I am trying to analytically solve the following recurrence equation
$(\lambda + \gamma) f_n = \gamma (1-p)^{n-1} f_{n+1} + \gamma [1-(1-p)^{n-1}] f_{n+2} + \lambda f_{n-1}\,,$
Under constraints of the type $f_{-1}=f_{-2}=0$ and $\sum_{n=0}^\infty f_n = 1$ for $n\geq 0$ and $p \in (0,1), \lambda,\gamma > 0$.
I tried Mathematica's RSolve without success.
1) Is there any hope for an analytical solution? 2) Could you recommend any material that would help me understanding this equation better.
Thanks a lot!
To make it easier to key in, I will write it as $(a+b) f_n = b q^{n-1} f_{n+1} + b [1-q^{n-1}] f_{n+2} + af_{n-1} $, where $q = 1-p$. Getting the highest subscript on the left, this becomes $b [1-q^{n-1}] f_{n+2} =-b q^{n-1} f_{n+1}+(a+b) f_n-af_{n-1} $. Shifting the indices up by one, this becomes $b [1-q^{n}] f_{n+3} =-b q^{n} f_{n+2}+(a+b) f_{n+1}-af_{n} $.
If we try $f_n = r^n$, then $b [1-q^{n}] r^{n+3} =-b q^{n} r^{n+2}+(a+b) r^{n+1}-ar^{n} $ or $b [1-q^{n}] r^{3} =-b q^{n} r^{2}+(a+b) r-a $ which doesn't work.
Using the next simplest approach, let $F(x) =\sum_{n=0}^{\infty} f_n x^n $.
Starting on the $f_n$ term:
$\sum_{n=0}^{\infty} af_n x^n =-aF(x) $.
$\sum_{n=0}^{\infty}(a+b) f_{n+1}x^n =\frac{a+b}{x}\sum_{n=0}^{\infty}(a+b) f_{n+1}x^{n+1} =\frac{a+b}{x}\sum_{n=1}^{\infty}(a+b) f_{n}x^{n} =\frac{a+b}{x}(F(x)-f_0) $.
$\sum_{n=0}^{\infty}b q^{n} f_{n+2} x^n =\frac{b}{q^2x^2}\sum_{n=0}^{\infty}q^{n+2} f_{n+2} x^{n+2} =\frac{b}{q^2x^2}\sum_{n=2}^{\infty} f_{n} (qx)^{n} =\frac{b}{q^2x^2}(F(qx)-f_0-qxf_1) $.
$\begin{array}\\ \sum_{n=0}^{\infty}b [1-q^{n}] f_{n+3}x^n &=\frac{b}{x^3}\sum_{n=0}^{\infty} [1-q^{n}] f_{n+3}x^{n+3}\\ &=\frac{b}{x^3}\sum_{n=3}^{\infty} [1-q^{n-3}] f_{n}x^{n}\\ &=\frac{b}{x^3}\left(\sum_{n=3}^{\infty}f_{n}x^{n} -\sum_{n=3}^{\infty} q^{n-3} f_{n}x^{n}\right)\\ &=\frac{b}{x^3}\left((F(x)-f_0-xf_1-x^2f_2) -\frac1{q^3}\sum_{n=3}^{\infty} f_{n}(qx)^{n}\right)\\ &=\frac{b}{x^3}\left((F(x)-f_0-xf_1-x^2f_2) -\frac1{q^3}(F(qx)-f_0-qxf_1-q^2x^2f_2))\right)\\ &=\frac{b}{x^3}(F(x)-\frac1{q^3}(F(qx)) +\frac{b}{x^3}\left((-f_0-xf_1-x^2f_2) -\frac1{q^3}(-f_0-qxf_1-q^2x^2f_2))\right)\\ &=\frac{b}{x^3}F(x)-\frac{b}{q^3x^3}F(qx) +\frac{b}{x^3}\left((\frac1{q^3}-1)f_0+x(\frac1{q^2}-1)f_1-x^2(\frac1{q}-1)f_2 \right)\\ \end{array} $
Therefore, $\frac{b}{x^3}F(x)-\frac{b}{q^3x^3}F(qx) +\frac{b}{x^3}\left((\frac1{q^3}-1)f_0+x(\frac1{q^2}-1)f_1-x^2(\frac1{q}-1)f_2 \right) =\frac{b}{q^2x^2}(F(qx)-f_0-qxf_1) +\frac{a+b}{x}(F(x)-f_0) -aF(x) $
or $F(x)(\frac{b}{x^3}-\frac{a+b}{x}+a) -F(qx)(\frac{b}{q^3x^3}+\frac{b}{q^2x^2}) =f_0(-\frac{b}{x^3}(\frac1{q^3}-1)-\frac{b}{q^2x^2}+\frac{a+b}{x}) +xf_1(-\frac{b}{x^3}+q\frac{b}{q^2x^2}) +x^2f_2\frac{b}{x^3}(\frac1{q}-1) $.
This can be simplified a little, but the main thing is that $F(x)$ is equal to a function of $F(qx)$ plus some other function. If $|q| < 1$, then, presumably, we can iterate, getting $F(x)$ in terms of $F(q^2x)$, then $F(q^3x)$, and so on, finally getting $F(x)$ in terms of $F(q^mx)$ for arbitrary $m$.
Letting $m \to \infty$, this gives $F(x)$ in terms of $F(0)$ and a real mess (to use technical terminology).
I'll leave it at this, since this was annoying enough.