How to solve Real Analysis Find the Fouries series of the function given below
$$
f(x) = \lvert\cos t\rvert \quad\hbox{for all $t$}.
$$
What does it means for all $x$? Can someone show me some  work done so that I can understand better?
 A: We treat the function $f$ as a $2\pi$-periodic function, so that we can apply the most basic formulae for the Fourier coefficients. The function $f$ is even; whence all sine coefficients $b_k$ are $=0$, and
$$a_k={1\over\pi}\int_{-\pi}^\pi f(x)\ \cos(k x)\ dx={2\over\pi}\int_0^\pi f(x)\ \cos(k x)\ dx\qquad(k\geq0)\ .$$
In addition, $f$ is even with respect to the point $x={\pi\over2}$, and this implies that $a_k$ is $=0$ for odd $k$ (to see this draw the graphs of $f$ and of $x\mapsto\cos\bigl((2j+1)x\bigr)$ in the same figure). In the interval $\bigl[0,{\pi\over2}\bigr]$ our function is just $x\mapsto\cos x$. Therefore we get
$$\eqalign{a_{2j}&={4\over\pi}\int_0^{\pi/2}\cos x\cos(2j x)\ dx={2\over\pi}\int_0^{\pi/2}\bigl(\cos((2j+1)x)+\cos((2j-1))x\bigr)\ dx\cr
&={2\over\pi}\left.\biggl({1\over2j+1}\sin((2j+1)x)+{1\over2j-1}\sin((2j-1)x)\biggr)\right|_0^{\pi/2} \cr
&={2\over\pi}\biggl({(-1)^j\over 2j+1} -{(-1)^j\over 2j-1}\biggr)={4\over\pi}{(-1)^{j-1}\over 4j^2-1}\ .\cr}$$
In particular $a_0={4\over\pi}$. All in all we obtain
$$|\cos x|={a_0\over2}+\sum_{j=1}^\infty a_{2j}\cos(2j x)={2\over\pi}-{4\over\pi}\sum_{j=1}^\infty{(-1)^j\over 4j^2-1}\cos(2jx)\ ,$$
and the resulting series is uniformly convergent on ${\mathbb R}$.
A: The Fourier series of a function $f(x)$ is a series on the form
$${1\over 2}a_0 +\sum_{n=1}^\infty \bigl(a_n \cos(nx) + b_n \sin(nx)\bigr)$$
where the coefficients $a_n$ and $b_n$ are given by the formulas $$a_n ={1\over\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,{\rm d}x\qquad n=0,1,2,\dots$$ $$b_n ={1\over\pi}\int_{-\pi}^\pi f(x)\sin(nx)\,{\rm d}x\qquad n=1,2,\dots$$
Since your function $f(x)=|\cos(x)|$ is an even function, all $b_n=0$. (You are integrating an odd function over a symmetric interval about 0.) 
For the same reason, you get that
$$a_n ={2\over\pi}\int_0^\pi f(x)\cos(nx)\,{\rm d}x$$
Your function $f(x)$ is defined by $f(x)=\cos(x)$ for $0\leq x\leq {\pi\over 2}$, and by $f(x)=-\cos(x)$ for ${\pi\over 2}\leq x\leq \pi$, so it might be wise to divide this integral in two parts when computing it. 
When you have computed the Fourier coefficients, try to plot $f(x)$ and the series you get by using the first few terms of the Fourier series. You should get two reasonably similar-looking graphs.
A: Note that the Fourier coefficients of a function $f(x)$ are given by
$$c_n=\frac{1}{2\pi}\int_0^{2\pi}e^{-inx}f(x)dx.$$
For $f(x)=|\cos x|$ we have
$$\begin{align}
c_n &=\frac{1}{2\pi}\int_0^{2\pi}e^{-inx}|\cos x|dx\\
&=\frac{1}{2\pi}\int_0^{\pi}e^{-inx}|\cos x|dx+\frac{1}{2\pi}\int_{\pi}^{2\pi}e^{-inx}|\cos x|dx\\
\end{align}$$
and if $n$ is odd we have $e^{-in(x+\pi)}=-e^{-inx}$ so the two integrals cancel. If $n$ is even then $e^{-in(x+\pi)}=e^{-inx}$ so both integrals agree and 
$$\begin{align}
c_n&=\frac{1}{\pi}\int_0^{\pi}e^{-inx}|\cos x|dx\\
&=\frac{1}{\pi}\int_0^{\pi/2}e^{-inx}\cos xdx+\frac{1}{\pi}\int_{\pi/2}^{\pi}e^{-inx}\cos xdx\\
\end{align}$$
and if $\frac{n}{2}$ is odd we have $e^{-in(x+\pi/2)}=e^{-inx+(n/2)\pi}=-e^{-inx}$ so the integrals again cancel. Thus if $n\neq 4m$ the Fourier coefficient $c_n$ is $0$. If $n=4m$ then we have $e^{-in(x+\pi/2)}=e^{-inx}$ so both integrals agree and
$$\begin{align}
c_n&=\frac{2}{\pi}\int_0^{\pi/2}e^{-inx}\cos xdx\\
&=\frac{2}{\pi}\int_0^{\pi/2}\cos nx\cos xdx+\frac{2i}{\pi}\int_0^{\pi/2}\sin nx\cos xdx\\
\end{align}$$
and both of these integrals can be found in many tables, or on Wolfram|Alpha.
A: One can use Fourier series on a general interval $[a, a + L]$,
$$ \sum _{n=-\infty }^{\infty }{\it c\_n}\,{{\rm e}^{{\frac {2\,inx\pi }{
L}}}} \,,$$
where $c_n$ are given by,
$$ c_n = \int _{a}^{a+L}\!f \left( x \right) {{\rm e}^{{\frac {-2\,in\pi \,x}{L
}}}}{dx}$$
Applying this to our problem, first we note that |cos(x)| is a periodic function of period $\pi$ and $\cos(x)$ is positive on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$,
$$ c_n = \int _{-\frac{\pi}{2}}^{-\frac{\pi}{2}+\pi} \cos(x) {{\rm e}^{{\frac {-2\,in\pi \,x}{L
}}}}{dx} = 2\,{\frac { \left( -1 \right) ^{1+n}}{\pi \, \left( 4\,{n}^{2}-1
 \right) }}
\,.$$
Then the Fourier series of $|\cos(x)|$ is given by,
$$ |\cos(x)| =  \sum _{n=-\infty }^{\infty }2\,{\frac { \left( -1 \right) ^{1+n}}{\pi \, \left( 4\,{n}^{2}-1\right) }}\,{{\rm e}^{{\frac {2\,inx\pi }{L}}}}\,. $$
Writing the sum as,
$$ |\cos(x)| =  2\,{\pi }^{-1}+\sum _{n=1}^{\infty}2\,{\frac { \left( -1 \right) ^{1+n}
 \left( {{\rm e}^{2\,inx}}+{{\rm e}^{-2\,inx}} \right) }{\pi \,
 \left( 4\,{n}^{2}-1 \right) }}
= 2\,{\pi }^{-1}+\sum _{n=1}^{\infty}4\,{\frac { \left( -1 \right) ^{1+n}
\cos \left( 2\,nx \right) }{\pi \, \left( 4\,{n}^{2}-1 \right) }}
\,, $$
which gives the desired result.
