I'm stuck on this dam question:

http://i933.photobucket.com/albums/a...8/Analysis.png

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.

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- Jan 11th 2010, 03:00 AMgebauer1988Power Series
I'm stuck on this dam question:

http://i933.photobucket.com/albums/a...8/Analysis.png

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. - Jan 11th 2010, 06:54 AMHallsofIvy
- Jan 11th 2010, 07:06 AMgebauer1988
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)

- Jan 11th 2010, 07:41 AMtonio

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

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

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

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

You do the rest.

Tonio - Jan 11th 2010, 07:46 AMgebauer1988
Superb, thanks for the explanation