Trigonometric Fourier Series of Even and Odd Functions

Trigonometric Fourier Series of Even and Odd Functions

We will now state two important theorems which will allow us to evaluate trigonometric Fourier series more easily in many circumstances.

 Theorem 1: Let $f \in L([-\pi, \pi])$ be a $2\pi$-periodic function. If $f$ is an even function ($f(x) = f(-x)$ for all $x \in [-\pi, \pi]$) then the trigonometric Fourier series of $f$ has the form $\displaystyle{f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} c_n \cos nx}$ where $\displaystyle{c_n = \frac{2}{\pi} \int_0^{\pi} f(t) \cos nt \: dt}$.
• Proof: Suppose that $f(x) = f(-x)$ for all $x \in [-\pi, \pi]$. Consider the trigonometric Fourier series of $f$:
(1)
\begin{align} \quad f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) \end{align}
• Where $\displaystyle{a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \: dt}$ and $\displaystyle{b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt \: dt}$.
• Since $f$ is an even function and $\sin nt$ is an odd function for all $n \in \mathbb{N}$ we have that $f(t) \sin nt$ is an odd function. But $f(t) \sin nt$ is an odd function on the symmetric interval $[-\pi, \pi]$ and so $\displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt \: dt = 0}$ for all $n \in \mathbb{N}$. Thus:
(2)
\begin{align} \quad f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx \end{align}
• Also, since $f$ is an even function and $\cos nt$ is an even function for all $n \in \{ 0, 1, 2, ... \}$ we have that $f(t) \cos nt$ is an even function. But $f(t) \cos nt$ is an even function on the symmetric interval $[-\pi, \pi]$ and so $\displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \: dt = \frac{2}{\pi} \int_{0}^{\pi} f(t) \cos nt \: dt = c_n}$ for all $n \in \{ 0, 1, 2, ... \}$. Thus:
(3)
\begin{align} \quad f(x) \sim \frac{c_n}{2} + \sum_{n=1}^{\infty} c_n \cos nx \quad \blacksquare \end{align}

The following theorem is an analogous to Theorem 1 above for odd functions.

 Theorem 2: Let $f \in L([-\pi, \pi])$ be a $2\pi$-periodic function. If $f$ is an odd function ($-f(x) = f(-x)$ for all $x \in [-\pi, \pi]$) then the trigonometric Fourier series of $f$ has the form $\displaystyle{f(x) \sim \sum_{n=1}^{\infty} d_n \sin nx}$ where $\displaystyle{d_n = \frac{2}{\pi} \int_0^{\pi} f(t) \sin nt \: dt}$.
• Proof: Suppose that $-f(x) = f(-x)$ for all $x \in [-\pi, \pi]$. Consider the trigonometric Fourier series of $f$:
(4)
\begin{align} \quad f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) \end{align}
• Where $\displaystyle{a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \: dt}$ and $\displaystyle{b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt \: dt}$.
• Since $f$ is an odd function and $\cos nt$ is an even function for all $n \in \mathbb{N}$ we have that $f(t) \cos nt$ is an odd function. But $f(t) \cos nt$ is an odd function on the symmetric interval $[-\pi, \pi]$ and so $\displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \: dt = 0}$ for all $n \in \mathbb{N}$. Thus:
(5)
\begin{align} \quad f(x) \sim \sum_{n=1}^{\infty} b_n \cos nx \end{align}
• Also, since $f$ is an odd function and $\sin nt$ is an odd function for all $n \in \{ 0, 1, 2, ... \}$ we have that $f(t) \sin nt$ is an even function. But $f(t) \sin nt$ is an even function on the symmetric interval $[-\pi, \pi]$ and so $\displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt \: dt = \frac{2}{\pi} \int_{0}^{\pi} f(t) \sin nt \: dt = d_n}$ for all $n \in \{ 0, 1, 2, ... \}$. Thus:
(6)