# solve recursion equation with condition

How can I find $f(x)$ in terms of $x$ in the following equation?

$-k\, x \,f(x) + k \, (x + 1)\, f(x + 1) - r \, f(x) + r\, f(x - 1) =0$

$r$ and $k$ are constants and $f(x)$ should satisfy:

$\sum_0^{\infty} f(x)=1$

Solve just find $f(x)$ in terms of other statements:

Solve[-k x f[x] + k (x + 1) f[x + 1] - r f[x] + r f[x - 1] == 0, f[x]]

{{f[x] -> (2 f[-1 + x] + f[1 + x] + x f[1 + x])/(2 + x)}}


What you have is basically a nonlinear second order recursion, and in this case it can be solved by:

sol = RSolve[-k x f[x] + k (x + 1) f[x + 1] - r f[x] + r f[x - 1] == 0, f[x], x]


The answer is fairly large, and besides having variables r and k, it also has two constants C and C, so there may be enough flexibility to enforce your desired constraint. The two constants are basically caused by the initial conditions of the recursion. For example, if you let f and f both be zero you can enforce a very simple solution

RSolve[-k x f[x] + k (x + 1) f[x + 1] - r f[x] + r f[x - 1] == 0
&& f == 0 && f == 0, f[x], x]


that unfortunately violates your constraint. Nonetheless, since we are free to choose the two constants, let's make our lives easy and choose C=0. In this case the solution is summable:

Sum[sol[[1, 1, 2]] //. C -> 0, {x, 0, Infinity}]

(2 E^(r/k) k C)/r


So now, for any specified r and k, you just need to pick C so that this is 1. Thus

Solve[(2 E^(r/k) k C)/r == 1, C]

C -> (E^(-(r/k)) r)/(2 k)

• Thanks bill, but is it possible to put the mentioned constraint and then find the solution? Jan 30, 2015 at 13:43
• Yes, by judicious choice of the constants C and C. See update. Jan 30, 2015 at 13:54