1. ## Power Series

I'm stuck on this dam question:

I mean the first parts are pretty simple but I don't see how it translates to the series, any help from you old chaps would be greatly appreciated.

2. Originally Posted by gebauer1988
I'm stuck on this dam question:

I mean the first parts are pretty simple but I don't see how it translates to the series, any help from you old chaps would be greatly appreciated.
Okay, what did you get for the power series for $e^z+ e^{\omega z}+ e^{2\omega z}$? Look at the real and imaginary parts of that.

Do you see that $\omega^3= 1$ and so $\omega^n= \omega^k$ where k= n (mod 3)?

3. Thanks for that top notch response, now I am beginning to question whether I have my power series right you see. I have the sum of (3/n)e^((w^n)z)

4. Originally Posted by gebauer1988
I'm stuck on this dam question:

I mean the first parts are pretty simple but I don't see how it translates to the series, any help from you old chaps would be greatly appreciated.

I suppose, and hope, you know that $\sum\limits_{n=0}^\infty\frac{x^n}{n!}=e^x$ , and for n not a multiple of 3 we get $1+w^n+w^{2n}=1+e^{2\pi n\slash 3}+e^{4\pi n\slash 3}=\frac{e^{3\cdot 2\pi n\slash 3}-1}{e^{2\pi n\slash 3}-1}=0$ , and $1+w^n+w^{2n}=3$ , for n a multiple of 3 , so we get:

$e^z+e^{wz}+e^{w^2z}=\sum\limits_{n=0}^\infty\frac{ z^n+(wz)^n+(w^2z)^n}{n!}$ $=\sum\limits_{n=0}^\infty \frac{z^n}{n!}\left(1+w^n+(w^n)^2\right)=3\sum\lim its_{n=0}^\infty\frac{z^{3n}}{(3n)!}$

From the above we get $\sum\limits_{n=0}^\infty\frac{8^n}{(3n)!}=\sum\lim its_{n=0}^\infty\frac{2^{3n}}{(3n)!}=\frac{1}{3}\l eft(e^2+e^{2w}+e^{2w^2}\right)$ $=\frac{1}{3}\left(e^2+e^{2\displaystyle{e^{2\pi \slash 3}}}+e^{2\displaystyle{e^{4\pi\slash 3}}}\right)$ , and:

$e^{2e^{2\pi\slash 3}}+e^{2e^{4\pi\slash 3}}=e^{-1+\sqrt{3}i}+e^{-1-\sqrt{3}i}=2e^{-1}\cos\sqrt{3}\Longrightarrow\sum\limits_{n=0}^\in fty\frac{8^n}{(3n)!}=\frac{e^2+2e^{-1}\cos\sqrt{3}}{3}$ $\cong 2.4236$

You do the rest.

Tonio

5. Superb, thanks for the explanation