Fourier sine series, Fourier cosine series

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When a function is an odd function, its Fourier series contains only the sine functions. Such a Fourier series is called a Fourier sine series. When a function is an even function, its Fourier series contains only the cosine functions (and possibly a constant). Such a Fourier series is called a Fourier cosine series.



Example. Consider the function \(f \in \mathcal{P}_{2\pi}\) defined by

\[f(x) = x ~~~ (-\pi < x < \pi).\]

Note that \(f\) is an odd function (i.e., \(f(-x) = -f(x)\)). The Fourier coefficients are obtained as (exercise!)

\[\begin{eqnarray*} a_n &=& \frac{1}{\pi}\int_0^{2\pi}f(x)\cos(nx)dx =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\\ &=& 0 \\ b_n &=& \frac{1}{\pi}\int_0^{2\pi}f(x)\sin(nx)dx =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx.\\ &=& (-1)^{n+1}\frac{2}{n}. \end{eqnarray*}\]

Thus, we have

\[x \sim 2\left[\frac{\sin x}{1} - \frac{\sin(2x)}{2} + \frac{\sin(3x)}{3} -\cdots +(-1)^{n+1}\frac{\sin(nx)}{n} + \cdots\right]\]

for \(-\pi < x < \pi\). □

More generally, if \(f\in \mathcal{R}_{2\pi}^{1}\) is an odd function, the coefficients of \(\cos(nx)\) are all 0 so \(S[f]\) consists of only the \(\sin(nx)\) terms,

\[f \sim S[f] = \sum_{n=1}^{\infty}b_n\sin(nx).\]

Such a Fourier series is called a Fourier sine series.

In contrast, if \(f\in \mathcal{R}_{2\pi}^{1}\) is an even function, the coefficients of \(\sin(nx)\) are all 0 so \(S[f]\) consists of only the \(\cos(nx)\) terms,

\[f \sim S[f] = \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos(nx).\]

Such a Fourier series is called a Fourier cosine series.

Example (evenfunc). Consider the function \(f \in \mathcal{P}_{2\pi}\) defined by

\[f(x) = 1 - \frac{|x|}{\pi} ~~~ (|x| \leq \pi).\]

\(f\) is a continuous, even, and periodic function. Clearly, \(b_n = 0\) (verify!). It is easy to see \(a_0 = 1\). For \(n = 1, 2, \cdots\),

\[\begin{eqnarray*} \pi a_n &=& \int_{-\pi}^{\pi}f(x)\cos(nx)dx = 2\int_{0}^{\pi}\left(1 - \frac{x}{\pi}\right)\cos(nx)dx\\ &=& \frac{2}{n^2\pi}\{1 - (-1)^n\} ~~~ (n = 1, 2, \cdots). \end{eqnarray*}\]

That is,

\[a_n = \left\{ \begin{array}{cc} 1 & (n = 0),\\ \frac{4}{\pi^2 n^2} & (n = 1, 3, 5, \cdots),\\ 0 & (n = 2, 4, 6, \cdots). \end{array} \right.\]

Therefore,

\[1 - \frac{|x|}{\pi} \sim \frac{1}{2} + \frac{4}{\pi^2}\left(\frac{\cos x}{1^2} + \frac{\cos(3x)}{3^2} + \frac{\cos(5x)}{5^2} + \cdots\right) ~~~ (-\pi < x < \pi).\]

□

Now, suppose we have a function \(g(x)\) on \(I_0 = (0, \pi)\) that is integrable and continuous on \(I_0\) except for finitely many points. Based on this function, let us define the following odd function \(f_o\):

\[\begin{eqnarray*} f_o(x) &=& g(x) \text{ for } x \in (0, \pi),\\ f_o(x) &=& -g(-x) \text{ for } x \in (-\pi, 0),\\ f_o(x + 2n\pi) &=& f_o(x) \text{ for } x \in (-\pi, \pi)\setminus \{0\}, n\in \mathbb{N}. \end{eqnarray*}\]

Similarly, we can define the following even function \(f_e\):

\[\begin{eqnarray*} f_e(x) &=& g(x) \text{ for } x \in (0, \pi),\\ f_e(x) &=& g(-x) \text{ for } x \in (-\pi, 0),\\ f_e(x + 2n\pi) &=& f_e(x) \text{ for } x \in (-\pi, \pi)\setminus \{0\}, n\in \mathbb{N}. \end{eqnarray*}\]

The corresponding Fourier series are of the forms

\[\begin{eqnarray*} f_o &\sim & \sum_{n=1}^{\infty}b_n\sin(nx),\\ f_e &\sim & \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos(nx), \end{eqnarray*}\]

where

\[\begin{eqnarray*} a_n &=& \frac{1}{\pi}\int_{-\pi}^{\pi}f_e(x)\cos(nx)dx = \frac{2}{\pi}\int_{0}^{\pi}g(x)\cos(nx)dx,\\ b_n &=& \frac{1}{\pi}\int_{-\pi}^{\pi}f_o(x)\sin(nx)dx = \frac{2}{\pi}\int_{0}^{\pi}g(x)\sin(nx)dx. \end{eqnarray*}\]

Using these coefficients, \(g(x)\) on \(I_0 = (0, \pi)\) can be expressed in two ways:

\[\begin{eqnarray*} g(x) &\sim& \sum_{n=1}^{\infty}b_n\sin(nx),\\ g(x) &\sim & \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos(nx). \end{eqnarray*}\]

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