How to Find the Fourier Series of a Function

Last Updated: January 6, 2023

In Fourier analysis, a Fourier series is a method of representing a function in terms of trigonometric functions. Fourier series are extremely prominent in signal analysis and in the study of partial differential equations, where they appear in solutions to Laplace's equation and the wave equation.

Preliminaries

Let $f(x)$ $f(x)$ be a piecewise continuous function defined on $[-L,L].$ $[-L,L].$ Then the function may be written in terms of its Fourier series. We note that the sums begin with $n=0,$ $n=0,$ but because $\cos 0x=1$ $\cos 0x=1$ and $\sin 0x=0,$ $\sin 0x=0,$ we may write out the constant term separately and begin both sums with $n=1.$ $n=1.$
- $f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left(a_{n}\cos {\frac {n\pi x}{L}}+b_{n}\sin {\frac {n\pi x}{L}}\right)$
The coefficients $a_{n}$ $a_{{n}}$ and $b_{n}$ $b_{{n}}$ are known as Fourier coefficients. To decompose a function into its Fourier series, we must find these coefficients.
- To recognize what they are, we write out the function $f(x)=|f\rangle$ $f(x)=|f\rangle$ in terms of a basis $g_{n}.$ $g_{{n}}.$ In order for this basis to be useful, it must be orthonormal so that $\langle g_{n}|g_{m}\rangle =\delta _{nm},$ $\langle g_{{n}}|g_{{m}}\rangle =\delta _{{nm}},$ the Kronecker delta that equals $1$ $1$ if $n=m$ $n=m$ and $0$ $0$ otherwise. The expression below simply means that we are projecting $f$ $f$ onto $g_{n}.$ $g_{{n}}.$
  - $|f\rangle =\sum _{n}|g_{n}\rangle \langle g_{n}|f\rangle$
- For functions defined on the interval $[-L,L],$ $[-L,L],$ we define the following inner product. Notice that this inner product is normalized. The ${}^{*}$ ${}^{{*}}$ symbol denotes the complex conjugate.
  - $\langle g_{n}|f\rangle ={\frac {1}{L}}\int _{-L}^{L}g_{n}(x)^{*}f(x)\mathrm {d} x$
- The functions $\cos {\frac {n\pi x}{L}}$ $\cos {\frac {n\pi x}{L}}$ and $\sin {\frac {n\pi x}{L}}$ $\sin {\frac {n\pi x}{L}}$ comprise the Fourier basis. With this in mind, we may write the Fourier coefficients below. When one substitutes $f(x)$ $f(x)$ with an element of the Fourier basis, the coefficient goes to unity. Hence the basis elements under this inner product form an orthonormal set.
  - $a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)\cos {\frac {n\pi x}{L}}\mathrm {d} x$
  - $b_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)\sin {\frac {n\pi x}{L}}\mathrm {d} x$
- What is the interpretation of the constant term $a_{0},$ $a_{{0}},$ and why do we need an extra ${\frac {1}{2}}$ ${\frac {1}{2}}$ in the expression? This expression is in fact the average value of $f(x)$ $f(x)$ over the interval. (If the function is periodic, then it is the average value of the function over the entire domain.) The extra ${\frac {1}{2}}$ ${\frac {1}{2}}$ is there because of the boundaries and compensates for the fact that we are integrating over an interval with length $2L.$ $2L.$
  - ${\frac {a_{0}}{2}}={\frac {1}{2L}}\int _{-L}^{L}f(x)\mathrm {d} x$

Steps

1
Decompose the following function in terms of its Fourier series. Generally speaking, we may find the Fourier series of any (piecewise continuous - see the tips) function on a finite interval. If the function is periodic, then the behavior of the function in that interval allows us to find the Fourier series of the function on the entire domain.
- $f(x)=x^{2}-2x+1:\ [-1,1]$
2
Identify the even and odd parts of the function. Every function may be decomposed into a linear combination of even and odd functions. The Fourier basis is convenient for us in that this series already separates these components. Therefore, by careful observation of which parts of the function are even and which are odd, we can do the integrals separately knowing which terms vanish and which do not.
- For our function, $x^{2}+1$ is even and $-2x$ is odd. This means that $b_{n}=0$ for $x^{2}+1$ and $a_{n}=0$ for $-2x.$
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3
Evaluate the constant term. The constant term $a_{0}$ $a_{{0}}$ is actually the $\cos {\frac {0\pi x}{L}}=1$ $\cos {\frac {0\pi x}{L}}=1$ term of the cosines. Note that $-2x$ $-2x$ does not contribute to the integral because any constant function is even.
- $a_{0}=\int _{-1}^{1}(x^{2}+1)\mathrm {d} x={\frac {8}{3}}$
4
Evaluate the Fourier coefficients. Here, we may evaluate by way of integration by parts. It is useful to recognize that $\cos n\pi =(-1)^{n}$ $\cos n\pi =(-1)^{{n}}$ and $\sin n\pi =0.$ $\sin n\pi =0.$ It is also worth noting that the integral of a trigonometric function over one period vanishes.
- $a_{n}=\int _{-1}^{1}(x^{2}+1)\cos n\pi x\mathrm {d} x={\frac {4(-1)^{n}}{n^{2}\pi ^{2}}}$
- $b_{n}=\int _{-1}^{1}-2x\sin n\pi x\mathrm {d} x={\frac {4(-1)^{n}}{n\pi }}$
5
Write out the function in terms of its Fourier series. This series converges on the interval $(-1,1).$ $(-1,1).$ Because the function is not periodic, the series does not hold on the whole interval, but rather in the neighborhood of any interior point (point-wise convergence as opposed to uniform convergence).
- $x^{2}-2x+1={\frac {4}{3}}+\sum _{n=1}^{\infty }\left[{\frac {4(-1)^{n}}{n^{2}\pi ^{2}}}\cos n\pi x+{\frac {4(-1)^{n}}{n\pi }}\sin n\pi x\right]$
- The image shows the Fourier series up to $n=3,$ $n=15,$ and $n=100.$ We can clearly see the convergence here, as well as the overshoots near the boundaries that do not seem to vanish at higher $n.$ This is the Gibbs phenomenon, which is the result of the failure of the series to converge uniformly on the prescribed interval.