In this article, we will be going through the process of solving the coefficients of Fourier Series.
Brief Introduction
In simple language, a periodic function, $f(x)$, is defined as a function in which, a value $P$ exists such that $f(P + x) = f(x)$. This periodic function can be represented as a Fourier series, namely the series of harmonically-related sines and cosines, as follows:
\(\begin{align} f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} + b_n \sin{nx}\right\} \label{fourier} \end{align}\) where $a_0$, $a_n$, and $b_n$ are constant values called Fourier coefficients.
In other references, the Fourier series of $f(x)$ is also defined as follows:
\[\begin{align*} f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{\frac{2\pi nx}{P}} + b_n \sin{\frac{2\pi nx}{P}}\right\} \end{align*}\]which is indeed the same representation as \eqref{fourier}, given the assumption $P = 2\pi$. However, note that we will be using the representation in \eqref{fourier} for our purpose of solving Fourier coefficients.
There are other ways of representing the Fourier series of a function. For example, when we define new variables $d_n$, $\varphi_n$ and $\theta_n$ such that
\[\begin{align*} d_n = \sqrt{a_n^2 + b_n^2}, &{}& \tan{\varphi_n} = \frac{a_n}{b_n}, &{}& \tan{\theta_n} = \frac{b_n}{a_n} \end{align*}\]the following are valid forms of Fourier series.
\[\begin{align*} f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} d_n \sin{nx + \varphi_n} &{}& \text{or} &{}& f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} d_n \cos{nx + \theta_n} \end{align*}\]Note that a periodic function must satisfy certain conditions so that it can be represented as a Fourier series (there are many resources regarding this topic on the Internet).
In the upcoming sections, we would like to solve the Fourier coefficients (namely $a_0$, $a_n$ and $b_n$) in terms of $x$.
Fourier coefficient $a_0$
First, we integrate both sides of \eqref{fourier} to the range of $[-\pi, \pi]$.
\[\begin{align} \int_{-\pi}^{\pi} f(x) \mathop{dx} &= \int_{-\pi}^{\pi} \frac{a_0}{2} \mathop{dx} + \int_{-\pi}^{\pi} \left\{\sum_{n=1}^{\infty} \left\{a_n \cos nx + b_n \sin nx\right\}\right\} \mathop{dx} \notag \\ &= \frac{a_0}{2} x\Big|_{-\pi}^{\pi} + \sum_{n=1}^{\infty} \left\{a_n \frac{\sin nx}{n} \Big|_{-\pi}^{\pi} + b_n \frac{-\cos nx}{n} \Big|_{-\pi}^{\pi} \right\} \label{fourier:a_0} \end{align}\]In fact, we can verify that the summation at the right hand side of \eqref{fourier:a_0} equals to 0 because
\[\begin{align*} \frac{\sin{nx}}{n} \Big\rvert_{-\pi}^{\pi} &= \frac{\sin{n\pi}}{n} - \frac{\sin{n(-\pi)}}{n} \\ &= 0 - 0 \\ &= 0 \end{align*}\]and
\[\begin{align*} \frac{\cos{nx}}{n} \Big\rvert_{-\pi}^{\pi} &= \frac{\cos{n\pi}}{n} - \frac{\cos{n(-\pi)}}{n} \\ &= \frac{\cos{n\pi}}{n} - \frac{\cos{n\pi}}{n} \\ &= 0 \end{align*}\]Therefore, \eqref{fourier:a_0} can be rewritten as follows:
\[\begin{align*} \int_{-\pi}^{\pi} f(x) \mathop{dx} &= \frac{a_0}{2} x\Big|_{-\pi}^{\pi} \\ &= \frac{a_0}{2} (2 \pi) = a_0 \pi \end{align*}\]which can be rearranged to obtain the expression of Fourier coefficients $a_0$ that we are interested in.
\[\begin{equation} a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \mathop{dx} \end{equation}\]Fourier coefficient $a_n$
To solve $a_n$, i.e. to transform \eqref{fourier} into an expression of $a_n$ in terms of $x$, we will first multiply both sides of \eqref{fourier} by $\cos mx$, where $m \in \mathbf{Z^+}$. Then, we integrate both sides of \eqref{fourier} to the range of $[-\pi, \pi]$, as follows:
\[\begin{align} \int_{-\pi}^{\pi} f(x) \cos mx \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \cos mx \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \left\{\sum_{n=1}^{\infty} \left\{a_n \cos nx \cos mx + b_n \sin nx \cos mx\right\}\right\} \mathop{dx} \label{fourier:a_n_1} \end{align}\]The next part is a bit tricky. Take note that we apply trigonometric identities to convert the terms enclosed within the summation as such:
\[\begin{align} \cos nx \cos mx &= \frac{1}{2} \left[\cos{(n+m)x} + \cos{(n-m)x}\right] \label{cosAndCos} \\ \sin nx \cos mx &= \frac{1}{2} \left[\sin{(n+m)x} + \sin{(n-m)x}\right] \label{sinAndCos} \end{align}\]Need some quick refreshers on trigonometric identities? Click me. 😀
Some refreshers on the trigonometric identities which are applied here:
\[\begin{align*} \cos a \cos b &= \frac{1}{2} \left[\cos{a} \cos{b} - \sin{a} \sin{b} + \cos{a} \cos{b} + \sin{a} \sin{b}\right] \\ &= \frac{1}{2} \left[\cos{(a+b)} + \cos (a - b)\right] \\ \sin a \cos b &= \frac{1}{2} \left[\sin{a} \cos{b} + \cos{a} \sin{b} + \sin{a} \cos{b} - \cos{a} \sin{b}\right] \\ &= \frac{1}{2} \left[\sin{(a+b)} + \sin{(a-b)}\right] \end{align*}\]By substituting \eqref{cosAndCos} and \eqref{sinAndCos} into the right hand side of \eqref{fourier:a_n_1}, we have
\[\begin{align} \int_{-\pi}^{\pi} f(x) \cos mx \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \cos mx \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \frac{1}{2} \Biggl\{ \sum_{n=1}^{\infty} \Bigl\{ a_n \left(\cos{(m+n)x} + \cos{(m-n)x} \right) \notag \\ & \hspace{16mm}+ b_n \left(\sin{(m+n)x - \sin{(m-n)x}}\right)\Bigr\} \Biggr\} \mathop{dx} \label{fourier:a_n_2} \end{align}\]First, we consider the case where $m \neq n$ for \eqref{fourier:a_n_1}. We can verify that
\[\begin{align*} \int_{-\pi}^{\pi} \cos{(m+n)x} \mathop{dx} &= \left.\frac{\sin{(m+n)}x}{m+n}\right|_{-\pi}^{\pi} \\ &= \left[\frac{\sin{(m+n)\pi}}{m+n}\right] - \left[\frac{\sin{(m+n)(-\pi)}}{m+n}\right] \\ &= \left[\frac{\sin{(m+n)\pi}}{m+n}\right] - \left[-\frac{\sin{(m+n)\pi}}{m+n}\right] \\ &= 0 + 0 \\ &= 0 \end{align*}\]and
\[\begin{align*} \int_{-\pi}^{\pi} \sin{(m+n)x} \mathop{dx} &= \left.\frac{-\cos{(m+n)}x}{m+n}\right|_{-\pi}^{\pi} \\ &= \left[\frac{-\cos{(m+n)\pi}}{m+n}\right] - \left[\frac{-\cos{(m+n)(-\pi)}}{m+n}\right] \\ &= \left[\frac{-\cos{(m+n)\pi}}{m+n}\right] - \left[\frac{-\cos{(m+n)\pi}}{m+n}\right] \\ &= 0 \end{align*}\]because $m+n \in \mathbf{Z}$.
Therefore, we can also verify that
\[\begin{align*} \int_{-\pi}^{\pi} \cos{(m-n)x} \mathop{dx} = 0 &{}& \text{and} &{}& \int_{-\pi}^{\pi} \sin{(m-n)x} \mathop{dx} = 0 \end{align*}\]given $m-n \in \mathbf{Z}$.
This reduces the summation in \eqref{fourier:a_n_1} to only terms where $m=n$, as follows:
\[\begin{align} \int_{-\pi}^{\pi} f(x) \cos mx \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \cos mx \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \frac{1}{2} \bigl\{ a_m \left(\cos{2mx} + \cos{0} \right) + b_m \left(\sin{2mx} + \sin{0}\right)\bigr\} \mathop{dx} \label{fourier:a_n_3} \end{align}\]From \eqref{fourier:a_n_3}, we can verify that
\[\begin{align*} \int_{-\pi}^{\pi} \cos{(2m)x} \mathop{dx} = 0 &{}& \int_{-\pi}^{\pi} \cos{0} \mathop{dx} &= \int_{-\pi}^{\pi} \mathop{dx} = 2\pi \\ \int_{-\pi}^{\pi} \sin{(2m)x} \mathop{dx} = 0 &{}& \int_{-\pi}^{\pi} \sin{0} \mathop{dx} &= \int_{-\pi}^{\pi} 0 \mathop{dx} = 0 \end{align*}\]Therefore, we can further reduce \eqref{fourier:a_n_3} to the following:
\[\begin{align} \int_{-\pi}^{\pi} f(x) \cos mx \mathop{dx} = \frac{a_m}{2} \int_{-\pi}^{\pi} \cos{0} \mathop{dx} = a_m \pi \label{fourier:a_n_4} \end{align}\]There is still a last step before completing the expression of Fourier equation in terms of Fourier coefficient $a_n$. By rearranging the term in \eqref{fourier:a_n_4}, and then replacing $m$ with $n$, we obtain
\[\begin{align} a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \mathop{dx} \end{align}\]Fourier coefficient $b_n$
We use the similar approach as previous section to transform \eqref{fourier} into the expression of $b_n$ in terms of $x$. The difference here is that, we will first multiply both sides of \eqref{fourier} by $\sin mx$ instead, where $m \in \mathbf{Z^+}$. Then, we integrate both sides of \eqref{fourier} to the range of $[-\pi, \pi]$, as follows:
\[\begin{align} \int_{-\pi}^{\pi} f(x) \sin{mx} \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \sin{mx} \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \left\{\sum_{n=1}^{\infty} \left\{a_n \sin{mx} \cos{nx} + b_n \sin{mx} \sin{nx}\right\}\right\} \mathop{dx} \label{fourier:b_n_1} \end{align}\]The trigonometric identities used here are slightly different from \eqref{fourier:a_n_1}, as follows:
\[\begin{align} \sin{mx} \cos{nx} &= \frac{1}{2} \left[\sin{(m-n)x} + \sin{(m+n)x}\right] \label{cosAndSin} \\ \sin{mx} \sin{nx} &= \frac{1}{2} \left[\cos{(m-n)x} - \cos{(m+n)x}\right] \label{sinAndSin} \end{align}\]Now, substitute \eqref{cosAndSin} and \eqref{sinAndSin} into \eqref{fourier:b_n_1}.
\[\begin{align} \int_{-\pi}^{\pi} f(x) \sin{mx} \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \sin{mx} \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \frac{1}{2} \Biggl\{ \sum_{n=1}^{\infty} \bigl\{ a_n \left(\sin{(m-n)x} + \sin{(m+n)x} \right) \notag \\ & \hspace{16mm}+ b_n \left(\cos{(m-n)x - \cos{(m+n)x}}\right)\Bigr\} \Biggr\} \mathop{dx} \label{fourier:b_n_2} \end{align}\]Considering case where $m \neq n$, we have already verified the following from previous sections.
\[\begin{align*} \int_{-\pi}^{\pi} \sin{(m-n)x} \mathop{dx} = 0 &{}& \int_{-\pi}^{\pi} \sin{(m+n)x} \mathop{dx} = 0 \\ \int_{-\pi}^{\pi} \cos{(m-n)x} \mathop{dx} = 0 &{}& \int_{-\pi}^{\pi} \cos{(m+n)x} \mathop{dx} = 0 \end{align*}\]Therefore, the summation in \eqref{fourier:b_n_2} can be reduced to only terms where $m = n$ as follows:
\[\begin{align} \int_{-\pi}^{\pi} f(x) \sin{mx} \mathop{dx} =& \int_{-\pi}^{\pi} \frac{a_0}{2} \sin{mx} \mathop{dx} \notag \\ &+ \int_{-\pi}^{\pi} \frac{1}{2} \Bigl\{ a_m \left(\sin{0} + \sin{2mx} \right) + b_m \left(\cos{0} - \cos{2mx}\right)\bigr\} \mathop{dx} \label{fourier:b_n_3} \end{align}\]Again from previous section, recall that
\[\begin{align*} \int_{-\pi}^{\pi} \sin{0} \mathop{dx} &= \int_{-\pi}^{\pi} 0 \mathop{dx} = 0 &{}& \int_{-\pi}^{\pi} \sin{(2m)x} \mathop{dx} = 0 \\ \int_{-\pi}^{\pi} \cos{0} \mathop{dx} &= \int_{-\pi}^{\pi} \mathop{dx} = 2\pi &{}& \int_{-\pi}^{\pi} \cos{(2m)x} \mathop{dx} = 0 \end{align*}\]Therefore, \eqref{fourier:b_n_3} can be further reduced to
\[\begin{align} \int_{-\pi}^{\pi} f(x) \sin{mx} \mathop{dx} = \frac{b_m}{2} \int_{-\pi}^{\pi} \cos{0} \mathop{dx} = b_m \pi \label{fourier:b_n_4} \end{align}\]Finally, we rearrange the terms in \eqref{fourier:b_n_4}, then replace $m$ with $n$ to obtain the expression of Fourier coefficient, $b_n$ in terms of $x$.
\[\begin{align} b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin{nx} \mathop{dx} \end{align}\]Bonus: Fourier Series for Odd and Even Functions
A function $f$ of variable $x$, denoted by $f(x)$, is defined as an even function if the following condition is satisfied.
\[\begin{equation*} f(-x) = f(x) \end{equation*}\]In other words, when we plot the graph of $f(x)$ against $x$ on the Cartesian plane (or $xy$-plane), the value of $f(x)$ is symmetric about the $y$-axis. Some examples of even function are $f(x) = \lvert x \rvert$ and $f(x) = x^2$.
A function $f$ of variable $x$, denoted by $f(x)$, is defined as an odd function if the following condition is satisfied.
\[\begin{equation*} f(-x) = -f(x) \end{equation*}\]Unlike an even function, when we plot the graph of $f(x)$ against $x$ on the same plane, the value of $f(x)$ is symmetric about the line $y = -x$. Some examples are $f(x) = x^n$ where $n$ is odd.
Recall that some periodic functions can be represented in a Fourier series \eqref{fourier} as follows:
\[\begin{equation*} f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} + b_n \sin{nx} \right\} \end{equation*}\]In fact, we have implicitly applied the concept of odd and even functions when solving the coefficients of the Fourier series of a periodic function, $f(x)$ in previous sections. To be more concise, the sine function, $\sin{x}$ is an odd function and the cosine function $\cos{x}$ is an even function, as such:
\[\begin{align*} \sin{(-x)} = -\sin{x} &{}& \cos{(-x)} = \cos{x} \end{align*}\]In this section, the same concept is applied to find the Fourier series representation of odd or even periodic functions. The first step is to identify the Fourier series representation of $f(-x)$.
\[\begin{align} f(-x) &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{(-nx)} + b_n \sin{(-nx)} \right\} \notag \\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} - b_n \sin{nx} \right\} \label{fourier:neg} \end{align}\]If $f(x)$ is odd, then we can find the Fourier series representation of $f(x)$ by manipulating \eqref{fourier} and \eqref{fourier:neg} as follows:
\[\begin{align} f(x) &= \frac{1}{2} \left\{f(x) - f(-x)\right\} \notag \\ &= \frac{1}{2} \Biggl\{ \Bigl\{ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} + b_n \sin{nx} \right\} \Bigr\} \notag \\ & \hspace{11mm} - \Bigl\{ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} - b_n \sin{nx} \right\} \Bigr\} \Biggr\} \notag \\ &= \sum_{n=1}^{\infty} \left\{b_n \sin{nx}\right\} \end{align}\]Similarly, we can also find the Fourier series representation of an even function, $f(x)$ by manipulating \eqref{fourier} and \eqref{fourier:neg} as follows:
\[\begin{align} f(x) &= \frac{1}{2} \left\{f(x) + f(-x)\right\} \notag \\ &= \frac{1}{2} \Biggl\{ \Bigl\{ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} + b_n \sin{nx} \right\} \Bigr\} \notag \\ & \hspace{11mm} + \Bigl\{ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx} - b_n \sin{nx} \right\} \Bigr\} \Biggr\} \notag \\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \left\{a_n \cos{nx}\right\} \end{align}\]References
-
“Definition of Fourier Series and Typical Examples,” Math24, 26-Apr-2020. [Online]. Available: https://www.math24.net/fourier-series-definition-typical-examples/. [Accessed: 28-Aug-2020].
-
“Fourier Series,” Brilliant Math & Science Wiki. [Online]. Available: https://brilliant.org/wiki/fourier-series/. [Accessed: 28-Aug-2020].
-
Libretexts, “4.6: Fourier series for even and odd functions,” Mathematics LibreTexts, 14-May-2020. [Online]. Available: https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Partial_Differential_Equations_(Walet)/04%3A_Fourier_Series/4.06%3A_Fourier_series_for_even_and_odd_functions. [Accessed: 28-Aug-2020].