I have an equation $f(x) = \alpha f(\frac{x}{\beta}) $ which I am trying to solve with generating functions, but can only find the trivial solution $ f(x)=0 $. I have tried:
$$\begin{align*}
f(x)&=\sum_{n=0}^{\infty}{a_{n}x^{n}}\\
\sum_{n=0}^{\infty}{a_{n}x^{n}}&=\alpha \sum_{n=0}^{\infty}x^{n}{\frac{a_{n}}{b^{n}}}\\
a_{n}\left(1-\frac{\alpha}{\beta^{n}}\right)&=0\\
a_{n}&=0\\
f(x)&=0\\
\end{align*}$$
But there is also a solution $c \alpha^{\log_{\beta}(x) – 1}$ which this method dosen't seem to give. Have I made a mistake anywhere?
Best Answer
As stated in the comment section, the equation $a_n(\beta^n-\alpha) = 0$ imply the subequations $a_n = 0$ or $\beta^n-\alpha = 0$. In other words, when the fixed parameters $\alpha,\beta$ are such that $\beta^n-\alpha \neq 0$, then you can conclude indeed that $a_n = 0$ $\forall n$ and $f \equiv 0$. However, when $\beta^n-\alpha = 0$ for a given $n$, then $a_m = 0$ $\forall m \neq n$ and the coefficient $a_n$ remains free, hence the monomial solution $f(x) = a_nx^n$, mentioned by Qiaochu Yuan.
Nevertheless, in this context, you are limiting yourself to "nice" solutions admitting a Taylor expansion. In the same spirit, we may consider non-integer exponents $n$ $-$ it may viewed as a Frobenius method applied to functional equations. Then the subequation $\beta^n-\alpha = 0$ is always solved by $n = \log_\beta\alpha$ (which be complex notwithstanding), hence $f(x) \propto x^{\log_\beta\alpha} \propto \alpha^{\log_\beta x}$, also mentioned in the comment section.
It is to be noted that the present functional equation is linear. In consequence, the superposition principle permits to construct linear combinations of several solutions, for instance by considering different branches of the logarithm or in the case where $\alpha = \pm1$ and $\beta = -1$, so that all even-, respectively, odd-indexed coefficients are free parameters, whose linear combinations generates all the even/odd functions, as well as (anti-)periodic functions by composing them with logarithm, as mentioned by Ryszard Szwarc.
In a broader perspective, this functional equation, rewritten as $f(\beta x) = \alpha f(x)$, can be seen as an eigenvalue problem for a (linear) composition operator $C_\beta(f)(x) := f(\beta x)$; the solutions are then the eigenfunctions associated to the eigenvalue $\alpha$. You can then follow this answer of mine, but it may turn out to be a quite overcomplicated method in the present case.
Finally, some distributional solutions may be even considered in certain cases, such as $f(x) = \delta(\alpha x)$ for $\alpha > 0$, $\beta = 1$, where $\delta$ is the Dirac delta function.