Rewriting power series $\sum\limits_0^{\infty}{\frac{\pi^nz^{2n+1}}{2n +2}}$ as $\sum\limits_0^{\infty}{a_n(z-z_0)^n}$

Question

I'm suppossed to bring this power series $$\sum_0^{\infty}{\frac{\pi^nz^{2n+1}}{2n +2}}$$

into the standard form $$\sum_0^{\infty}{a_n(z-z_0)^n}$$

Now, I know that $$a_n=\frac{\pi^n}{2n +2}$$ and $$z_0=0$$

but have absolutely no idea how I'm suppossed to rewrite $z^{2n+1}$ as some singular term $z^n$. I could define a new variable over the interval of uneven numbers, but I don't think that's the point here. Maybe I'm also misunderstanding something.

Use $$a_{2n}=0\qquad a_{2n+1}=\frac{\pi^n}{2n +2}$$ then indeed the series is $$\sum_{n=0}^\infty a_nz^n$$ — Did, May 09 '17 at 13:43
Alternatively, in the standard form equation, you can set $a_n = 0$ if $n$ is even and $a_n = \frac{\pi^{(n-1)/2}}{n+1}$ if $n$ is odd. — Paul Aljabar, May 09 '17 at 13:47
@Did But wouldn't this practically be the same as defininf n as beinf from the set of uneven integers? — Skydiver, May 09 '17 at 13:47
It might help to use different letters to denote the index in each of the summations. — Paul Aljabar, May 09 '17 at 13:49

score 1 · Accepted Answer · answered May 09 '17 at 14:23

1

Your series is

$$\sum a_n (z-z_0)^n=$$ $$\sum a_{2n}(z-z_0)^{2n}+a_{2n+1}(z-z_0)^{2n+1} $$

thus

$$a_{2n}=0$$ $$a_{2n+1}=\frac {\pi^n}{2n+2} $$ $$z_0=0$$

answered May 09 '17 at 14:23

hamam_Abdallah

62,951

Rewriting power series $\sum\limits_0^{\infty}{\frac{\pi^nz^{2n+1}}{2n +2}}$ as $\sum\limits_0^{\infty}{a_n(z-z_0)^n}$

1 Answers1