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I found this on Priestley's Complex Analysis in the Laplace transforms bit.

Suppose $f$ satisfies $f'(t)=f(kt)$ for $t>0$, where $0<k<1$ and $f(0)=1$. Prove that $$f(t)=\sum_{n=0}^{\infty}\frac{k^{n(n-1)/2}}{n!}t^n$$

Applying the Laplace transform directly to $f'(t)=f(kt)$ gives a functional equation but I'm unsure how to solve it. Any hints?

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I do not know if the application of the Laplace transform would lead to a quicker solution, however, effectively, it is required to prove that $$f^{(n)}(0)=k^{\frac{n(n-1)}{2}}$$ In order to show this we can prove that $$f^{(n)}(t)=k^{\frac{n(n-1)}{2}}f(k^nt)$$ That it is true for $n=0$ and $n=1$ is easily verified. Now for $n\ge 1$ $$f^{(n+1)}(0)=k^{\frac{n(n-1)}{2}}\left(f(k^nt)\right)'=k^{\frac{n(n-1)}{2}+n}f(k^{n+1}t)=k^{\frac{n(n+1)}{2}}f(k^{n+1}t)$$ So the result holds by induction.

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I can see where you're going with this but with $f$ being a real function, how can I be sure it is representable by a power series? – E.Lim Nov 15 '12 at 8:43
well, obviously this is a MacLaurin expansion and the given relation allows to deduce existence of the derivatives recursively – Valentin Nov 16 '12 at 17:11
Back to this. I mean, not every infinitely differentiable function is equal to its Taylor series, example $f(x)=e^{-1/x^2}$ – E.Lim Dec 19 '12 at 13:51

Let $f(t)=\sum\limits_{n=0}^\infty a_nt^n$ ,

Then $\sum\limits_{n=0}^\infty na_nt^{n-1}=\sum\limits_{n=0}^\infty a_n(kt)^n$

$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=0}^\infty k^na_nt^n=0$

$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=1}^\infty k^{n-1}a_{n-1}t^{n-1}=0$


$\therefore na_n-k^{n-1}a_{n-1}=0$


$\therefore\begin{cases}a_0=a_0\\a_n=\dfrac{(k^0k^1k^2......k^{n-1})a_0}{1\times2\times3\times......n}\forall n\in\mathbb{N}\end{cases}$

$\begin{cases}a_0=a_0\\a_n=\dfrac{k^{0+1+2+......+(n-1)}a_0}{n!}\forall n\in\mathbb{N}\end{cases}$

$\begin{cases}a_0=a_0\\a_n=\dfrac{k^{\frac{n(n-1)}{2}}a_0}{n!}\forall n\in\mathbb{N}\end{cases}$

$a_n=\dfrac{k^{\frac{n(n-1)}{2}}a_0}{n!}\forall n\in\mathbb{Z}^*$

$\therefore f(t)=C\sum\limits_{n=0}^\infty\dfrac{k^{\frac{n(n-1)}{2}}t^n}{n!}$

$f(0)=1$ :


$\therefore f(t)=\sum\limits_{n=0}^\infty\dfrac{k^{\frac{n(n-1)}{2}}t^n}{n!}$

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