1. Jun 14, 2010

### yungman

1. The problem statement, all variables and given/known data
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 [itex] -\pi < x < \pi \hbox{ and a is real and not equal 0} [/tex]

2. Relevant equations

Given:

$$\int^{\pi}_{\pi} e^{ax} e^{-jnx}dx = \int^{\pi}_{\pi} C_n e^{jnx} e^{-jnx} dx = 2\pi C_n$$

$$C_n = \frac{1}{2\pi} \int^{\pi}_{\pi} e^{ax} e^{-jnx}dx$$

$$f(x)=e^{ax} = \sum_{n=-\infty}^{\infty} C_n e^{jnx}$$

$$e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx}$$

3. The attempt at a solution

$$e^{ax} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a-jn} e^{jnx} = \frac{sinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}(a+jn)[cos(nx) + jsin(nx)]$$

$$(-1)^{-n}=(-1)^n \Rightarrow \sum_{n=-\infty}^{\infty}\frac{j(-1)^n n}{a^2+ n^2} = 0$$

$$\Rightarrow e^{ax} = \frac{asinh(\pi a)}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{a^2+ n^2}[cos(nx) + jsin(nx)]$$

I cannot see how to get from this to this:

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

Thanks

Alan

Last edited: Jun 14, 2010
2. Jun 14, 2010

### vela

Staff Emeritus
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.

3. Jun 14, 2010

### yungman

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

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

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

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

4. Jun 15, 2010

### vela

Staff Emeritus
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.

Last edited: Jun 15, 2010
5. Jun 15, 2010

### 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

Last edited: Jun 15, 2010