Finding Trigonometric Fourier Series of Functions of Period 2L

Finding Trigonometric Fourier Series of Functions of Period 2L

So far, all of the formulas we have looked at regarding trigonometric Fourier series have been for functions $f$ that are defined on an interval of length $2\pi$ and/or are $2\pi$-periodic. We would like to generalize obtaining a trigonometric Fourier series for functions that are instead of period $2L$. Fortunately, this is rather simple.

Suppose that $f$ is $2L$-periodic ($L > 0$), i.e., $f(x + 2L) = f(x)$ for all $x \in \mathbb{R}$. Then the trigonometric Fourier series of $f$ is given by:

(1)
\begin{align} \quad f(x) \sim \frac{c_0}{2} + \sum_{n=1}^{\infty} \left [ c_n \cos \left ( \frac{n \pi x}{L} \right ) + d_n \sin \left ( \frac{n \pi x}{L} \right ) \right ] \end{align}

Where $\displaystyle{c_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos \left ( \frac{n \pi t}{L} \right ) \: dt}$ and $\displaystyle{d_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin \left ( \frac{n \pi t}{L} \right ) \: dt }$.

For example, suppose that we want to find the Fourier series of the function $f(x) = \mid x \mid$ for $-1 \leq x \leq 1$ and $f(x + 2) = f(x)$ for all $x \in \mathbb{R}$ ($f$ is $2$-periodic). Then $L = 1$ and:

(2)
\begin{align} \quad a_0 = \frac{1}{L} \int_{-L}^{L} f(t) \: dt = \int_{-1}^{1} \mid L \mid \: dt = \int_{-1}^{0} -t \: dx + \int_0^{1} t \: dx = -\frac{t^2}{2} \biggr \lvert_{-1}^{0} + \frac{t^2}{2} \biggr \lvert_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \end{align}
(3)
\begin{align} \quad c_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos \left ( \frac{n \pi t}{L} \right ) \: dt = \int_{-1}^{1} \mid t \mid \cos n \pi t \: dt = \int_{-1}^{0} -t \cos n\pi t \: dt + \int_0^1 t \cos n \pi t \: dt \quad (*) \end{align}

We use integration by parts to evaluate the integrals above. Let $u = t$ and let $dv = \cos n \pi t$. Then $du = dt$ and $\displaystyle{v = \frac{\sin n \pi t}{n \pi}}$, so:

(4)
\begin{align} \quad \int t \cos n \pi t \: dt = t \frac{\sin n \pi t}{n \pi} - \int \frac{\sin n \pi t}{n \pi} \: dt = t\frac{\sin n \pi t}{n \pi} + \frac{\cos n \pi t}{n^2 \pi^2} \end{align}

Evaluating the righthand side of the equation above at $t = 0$ and $t = 1$ yields:

(5)
\begin{align} \quad \int_0^1 t \cos n \pi t \: dt = \frac{\cos n \pi}{n^2 \pi^2} - \frac{1}{n^2 \pi^2} = \frac{(\cos n \pi - 1)}{n^2 \pi^2} \quad (**) \end{align}

We similarly find that:

(6)
\begin{align} \quad \int_{-1}^{0} t \cos n \pi t \: dt = \frac{(\cos n \pi - 1)}{n^2 \pi^2} \quad (***) \end{align}

Thus, by plugging $(**)$ and $(***)$ into $(*)$ we see that:

(7)
\begin{align} \quad c_n = \frac{2(\cos n \pi - 1)}{n^2 \pi^2} \end{align}

Furthermore we have that:

(8)
\begin{align} \quad d_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin \left ( \frac{n \pi t}{L} \right ) \: dt = \int_{-1}^{1} \mid t \mid \sin n \pi t \: dt \quad (****) \end{align}

Note that $\mid t \mid$ is an even function and $\sin n \pi t$ is an odd function on $[-L, L] = [-1, 1]$, so $\mid t \mid \sin n \pi t$ is an odd function on the symmetric interval $[-1, 1]$ so the integral $(****)$ above is equal to $0$ and the trigonometric Fourier series of $f$ is therefore:

(9)
\begin{align} \quad f(x) & \sim \frac{1}{2} + \sum_{n=1}^{\infty} \frac{2(\cos n \pi - 1)}{n^2 \pi^2} \cos n \pi x \\ & \sim \frac{1}{2} + \frac{-4}{\pi^2} \cos \pi x + \frac{-4}{9 \pi^2} \cos 3\pi x + \frac{-4}{25 \pi^2} \cos 5 \pi x + ... \\ & \sim \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4}{(2n - 1)^2 \pi^2} \cos (2n - 1)\pi x \end{align}

The graph of the first three approximations of $f$ (in red, yellow, and green) are graphed below alongside with $f$ itself (in blue):

Screen%20Shot%202016-04-23%20at%206.16.36%20PM.png

Clearly the Fourier series of $f$ converges to $f$ on all of $\mathbb{R}$. In particular, the series converges at $x = 1$ to $f(1)$, and consequentially:

(10)
\begin{align} \quad 1 = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4}{(2n - 1)^2 \pi^2} \cos (2n - 1)\pi \end{align}

Therefore:

(11)
\begin{align} \sum_{n=1}^{\infty} \frac{4}{(2n - 1)^2 \pi^2} \cos (2n - 1)\pi = -\frac{1}{2} \end{align}

Finding the sum of the series above would have been much more difficult without the use of Fourier series.

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