Fourier Series Reference Cheat Sheet

A printable reference covering Fourier coefficients, sine and cosine series, convergence, Parseval's identity, and common waveform expansions for college.

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Fourier series represent periodic functions as sums of sines and cosines. This cheat sheet helps college students quickly identify the correct coefficient formulas, symmetry shortcuts, and convergence rules. It is useful in calculus, differential equations, signal processing, physics, and engineering courses. The goal is to make setup and interpretation faster without losing mathematical precision. The core idea is that a function with period $2L$ can be expanded using orthogonal basis functions $\cos\left(\frac{n\pi x}{L}\right)$ and $\sin\left(\frac{n\pi x}{L}\right)$ . Coefficients are found by integrating the function against these basis functions over one full period. Even functions produce cosine series, odd functions produce sine series, and piecewise smooth functions converge to midpoint values at jumps. Energy relationships such as Parseval's identity connect the function's squared size to the sum of squared Fourier coefficients.

Key Facts

For a function with period $2L$ , the Fourier series is $f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n\cos\left(\frac{n\pi x}{L}\right) + b_n\sin\left(\frac{n\pi x}{L}\right)\right]$ .
The constant coefficient is $a_0 = \frac{1}{L}\int_{-L}^{L} f(x)\,dx$ .
The cosine coefficients are $a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx$ for $n \ge 1$ .
The sine coefficients are $b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx$ for $n \ge 1$ .
If $f$ is even, then $b_n = 0$ and $a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx$ .
If $f$ is odd, then $a_0 = 0$ , $a_n = 0$ , and $b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx$ .
At a jump discontinuity, the Fourier series converges to $\frac{f(x^-) + f(x^+)}{2}$ rather than to either one-sided value.
Parseval's identity for period $2L$ is $\frac{1}{L}\int_{-L}^{L} |f(x)|^2\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right)$ .

Vocabulary

Fourier series: A representation of a periodic function as an infinite sum of sine and cosine terms.
Period: The positive length $T$ such that $f(x+T)=f(x)$ for all relevant values of $x$ .
Fourier coefficient: A number such as $a_n$ or $b_n$ that measures how much of a specific cosine or sine mode appears in the function.
Orthogonality: A property where integrals of different basis functions over a full period equal $0$ , allowing coefficients to be isolated.
Half-range expansion: A sine-only or cosine-only Fourier series built from a function originally defined on an interval such as $0 < x < L$ .
Gibbs phenomenon: The persistent overshoot near a jump discontinuity that remains even as more Fourier terms are added.

Common Mistakes to Avoid

Using the wrong period in the basis functions, because $\cos(nx)$ and $\sin(nx)$ apply directly only when the period is $2\pi$ .
Forgetting the factor $\frac{1}{L}$ or $\frac{2}{L}$ in coefficient formulas, which changes every coefficient by a constant scale factor.
Treating a discontinuity value as the Fourier series limit, because the series converges to $\frac{f(x^-) + f(x^+)}{2}$ at a jump.
Ignoring symmetry, which leads to unnecessary integration and can produce incorrect nonzero coefficients for terms that should vanish.
Confusing sine and cosine half-range extensions, because a sine series corresponds to an odd extension while a cosine series corresponds to an even extension.

Practice Questions

1 Find the Fourier series coefficients for $f(x)=x$ on $-\pi < x < \pi$ with period $2\pi$ .
2 Compute the cosine series for $f(x)=x^2$ on $-\pi < x < \pi$ with period $2\pi$ .
3 For the square wave $f(x)=1$ on $0<x<\pi$ and $f(x)=-1$ on $-\pi<x<0$ , determine which coefficients $a_0$ , $a_n$ , and $b_n$ are zero.
4 Explain why a Fourier series may fail to equal $f(x)$ exactly at a jump discontinuity, even when the function is otherwise well behaved.

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