Why does this method fail for finding the Fourier series for $\cos\left(\frac{x}{2}\right)$ on the interval $-\pi \lt x \lt \pi$?

Question

This question is strongly related to this question (that does not have an answer).

From "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 417

$$a_n=\frac{2}{L}\int_{x_0}^{x_0+L}f(x)\cos\left(\frac{2\pi nx}{L}\right)dx\tag{1}$$

similarly $$b_n=\frac{2}{L}\int_{x_0}^{x_0+L}f(x)\sin\left(\frac{2\pi nx}{L}\right)dx\tag{2}$$ where $x_0$ is arbitrary but is often taken as $0$ or $-\frac{L}{2}$ and $L$ is the period of $f(x)$.

The following question I asked on "Chegg Study" website is here:

Show that the Fourier series of $f(x)=\cos\left(\frac{x}{2}\right)$ for $-\pi\lt x \lt \pi$ is given by $$f(x)=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cos(nx)}{4n^2 - 1}$$

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So my attempt at the finding the Fourier series for $\cos\left(\frac{x}{2}\right)$ on the interval $-\pi \lt x \lt \pi$ is by first noting that since $\cos\left(\frac{x}{2}\right)$ is not periodic on $-\pi \lt x \lt \pi$,$\,$ I must therefore make the function periodic by extending the interval to $-2\pi \lt x \lt 2\pi$. By the Dirichlet conditions for convergence the function must be periodic.

Now that the function is symmetric and periodic about $x=0$ the $b_n=0 \,\forall \,n \in \mathbb N$

A plot of the extended function is shown below:

Now by $(1)$, $$\begin{align}a_n &=\frac{2}{L}\int_{x_0}^{x_0+L}f(x)\cos\left(\frac{2\pi nx}{L}\right)dx \\&=\frac{2}{4\pi}\int_{-2\pi}^{2\pi}\cos\left(\frac{x}{2}\right)\cos\left(\frac{2\pi nx}{4\pi}\right)dx\\&=\frac{1}{2\pi}\int_{-2\pi}^{2\pi}\cos\left(\frac{x}{2}\right)\cos\left(\frac{nx}{2}\right)dx\\&=0\end{align}$$

and this has been verified here by Wolfram Alpha.

All I did was substitute the value of the period $L$ into the argument of the cosine, the limits and the denominator in front of the integral.

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Why is this giving me the wrong answer when it clearly works for other functions like $x^2$ on $0 \lt x \le 2$ as can be seen here "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 422 and 423?

The method used in the $x^2$ and $\cos(x/2)$ case was general and did not specify any restrictions on the function being extended.

So, put in another way, is there some kind of rule that says extending a trigonometric function to a full period is always forbidden since integrating over the period will be zero?

Answer 1

You want to expand $\cos(x/2)$ on $[-\pi,\pi]$ in a Fourier series. The resulting series will converge to the periodic extension of $\cos(x/2)$ from $[-\pi,\pi]$ to $\mathbb{R}$ with period $2\pi$. The Fourier coefficients are $$ a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\cos(x/2)dx \\ a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\cos(x/2)\cos(n x)dx,\;\; n \ge 1 \\ b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\cos(x/2)\sin(n x)dx=0,\;\; n \ge 1. $$ $b_n=0$ because the integrand is odd; $a_n$ can be computed using the identity $$ \cos(x/2)\cos(n x)=\frac{1}{2}\{\cos(x/2+n x)+\cos(x/2-n x)\} \\ %% \cos(x/2)\sin(n x)=\frac{1}{2}\{\sin(n x+x/2)+\sin(n x-x/2)\}. $$ Using this, $$ a_n = \left.\frac{1}{2\pi}\left\{\frac{\sin(x/2+nx)}{1/2+n}+\frac{\sin(x/2-nx)}{1/2-n}\right\}\right|_{x=-\pi}^{\pi} \\ = \frac{(-1)^n}{2\pi}\left[\frac{1}{n+1/2}-\frac{1}{n-1/2}\right] \\ = \frac{(-1)^{n+1}}{2\pi}\frac{1}{n^2-1/4}. $$

Answer 2

$a_1=\frac{1}{2\pi}\int_{-2\pi}^{2\pi}cos^2(\frac{x}{2})dx=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}1dx=1$.

Answer 3

If you "extend the period" you get a rather different looking curve

The question that they are asking seem to be to evaluate.

$a_n = \frac 1{2\pi}\int_{-\pi}^{\pi} \cos \frac x2 \cos nx\ dx\\ \frac {1}{4\pi} \int_{-\pi}^{\pi} \cos (n+\frac 12)x + \cos (n-\frac 12)x \ dx\\ $

Answer 4

This was generated with gnuplot - it shows a function that is defined by

$$f(x) = cos(x/2) \quad -\pi<x<\pi$$ and $$f(x+2\pi)=f(x)$$

I think you need to stick to the range of $-\pi<x<\pi$ - which is what @DougM is saying in his answer

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Edit about $x^2$ range extension...

The picture in the reference you gave is helpful here - where it shows the $x^2$ function being 'mirrored' or reflected in the $y$-axis.

So $x^2$ is an even function and so in the Fourier series we will only have contributions from $\cos(kx)$ and not $\sin(kx)$

This means that the critical integrals of $$\int x^2 \cos(kx) dx $$ are even - or at least the $x^2 \cos(kx)$ function inside is even.

Now if the function is even then integrating over the range $[0,2]$ will give exactly half the value that you get integrating over $[-2,2]$ - because everything is completely symmetrical about the $y$-axis or the line $x=0$.

In your case of the cos$(x/2)$ you also have an even function, but you are not expanding into the mirror image on the other side of the $y$-axis... instead you are making the range wider and the $\cos(x/2)$ function now is negative and things change.

For example - with the extended range the $a_0$ term will be zero, but in the figure above everything is above the $x$-axis so $a_0$ will be positive and non-zero.