Expand the function f(x) = x^3 in a Fourier sine series

[Ylle]

Homework Statement

Expand the function f(x) = x³ in a Fourier sine series on the interval 0 <= x <= 1

Homework Equations

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

and

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

The Attempt at a Solution

Well, I first calculate b_k 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

 

[D H]

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

 

[zcd]

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

 

[Ylle]

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

 

[D H]

No, you misunderstood my question. What is

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

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

is not correct.

 

[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 ?

 

[zcd]

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

 

[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 b_k in the first equation I posted.

 

[D H]

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

 

[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 ? :)

 

[D H]

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.

 

[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 :)

 

[Ylle]

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

 

[D H]

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

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

 

[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

 

[D H]

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 {b_k}s. These will just be a bunch of numbers. In this case, each b_k will be non-zero.
2. Then you form the sine series using the {b_k}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 b_k, 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.

 

[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 :)

 

[zcd]

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

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 b_n term in general.

 

[D H]

So the definite integral is still wrong?:

Yes.

$$\[\int{{{x}^{3}}\sin \left( k\pi x \right)dx=}\left( \left( \frac{3{{x}^{2}}}{{{\left( k\pi \right)}^{2}}}-\frac{6}{{{\left( k\pi \right)}^{4}}} \right)\sin \left( k\pi x \right)+\left( \frac{6x}{{{\left( k\pi \right)}^{3}}}-\frac{{{x}^{3}}}{k\pi } \right)\cos \left( k\pi x \right) \right)\]$$

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 $\cos(k\pi)=(-1)^k$.

 

[Ylle]

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