Fourier series for even and odd functions

Fourier series: Coefficients of Fourier series

Fourier series for even and odd functions

In the Fourier series of odd and even functions about half the coefficients will be zero.

Let $L$ be a positive number and let $f$ be a $2L$ -periodic function. Suppose that $f$ has Fourier series

$\begin{array}{rcl}\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos \left(\frac{\pi n x}{L}\right) + b_n \sin \left(\frac{\pi n x}{L}\right) \end{array}$

If $f$ is even, then $b_n=0$ for $n=1,2,\ldots$
If $f$ is odd, then $a_n=0$ for $n=0,1,2\ldots$

Suppose that $f$ is even. Since $\cos\left(\frac{\pi n x}{L}\right)$ is an even function, the product $f(x)\cdot \cos\left(\frac{\pi n x}{L}\right)$ is also even. At the same time, $\sin\left(\frac{\pi n x}{L}\right)$ is odd, so $f(x)\cdot \sin\left(\frac{\pi n x}{L}\right)$ is also odd. Using the Euler formulas for the coefficients, we can calculate the coefficients for $f$ .

$\begin{array}{rcl}\displaystyle a_n &=&\displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{\pi n x}{L}\right) \dd x = \frac{2}{L} \int_0^{L} f(x)\cos\left(\frac{\pi n x}{L}\right) \dd x \quad \text{for } n=0,1,\ldots \\ b_n &=& \displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{\pi n x}{L}\right) \dd x = 0 \quad \text{for } n=1,2,\ldots \end{array}$

So the Fourier series of an even function contains only cosines and can be written as stated.

Now let $f(x)$ be an odd function. Since $\cos\left(\frac{\pi n x}{L}\right)$ is clearly an even function for all non-negative $n$ , the product $f(x)\cos\left(\frac{\pi n x}{L}\right)$ is odd. At the same time, $\sin\left(\frac{\pi n x}{L}\right)$ is odd so that $f(x)\sin\left(\frac{\pi n x}{L}\right)$ is even. Then, again using the Euler formulas, we can calculate the coefficients of the Fourier series for $f$ :

$\begin{array}{rcl}\displaystyle a_n &=&\displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{\pi n x}{L}\right) \dd x = 0 \quad \text{for } n=0,1,\ldots \\ b_n &=& \displaystyle\frac{1}{L} \int_{-L}^{L} f(x)\sin\left(\frac{\pi n x}{L}\right) \dd x = \displaystyle\frac{2}{L} \int_0^{L} f(x)\sin\left(\frac{\pi n x}{L}\right) \dd x \quad \text{for } n=1,2,\ldots \end{array}$

Thus, the Fourier series of a periodic even function contains only terms with sines and can be written as

$\begin{array}{c}\displaystyle \sum_{n=1}^{\infty} b_n \sin\left(\frac{\pi n x}{L}\right) \end{array}$

Piecewise smooth

Even in the case where $f$ is periodic and piecewise smooth and behaves like an even or odd function at the point where $f$ and $f'$ are smooth, the theorem holds.

For instance, the $2$ -periodic function $f$ that is $-1$ on $\ivco{-1}{0}$ and $1$ on $\ivco{0}{1}$ is odd outside the points $2k$ for integers $k$ and the coefficient $a_0$ of the Fourier series is $0$ .

Consider the $2$ -periodic function that is determined by $f(x)=\begin{cases}-3 x ^2 &\text{if } -1 \leq x\lt 0\\ 3 x ^2 &\text{if } 0\leq x\lt 1\end{cases}$

Its Fourier series has the form $f(x)=\frac{a_0}{2}+\sum_{m=1}^{\infty}\left(a_m\cdot\cos\left(\frac{m\cdot\pi\cdot x}{L}\right)+b_m\cdot\sin\left(\frac{m\cdot\pi\cdot x}{L}\right)\right)$ where $L=1$ is the half-period of $f$ .

Of which of the coefficients of the Fourier series can we tell that they are zero without further computation?

$a_0 =0$
$a_m=0$ for $m=1,2,3,\dots$

The function is odd, so all coefficients $a_m$ are zero.

The graph of the function $f(x)$ has been plotted over three periods in the figure below.

New example

Fourier series: Coefficients of Fourier series

Fourier series for even and odd functions

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