# Using Fourier Sine basis to write x(L-x) [0,L]

1. Oct 21, 2013

### Hakkinen

1. The problem statement, all variables and given/known data
A function $F(x) = x(L-x)$ between zero and L. Use the basis of the preceding problem to write this vector in terms of its components:
$F(x)= \sum_{n=1}^{\infty}\alpha _{n}\vec{e_{n}}$
If you take the result of using this basis and write the resulting function outside the interval $0<x<L$, graph the result

2. Relevant equations
The basis as mentioned above is $\vec{e_{n}}=\sqrt{\frac{2}{L}}\sin {\frac{n\pi x}{L}}$

3. The attempt at a solution

I don't really know where to start, do I just need to write F(x) as a fourier series and then compute the coefficients using the standard "trick" to do so, ie sticking F(x) in the integral and so on?

The previous problem was just showing that the basis is orthonormal.

2. Oct 21, 2013

### LCKurtz

Orthonormal with respect to what inner product? What happens if, in the equation$$x(L-x)= \sum_{n=1}^{\infty}\alpha _{n}\vec{e_{n}}$$you take the inner product of both sides with $e_k$? You need to show some work.

Last edited: Oct 21, 2013
3. Oct 21, 2013

### Hakkinen

Sorry, it's orthonormal with respect to the inner product defined as $< f^{*},g> = \int_{0}^{L}f^{*}(x)g(x)dx$

EDIT:

So, taking the inner product of both sides of that equation

$\sqrt{\frac{2}{L}}\int_{0}^{L}x(L-x)\sin {\frac{k\pi x}{L}}dx=\sum_{n=1}^{\infty}a_{n}\delta _{nk}$
After evaluating the integral I should be able to get the coefficients $a_{n}$ ?

Last edited: Oct 21, 2013
4. Oct 21, 2013

### LCKurtz

Notice I edited my post to put $x(L-x)$ for $F(x)$. You need to write out the both sides and show me what you get. You have both $k$ and $n$ variables on the RHS.

5. Oct 21, 2013

### Hakkinen

Okay, so the RHS should actually just be $a_{k}$ because of the delta term?

Thanks for your help btw, this forum is excellent and it's people like you that make it what it is!

6. Oct 21, 2013

### LCKurtz

Yes, so you have a formula you can work out for $a_k$. So what do you finally get for the answer to your problem?

7. Oct 21, 2013

### Hakkinen

Well assuming I integrated everything correctly, I got some nice cancellations and ended up with
$a_{k}=2\sqrt{\frac{2}{L}}(\frac{L}{k\pi})^3\cos \frac{k\pi x}{L}$ all evaluated at L and 0.

I'm unsure of how to proceed and write out the full expression, with the $(\frac{1}{k})^3$ term in the expression...

8. Oct 21, 2013

### LCKurtz

Wouldn't you expect $a_k$ to depend on $k$? You have to put in the limits 0 and L and simplify it, because you are going eventually to put those $a_k$'s in the FS for $x(L-x)$.

9. Oct 22, 2013

### vanhees71

It's important to note that the correct definition for the scalar product on the Hilbert space $\mathrm{L}^2([-L,L],\mathbb{C})$ is
$$\langle f|g \rangle=\int_{-L}^{L} \mathrm{d} x \; f^*(x) g(x).$$
Further the orthonormal function system $|e_k \rangle$ given by
$$e_k(x)=\sqrt{2}{L} \sin \left (\frac{n \pi x}{L} \right)$$
spans all odd functions in this Hilbert space.

Now you simply calculate the Fourier coefficients of the function which is continued from the interval $[0,L]$ to the complete interval $[-L,L]$ as an odd function and evaluate the Fourier coefficients
$$F_n=\langle e_n | F \rangle,$$
using the above given definition of the scalar product!

10. Oct 22, 2013

### LCKurtz

@vanhees71: While that may all be true, since Hakkineen didn't give any context for his question, how are you to say the question doesn't arise from a Sturm-Liouville problem$$y'' +\lambda y = 0$$ $$y(0) = 0,~y(L) = 0$$That gives rise to precisely the given orthonormal set with everything defined only on $[0,L]$.

11. Oct 22, 2013

### Hakkinen

Thanks for the clarification. The problem didn't really have any context other than what I stated, it's from a math methods book so hopefully when I cover Sturm-Liouville problems soon in my PDE class I can see some relevant examples.

So for the final expression of F(x), I have:
$F(x)=\sum_{k=1}^{\infty}a_{k}\hat{e}_{k}$

where a_k is given by
$a_{k}=2\sqrt{\frac{2}{L}}(\frac{L}{k\pi })^3[\cos k\pi -1]$
Which equals 0 when k is even, and
$-4\sqrt{\frac{2}{L}}(\frac{L}{k\pi })^3$ when k is odd.

12. Oct 23, 2013

### LCKurtz

That looks OK to me.