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

−kxf(x)+k(x+1)f(x+1)−rf(x)+rf(x−1)=0 -k\, x \,f(x) + k \, (x + 1)\, f(x + 1) – r \, f(x) + r\, f(x – 1) =0

rr and kk are constants and f(x)f(x) should satisfy:

∑∞0f(x)=1\sum_0^{\infty} f(x)=1

Solve just find f(x)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)}}

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1 Answer

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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[1] and C[2], 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[0] and f[1] 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] == 0 && f[1] == 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[2]=0. In this case the solution is summable:

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

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

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

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

C[1] -> (E^(-(r/k)) r)/(2 k)

Thanks bill, but is it possible to put the mentioned constraint and then find the solution?

– Soodeh Z.

Jan 30 ’15 at 13:43

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Yes, by judicious choice of the constants C[1] and C[2]. See update.

– bill s

Jan 30 ’15 at 13:54

thanks bill. Nice answer.

– Soodeh Z.

Jan 30 ’15 at 13:59