I am working on the following:

Let $f : \mathbb C \to \mathbb C$ be analytic. Suppose for all $z \in \mathbb C$ hold $f(2z) = 4f(z)$ and $f(1) = 1$. Then $f(z) = z^2$ for all $z \in \mathbb C$.

I tried to solve this with the Taylor Series Theorem. We know that $f$ is analytic and so it has a valid expansion like \begin{align*} f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(0)}{n!} z^n \end{align*} for any $B_R(0) \subset \mathbb C$. Maybe it is better zu expand at $z_0 = 1$ since we know that $f(1) = 1$. If we do that we have instead \begin{align*} f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(1)}{n!} (z-1)^n = f(1) + \sum_{n = 1}^\infty \frac{f^{(n)}(1)}{n!} (z-1)^n = 1+\sum_{n = 1}^\infty \frac{f^{(n)}(1)}{n!} (z-1)^n. \end{align*} How can I go on?

We had the following version of the taylor series theorem:

Let $D \subset \mathbb C$ be a domain and $f : D \to \mathbb C$ differentiable. Then $f$ is analytic in $D$ and for any ball $B_R(z_0) \subset D$ the power series expansion \begin{align*} f(z) = \sum_{n = 0}^\infty \frac{f^{(n)(z_0)}}{n!}(z-z_0)^n, \quad |z-z_0| < R \end{align*} is valid. Further if $r \in (0,R)$ then \begin{align*} f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{S_r^+(z_0)} \frac{f(w)}{(w-z_0)^{n+1}} \, dw. \end{align*}

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    $\begingroup$ I would not use Taylor series. Use the identity theorem. $\endgroup$ – Daniel Fischer Dec 31 '13 at 14:03
  • $\begingroup$ I am currently practicing for exams and although I strongly believe that it is easier with the identity theorem, this problem occured in the section of the taylor series theorem, so I would like to know how to solve it that way, please. $\endgroup$ – numerion Dec 31 '13 at 14:10
  • $\begingroup$ What does the "Taylor series theorem" say? Is it just the theorem that a holomorphic function is represented by its Taylor series, or does it say anything else? $\endgroup$ – Daniel Fischer Dec 31 '13 at 14:15
  • $\begingroup$ I edited the version we had in the lecture of the theorem. $\endgroup$ – numerion Dec 31 '13 at 14:25

Let $a_n:=\displaystyle\frac{f^{(n)}(0)}{n!}$ so that $f(z)=\sum_{n\ge 0}a_nz^n$. Now plug in the given condition $4f(z)=f(2z)$: $$\sum_{n\ge 0}4a_nz^n=\sum_{n\ge 0}a_n2^nz^n $$ for all $z$. But this tells us that all coefficients on both sides must coincide, i.e. $4a_n=2^na_n$, that is, $(4-2^n)a_n=0$, so either $a_n=0$ or $n=2$, resulting in $f(z)=a_2z^2$.
Finally, condition $f(1)=1$ implies $a_2=1$.


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