Müntz–Szász Theorem

First compute $\sum 1/n^2 = \pi^2/6$ (complete proof). Take $f(x) = x$ on $(-\pi, \pi)$. Since $f$ is continuous, $f \in L^2(-\pi, \pi)$ (indeed $\int_{-\pi}^{\pi} x^2\,dx < \infty$), so Parseval's identity applies to $f$.

Since $f$ is odd, every cosine coefficient vanishes: $a_0 = 0$ and $a_k = 0$ for all $k$, and only the sine coefficients $b_k = \frac{1}{\pi}\int_{-\pi}^{\pi} x \sin(kx)\,dx = \frac{2(-1)^{k+1}}{k}$ (integration by parts, next step) survive. Parseval in the form $\frac{1}{\pi}\int_{-\pi}^{\pi} f^2\,dx = \frac{a_0^2}{2} + \sum_{k=1}^{\infty} (a_k^2 + b_k^2)$ therefore reads $\frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \sum_{k=1}^{\infty} b_k^2$. The left side is $\frac{2\pi^2}{3}$, and $b_k^2 = \frac{4}{k^2}$, so $4\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{2\pi^2}{3}$, i.e. $\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.6449 < \infty$.

Compute the coefficient $b_k$ by explicit integration by parts. Since $x\sin(kx)$ is even, $b_k = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(kx)\,dx = \frac{2}{\pi}\int_{0}^{\pi} x\sin(kx)\,dx$. With $u = x$, $dv = \sin(kx)\,dx$ we get $du = dx$, $v = -\frac{\cos(kx)}{k}$, so the antiderivative is $\int x\sin(kx)\,dx = -\frac{x\cos(kx)}{k} + \int \frac{\cos(kx)}{k}\,dx = -\frac{x\cos(kx)}{k} + \frac{\sin(kx)}{k^2}$.

Evaluating from $0$ to $\pi$ gives $\left[-\frac{\pi\cos(k\pi)}{k} + \frac{\sin(k\pi)}{k^2}\right] - 0 = -\frac{\pi(-1)^k}{k}$ (since $\sin(k\pi) = 0$). Hence $b_k = \frac{2}{\pi}\cdot\left(-\frac{\pi(-1)^k}{k}\right) = -\frac{2(-1)^k}{k} = \frac{2(-1)^{k+1}}{k}$.

Now quote the necessity direction of the Müntz–Szász theorem: if $\sum \frac{1}{\lambda_n} < \infty$ then $\operatorname{span}\{1, x^{\lambda_1}, x^{\lambda_2}, \ldots\}$ is not dense in $C[0,1]$. [Sketch marker: the full proof of this direction needs complex analysis — convergence of $\sum 1/\lambda_n$ makes a Blaschke-type infinite product converge, producing a nonzero analytic function orthogonal (in a weighted $L^2$ inner product) to every $x^{\lambda_n}$; Hahn–Banach then separates a continuous function at fixed positive distance from the span. We use only the conclusion here.]

For $\lambda_n = n^2$ we have $\sum \frac{1}{\lambda_n} = \frac{\pi^2}{6} < \infty$, so $\operatorname{span}\{1, x, x^4, x^9, \ldots\}$ is not dense in $C[0,1]$ — which is why the sparse basis in the scene stays stuck at error $\approx 0.16$.

Let $p_n$ be even polynomials with $p_n \to f$ uniformly on $[-1, 1]$. Uniform convergence implies pointwise convergence, so $f(x) = \lim_{n \to \infty} p_n(x)$ for each $x \in [-1, 1]$. Each $p_n$ is even, so $p_n(x) = p_n(-x)$, and hence $f(x) = \lim_{n \to \infty} p_n(x) = \lim_{n \to \infty} p_n(-x) = f(-x)$.

So $f$ is even — a uniform limit of even polynomials is always even. (End of proof.)

$g(x) = x$ is continuous on $[-1, 1]$ but not even. If $\operatorname{span}\{1, x^2, x^4, \ldots\}$ were dense in $C[-1, 1]$, some sequence of even polynomials would converge uniformly to $g$, and by what we just proved the limit $g$ would have to be even — a contradiction. Hence $\operatorname{span}\{1, x^2, x^4, \ldots\}$ is not dense in $C[-1, 1]$.

(Note: this is a $C[-1,1]$ statement, distinct from the scene's sparse basis $\{1, x, x^4, \ldots\}$ on $C[0,1]$ — the non-density of the latter is the job of the Müntz–Szász necessity direction in the first problem.)

The harmonic series diverges: for $H_n = \sum_{k=1}^{n} \frac{1}{k}$, each term satisfies $\frac{1}{k} \ge \int_{k}^{k+1} \frac{dx}{x}$, so $H_n \ge \int_{1}^{n+1} \frac{dx}{x} = \ln(n+1) \to \infty$. Hence for $\lambda_n = n$ we have $\sum \frac{1}{\lambda_n} = \infty$.

(Numerical witness: $H_{1000} \approx 7.4855 > \ln(1001) \approx 6.9088$.)

For $\lambda_n = n$, $\operatorname{span}\{1, x, x^2, \ldots\}$ is exactly the set of all (real-coefficient) polynomials — $x^k$ is itself the $k$-th basis element, so every polynomial of degree at most $d$ is a linear combination of the first $d+1$ basis elements. By the Weierstrass approximation theorem, polynomials are uniformly dense in $C[0,1]$, so this span is dense too.

No need to quote the sufficiency direction of Müntz–Szász at all — Weierstrass handles this case directly.

Concrete check: for $N \ge 2$, $x^2$ is itself a basis element, so it trivially lies in $\operatorname{span}\{1, x, \ldots, x^N\}$ — exact membership, not approximation. The scene shows this density working in practice even on a function with a cusp like $\sqrt{x}$: the dense basis's least-squares error shrinks $0.1003 \to 0.0708 \to 0.0544$ for $N = 4, 6, 8$.

(End of proof.)