# Power series and exponential excercise

The power series $f(z):= \sum_{\nu \geq 0}a_{\nu}z^{\nu}$ converges in a disc $B$ centered at $0$. For every $z\in B$ such that $2z$ is also in $B$, $f$ satisfies $f(2z) = (f(z))^{2}$. Show that if $f(0)\neq 0$, then $f(z)=\mathrm{exp} bz$, with $b:=f'(0)=a_{1}.$

I'm completely lost with this problem. Could someone give me a hint?

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I'm also lost. Shouldn't the series for $f(z)$ start with $\nu=0$ in order that $f(0)\neq 0$? – Fabian Mar 29 '11 at 8:03
yep I'm sorry that's a typo. I'll fix it immediatly. – Sak Mar 29 '11 at 8:06
I think the course of action would be to prove that the power series is holomorphic and then calculate it's derivatives to see if it satisties f' = f. But I'm kind of confused with the propperties of power series. It would be great if someone could give me some steps to follow and then I can fill the gaps as to how to use power series. – Sak Mar 29 '11 at 8:08
@Chu: I think it is obvious that the power series is holomorphic (inside the convergence radius), see en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions. If it is not obvious, what do you already know about holomorphic functions? – Fabian Mar 29 '11 at 8:14
In principle, you can plug in the power series into $f(2z) = (f(z))^2$ and compare coefficients order by order in $z$. – Fabian Mar 29 '11 at 8:15

First you can plug in $z = 0$ to get that $a_0 = 1$. Since $\log(1 + z)$ is analytic on $|z| < 1$, the composition $\log(1 + \sum_{n=1}^{\infty}a_nz^n) = \log(f(z))$ will also be analytic on some disk $D$ centered at $z = 0$. Let $g(z) = \log(f(z))$. Your conditions say that $g(2z) = 2g(z)$ now. Look at the power series of $g(z)$ to show that this equation forces $g(z)$ to be of the form $cz$, and you have enough info to figure out what $c$ is.
Hint: take the logarithm of your equation and arrive at an equation for $g(z) = \log f(z)$.