How to show convergence of my generalized Fourier Series to the values I specify in the body of this question.

Question

This question is related to another one I asked earlier here.

For reference, I asked for help writing a generalized Fourier Series for the function $f(x) = 1$ for $0<x<1$, in terms of the eigenfunctions $\displaystyle w_{n} = \cos \left[\left( n - \frac{1}{2}\right)\pi x \right]$, $n = 0,1,2, \cdots$ for the Sturm-Liouville problem:

$\begin{matrix} - w^{\prime\prime} (x) = \mu w(x) \\ w^{\prime}(0) = w(1) = 0. \end{matrix}$

With the help of the person who answered me, I was able to successfully show that $\displaystyle \tilde{f}(x) = \tilde{f}_{E}(x) \sim \frac{1}{2}a_{0} + \sum_{n=1}^{\infty}a_{n} \cos \left[ \left( n - \frac{1}{2}\right)\pi x \right] \\ \displaystyle = \frac{1}{2(1)}\int_{-1}^{1}1\, dx +\sum_{n=1}^{\infty}a_{n} \cos \left[ \left( n - \frac{1}{2}\right)\pi x \right] \\ \displaystyle = 1 + \sum_{n=1}^{\infty}a_{n} \cos \left[ \left( n - \frac{1}{2}\right)\pi x \right] \\ \displaystyle = 1 + \frac{4}{\pi}\sum_{n=1}^{\infty} (-1)^{n+1}(2n-1)^{-1}\cdot \cos \left[ \left( n - \frac{1}{2}\right) \pi x \right]$

is the generalized Fourier Series I sought.

Now, however, I am having trouble showing the convergence

According to the back of my book, this series should converge pointwise to

$\tilde{f}(x) = \begin{cases} 1 & \text{if}\, |x|<1 \\ 0 & \text{if}\, 1 < |x|<2 \end{cases}$

$\tilde{f}(x) = \tilde{f}(x+4)$ for all $x$.

Now, there's a theorem in my book that says that if $f$ and $f^{\prime}$ are both sectionally continuous on $(a,b)$, then the series converges pointwise to $\displaystyle \frac{1}{2}\left[f(x^{+}) + f(x^{-1}) \right]$ at each $x \in (a,b)$ (where $f(x^{+})$ is the limit of $f$ as we approach $x$ from the right, and $f(x^{-})$ is the limit of $f$ as we approach $x$ from the left)

Since $f(x) = 1$ on $0 < x < 1$, by this theorem, it makes sense that $\tilde{f}$ should converge pointwise to $\displaystyle \frac{1}{2}\left[ 1 + 1 \right] = \frac{1}{2}(2) = 1$ at each $x \in (0, 1)$.

But, where does the $|x|<1$ come into play? And also, why does $\tilde{f}$ converge pointwise to $0$ for $1 < |x| < 2$? Why did we start caring about what happens between $1$ and $2$? Does this have something to do with the $2L$-periodic extension (I suppose here, then, since $L = 1$, it would be the $2$-periodic extension), and if so, how?

Furthermore, where does the $\tilde{f} = \tilde{f} (x+4)$ come from?

I ask you to please be patient with me, and not get frustrated that I don't know these things, even if they are very basic to the study of Fourier Series. I'm just trying to learn and getting very confused in the process. By helping me get un-confused, you're doing a very good deed! Thank you.

Answer 1

To show that the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}\cos((n-1/2)\pi x)}{n}$ converges for $|x|<1$, we can simply apply Dircichlet's Test.

To do so, we need only show that there exists a number $L$ such that for any $N$, we have for any fixed $x_0 \in (0,1)$ (or $x_0\in (-1,0)$)

$$\begin{align} \left|\sum_{n=1}^N (-1)^{n+1}\cos((n-1/2)\pi x_0)\right|&=\left|\sec(\pi x_0/2)\sin^2\left(\frac{N\pi (x_0-1)}{2}\right)\right| \tag 1\\\\ &\le \sec(\pi x_0/2)\\\\ &=L \end{align}$$

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To arrive at $(1)$, we use the sum angle formula $\cos(x)\cos(y)=\frac{\cos(x+y)+\cos(x-y)}{2}$ with $x=\pi x_0/2$ and $y=(n-1/2)\pi x_0$. Then, we can write

$$\begin{align} \cos(\pi x_0/2)\sum_{n=1}^N (-1)^{n+1}\cos((n-1/2)\pi x_0)&= \sum_{n=1}^N \frac{\left((-1)^{n+1}\cos(n\pi x_0)-(-1)^n\cos((n-1)\pi x_0)\right)}{2}\\\\ &=\frac12\left((-1)^{N+1}\cos(N\pi x_o)+1\right)\\\\ &=\frac12 \left(1-\cos(N\pi(x_0-1)) \right)\\\\ &=\sin^2\left(\frac{N\pi(x_0-1)}{2}\right) \end{align}$$

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It was established in THIS ANSWER that the eigenfunctions $\cos((n-1/2)\pi x)$ form a complete orthogonal set on $(0,1)$, and that we can therefore represent the function $f(x)=1$ by the series

$$1\sim \frac4\pi \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi x)$$

for $x\in (0,1)$.

Now, note that $\cos((n-1/2)\pi x)$ has period $4$. To do this, we write

$$\begin{align} \cos((n-1/2)\pi (x+4))&=\cos((n-1/2)\pi x)\cos((n-1/2)4\pi)-\sin((n-1/2)\pi x)\sin((n-1/2)4\pi)\\\\ &=\color{blue}{\cos((n-1/2)\pi x)}\color{red}{\cos((2n-1)2\pi)}-\color{green}{\sin((n-1/2)\pi x)}\color{purple}{\sin((2n-1)2\pi)}\\\\ &=\color{blue}{\cos((n-1/2)\pi x)}\color{red}{1}-\color{green}{\sin((n-1/2)\pi x)}\color{purple}{0}\\\\ &=\cos((n-1/2)\pi x) \end{align}$$

Similarly, we see that $\cos((n-1/2)\pi x)=\cos((n-1/2)\pi (2-x-2))=-\cos((n-1/2)(2-x))$. Therefore, for $x\in (1,2)$,

$$\begin{align}1 & \sim \frac4\pi\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi (2-x))\\\\ &=-\frac4\pi\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi x) \end{align}$$

and hence for $x\in (1,2)$, we have

$$\frac4\pi\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi x) \sim -1$$

Putting it together, we see that

$$\frac4\pi\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi x)=\begin{cases}1&,0<x<1\\\\-1&,1<x<2\end{cases}$$

Finally, adding $1$ and dividing by $2$ yields

$$\frac12 +\frac2\pi\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\cos((n-1/2)\pi x)=\begin{cases}1&,0<x<1\\\\0&,1<x<2\end{cases}$$

Answer 2

I am sure @Dr. MV already answered your question, just want to add something to the discussion that may help. Consider the function

$$ f(x)=\begin{cases} 1\quad \mbox{for}\quad |x| < 1 \\ 0\quad \mbox{for}\quad 1<|x|<2 \end{cases} $$

I would like to make a Fourier expansion of $f(x)$. Since the series is periodic in nature we need to make a periodic extension of $f$ and because $f$ is even in its domain, the natural choice for the extension $\tilde{f}$ is to make it even as well with period $L$

$$ \tilde{f}(x + L)= \tilde{f}(x) $$

$L=4$ comes from the $\tilde{f}(x)$ has to be $f(x)$ for $|x|<2$

The Fourier series for $\tilde{f}$ is simply

$$ \tilde{f}(x) = \frac{1}{2}a_0 + \sum_{n = 1}^{+\infty}a_n \cos \frac{2\pi n x}{L} + b_n \sin \frac{2\pi n x}{L} \tag{1} $$

where

$$ a_0 = \frac{2}{L}\int_{-L/2}^{L/2}{\rm d}x\; \tilde{f}(x) = \frac{1}{2}\int_{-2}^{2}{\rm d}x\; f(x) = 1 $$

$$ a_n = \frac{2}{L}\int_{-L/2}^{L/2}{\rm d}x\; \tilde{f}(x) \cos\frac{2\pi n x}{L}= \frac{1}{2}\int_{-2}^{2}{\rm d}x\; f(x)\cos\frac{\pi n x}{2} = \frac{2}{n\pi}\sin\frac{n\pi}{2} \quad n =1,2,\cdots $$

and

$$ b_n = \frac{2}{L}\int_{-L/2}^{L/2}{\rm d}x\; \tilde{f}(x) \sin\frac{2\pi n x}{L}= \frac{1}{2}\int_{-2}^{2}{\rm d}x\; f(x)\sin\frac{\pi n x}{2} = 0 \quad n =1,2,\cdots $$

Note that

$$ a_{2k} = 0 \quad\mbox{and}\quad a_{2k - 1} = \frac{2}{(2k-1)\pi}(-1)^k \qquad k =1,2,\cdots $$

Eq. (1) then becomes

$$ \tilde{f}(x) = \frac{1}{2} - \frac{2}{\pi}\sum_{k = 1}^{+\infty}\frac{(-1)^k}{2k-1}\cos \left((k - 1/2)\pi x\right) \tag{2} $$

Which is basically what you have before. The figure below shows $f(x)$ and $\tilde{f}(x)$ truncated with 20 terms in the sum