Please help on complex fourier expansion.

[yungman]

Homework Statement

To prove:

e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+n^2}(a cos(nx) + n sin(nx))

For -\pi &lt; x &lt; \pi \hbox{ and a is real and not equal 0} [/tex]<br /> <br /> <h2>Homework Equations</h2><br /> <br /> Given:<br /> <br /> \int^{\pi}_{\pi} e^{ax} e^{-jnx}dx = \int^{\pi}_{\pi} C_n e^{jnx} e^{-jnx} dx = 2\pi C_n<br /> <br /> C_n = \frac{1}{2\pi} \int^{\pi}_{\pi} e^{ax} e^{-jnx}dx<br /> <br /> f(x)=e^{ax} = \sum_{n=-\infty}^{\infty} C_n e^{jnx}<br /> <br /> e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx}<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> 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)]<br /> <br /> (-1)^{-n}=(-1)^n \Rightarrow \sum_{n=-\infty}^{\infty}\frac{j(-1)^n n}{a^2+ n^2} = 0<br /> <br /> \Rightarrow e^{ax} = \frac{asinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}[cos(nx) + jsin(nx)]<br /> <br /> I cannot see how to get from this to this:<br /> <br /> e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+n^2}(a cos(nx) + n sin(nx))<br /> <br /> Please help me.<br /> <br /> Thanks<br /> <br /> Alan

 

[vela]

If you group the +n and -n terms together, you can write the sum as

\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx} = \frac{1}{a}+\sum_{n=1}^\infty \left[\frac{(-1)^n}{a-jn} e^{jnx} + \frac{(-1)^{-n}}{a-j(-n)} e^{j(-n)x}\right]

Try simplifying what's inside the square brackets.

 

[yungman]

If you group the +n and -n terms together, you can write the sum as

\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx} = \frac{1}{a}+\sum_{n=1}^\infty \left[\frac{(-1)^n}{a-jn} e^{jnx} + \frac{(-1)^{-n}}{a-j(-n)} e^{j(-n)x}\right]

Try simplifying what's inside the square brackets.

Thanks for your reply. I worked on this and I still going nowhere. THis is what I have:

\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx} = \frac{1}{a}+\sum_{n=1}^\infty [\frac{(-1)^n}{a-jn} e^{jnx} + \frac{(-1)^{-n}}{a-j(-n)} e^{j(-n)x}] = <br /> <br /> \frac{1}{a}+\sum_{n=1}^\infty \frac{(-1)^n}{a^2+n^2} [a(e^{jnx} + e^{-jnx}) + jn(e^{jnx} + e^{-jnx})]

= \frac{1}{a}+\sum_{n=1}^\infty \frac{(-1)^n}{a^2+n^2} [2a(cos(nx)) + j2n(sin(nx))]

Still with this, it is no where close to the answer. The important thing is the answer contain sum from n=-\infty \hbox{ to } n=\infty[/tex]

 

[vela]

You're close. Remember that sin(nx) has a 2j on the bottom, not just a 2.EDIT: I checked the formula you're trying to prove, and there's a mistake in it. It should be

e^{ax} = \frac{\sinh \pi a}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+n^2}(a \cos nx - n \sin nx)

The sign of the sine term is wrong in the original.

 

[yungman]

I went through my derivation again and I found out what I did wrong. This is my new finding:

\sum_{n=-\infty}^{\infty} (-1)^n n[cos(nx)] = ... (-1)^{-2}(-2)cos(-2x) + (-1)^{-1}(-1)cos(-x) + 0 + (-1)^1(1)cos(x) + (-1)^2(2)cos(2x)...

= ...-2cos(2x) +cos(x) + 0 - cos(x) + 2cos(2x)...= 0

\sum_{n=-\infty}^{\infty} (-1)^n a[sin(nx)] = ... (-1)^{-2}(a)sin(-2x) + (-1)^{-1}(a)sin(-x) + 0 + (-1)^1(a)sin(x) + (-1)^2(a)sin(2x)...

= ...-a[sin(2x)] +a[sin(x)] + 0 - a[sin(x)] + a[sin(2x)]...= 0

With that, I go on to substitude into the original equation:

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)]

= \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}[a(cos(nx)) + jn(cos(nx)) +ja(sin(nx))-n(sin(nx))]

= \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}[a(cos(nx)) -n(sin(nx))]

My original assumption was wrong and now I got the answer. Thanks for your help anyway.

Alan