Convergence of Fourier Series, calculus lesson

A square wave has jumps, yet its Fourier series is built from perfectly smooth sines. Something has to give at the jumps. The series makes a fair compromise there and picks the midpoint of the jump, but near the jump the partial sums overshoot by a fixed amount no matter how many terms you take. Knowing when the series converges, and how, tells you how far to trust a truncated Fourier series.

Key terms

Piecewise smooth periodic functions

Definition A function $f$ with period $2L$ is piecewise $C^1$ (piecewise smooth) if there is a partition $-L = x_0 < x_1 < \cdots < x_k = L$ such that on each open piece $(x_{j-1}, x_j)$ the function $f$ is continuously differentiable, and $f$ and $f'$ restricted to that piece extend to continuous functions on the closed interval $[x_{j-1}, x_j]$ . The values of $f$ at the finitely many partition points do not matter.

In plain words Cut one period into finitely many pieces. On each piece the graph is a smooth curve that stays smooth right up to both ends, with no infinite slopes and no wild oscillation.

Example The square wave and the sawtooth $f(x) = x$ on $(-\pi, \pi)$ , repeated with period $2\pi$ , are piecewise $C^1$ . The function $\sqrt{\lvert x \rvert}$ on $[-\pi, \pi]$ is not, because its derivative $\frac{1}{2\sqrt{x}}$ becomes unbounded as $x \to 0^+$ .

One-sided limits of the function and its derivative

Definition At each point $x$ write

f(x^+) = \lim_{t \to x^+} f(t), \qquad f(x^-) = \lim_{t \to x^-} f(t),

and similarly $f'(x^+)$ and $f'(x^-)$ for the one-sided limits of $f'$ . For a piecewise $C^1$ function all four limits exist and are finite at every $x$ . The function is continuous at $x$ exactly when $f(x^+) = f(x^-) = f(x)$ , and at a jump the jump size is $f(x^+) - f(x^-)$ .

In plain words Approach the point from the right and from the left, and record where the function and its slope are heading. For a piecewise smooth function both sides always give finite answers.

Example For the sawtooth at $x = \pi$ , the left limit is $f(\pi^-) = \pi$ . To the right of $\pi$ the graph has restarted, $f(\pi + t) = -\pi + t$ for small $t > 0$ , so $f(\pi^+) = -\pi$ . The jump size is $-2\pi$ . The derivative is $1$ on both sides, so $f'(\pi^+) = f'(\pi^-) = 1$ .

Convergence to the midpoint

Definition Let $f$ be $2L$ -periodic and piecewise $C^1$ . Then for every real $x$ its Fourier series converges, and

\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) = \frac{f(x^+) + f(x^-)}{2}.

In particular the series converges to $f(x)$ at every point where $f$ is continuous.

In plain words Where the graph is unbroken the series gives the right value. At a jump it gives the average of the two sides, whatever value $f$ happens to have there.

Example The sawtooth has Fourier series

2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx).

At $x = \pi$ every term is $0$ , and indeed the midpoint is $\frac{\pi + (-\pi)}{2} = 0$ . At $x = \frac{\pi}{2}$ , where $f$ is continuous, the series gives

2\left(1 - \frac{1}{3} + \frac{1}{5} - \cdots\right) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} = f\left(\frac{\pi}{2}\right).

Periodic extension and the endpoints

Definition A function given only on one interval of length $2L$ , say on $(-L, L]$ , is turned into a $2L$ -periodic function by its periodic extension, $F(x + 2kL) = f(x)$ for $x \in (-L, L]$ and every integer $k$ . The Fourier series of $f$ is the Fourier series of $F$ . At $x = \pm L$ it converges, for piecewise $C^1$ data, to

\frac{f(-L^+) + f(L^-)}{2}.

If $f(-L^+) \neq f(L^-)$ , the extension has a jump at $\pm L$ even when $f$ is smooth on the open interval.

In plain words The series always treats your function as repeating. If the right end does not meet the left end, the repetition creates a jump there, and the series picks its midpoint.

Example For $f(x) = x^2$ on $[-\pi, \pi]$ the ends match, since $f(-\pi) = f(\pi) = \pi^2$ , so the extension is continuous and the series equals $\pi^2$ at $x = \pi$ . Put $x = \pi$ into the Fourier series of $x^2$ and use $\cos(n\pi) = (-1)^n$ . This gives

\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{1}{n^2}, \qquad \text{so} \qquad \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.

Gibbs phenomenon

Definition Let $f$ be piecewise $C^1$ with a jump of size $J = f(x_0^+) - f(x_0^-) \neq 0$ at $x_0$ . Then the partial sums $S_N$ overshoot the jump near $x_0$ , and the overshoot does not shrink as $N \to \infty$ . The largest value of $S_N$ just to the right of an upward jump tends to

f(x_0^+) + J\left(\frac{1}{\pi}\int_0^{\pi} \frac{\sin t}{t} \, dt - \frac{1}{2}\right) \approx f(x_0^+) + 0.0895\, J,

and there is a matching undershoot on the other side. The overshoot moves closer to $x_0$ as $N$ grows, but its height stays near $9$ percent of the jump.

In plain words Next to a jump the partial sums always poke out beyond the function by about $9$ percent of the jump. More terms make the spike narrower, not shorter.

Example The square wave jumps from $-1$ to $1$ at $0$ , so $J = 2$ . The peak of $S_N$ just right of $0$ sits at $x = \frac{\pi}{N + 1}$ for odd $N$ .

$N$	Peak of $S_N$
$9$	$1.1823$
$49$	$1.1791$
$199$	$1.1790$
limit	$1.1790$

The overshoot stays near $0.179 = 0.0895 \cdot 2$ .

Pointwise versus uniform behavior

Definition Let $f$ be $2L$ -periodic and piecewise $C^1$ .

If $f$ is continuous on all of $\mathbb{R}$ , so that no jumps occur even at $\pm L$ after periodic extension, then its Fourier series converges to $f$ absolutely and uniformly on $\mathbb{R}$ .
If $f$ has a jump at $x_0$ , the convergence is pointwise but not uniform on any interval containing $x_0$ . The partial sums are continuous, and a uniform limit of continuous functions would be continuous.

In plain words A continuous, piecewise smooth periodic function is matched everywhere at once. A function with a jump is matched only point by point, and the Gibbs overshoot keeps the worst error from shrinking.

Example For $\lvert x \rvert$ on $[-\pi, \pi]$ the extension is continuous, and the coefficients satisfy $\lvert a_n \rvert \le \frac{4}{\pi n^2}$ . By the Weierstrass M-test the series converges uniformly. For the square wave the worst error $\sup_x \lvert S_N(x) - f(x) \rvert$ stays above $0.17$ for every large $N$ , because of the overshoot, so the convergence is not uniform.

Check yourself

To what value does the Fourier series of the square wave converge at $x = \pi$ ?

Show the answer

There the wave jumps from $1$ to $-1$ , so the series converges to the midpoint $\frac{1 + (-1)}{2} = 0$ . Every term $\frac{4}{n\pi}\sin(n\pi)$ is indeed $0$ .

Let $f(x) = e^x$ on $(-\pi, \pi]$ , extended with period $2\pi$ . What is the sum of its Fourier series at $x = \pi$ ?

Show the answer

The extension jumps at $\pi$ from $e^{\pi}$ to $e^{-\pi}$ , so the sum is the midpoint

\frac{e^{\pi} + e^{-\pi}}{2} = \cosh \pi.

Evaluate the Fourier series of $x^2$ at $x = 0$ to find $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ .

Show the answer

The extension is continuous at $0$ , so the series equals $0^2 = 0$ there and

0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}.

Hence

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}.

Convergence of Fourier Series

Key terms

Piecewise smooth periodic functions

One-sided limits of the function and its derivative

Convergence to the midpoint

Periodic extension and the endpoints

Gibbs phenomenon

Pointwise versus uniform behavior

Check yourself

How this lesson connects

Builds on