Weierstrass Function¶

Definition¶

The Weierstrass function is defined as

f(x) = \sum_{n=1}^\infty a^n \cos(b^n x)

where $0<a<1$ and $ab>1\text{.}$

Each summand uses a Trigonometric Functions cosine with increasing frequency. The Infinite Series of Functions converges uniformly (by the Weierstrass $M$ -test) because $|a^n\cos(b^nx)|\le a^n$ and the geometric series converges for $0<a<1$ $1$ .

Interpretation¶

The Weierstrass function is one of the best-known examples of a function that is continuous everywhere but differentiable nowhere. It shows that continuity, however it is pictured, does not force a graph to have a tangent at even a single point.

A partial sum of the Weierstrass function, a jagged curve, with a small window marked; the zoomed view of the window shows the same roughness.

The figure plots the finite partial sum $\displaystyle \sum_{n=1}^{8}a^n\cos(b^nx)$ with $a=1/2$ and $b=3$ $3$ . The right panel magnifies the red window and illustrates the repeated roughness; the claim about the infinite limit function is stated above.

The two conditions on $a$ and $b$ play opposite roles. The condition $0<a<1$ makes the amplitudes $a^n$ summable, so the Weierstrass $M$ -test gives uniform convergence, and the uniform limit of the continuous partial sums is continuous. The condition $ab>1$ concerns slopes: the $n$ -th term has derivative $-(ab)^n\sin(b^nx)$ , so each new term is smaller in height but steeper, and the steepness keeps increasing at ever smaller scales. The terms of the differentiated series reach size $(ab)^n\to\infty$ $\infty$ , so that series does not converge uniformly and term-by-term differentiation is not available; this explains why differentiability can fail, although the proof that it fails at every point takes more work. If instead $0<ab<1$ $1$ , the derivative series converges uniformly by the $M$ -test and $f$ is differentiable.

In Weierstrass's original paper the function is $\displaystyle \sum_{n=1}^\infty a^n\cos(b^n\pi x)$ with $0<a<1$ $1$ , $b$ a positive odd integer and $ab>1+\tfrac{3\pi}{2}$ $\frac{3 π}{2}$ . Riemann's series $\displaystyle \sum_{n=1}^\infty\sin(n^2x)/n^2$ is sometimes mentioned in the same context; it is continuous by the same $M$ -test argument, but it is differentiable at some points, so it is not a nowhere differentiable function.

A common slip is to read the plotted curve as the function itself. Any plot shows a finite partial sum, which is differentiable everywhere; the roughness at every scale belongs only to the infinite sum.

Key Results¶

The Weierstrass function is continuous everywhere as long as $0<a<1\text{.}$
The Weierstrass function is nowhere differentiable if $ab>1\text{.}$

Examples¶

$\displaystyle f(x) = \sum_{n=1}^\infty \frac{1}{2^n} \cos(3^n x)$ is a particular Weierstrass function.

Prerequisites¶

Weierstrass $M$ -test

Sources¶

Math III: Analysis · Lecture 2

Symbols: infinity, pi, sum

Last updated: 27 October 2026 11:52