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

[Hakkinen]

Homework Statement

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

Homework Equations

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

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.

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

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.

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.

 

[Hakkinen]

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

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} ?

 

[LCKurtz]

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

In response to your question, the RHS should just be zero correct? as you're taking the inner product of two basis vectors that are known to be orthogonal.

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.

 

[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!

 

[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?

 

[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...

 

[LCKurtz]

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...

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 $x(L-x)$.

 

[vanhees71]

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} ?

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!

 

[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 $[0,L]$.

 

[Hakkinen]

@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 $[0,L]$.

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:
<br /> F(x)=\sum_{k=1}^{\infty}a_{k}\hat{e}_{k}<br />

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

 

[LCKurtz]

That looks OK to me.