# Homework Help: Expand the function f(x) = x^3 in a Fourier sine series

1. Sep 3, 2009

### Ylle

1. The problem statement, all variables and given/known data
Expand the function f(x) = x3 in a Fourier sine series on the interval 0 <= x <= 1

2. Relevant equations
$$$f\left( x \right)=\sum\limits_{k=1}^{\infty }{{{b}_{k}}\sin \left( k\pi x/a \right)}0<x\le a$$$

and

$$${{b}_{k}}=\frac{2}{a}\int_{0}^{a}{f\left( x \right)\sin \left( k\pi x/a \right)dx}$$$

3. The attempt at a solution

Well, I first calculate bk which becomes:

$$${{b}_{k}}=2\left( \left( \frac{3}{{{\left( k\pi \right)}^{2}}}-\frac{6}{{{\left( k\pi \right)}^{4}}} \right)\sin \left( k\pi \right) \right)$$$,
since x3 is an odd equation which means that the sine parts are the only ones who are non-zero => Cosine parts disappear.

And then I just insert that in the first equation. But the I start thinking, sin(pi) = 0, så everything ends up being 0 here :S

What have I done wrong ?

Regards

2. Sep 3, 2009

### D H

Staff Emeritus
Show your integration. Hint: What happened to the cosine terms?

3. Sep 3, 2009

### zcd

I think you're misunderstanding the part about cos parts disappearing. That only applies for $$a_{n}=\int_{-T}^{T} f(x)\cos(\frac{n\pi x}{T})\,dx$$.

4. Sep 3, 2009

### Ylle

Ahhhh... I've missed that part.
So I just integrate as normal, without removing the cosine part ?

5. Sep 3, 2009

### D H

Staff Emeritus
No, you misunderstood my question. What is

$$${{b}_{k}}=\frac{2}{a}\int_{0}^{a}{f\left( x \right)\sin \left( k\pi x/a \right)dx}$$$

$$${{b}_{k}}=2\left( \left( \frac{3}{{{\left( k\pi \right)}^{2}}}-\frac{6}{{{\left( k\pi \right)}^{4}}} \right)\sin \left( k\pi \right) \right)$$$

is not correct.

6. Sep 3, 2009

### Ylle

Well, as I thought I just cut out the cosine part I just gave that answer, but with the cosine it becomes:

$$${{b}_{k}}=2\left( \left( \frac{3}{{{\left( k\pi \right)}^{2}}}-\frac{6}{{{\left( k\pi \right)}^{4}}} \right)\sin \left( k\pi \right)+\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right)\cos \left( k\pi \right) \right)$$$

That is for [0, 1].

Or am I wrong ?

Last edited: Sep 3, 2009
7. Sep 3, 2009

### zcd

You can't cut that part out. For the odd function cutting out cos terms, that only applies for the entire an.

8. Sep 3, 2009

### Ylle

Yeah, I understand that now. But wouldn't that mean that the integral would be the one as I posted above ? And then just add that to the bk in the first equation I posted.

9. Sep 3, 2009

### D H

Staff Emeritus
Better, but still not correct. Double check your integration.

10. Sep 3, 2009

### Ylle

Ahhh, didn't see that.
sin(pi) = 0, so all the sine parts disappears, and cos(0) = 1 and cos(pi) = -1
So we get:

$$${{b}_{k}}=2\left( -\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right) \right)=-2\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right)$$$

Or am I wrong again ? :)

Last edited: Sep 3, 2009
11. Sep 3, 2009

### D H

Staff Emeritus
Still wrong. Your indefinite integral is wrong (well, it was wrong before you edited it away), which made your definite integral wrong as well.

Show your work. When you get a hairy integral, it is always a good idea to check your work by differentiating to see if you obtain the original expression.

That or ask a computer to do the integration for you, but that's no fun.

12. Sep 3, 2009

### Ylle

So this is the wrong definite integral ?:

$$${{b}_{k}}=2\left( -\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right) \right)=-2\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right)$$$

And I see now.
I just thought that x could only be natural numbers. But I see now :)

Thank you very much DH :)

Last edited: Sep 3, 2009
13. Sep 3, 2009

### Ylle

But, anyways, even though the integral is wrong. Wouldn't the whole thing become zero anyways, after I put bk into the equation where it's multiplied by sin(k*x*pi) ?

Last edited: Sep 3, 2009
14. Sep 3, 2009

### D H

Staff Emeritus
First question: What is your indefinite integral?
Second question: Did you double-check your integration by differentiating the result?

To answer your question, the {bk} will not be zero. $\sin(k\pi x)$ is zero only for a few special values of k and x.

15. Sep 3, 2009

### Ylle

1. My indefinte is: $$$\int{{{x}^{3}}\sin \left( k\pi x \right)dx=}\left( \left( \frac{3{{x}^{2}}}{{{\left( k\pi \right)}^{2}}}-\frac{6}{{{\left( k\pi \right)}^{4}}} \right)\sin \left( k\pi x \right)+\left( \frac{6x}{{{\left( k\pi \right)}^{3}}}-\frac{{{x}^{3}}}{k\pi } \right)\cos \left( k\pi x \right) \right)$$$

2. Yes, I would say I've doublechecked :)

3. And I don't understand. No matter what number I put instead of k and x in sin(x*pi*k) it becomes zero :S

16. Sep 3, 2009

### D H

Staff Emeritus
That looks much better. How did you go from that (correct) indefinite integral to the (incorrect) definite integral?

You apparently don't understand the basic concept.
1. First you evaluate the {bk}s. These will just be a bunch of numbers. In this case, each bk will be non-zero.
2. Then you form the sine series using the {bk}s. You do not evaluate this because you don't know x. Yet. This series is just another way of writing f(x).

You can find the value of the series for some particular x by plugging in that particular value of x. All you are doing here is evaluating the $\sin(k\pi x)$, multiplying by the already known bk, and summing.

There is no integration in this step. For example, for x=1/1000 you won't find a zero term until k=1000.

17. Sep 3, 2009

### Ylle

So the definite integral is still wrong?:

$$${{b}_{k}}=2\left( -\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right) \right)=-2\left( \frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right)$$$

And I see now. I just thought that x could only be natural numbers. But I see now :)

Thank you very much DH :)

18. Sep 3, 2009

### zcd

You have it right here. Only the sin parts go to 0, the cos values alternate + and - because $$\cos(n\pix)$$ is 1 for n even and -1 for n odd. Evaluate the limits of the integral and don't forget the constant multiplier, and then you have the bn term in general.

19. Sep 3, 2009

### D H

Staff Emeritus
Yes.

That is correct.

The sine term vanishes at both x=0 and x=1, and the cosine term vanishes at x=0. All that is left is the cosine term at x=1:

$$\int_0^1x^3\sin( k\pi x)dx = \left(\frac{6}{{{\left( k\pi \right)}^{3}}}-\frac{1}{k\pi } \right)\cos(k\pi)$$

This simplifies a bit more as zcd hinted. Use $\cos(k\pi)=(-1)^k$.

20. Sep 4, 2009

### Ylle

Ahhh, I just forgot to write the cosine part with my solution. My bad...
But again, ty very much.