The Weierstrass Function

Each term $u_n(x) = a^n \cos(b^n \pi x)$ is continuous.

Since $|\cos| \le 1$, $|u_n(x)| \le a^n = \left(\frac12\right)^n =: M_n$ for every $x$, and $\sum_{n=0}^{\infty} M_n = \sum_{n=0}^{\infty} \left(\frac12\right)^n = 2 < \infty$.

By the Weierstrass M-test (the m-test scene in the series chapter), $\sum u_n$ converges uniformly on $\mathbb{R}$.

A uniform limit of continuous functions is continuous: fix $x_0$; then $|W(x) - W(x_0)| \le |W(x) - W_N(x)| + |W_N(x) - W_N(x_0)| + |W_N(x_0) - W(x_0)|$.

The first and last terms are small by the tail bound $(1/2)^N$, the middle one by the continuity of the finite sum $W_N$.

Hence $W$ is continuous at $x_0$, and $x_0$ was arbitrary, so $W$ is continuous on all of $\mathbb{R}$.

Check the numbers: at $N = 6$, $|W(x) - W_6(x)| \le (1/2)^6 = 0.015625$ for every $x$ — the curve on screen sits within 0.016 of the true $W$. Raising $N$ drives the tail bound $(1/2)^N$ to 0, so the on-screen $W_N$ approaches $W$ uniformly.

Theorem (Weierstrass–Hardy): if $b$ is an odd integer and $ab > 1 + \frac{3\pi}{2}$, then $W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ is differentiable at no point. We use $a = \frac12$, $b = 13$, so $ab = 6.5 > 1 + \frac{3\pi}{2} \approx 5.712$ — the condition holds.

The mechanism — zooming in at scale $b^{-n}$: the $n$-th harmonic $\cos(b^n \pi x)$ completes a full period over every interval of length about $b^{-n}$, so near any point $x_0$ there is a spike point $x_n$ with $|x_n - x_0| \lesssim b^{-n}$ where the $n$-th cosine swings nearly trough-to-peak (about $2$). Across this spike, the $n$-th term's difference quotient is $a^n \cdot (\approx 2) / (\approx b^{-n}) \approx 2(ab)^n = 2 \cdot 6.5^n \to \infty$ (about $6.37 \times 10^6$ at $n = 8$).

The remaining terms cannot beat the spike: the terms with $k < n$ are nearly flat at this scale (differentiating brings out $a^k b^k \pi$, a geometric sum with ratio $ab$ and a strictly smaller constant than the spike's), and the terms with $k > n$ are uniformly tiny by the tail bound $2a^{n+1}/(1-a)$. The classical proof chooses the spike points delicately, tracks all three constants, and shows that $ab > 1 + \frac{3\pi}{2}$ is exactly the inequality saying "the spike's constant wins."

Honest label: this is an idea-level sketch.

The full proof is the classical Weierstrass–Hardy construction — building the spike points and tracking all three constants to the end — which we do not reproduce line by line here.

The takeaway it conveys is this: the difference quotient blows up to infinity along the spike sequence $x_n \to x_0$, so no finite number can serve as $W'(x_0)$; $x_0$ was arbitrary, so $W$ is differentiable nowhere.

Each $W_N$ is a finite sum, so termwise differentiation is legitimate: $W_N'(x) = -\sum_{n=0}^{N} a^n b^n \pi \sin(b^n \pi x)$ exists at every $x$.

For instance $W_2'(x) = -\pi\sin(\pi x) - \frac{13\pi}{2}\sin(13\pi x) - \frac{169\pi}{4}\sin(169\pi x)$ (since $a^1 b^1 \pi = \frac{13\pi}{2}$ and $a^2 b^2 \pi = \frac{169\pi}{4}$).

Every curve on screen is a smooth function.

But by problem 1's tail bound $|W - W_N| \le (1/2)^N \to 0$, we have $W_N \to W$ uniformly on $\mathbb{R}$. And by problem 2, the limit $W$ is differentiable at no point.

Hence a uniform limit of differentiable functions need not be differentiable — continuity is preserved under uniform limits (problem 1's argument), but differentiability is not. This is why analysis demands stronger convergence (e.g. uniform convergence of the derivatives) before differentiating termwise.