Analysis

Fourier series

Almost any periodic wave, however jagged, can be built by adding up pure sine and cosine tones of increasing frequency. The subtle part is not the construction — it is knowing in what sense the infinite sum comes back to the original wave.

IntuitionBuilding a wave out of pure tones

A musical tone from a tuning fork is a pure sine wave. A tone from a violin, with the same pitch, sounds completely different — because it is really a sum of the fundamental frequency plus quieter harmonics at $2, 3, 4, \dots$ times that frequency, in proportions that give the violin its timbre. Joseph Fourier's radical idea (1807) was that this is not special to musical instruments: almost any periodic signal, including one with sharp corners or jumps like a square wave, can be written as a sum of sines and cosines whose frequencies are whole-number multiples of a single base frequency.

A square wave approximated by just

3

sine harmonics. Each added harmonic bends the wiggly curve a little closer to the sharp corners of the square wave.

UndergraduateFourier coefficients: finding the amplitudes

Definition: Fourier series

Let $f$ be a function with period $2\pi$ (so $f(x+2\pi)=f(x)$ for all $x$ ), integrable on $[-\pi,\pi]$ . Its Fourier series is $f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos nx + b_n\sin nx\right)$ , and the partial sum $S_N f(x) = \frac{a_0}{2} + \sum_{n=1}^{N}\left(a_n\cos nx + b_n\sin nx\right)$ is the best approximation to $f$ built from only the first $N$ harmonics. The symbol $\sim$ is deliberate: whether the series actually converges back to $f(x)$ is a separate question, addressed further down.

a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx\ (n\ge 0), \qquad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx\ (n\ge 1)

These formulas are not pulled out of thin air. The functions $1,\cos x,\sin x,\cos 2x,\sin 2x,\dots$ are orthogonal on $[-\pi,\pi]$ : the integral of the product of two different ones is $0$ , while $\int_{-\pi}^{\pi}\cos^2 nx\,dx = \int_{-\pi}^{\pi}\sin^2 nx\,dx = \pi$ for $n\ge1$ . Multiplying the series for $f$ by $\cos mx$ and integrating term by term makes every term vanish except the one matching $m$ , which isolates $a_m$ — exactly the trick used to find coordinates with respect to an orthonormal basis in a vector space, except the "vectors" here are functions.

f(x) \sim \sum_{n=-\infty}^{\infty} \hat f(n)\, e^{inx}, \qquad \hat f(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\, e^{-inx}\, dx

Example: The Fourier series of a square wave

Find the Fourier series of the $2\pi$ -periodic square wave $f(x) = 1$ for $0 < x < \pi$ and $f(x) = -1$ for $-\pi < x < 0$ .

Solution

Since $f$ is odd, $a_n=0$ for all $n$ . For $b_n$ : $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx = \frac{2}{\pi}\int_0^\pi \sin nx\,dx = \frac{2}{n\pi}\left(1-\cos n\pi\right)$ . This is $0$ when $n$ is even, and $\frac{4}{n\pi}$ when $n$ is odd. So $f(x) \sim \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\left((2k+1)x\right)}{2k+1} = \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)$ .

The same square wave with

25

harmonics. The approximation is much sharper, but right next to the jump it still overshoots the target value by about

9\%

of the jump's height — no matter how many harmonics are added. This persistent overshoot is the Gibbs phenomenon.

UndergraduateParseval's identity: conservation of energy

Theorem: Parseval's identity

If $f$ has Fourier coefficients $a_n, b_n$ , then $\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right)$ .

Why is it true?

This is exactly the Pythagorean theorem in the infinite-dimensional space of periodic functions: since $1,\cos nx,\sin nx$ are orthogonal, the "squared length" $\int |f|^2$ of $f$ splits into the sum of the squared lengths of its components along each direction, with no cross terms surviving the integration.

Proof

Expand $|f(x)|^2$ using the series for $f$ and integrate term by term over $[-\pi,\pi]$ . By orthogonality, $\int \cos nx\cos mx = \int \sin nx\sin mx = 0$ for $n\ne m$ and $\int \cos nx\sin mx = 0$ always, so only the "diagonal" terms $a_n^2\int\cos^2 nx = a_n^2\pi$ , $b_n^2\int\sin^2 nx = b_n^2\pi$ , and $(a_0/2)^2\int 1 = a_0^2\pi/2$ survive; dividing by $\pi$ gives the stated formula.

Example: The Basel problem from Parseval's identity

Apply Parseval's identity to the sawtooth wave $f(x) = x$ on $(-\pi, \pi]$ , extended periodically, to evaluate $\sum_{n=1}^{\infty} \frac{1}{n^2}$ — the [Basel problem](/problems/basel-problem).

Solution

Since $f$ is odd, $a_n=0$ ; a short computation gives $b_n = \frac{2}{\pi}\int_0^\pi x\sin nx\,dx = \frac{2(-1)^{n+1}}{n}$ . Parseval's identity reads $\frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \sum_{n=1}^{\infty} b_n^2$ . The left side is $\frac{1}{\pi}\cdot\frac{2\pi^3}{3} = \frac{2\pi^2}{3}$ , and the right side is $\sum_{n=1}^{\infty} \frac{4}{n^2} = 4\sum_{n=1}^{\infty}\frac{1}{n^2}$ . Equating and solving gives $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ — the value Leonhard Euler first announced in 1735 by a completely different, more delicate argument using the infinite product for $\sin x$ ; this Fourier-analytic derivation came later, once the theory was made rigorous.

AdvancedDoes the series actually converge back to $f$ ?

In the sense of Parseval's identity — convergence in average squared error, called $L^2$ convergence — the answer is always yes, for any $f$ with $\int |f|^2 < \infty$ . Pointwise convergence, at a specific $x$ , is far more delicate. Dirichlet proved in 1829 that $S_N f(x) \to f(x)$ whenever $f$ is piecewise smooth (and at a jump, the series converges to the average of the left and right limits). It was widely believed this should extend to every continuous function, until du Bois-Reymond constructed a continuous function whose Fourier series diverges at a single point (1873). Kolmogorov went further in 1923, building an integrable function whose Fourier series diverges everywhere.

Theorem: Carleson's theorem

If $f \in L^2(-\pi,\pi)$ , then $S_N f(x) \to f(x)$ for almost every $x$ (Hunt extended this in 1968 to every $f \in L^p$ with $p > 1$ ).

Why is it true?

Kolmogorov's counterexample lives in $L^1$ , the largest space where the Fourier coefficients still make sense; Carleson's theorem (1966) shows the trouble genuinely disappears as soon as $f$ has finite energy ( $L^2$ ). The proof does not exhibit convergence directly — it controls the maximal partial sum $\sup_N |S_N f(x)|$ using a delicate time-frequency decomposition of the function into localized wave packets.

Proof

The full argument is considered one of the hardest proofs in twentieth-century analysis and is well beyond this page; Charles Fefferman gave a more transparent (though still technical) version in 1973 using a combinatorial argument on time-frequency "tiles". The topic [Harmonic analysis](/giai-tich-dieu-hoa) sketches the underlying picture.

Historical note

Trigonometric series appeared decades before Fourier, in an 18th-century argument over the vibrating string. Leonhard Euler and Jean le Rond d'Alembert found traveling-wave solutions to the string equation, but disagreed with Daniel Bernoulli, who claimed in 1753 that every possible shape of a plucked string could be written as a sum of sine modes $\sin(nx)$ — a claim Euler thought absurd for shapes with corners. Bernoulli turned out to be right in spirit. Joseph Fourier revived the idea in a 1807 memoir on heat conduction (initially rejected by the Paris Academy for lack of rigor) and a full 1822 treatise, claiming the expansion works for essentially arbitrary functions; making that claim precise occupied analysts for the next century and a half.

Leonhard Euler Joseph Fourier

For a function of period $2\pi$ , the Fourier coefficient $a_0$ is defined as

If $f$ is an odd function (that is, $f(-x)=-f(x)$ ), which Fourier coefficients automatically vanish?

The Gibbs phenomenon says that near a jump discontinuity, the partial sums $S_N f$

Using Parseval's identity on the Fourier series of $f(x)=x$ on $(-\pi,\pi]$ , the sum $\sum_{n=1}^{\infty} \frac{1}{n^2}$ equals

References

Joseph Fourier (1822). Théorie analytique de la chaleur · DOI:10.1017/cbo9780511693229
Elias M. Stein, Rami Shakarchi (2003). Fourier Analysis: An Introduction
T. W. Körner (1988). Fourier Analysis
Lennart Carleson (1966). On convergence and growth of partial sums of Fourier series