Even and Odd Fourier Series for f(x) = π − x on (0, π)

For a function specified only on $0<x<\pi$, “the even and odd Fourier series” usually means its two half-range expansions. The cosine series comes from reflecting the function evenly across $x=0$; the sine series comes from reflecting it oddly. The coefficients—and the periodic functions represented—are therefore different.

For $f(x)=\pi-x$, the even extension produces a cosine series with only odd-indexed cosine terms. The odd extension produces a sine series with every positive integer frequency. Both equal $\pi-x$ at each point strictly inside the original interval.

Half-range coefficient formulas

A cosine half-range expansion on $(0,\pi)$ has the form

$$f(x)\sim \frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx),$$

where

$$a_0=\frac{2}{\pi}\int_0^\pi f(x)\,dx, \qquad a_n=\frac{2}{\pi}\int_0^\pi f(x)\cos(nx)\,dx.$$

A sine half-range expansion has the form

$$f(x)\sim\sum_{n=1}^{\infty}b_n\sin(nx), \qquad b_n=\frac{2}{\pi}\int_0^\pi f(x)\sin(nx)\,dx.$$

The factor $2/\pi$ is the half-range normalization for an interval of length $\pi$. The constant term in the cosine series is $a_0/2$, not $a_0$.

Even extension: cosine series

First calculate the constant coefficient:

$$\begin{aligned} a_0&=\frac{2}{\pi}\int_0^\pi(\pi-x)\,dx\\ &=\frac{2}{\pi}\left[\pi x-\frac{x^2}{2}\right]_0^\pi =\pi. \end{aligned}$$

For $n\ge 1$, integration by parts gives

$$\begin{aligned} a_n&=\frac{2}{\pi}\int_0^\pi(\pi-x)\cos(nx)\,dx\\ &=\frac{2}{\pi}\frac{1-\cos(n\pi)}{n^2}\\ &=\frac{2\bigl(1-(-1)^n\bigr)}{\pi n^2}. \end{aligned}$$

If $n$ is even, this coefficient is zero. If $n$ is odd, it is $4/(\pi n^2)$. Thus the cosine expansion is

$$\boxed{\pi-x=\frac{\pi}{2}+\frac{4}{\pi}\left(\cos x+\frac{\cos 3x}{3^2}+\frac{\cos 5x}{5^2}+\cdots\right),\quad 0<x<\pi.}$$

Equivalently, the sum can be written as $\frac{\pi}{2}+\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\cos((2k+1)x)}{(2k+1)^2}$.

This series represents the even periodic extension, which on $[-\pi,\pi]$ is $\pi-|x|$. It is continuous at the endpoints when repeated with period $2\pi$, and its cosine coefficients decay as $1/n^2$.

Odd extension: sine series

For the sine coefficients,

$$\begin{aligned} b_n&=\frac{2}{\pi}\int_0^\pi(\pi-x)\sin(nx)\,dx\\ &=\frac{2}{\pi}\left(\frac{\pi(1-(-1)^n)}{n}+\frac{\pi(-1)^n}{n}\right)\\ &=\frac{2}{n}. \end{aligned}$$

So the odd half-range expansion is

$$\boxed{\pi-x=2\left(\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots\right),\quad 0<x<\pi.}$$

Here the odd periodic extension on $(-\pi,\pi)$ equals $\pi-x$ for positive $x$ and $-\pi-x$ for negative $x$. It has a jump at $x=0$, which explains why its sine coefficients decrease only as $1/n$.

A quick check at $x=\pi/2$

The function value is $f(\pi/2)=\pi/2$. In the cosine series, every term vanishes because $\cos((2k+1)\pi/2)=0$, leaving $\pi/2$. In the sine series, the even-indexed terms vanish and the odd-indexed terms alternate:

$$2\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\right)=\frac{\pi}{2},$$

by the alternating arctangent series. The two expansions agree with the function, despite having different terms and convergence behavior.

Convergence and endpoint values

On every point $0<x<\pi$, both Fourier series converge to $\pi-x$. The cosine series also converges to the even extension’s endpoint values: at $x=0$ it gives $\pi$, and at $x=\pi$ it gives $0$.

For the odd extension, the periodic function jumps at $x=0$. A Fourier series at a jump converges to the average of the left- and right-hand limits, so the sine series has value $0$ there—not $\pi$. At $x=\pi$, the periodic odd extension is continuous with value $0$, and the series converges to $0$.

Common mistakes

• Using the full-range coefficient factor: For these half-range formulas, use $2/\pi$ in the integrals, not $1/\pi$.
• Forgetting the half in the constant term: The cosine series begins with $a_0/2=\pi/2$.
• Dropping parity-dependent terms incorrectly: In the cosine series, even indices vanish and odd indices remain. In the sine series, every coefficient is $2/n$.
• Treating the extensions as identical: A cosine series and a sine series represent different periodic extensions; they happen to match the original function on $(0,\pi)$.
• Assuming the sine series equals the endpoint value of the original formula: At $x=0$, its value is determined by the odd extension’s jump and is the midpoint, $0$.