Exponential Fourier Series - Trouble solving

**4Math** · August 20th 2010, 02:15 AM

Hello,

I have troubling simplifying the following problem. It is the second integral solving for c(k) of f involving [i] that is confusing.

I also was wondering if there is simpler way to solve for the exponetial fourier series for the given problem, than the way I done it. I would greatly appriciate guidance.

Thank you

**Ackbeet** · August 20th 2010, 02:33 AM

Part of your problem is that for the $c_{k}$ integral, you can't use half the interval and multiply by two like you could for the $c_{0}$ case. You can only do that, in general, when your complete integrand is even. But the presence of that particular exponential function in the integrand precludes that. So you have to keep the full $[-\pi,\pi]$ interval for the limits in the $c_{k}$ integration, I'm afraid.

I don't know of any way to integrate $x^{2}e^{-ikx}$ other than by parts twice. I'm not sure I would recommend breaking the integrals up into the trig functions, because then you're going to have to do integration by parts twice on two integrals. There's no need, when working with the exponential Fourier series, to look at sin and cos individually. It's an easier integration with just the exponential in there.

So I would carry these changes through, and then we'll see what happens at the end. Sound good?

**4Math** · August 20th 2010, 07:02 AM

Originally Posted by Ackbeet

Part of your problem is that for the $c_{k}$ integral, you can't use half the interval and multiply by two like you could for the $c_{0}$ case. You can only do that, in general, when your complete integrand is even. But the presence of that particular exponential function in the integrand precludes that. So you have to keep the full $[-\pi,\pi]$ interval for the limits in the $c_{k}$ integration, I'm afraid.

I don't know of any way to integrate $x^{2}e^{-ikx}$ other than by parts twice. I'm not sure I would recommend breaking the integrals up into the trig functions, because then you're going to have to do integration by parts twice on two integrals. There's no need, when working with the exponential Fourier series, to look at sin and cos individually. It's an easier integration with just the exponential in there.

So I would carry these changes through, and then we'll see what happens at the end. Sound good?

I got the following expression after integrating the integral $[-\pi,\pi]$ :

$c_{k} = \frac{1}{2\pi} ({-\frac{\pi^{2}}{ik}}e^{-ik\pi}}+{\frac{2\pi}{k^{2}}e^{-ik\pi}+{\frac{2}{ik^{3}}e^{-ik\pi}+\frac{\pi^{2}}{ik}}e^{ik\pi}+{\frac{2\pi}{k ^{2}}e^{ik\pi}-{\frac{2}{ik^{3}}e^{ik\pi})$

and I know that

$e^{ik\pi}=(-1)^{k}$

$\frac{1}{2\pi}(\frac{\pi^{2}}{ik}}e^{ik\pi}{-\frac{\pi^{2}}{ik}}e^{-ik\pi}})=0$

$\frac{1}{2\pi}({\frac{2}{ik^{3}}e^{-ik\pi}-{\frac{2}{ik^{3}}e^{ik\pi})=0$

$\frac{1}{2\pi}({\frac{2\pi}{k^{2}}e^{ik\pi}+{\frac {2\pi}{k^{2}}e^{-ik\pi})={\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}} e^{-ik\pi}$

$c_{k}$ suppose to be ${\frac{{2}(-1^{k})}{k^{2}}$

How do I get:

$({\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}}e^{-ik\pi})$

to be equal ${\frac{{2}(-1^{k})}{k^{2}}$

I would appriciate a response. Thank you

**Ackbeet** · August 20th 2010, 07:09 AM

Your k exponent should be outside the parentheses (otherwise it only applies to the 1). Otherwise, it looks fine.

**4Math** · August 20th 2010, 07:54 AM

Originally Posted by Ackbeet

Your k exponent should be outside the parentheses (otherwise it only applies to the 1). Otherwise, it looks fine.

Where does the 2 in ${\frac{{2}(-1)^{k}}{k^{2}}$ comes from. How do I get that. I just got it from the answer in the book but don't know where it comes from.

**Ackbeet** · August 20th 2010, 08:00 AM

$({\frac{1}{k^{2}}e^{ik\pi}+{\frac{1}{k^{2}}e^{-ik\pi})=\frac{1}{k^{2}}(e^{ikx}+e^{-ikx})=\frac{1}{k^{2}}(\cos(k\pi)+i\sin(k\pi)+\cos( k\pi)-i\sin(k\pi))$
$=\frac{2}{k^{2}}\,\cos(k\pi).$

But now $\cos(k\pi)=(-1)^{k}$ for all $k\in\mathbb{Z}.$

The answer follows.

Does that help? It's basically all from the Euler formula, and the even-ness of cosine and the odd-ness of sine:

$e^{i\theta}=\cos(\theta)+i\sin(\theta).$

**4Math** · September 13th 2010, 12:29 PM

I have a question regarding solving for the exponetial Fourier series for a given function. The general formula is the following (the way I understood):

$\displaystyle f(x)=\displaystyle \sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}$

there $\Omega T={2\pi}$

Now, to solve for the exponetial Fourier series for the given problem above :

$\displaystyle f(x) = {x^{2}}$ $,\,\,-\pi < x \leq \pi$

Why do we solve for $c_{0}$ ? Its not in the general fomula.
--------------------------------------------------------------------------
Where as I have a different problem where I have to solve for the exponetial Fourier series for a ${2\pi}$ - periodic function for the following given function:

$\displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}$

In this case we just solve for $c_{k}$ only and not for $c_{0}$ . Why is that?

I do appriciate for a explanation. Thank you

**Ackbeet** · September 14th 2010, 02:32 AM

Why do we solve for $c_{0}$ ? It's not in the general formula.

I don't understand. If you have the sum

$\displaystyle{f(x)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}},$

then the $k=0$ term has the expression $c_{0}$ in it. Does that qualify as "being in the general formula"?

In the problem you have there, I would think you would solve for $c_{0}$ . Both functions you're dealing with have a nonzero average value. As I recall, the expression for the $c_{0}$ term looks a lot like the average value of the function over its period. *looking it up* In fact, it is the average value of the function over a period. Now if you have a function like $f(x)=x$ over the interval $(-\pi,\pi),$ and you can tell at a glance that its average value is zero, well then, you have no need to explicitly compute it. But such examples should not be considered to be the usual.

Make sense?

**4Math** · September 14th 2010, 03:58 AM

Alright, it makes sense. Thanks.

How about for this function:

$\displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}$

Do I need to compute $c_{0}$ ? If not, why?

**Ackbeet** · September 14th 2010, 04:28 AM

Well, do you think the average value of the function over that interval is nonzero?

**4Math** · September 14th 2010, 05:42 AM

Originally Posted by Ackbeet

Well, do you think the average value of the function over that interval is nonzero?

I guess I am confused after all and don't understand.

Do you mean that:

$\displaystyle{f(x)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega x}}\iff {f(x)=c_{0} + \sum_ {k\neq 0}c_{k}e^{ik\Omega x}}$

**Ackbeet** · September 14th 2010, 06:39 AM

Your double implication is correct, if the understood range of $k$ values is the integers. All you're doing there is splitting off one term from the sum.

**4Math** · September 14th 2010, 11:26 AM

Originally Posted by Ackbeet

Now if you have a function like $f(x)=x$ over the interval $(-\pi,\pi),$ and you can tell at a glance that its average value is zero, well then, you have no need to explicitly compute it. But such examples should not be considered to be the usual.

If I compute $c_{0}$ for this function:

$f(x)=x$ over the interval $(-\pi,\pi)$

I would get the following:

$\displaystyle c_{0} =\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx =\frac{2}{2\pi}\int_{0}^{\pi}x dx =\frac{\pi}{2}$ the value is ${\neq 0}$

**Ackbeet** · September 14th 2010, 11:27 AM

No, the formula you used is valid for an even integrand. You have an odd integrand.

**4Math** · September 14th 2010, 11:51 AM

Originally Posted by Ackbeet

No, the formula you used is valid for an even integrand. You have an odd integrand.

Which formula do I use for an odd integrand? By the way thanks for your response.

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