I am working on the following problem:
Let $f(x) = 1$ for $0 < x < 1$. Write a generalized Fourier series for $f(x)$ in terms of the eigenfunctions of the following Sturm-Liouville problem. Discuss the convergence of this series:
$\begin{matrix}-u^{\prime}{\prime}(x) = \mu u(x) \\ u(0) = u^{\prime}(1)=0 \end{matrix}$
What I've done thus far:
I have found the eigenvalues of this Sturm-Liouville problem to be $\displaystyle \mu_{n}= \left(\frac{\pi(2n-1)}{2} \right)^{2}$, and the eigenfunctions to be $\displaystyle u_{n}(x)=\sin \left(\frac{\pi(2n-1)}{2} \right)x$, $n = 1,2,\cdots$.
Now, since the eigenfunctions for a complete orthogonal set on $[0,1]$, we can expand $f(x) = 1$ as $\displaystyle f(x) = 1 \sim \sum_{n=1}^{\infty}b_{n}\sin\left( \frac{\pi(2n-1)}{2}\right)\pi x$, where $\displaystyle b_{n} = \frac{2}{1}\int_{0}^{1} 1 \cdot \sin\left(\frac{\pi (2n-1)}{2} \right)x dx = -\frac{4}{\pi(2n-1)}\left[\cos \left(\frac{\pi(2n-1)x}{2} \right) \right]_{0}^{1} = -\frac{4}{\pi(2n-1)}\left[\cos\left(\pi n - \frac{\pi}{2}\right) - 1 \right] = -\frac{4}{\pi(2n-1)}\left[\cos(\pi n)\cos \left( \frac{\pi}{2}\right)+ \sin(\pi n) \sin\left( \frac{\pi}{1}\right) - 1 \right] = \frac{4}{\pi(2n-1)}$.
So, $\displaystyle f(x) = 1 \sim \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left(\frac{\pi(2n-1)}{2} \right)x$.
EDIT: Thanks to the help of an extremely patient person who answered a similar question of mine here on MSE, I have come up with the following solution, which I would like someone to take a look at, and tell me if it's correct. If it's not, what can I do to fix it? Thank you! :)
$\displaystyle \sin \left( n - \frac{1}{2} \right) \pi x$ has a period of $4$, since $\displaystyle \begin{align}\sin\left( \left( n - \frac{1}{2}\right) \pi \left( x + 4 \right)\right) \\= \sin \left( \left( n-\frac{1}{2}\right) \left(\pi x + 4 \pi \right) \right) \\= \sin \left( \left( n - \frac{1}{2}\right) \pi x\right) + \left( n - \frac{1}{2}\right) 4 \pi) \\= \sin\left[\left(n - \frac{1}{2}\right)\pi x \right]\cos \left[\left( n - \frac{1}{2}\right)4 \pi \right] + \cos \left[\left(n - \frac{1}{2} \right)\pi x\right]\sin \left[\left( n - \frac{1}{2}\right)4 \pi \right] \\= \sin\left[\left(n - \frac{1}{2}\right)\pi x \right]\cos \left[\left( 2n -1 \right)2 \pi \right] + \cos \left[\left(n - \frac{1}{2} \right)\pi x\right]\sin \left[\left( n - \frac{1}{2}\right)4 \pi \right] \\ = \sin\left[ \left(n - \frac{1}{2}\right) \pi x \right](1) = \cos \left[ \left( n - \frac{1}{2}\right) \pi x \right](0) \\ = \sin \left[ \left( n - \frac{1}{2}\right)\pi x \right]\end{align} $.
Also,
$\displaystyle \begin{align} \sin\left[ \left(n - \frac{1}{2}\right) \pi x \right] \\ = \sin\left[ \left( n - \frac{1}{2}\right)\pi (x + 1 - 1)\right] \\ = \sin \left[ \left( n - \frac{1}{2}\right) \left[(x+1) \pi - \pi \right]\right] \\ = \sin\left[ \left(n - \frac{1}{2}\right) (x+1) \pi - \left( n - \frac{1}{2}\right)\pi\right] \\ = \sin \left[ \left( n - \frac{1}{2}\right)(x+1) \pi \right] \cos\left[ \left( n - \frac{1}{2}\right)\pi\right] \\- \cos\left[\left(n - \frac{1}{2}\right)(x+1)\pi\right]\sin \left[ \left( n - \frac{1}{2}\right)\pi \right] \\ = \sin \left[ \left( n - \frac{1}{2}\right) (x+1) \pi \right] (-1) - \cos \left[\left( n - \frac{1}{2}\right)(x+1)\pi\right](0) \\ = -\sin \left[\left( n - \frac{1}{2}\right)(x+1)\pi\right] \end{align}$
Therefore, for $x \in (-1,0)$, $\displaystyle \sin \left[ \left( n - \frac{1}{2}\right) (x+1) \pi \right] = - \sin\left[ \left(n - \frac{1}{2}\right) \pi x \right]$.
And so, for $x \in (-1,0)$, $\displaystyle f(x) = 1 \sim \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left[ \left( n-\frac{1}{2} \right) \pi x \right] \\ \displaystyle = -\frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left[ \left( n-\frac{1}{2} \right) \pi (x+1) \right]$.
Hence, for $x \in (-1,0)$, we have $\displaystyle -1 \sim \frac{4}{\pi}\sum_{n=1}^{\infty}(2n-1)^{-1}\sin \left[ \left( n - \frac{1}{2}\right) \pi (x+1)\right]$, and putting everything together, we see that
$\displaystyle \frac{4}{\pi}\sum_{n=1}^{\infty}(2n-1)^{-1}\sin \left[\left(n - \frac{1}{2}\right)\pi x \right] = \begin{cases} 1, & x\in (0,1)\\ -1, & x \in (-1,0) \end{cases}$
Adding $1$ to both sides and dividing by $2$, we obtain both
$\displaystyle f(x) = 1 \sim \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi x \right)\, \implies \\ \begin{align} 2 \sim 1 + \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi x \right) \, \implies \\ 1 \sim \frac{1}{2} + \frac{2}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi x \right)\end{align} $
-And-
$\displaystyle -1 \sim \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi (x+1) \right)\, \implies \\ \begin{align} 0 \sim 1 + \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi (x+1) \right) \, \implies \\ 0 \sim \frac{1}{2} + \frac{2}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}\sin \left( \left( n - \frac{1}{2}\right) \pi (x+1) \right)\end{align} $
So, we have that
$\displaystyle \frac{1}{2}+\frac{2}{\pi}\sum_{n=1}^{\infty}(2n-1)^{-1}\sin\left[\left( n - \frac{1}{2}\right)\pi x \right] = \begin{cases} 1, & x \in (0,1) \\ 0, & x \in (-1,0) \end{cases}$
The reason I'm not sure this answer is right is because of the $+ \frac{1}{2}$ term, which I've only seen happen in cases where we're trying to do a Fourier expansion in terms of even functions. But, is it the fact that $f(x) = 1$, which is even, the reason why the $+ \frac{1}{2}$ is appearing? Or did I make a mistake?
Thank you! :)
There is a mistake in your argument,
$$ \sin\left[(n-\frac{1}{2})\pi x\right] \ne - \sin\left[(n-\frac{1}{2})\pi (x + 1)\right] $$
You can test that just by checking with $n = 1$:
LHS:
$$ {\rm LHS} = \sin\left(\frac{1}{2}\pi x\right) $$
RHS:
$$ {\rm RHS} = -\sin\left(\frac{1}{2}\pi (x + 1)\right) = -\sin\left(\frac{1}{2}\pi x + \frac{\pi}{2}\right) = -\cos\left(\frac{1}{2}\pi x\right) \ne {\rm LHS} $$
What you should do is realize that
$$ \sin\left[(n-\frac{1}{2})\pi (-x)\right] = -\sin\left[(n-\frac{1}{2})\pi x\right] $$
therefore
$$ -1 \sim \frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{1}{2n-1}\sin\left[(n-\frac{1}{2})\pi (-x)\right] ~~~\mbox{for}~~~ 0<x<1 $$
or equivalently
$$ -1 \sim \frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{1}{2n-1}\sin\left[(n-\frac{1}{2})\pi x\right] ~~~\mbox{for}~~~ -1<x<0 $$
In summary you have
$$ \frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{1}{2n-1}\sin\left[(n-\frac{1}{2})\pi x\right] \sim \begin{cases} 1, & 0 < x < 1 \\ - 1, & -1 < x < 0\end{cases} $$
After this, you can definitely shift and scale as you did, reaching to the conclusion
$$ \frac{1}{2}+\frac{2}{\pi}\sum_{n=1}^{+\infty}\frac{1}{2n-1}\sin\left[(n-\frac{1}{2})\pi x\right] \sim \begin{cases} 1, & 0 < x < 1 \\ 0, & -1 < x < 0\end{cases} $$
Do not get confused by the $1/2$ term, it only gives you information about the average value of $f(x)$