# Functional equation $(n+1) f(n+1)= (a n+b) f(n)$ for $n=0,1,...$

I am looking for a solution to the following functional equation: \begin{align} (n+1) f(n+1)= (a n+b) f(n), n=0,1,... \end{align} where $$a$$ and $$b$$ are some positive constants. Moreover, $$f(n)$$ is positive and \begin{align} 0

I was able to find a solution if $$a=b \le 1$$ \begin{align} f(n)= \frac{c}{(1+c)^{n+1}}, \, c=\frac{1-a}{a}. \end{align}

My questions:

1. Is there a systematic way to solve this equation?
2. Is the solution to this equation unique?

Any reference would also be appreciated.

One way is to use generating functions. Define \begin{align*} g(x) = \sum_{n=0}^{\infty} f(n) x^n \end{align*} Henceforth, I will use the notation \begin{align*} g(x) \leftrightarrow\{f(n)\} \end{align*} to indicate that $$g(x)$$ has coefficients of $$f(n)$$ for $$x^n$$ in its series expansion. Then we have \begin{align*} g'(x) \leftrightarrow\{(n+1)f(n+1)\} \end{align*} and \begin{align*} ax g'(x) + bg(x) \leftrightarrow \{(an+b)f(n)\} \end{align*} We can set equal the corresponding generating functions \begin{align*} g'(x) = axg'(x) + bg(x) \end{align*} which has solution \begin{align*} g(x) = c_1 (1 - ax)^{-b/a} = \sum_{n=0}^{\infty} c_1 (-a)^n \binom{-b/a}{n}x^n \leftrightarrow \left\{c_1 (-a)^n \binom{-b/a}{n}\right\} \end{align*} So, \begin{align*} f(n) = c_1 (-a)^n \binom{-b/a}{n} \end{align*} The constant $$c_1$$ can be found be evaluating \begin{align*} 1 = \sum_{n=0}^{\infty} f(n) = g(1) = c_1 (1 - a)^{-b/a} \implies c_1 = (1 - a)^{b/a} \end{align*} So we find \begin{align*} f(n) = (1 - a)^{b/a} (-a)^n \binom{-b/a}{n} \end{align*} It's worth noting that $$f(n)$$ is always positive, so we also have $$0 < f(n) \le 1$$.
• Thanks. Really cool. I have a few questions: 1) how do you know the solution to these generating functions? 2) Is the final solution for $f(n)$ unique? 3) Can you give me some reference on how to solve functional equations?
• One more question. In the above it seems like $a$ has to be less than 1, right?\$