Fourier coefficients for arbitrary periods

Fourier series: Coefficients of Fourier series

Fourier coefficients for arbitrary periods

Sometimes the need arises to find a Fourier series for a #2L#-periodic function for a number #L# distinct from #\pi#. This is achievable with minor adjustments to what we have already seen in the case of a #2\pi#-periodic function.

Euler formulas for arbitrary periodsLet #L# be a positive number and let #f# be a #2L#-periodic function.

The Fourier series of #f# is

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

where

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

By use of the substitution #x=\frac{ L t}{\pi}#, we scale #f# to the #2\pi#-periodic function #\varphi(t)#. Thus,

\[\varphi(t) = f\left(\frac{L t}{\pi}\right)\]

The Fourier series for #\varphi#, say

\[\begin{array}{rcl}\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos (nt) + b_n \sin (nt) \right)\end{array}\]

is determined by the Euler formulas. Therefore, for # n=0,1,\ldots# we have

\[\begin{array}{rcl}\displaystyle a_n&=&\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} \varphi(t) \cos( nt) \, \dd t \\ &&\phantom{xxx}\color{blue}{\text{Euler formula}}\\&=& \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f\left(\frac{L\cdot t}{\pi}\right) \cos( nt) \, \dd t\\ &&\phantom{xxx}\color{blue}{\text{definition }\varphi}\\ &=& \displaystyle \frac{1}{L} \int_{-L}^{L} f(x) \cos\left( \frac{\pi n x}{L}\right) \,\dd x\\ &&\phantom{xxx}\color{blue}{\text{substitution } t=\frac{\pi x}{L}}\\ \end{array}\]

Similarly, for #n=1,2,\ldots# we can derive the following.

\[\begin{array}{rcl}\displaystyle b_n&=&\displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \sin\left( \frac{\pi n x}{L}\right) \,\dd x \end{array}\]

By the reverse substitution #t=\frac{\pi x}{L}# in the Fourier series for #\varphi#, we find for #f(x) = \varphi\left(\frac{\pi x}{L}\right)# the Fourier series

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

Scaling

The proof tells us that, if #f(x)# is a #2L#-period function, then #\varphi(x) = f\left(\frac{L\cdot x}{\pi}\right)# is a #2\pi#-periodic function, and that, if #\varphi# has Fourier series

\[\varphi(t) = \begin{array}{rcl}\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos (nt) + b_n \sin (nt) \right)\end{array}\]

then the Fourier series of #f(x) = \varphi\left(\frac{\pi\cdot x}{L}\right)# can be obtained by substituting #\frac{\pi\cdot x}{L}# for #t# in the Fourier series of #\varphi#. \[f\left(x\right) = \begin{array}{rcl}\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos\left (\frac{n\cdot \pi\cdot x}{L}\right) + b_n \sin\left (\frac{n\cdot\pi\cdot x}{L}\right)\right) \end{array}\]

ConvergenceThe Convergence theorem, to which we already referred when discussing the #2\pi#-periodic case, is valid for all finite periods.

Calculate the Fourier series of \[f(x)=\left\{\begin{array}{l l}{{x}\over{2}}-1, & 0\leq x< 2\\ -1, & -2 \lt x<0 \end{array}\right. , \hspace{1.1cm}f(x+4)=f(x)\] Enter simplified expressions for the coefficients \(a_0\), \(a_m\), and \(b_m\), \(m=1,2,\dots\), where the Fourier series is given by \[f(x)=\frac{a_0}{2}+\sum_{m=1}^{\infty}\left(a_m\cdot\cos\left(\frac{m\cdot\pi\cdot x}{L}\right)+b_m\sin\left(\frac{m\cdot\pi\cdot x}{L}\right)\right)\] and \(L\) is the half-period of \(f\).

\(a_0 = -{{3}\over{2}}\)
\(a_m=\frac{(-1)^m-1}{m^2\cdot\pi^2}\) \(m=1,2,3,\dots\)
\(b_m=-\frac{(-1)^m}{m\cdot\pi}\) \(m=1,2,3\dots\)
\(f(x)=-{{3}\over{4}}+\sum_{m=1}^{\infty}\left(\frac{(-1)^m-1}{m^2\cdot\pi^2}\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)-\frac{(-1)^m}{m\cdot\pi}\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right) \)

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

Noting that the half-period of \(f\) is \(L=2\), we can caluclate the coefficient \(a_0\) as follows.
\[\begin{array}{rcl}a_0&=&\frac{1}{L}\int_{-L}^{L}f(x)\hspace{0.1cm}{\dd x}\\&=&{{1}\over{2}}\left\{\int_{-2}^0-1\hspace{0.1cm}{\dd x}+\int_0^{2}\left({{x}\over{2}}-1\right)\hspace{0.1cm}{\dd x}\right\}\\&=&{{1}\over{2}}\left\{\left[- x\right]_{-2}^{0}+\left[{{x^2}\over{4}}-x\right]_0^{2}\right\} \\&=&{{1}\over{2}}\left\{\left(0-(2)\right)+\left(-1-0\right)\right\}\\&=&{{1}\over{2}}\left(-3\right)\\&=&-{{3}\over{2}}\end{array}\] The coefficients \(a_m\) are calculated as follows.
\[\begin{array}{rcl}a_m&=&\frac{1}{L}\int_{-L}^{L}f(x)\cdot\cos\left(\frac{m\cdot\pi\cdot x}{L}\right){\dd x}\\&=&{{1}\over{2}}\left\{\int_{-2}^0 -\cos\left(\frac{m\cdot\pi\cdot x}{2}\right) {\dd x}+\int_0^{2}\left({{x}\over{2}}-1\right)\cdot \cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\right\}\\&=&{{1}\over{2}}\int_{-2}^0 -\cos\left(\frac{m\cdot\pi\cdot x}{2}\right) {\dd x}+{{1}\over{2}}\int_0^{2}\left({{x}\over{2}}-1\right)\cdot \cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\\&=&{{1}\over{2}}\left[-\frac{2}{m\cdot\pi}\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_{-2}^0+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\\&=&{{1}\over{2}}\left(-\frac{2}{m\cdot\pi}\cdot\left(\sin(0)-\sin(-m\cdot\pi)\right)\right)+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\\&=&{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x} \end{array}\] The integral above can be evaluated using integration by parts. Setting \(g(x)=x-2\) and \(\frac{{\dd f(x)}}{{\dd x}}=\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)\), we have \(\frac{{\dd g(x)}}{{\dd x}}=1\) and \(f(x)=\frac{2}{m\cdot\pi}\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\). Hence \[\begin{array}{rcl}a_m&=&{{1}\over{4}}\left\{\left[f(x)\cdot g(x)\right]_0^{2}-\int_0^{2}\frac{{\dd g(x)}}{{\dd x}}\cdot f(x){\dd x}\right\}\\&=&{{1}\over{4}}\left\{\left[\frac{2\cdot(x-2)}{m\cdot\pi}\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_0^{2}-\frac{2}{m\cdot\pi}\int_0^{2}\sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\right\}\\&=&{{1}\over{4}}\left\{\left(0-0\right)+\frac{4}{m^2\cdot\pi^2}\left[\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_0^{2} \right\}\\&=&{{1}\over{4}}\left\{\frac{4}{m^2\cdot\pi^2}\left(\cos\left(m\cdot\pi\right)-\cos\left(0\right)\right) \right\}\\&=&{{1}\over{4}}\left\{\frac{4}{m^2\cdot\pi^2}\left((-1)^m-1\right) \right\}\\&=&\frac{(-1)^m-1}{m^2\cdot\pi^2}\end{array}\] Note that we have used the fact that \(\sin(m\cdot\pi)=0\) and \(\cos(m\cdot\pi)=(-1)^m\) for \(m\in\Bbb{Z}\). The coefficients \(b_m\) are calculated as follows.
\[\begin{array}{rcl}b_m&=&\frac{1}{L}\int_{-L}^{L}f(x)\cdot\sin\left(\frac{m\cdot\pi\cdot x}{L}\right){\dd x}\\&=&{{1}\over{2}}\left\{\int_{-2}^0 -\sin\left(\frac{m\cdot\pi\cdot x}{2}\right) {\dd x}+\int_0^{2}\left({{x}\over{2}}-1\right)\cdot \sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\right\}\\&=&{{1}\over{2}}\int_{-2}^0 -\sin\left(\frac{m\cdot\pi\cdot x}{2}\right) {\dd x}+{{1}\over{2}}\int_0^{2}\left({{x}\over{2}}-1\right)\cdot \sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\\&=&{{1}\over{2}}\left[\frac{2}{m\cdot\pi}\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_{-2}^0+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x} \\&=&{{1}\over{2}}\left(\frac{2}{m\cdot\pi}\cdot\left(\cos(0)-\cos(m\cdot\pi)\right)\right)+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x} \\&=&{{1}\over{2}}\left(\frac{2}{m\cdot\pi}\cdot\left(1-(-1)^m\right)\right)+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\\&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\int_0^{2}\left(x-2\right)\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\end{array}\] The integral above can be evaluated using integration by parts. Setting \(g(x)=x-2\) and \(\frac{{\dd f(x)}}{{\dd x}}=\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\), we have \(\frac{{\dd g(x)}}{{\dd x}}=1\) and \(f(x)=-\frac{2}{m\cdot\pi}\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)\). Hence \[\begin{array}{rcl}b_m&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\left\{\left[f(x)\cdot g(x)\right]_0^{2}-\int_0^{2}\frac{{\dd g(x)}}{{\dd x}}\cdot f(x){\dd x}\right\}\\&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\left\{\left[\frac{2\cdot(2-x)}{m\cdot\pi}\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_0^{2}+\frac{2}{m\cdot\pi}\int_0^{2}\cos\left(\frac{m\cdot\pi\cdot x}{2}\right){\dd x}\right\}\\&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\left\{\left(0-\frac{4}{m\cdot\pi}\right)+\frac{4}{m^2\cdot\pi^2}\left[\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right]_0^{2} \right\}\\&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\left\{-\frac{4}{m\cdot\pi}+\frac{4}{m^2\cdot\pi^2}\left(\sin\left(m\cdot\pi\right)-\sin\left(0\right)\right) \right\}\\&=&\frac{1\cdot(1-(-1)^m)}{m\cdot\pi}+{{1}\over{4}}\left\{-\frac{4}{m\cdot\pi} \right\}\\&=&-\frac{(-1)^m}{m\cdot\pi}\end{array}\]Hence the Fourier series for \(f(x)\) is \[\begin{array}{rcl}f(x)&=&\frac{a_0}{2}+\sum_{m=1}^{\infty}\left(a_m\cdot\cos\left(\frac{m\cdot\pi\cdot x}{L}\right)+b_m\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right)\\&=&-{{3}\over{4}}+\sum_{m=1}^{\infty}\left(\frac{(-1)^m-1}{m^2\cdot\pi^2}\cdot\cos\left(\frac{m\cdot\pi\cdot x}{2}\right)-\frac{(-1)^m}{m\cdot\pi}\cdot\sin\left(\frac{m\cdot\pi\cdot x}{2}\right)\right) \end{array}\]

New example