Fourier Sine and Cosine Series

Recall that the Fourier series of f(x) is defined by

\begin{displaymath}a_0 + \sum \Big(a_n\cos(nx) + b_n\sin(nx)\Big)\end{displaymath}

where

\begin{displaymath}\left\{\begin{array}{lclr}
a_0 &=& \displaystyle \frac{1}{2\p...
..._{-\pi}^{\pi} f(x) \sin(nx)dx,& 1 \leq n.\\
\end{array}\right.\end{displaymath}

We have the following result:

Theorem. Let f(x) be a function defined and integrable on interval $[-\pi, \pi]$.

(1)
If f(x) is even, then we have

\begin{displaymath}b_n = 0 ,\;\;\; \mbox{for all } n \geq 1;\end{displaymath}

and

\begin{displaymath}a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx.\end{displaymath}

(2)
If f(x) is odd, then we have

\begin{displaymath}a_n = 0 ,\;\;\; \mbox{for all } n \geq 0;\end{displaymath}

and

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx)dx.\end{displaymath}

This Theorem helps define the Fourier series for functions defined only on the interval $[0,\pi]$. The main idea is to extend these functions to the interval $[-\pi, \pi]$ and then use the Fourier series definition.

Let f(x) be a function defined and integrable on $[0,\pi]$. Set

\begin{displaymath}f_1(x) = \left\{ \begin{array}{lrl}
-f(-x),& -\pi \leq x < 0\\
f(x), & 0 \leq x \leq \pi
\end{array} \right.\end{displaymath}

and

\begin{displaymath}f_2(x) = \left\{ \begin{array}{lrl}
f(-x),& -\pi \leq x < 0\\
f(x), & 0 \leq x \leq \pi
\end{array} \right.\end{displaymath}

Then f1 is odd and f2 is even. It is easy to check that these two functions are defined and integrable on $[-\pi, \pi]$ and are equal to f(x) on $[0,\pi]$. The function f1 is called the odd extension of f(x),
while f2 is called its even extension.

Definition. Let f(x), f1(x), and f2(x) be as defined above.

(1)
The Fourier series of f1(x) is called the Fourier Sine series of the function f(x), and is given by

\begin{displaymath}f(x) \sim \sum_{n=1}^{\infty} b_n\sin(nx).\end{displaymath}

where

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx)dx.\end{displaymath}

(2)
The Fourier series of f2(x) is called the Fourier Cosine series of the function f(x), and is given by

\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty} a_n\cos(nx).\end{displaymath}

where

\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx,\;\;\mbox{...
...}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx\;\;\mbox{for $n \geq 1$}.\end{displaymath}

Example. Find the Fourier Cosine series of f(x) = x for $ x \in [0,\pi]$.
Answer. We have

\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} xdx = \frac{\pi}{2},\end{displaymath}

and

\begin{displaymath}a_n = \frac{2}{\pi} \int_{0}^{\pi} x\cos(nx)dx =
\frac{2}{n^2\pi} (\cos(n\pi) - 1) = \frac{2}{n^2\pi} ((-1)^n - 1).\end{displaymath}

Therefore, we have

\begin{displaymath}f(x) \sim \frac{\pi}{2} - \frac{4}{\pi}\left(\cos(x) +
\frac{1}{9}\cos(3x) + \frac{1}{25}\cos(5x)+\ldots\right).\end{displaymath}

Example. Find the Fourier Sine series of the function f(x) = 1 for $ x \in [0,\pi]$.
Answer. We have

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(nx)dx =
\frac{2}{n\pi}(-\cos(n\pi) + 1) = \frac{2}{n\pi}(1-(-1)^n).\end{displaymath}

Hence

\begin{displaymath}f(x) \sim \frac{4}{\pi}\left(\sin(x) + \frac{1}{3}\sin(3x) +
\frac{1}{5}\sin(5x)...\right).\end{displaymath}

Example. Find the Fourier Sine series of the function $f(x) = \cos(x)$ for $ x \in [0,\pi]$.
Answer. We have

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi}\cos(x)\sin(nx)dx\end{displaymath}

which gives b1 = 0 and for n > 1, we obtain

\begin{displaymath}b_n = \frac{2n}{\pi}\left(\frac{1 + (-1)^n}{n^2 -1}\right).\end{displaymath}

Hence

\begin{displaymath}\cos(x) \sim \frac{8}{\pi} \sum_{n=1}^{\infty}
\frac{n\sin(2nx)}{4n^2-1}.\end{displaymath}

Special Case of 2L-periodic functions.
As we did for $2\pi$-periodic functions, we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. First, recall the Fourier series of f(x)

\begin{displaymath}f(t) \sim a_0 + \sum_{n=1}^{\infty} \left(a_n\cos\left(n\frac{\pi t}
{L}\right) + b_n\sin\left(n\frac{\pi t}{L}\right)\right)\end{displaymath}

where

\begin{displaymath}\left\{\begin{array}{lclr}
a_0 &=& \displaystyle \displaystyl...
...f(x) \sin\left(n\frac{\pi x}{L}\right)dx,\\
\end{array}\right.\end{displaymath}

for $1\leq n$.
1.
If f(x) is even, then bn = 0, for $n \geq 1$. Moreover, we have

\begin{displaymath}a_0 = \frac{1}{L} \int_{0}^{L} f(x) \cos\left(n\frac{\pi x}{L}\right)dx,\end{displaymath}

and

\begin{displaymath}a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos\left(n\frac{\pi x}{L}\right)dx.\end{displaymath}

Finally, we have

\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty}a_n\cos\left(n\frac{\pi x}{L}
\right).\end{displaymath}

2.
If f(x) is odd, then an = 0, for all $n \geq 0$, and

\begin{displaymath}b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(n\frac{\pi x}{L}\right)dx.\end{displaymath}

Finally, we have

\begin{displaymath}f(x) \sim \sum_{n=1}^{\infty} b_n\sin\left(n\frac{\pi x}{L}\right).\end{displaymath}

The definitions of Fourier Sine and Cosine may be extended in a similar way.

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