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<f(n) &\le 1,\\ \sum_{n=0}^\infty f(n)&=1. \end{align}

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.


1 Answer 1


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 $0 < f(n) \le 1$.

  • $\begingroup$ 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? $\endgroup$
    – Boby
    Feb 9, 2019 at 2:42
  • 1
    $\begingroup$ 1) Once we set up the generating function, it transfers the problem from a recurrence relation to (often) a differential equation. At that point, you just apply techniques of differential equations. 2) The uniqueness follows from uniqueness of the differential equation for which you are solving. Given the right initial conditions, it can be unique, as in your problem. You would need to check, depending on the problem. 3) generatingfunctionology by Wilf is a great reference. $\endgroup$
    – Tom Chen
    Feb 9, 2019 at 20:32
  • $\begingroup$ One more question. In the above it seems like $a$ has to be less than 1, right?$ $\endgroup$
    – Boby
    Feb 10, 2019 at 19:33

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