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 $f$:

\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:

\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:

\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 $f$:

\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:

\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:

\begin{align} \quad f(x) \sim \sum_{n=1}^{\infty} d_n \cos nx \quad \blacksquare \end{align}