code>, I found that:</p> <p><code>$G(z)=\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}$</code></p> <p>The problem is, how am I going to do series expansion on G? Doing a simple expansion of <code>$\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}=\sum _{j=0}^{\infty } \left(\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)\right){}^j$</code> won't help. Since the power is too difficult to expand out.</p> <p>And contour integration isn't helping as well, since it is too difficult to compute analytically or numerically too.</p> <p>Here I would like to ask about direction in obtaining analytical solution, or approximated one.</p> <p>And in my case, my function G is given by:</p> <p><code>$G(z)=\left(1+\frac{3i}{2r^3}\left(r^2\left(\ln \left(1-\frac{e^{i r}}{z}\right)+\ln \left(1-e^{i r}z\right)\right)\right)-i r\left(\text{Li}_2\left(\frac{e^{i r}}{z}\right)+\text{Li}_2\left(e^{i r}z\right)\right)+\text{Li}_3\left(\frac{e^{i r}}{z}\right)+\text{Li}_3\left(e^{i r}z\right)\right){}^{-1}$</code></p> <p>p.s.:I have posted the same problem in <a href="http://www.voofie.com/content/44/solving-recurrence-equation-with-indexes-from-negative-infinity-to-positive-infinity/" rel="nofollow">Voofie</a>.</p> http://mathoverflow.net/questions/20704/solving-recurrence-equation-with-indexes-from-negative-infinity-to-positive-infin/25900#25900 Answer by Julián Aguirre for Solving recurrence equation with indexes from negative infinity to positive infinity Julián Aguirre 2010-05-25T16:39:50Z 2010-05-25T16:39:50Z <p>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 $\sum_{j=1}^\infty|t_j|&lt;\infty$ 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 $\sum_{j=1}^\infty|t_j|&lt;\dfrac{1}{2}$, 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.</p> <p>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 $|t|&lt;\dfrac{1}{2}$ for this to make sense.</p> <p>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 $\sum_{j=1}^\infty|t_j|&lt;\infty$.</p>