Simplify $f(x) + b\cdot f(x+a) + b^2 \cdot f(x+2a)+...$

Question

Is it possible to simplify the following expression: $$ S = f(x) + b\cdot f(x+a) + b^2 \cdot f(x+2a)+... $$ under the assumption that $f$ is infinitely smooth and defined on compact support (in other words $f$ is non-zero only on some interval $[-d,d]$) and $a,b$ are real numbers? I am also interested in the case when $a,b$ are either very small $(\to 0)$ or very large $(\to \infty)$. For example, if $a$ is very small number, we can expand $f$ into Taylor series and get: \begin{align} S &\approx f(x) \left[ 1+b+b^2 + ... \right] + abf'(x)\left[1+2b+3b^2+...\right]\\ &= \frac{f(x)}{1-b} + \frac{abf'(x)}{(1-b)^2}+... \end{align}

Answer 1

Not an answer, but too long for a comment.

You might be able to analyze this with the Fourier transform. Let:

$$h(x)=\sum_{k=0}^\infty b^kf(x+ka)$$

Then, at least if $|b|<1,$

$$\begin{align} \hat h(\xi)&=\sum_{k=0}^\infty b^ke^{2\pi ka \xi i}\hat f(\xi)\\ &=\frac{\hat f(\xi)}{1-be^{2\pi a\xi i}} \end{align} $$ Doing an inverse Fourier transform on this is going to require a case-by-case analysis, I think.

You also have an equality:

$$h(x)=f(x)+bh(x+a)$$