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Applied Mathematics

Peter Staab

Section 6.3 Even and Odd Functions; Half-Range Expansions

We saw above that the periodic extension of \(f(x)=x\) on \([-1,1]\) in Example 6.2.10 resulted in a odd function and that only the sine terms of the Fourier Series was left. That is, all of the Fourier coefficients for the cosine terms were 0. In this section, we use this idea to produce only even and odd extensions which results in only sine expansions or cosine expansions.
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To begin, let’s clearly define an even- and odd-periodic extension.
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Definition 6.3.1.

Let \(f\) be defined on \([0,L]\) for some \(L \gt 0\text{.}\)
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  • The even periodic extension of \(f\) is defined as
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    \begin{equation*} F(x) = \begin{cases} f(x) \amp 0 \leq x \leq L \\ f(-x) \amp -L \leq x \lt 0 \\ f(x+2kL) \amp \text{otherwise for appropriate integer $k$} \end{cases} \end{equation*}
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  • The odd periodic extension of \(f\) is defined as
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    \begin{equation*} F(x) = \begin{cases} f(x) \amp 0 \leq x \leq L \\ -f(-x) \amp -L \leq x \lt 0 \\ f(x+2kL) \amp \text{otherwise for appropriate integer $k$} \end{cases} \end{equation*}
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Example 6.3.2.

Graph the even- and odd-periodic extension of
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\begin{equation*} f(x) = 1-x \end{equation*}
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defined on \([0,1]\text{.}\)
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Solution.
For the even extension, we first graph the function on \([0,1]\text{,}\) then make the even extension of it on \([-1,0]\text{.}\) The original function is shown below as a solid line and the even extension is dashed.
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Figure 6.3.3. The even extension of \(f(x)=1-x\)
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and then we produce the period extension of period 2.
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Figure 6.3.4. The even extension of \(f(x)=1-x\)
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To find the odd extension, flip the original function on \([0,1]\) around the origin to get:
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Figure 6.3.5. Odd extension of \(f(x)=1-x\)
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Then extend the function on \([-1,1]\) in a periodic way.
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Figure 6.3.6. Periodic version of the odd extension of \(f(x) = 1-x\text{.}\)
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Subsection 6.3.1 The Fourier Sine and Cosine Series

We now address the Fourier series of the even- and odd-periodic extensions of \(f\) on \([0,L]\text{.}\) As in Example 6.2.10, there are no cosine terms and the Fourier series of the odd periodic extension of \(f\) can be written
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\begin{align} f(x) \amp = \sum_{n=1}^{\infty} b_n \sin \biggl(\frac{n\pi x}{L} \biggr)\quad\text{where,}\tag{6.3.1}\\ b_n \amp = \frac{2}{L} \int_{0}^L f(x) \sin \biggl(\frac{n\pi x}{L} \biggr) \,dx \tag{6.3.2} \end{align}
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and this is often called the Fourier sine series.
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Similarly, the Fourier series of the even periodic extension of \(f\) is
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\begin{align} f(x) \amp= a_0 + \sum_{n=1}^{\infty} a_n \cos \biggl( \frac{n \pi x}{L} \biggr)\quad\text{where,}\tag{6.3.3}\\ a_0 \amp = \frac{1}{L} \int_{0}^L f(x) \, dx\tag{6.3.4}\\ a_n \amp = \frac{2}{L} \int_0^L f(x) \cos \biggl( \frac{n\pi x}{L} \biggr)\, dx\tag{6.3.5} \end{align}
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and is called the Fourier cosine series.
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Example 6.3.7.

Find the Fourier Cosine Series of \(f(x) = x\) for \(0 \leq x \leq 1\text{.}\)
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Solution.
For this, we need to find the coefficients in ((6.3.4)) and ((6.3.5)),
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\begin{align*} a_0 \amp = \frac{1}{1} \int_0^1 x \,dx = \frac{1}{2} \\ a_n \amp = \frac{2}{1} \int_0^1 x \cos n \pi x \, dx \\ \amp = \frac{2(\cos n\pi -1)}{n \pi} = \frac{2}{n\pi} \bigl((-1)^n-1\bigr) \end{align*}
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and use ((6.3.3)) to find
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\begin{equation*} f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{2}{n\pi} \bigl((-1)^n-1\bigr) \cos n \pi x \end{equation*}
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Example 6.3.8.

Find the odd periodic extension of \(f(x) = x^2\) on \([0,2]\text{.}\) A graph of this is:
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Figure 6.3.9. Graph of odd extension of \(f(x) = x^2\) on \([0,2]\)
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Solution.
For this, we need to find \(b_n\) from ((6.3.2)),
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\begin{align*} b_n \amp = \frac{2}{2} \int_0^2 x^2 \sin \biggl( \frac{n \pi x}{2} \biggr) \, dx\\ \amp\qquad \text{and again, this can be done using tabular integration,}\\ \amp = -\frac{2 \, x^{2}}{n \pi} \cos \biggl( \frac{n \pi x}{2} \biggr) + \frac{8 \, x}{n^2\pi^2} \sin \biggl( \frac{n \pi x}{2} \biggr) + \frac{16}{n^3\pi^3} \cos \biggl( \frac{n \pi x}{2} \biggr) \biggr \vert_0^2 \\ \amp = -\frac{8}{n \pi} \cos (n \pi) + \frac{16}{n^3\pi^3} \cos (n \pi) = \frac{16-8n^2\pi^2}{n^3 \pi^3} (-1)^n \end{align*}
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and then use this in ((6.3.1))
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\begin{equation*} f(x) = \sum_{n=1}^{\infty}\frac{16-8n^2\pi^2}{n^3 \pi^3} (-1)^n \sin \biggl(\frac{n \pi x}{2} \biggr) \end{equation*}
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