Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help.  The question asks

Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.

Here is what I have done so far.  Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$.  Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function?  Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.
 A: Your function can be expressed as a convergent infinite product
$$f(z) = (1-z)(1-z/2)(1-z/4)(1-z/8)\dots$$
It's easy to see directly from this expression that it satisfies the functional equation $f(2z) = (1-2z)f(z)$.
I don't think that it can be expressed in terms of elementary functions, but I may be wrong.
Its values at the points $z=2^{-n}$ are related to the values of the Dedekind eta function.
A: Observations:
1) $f(1)=f(2*0.5)=(1-2*0.5)f(0.5)=0$. Hence $f(2)=(1-2*1)f(1)=0$, and so, by induction, you have $f(2^k)=0$ for all $k\geq 0$.
2) If there is another zero, say $z_0\neq 2^k$, then $z_k=z_0/2^k$ will also be zeroes of $f(z)$, and hence $f(z)=0$. So:
3) If $f(z)$ is not identically $0$, then $f(z)$ vanishes at precisely $2^k, k\geq0$ 
A: Your function has zero at $z=1/2$. I think that your formulas should be easier to work with if you expand at that point.
A: If i am not mistaken : if $f(z)$ is entire and $f(2z) = (1-2z)f(z)$, then $ \int_{C} f(z)dz =0  $ for any closed $C$. 
$$ \int_{C} f(z)dz = \int_{C} \frac{f(2z)}{1-2z}dz = 0  $$
then by Cauchy integral formula, $f(2 \cdot \frac{1}{2}) = f(1) = 0$, then $f(2) = (1-2)f(1) = 0$ and $f(2^n) = 0$. Now put $z$ with form $z=2^n$ to your series form of $f(z)$, will get $a_{0} = 0$
A: Your approach with $a_0:=1$ leads to the following recursion for the coefficients $a_k$:
$$a_k=-{2a_{k-1}\over 2^k-1}\qquad(k\geq1)\ .$$
Since $$\left|{a_{k-1}\over a_k}\right|={2^k-1\over2}\to\infty\qquad(k\to\infty)$$
it follows that these $a_k$ are the Taylor coefficients of an entire function. In other words: Your procedure is not only formal, but involves a factual function $f:\ {\mathbb C}\to{\mathbb C}$.
