- #1
yungman
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Homework Statement
To prove:
[tex]e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+n^2}(a cos(nx) + n sin(nx))[/tex]
For [itex] -\pi < x < \pi \hbox{ and a is real and not equal 0} [/tex]
Homework Equations
Given:
[tex] \int^{\pi}_{\pi} e^{ax} e^{-jnx}dx = \int^{\pi}_{\pi} C_n e^{jnx} e^{-jnx} dx = 2\pi C_n [/tex]
[tex] C_n = \frac{1}{2\pi} \int^{\pi}_{\pi} e^{ax} e^{-jnx}dx [/tex]
[tex] f(x)=e^{ax} = \sum_{n=-\infty}^{\infty} C_n e^{jnx}[/tex]
[tex]e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx}[/tex]
The Attempt at a Solution
[tex]e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}(a+jn)[cos(nx) + jsin(nx)][/tex]
[tex] (-1)^{-n}=(-1)^n \Rightarrow \sum_{n=-\infty}^{\infty}\frac{j(-1)^n n}{a^2+ n^2} = 0 [/tex]
[tex] \Rightarrow e^{ax} = \frac{asinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}[cos(nx) + jsin(nx)][/tex]
I cannot see how to get from this to this:
[tex]e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+n^2}(a cos(nx) + n sin(nx))[/tex]
Please help me.
Thanks
Alan
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