1. ## power series derivative

Consider the powerseries:
$f(z)=\displaystyle \sum_{n=0}^{\infty} a_n z^n$
with the property that
$f(\frac{1}{a})=\frac{1}{a^3} \forall n\in\mathbb N$
Prove that
$f'(0)=0$

I had hint to use Uniqueness Theorem of Power Series, but I still have no idea how to do it. Please help me, thanks a lot.

2. You mean $f(\frac{1}{n})= \frac{1}{n^3}\forall n\in \mathbb N$.

For the power series $\displaystyle f(z)= \sum_{n=0}^\infty a_nz^n$, $f'(z)= \displaystyle\sum_{n=0}^\infty} na_nz^{n-1}$ and, in particular, $f'(0)= a_1$. You are really asked to prove that $a_1= 0$.

3. Originally Posted by tsang

Consider the powerseries:
$f(z)=\displaystyle \sum_{n=0}^{\infty} a_n z^n$
with the property that

$f(\frac{1}{a})=\frac{1}{a^3} \forall n\in\mathbb N$

Prove that

$f'(0)=0$

I had hint to use Uniqueness Theorem of Power Series, but I still have no idea how to do it. Please help me, thanks a lot.
If $f(\frac{1}{a})= \frac{1}{a^{3}}\ , \ \forall a \in \mathbb{N}$ that means that is...

$\displaystyle \sum_{n=0}^{\infty} \frac{a_{n}}{a^{n}} = \frac{1}{a^{3}}\ , \ \forall a \in \mathbb{N}$ (1)

... so that is...

$a_{n}=\left\{\begin{array}{ll}1 ,\,\, n=3\\{}\\0 ,\,\, n \ne 3\end{array}\right.$ (2)

... and from (2) You conclude that is $f^{'}(0)=0$...

Kind regards

$\chi$ $\sigma$