Full Series
A function \(f(x)\) of period \(2\pi\) is written as the trigonometric series
\[
f(x)=a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx),
\]
\begin{gather*}
a_0=\frac1{2\pi}\int_{-\pi}^{\pi}f(x)\,\dd x,\\
a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos nx\,\dd x,\qquad
b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,\dd x,\qquad n=1,2,\dots
\end{gather*}
Alternatively, so that every coefficient carries the same factor \(1/\pi\), the FS is written as
\[
f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx),
\]
with the Fourier coefficients given by the Euler formulas
\begin{gather*}
a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,\dd x,\\
a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos nx\,\dd x,\qquad
b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,\dd x,\qquad n=1,2,\dots
\end{gather*}
Example 2.2 Write the Fourier series for
\[
f(x)=\begin{cases}-k & \text{if } -\pi<x<0,\\ \phantom{-}k&\text{if } 0<x<\pi,\end{cases}
\qquad\text{and}\qquad f(x+2\pi)=f(x).
\]
Such functions occur as external forces on mechanical systems, EMFs in circuits, etc.
DiscontinuityThe value of \(f(x)\) at a single point does not affect the integral,
so \(f(x)\) may be left undefined at \(x=0\) and \(x=\pm\pi\).
Solution. \(f\) is odd, so \(a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,\dd x=0\) and
\begin{align*}
a_n&=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos nx\,\dd x=0 &&\bigl(f(x)\cos nx\ \text{is odd}\bigr),\\
b_n&=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,\dd x=\frac2\pi\int_0^\pi k\sin nx\,\dd x &&\bigl(f(x)\sin nx\ \text{is even}\bigr)\\
&=\frac{2k}{n\pi}(1-\cos n\pi).
\end{align*}
Hence
\[
b_1=\frac{4k}{\pi},\qquad b_2=0,\qquad b_3=\frac{4k}{3\pi},\qquad b_4=0,\ \dots,
\]
\[
f(x)=\frac{4k}{\pi}\Bigl(\sin x+\frac13\sin3x+\frac15\sin5x+\cdots\Bigr).
\]
Leibniz series (1674) Setting \(x=\pi/2\) gives
\[
f\Bigl(\frac\pi2\Bigr)=k=\frac{4k}\pi\Bigl(1-\frac13+\frac15-+\cdots\Bigr),
\qquad\text{thus}\qquad
1-\frac13+\frac15-\frac17+-\cdots=\frac\pi4.
\]
