The <polynomial Runge theorem> says that if $K\subset\mathbb C$ is <compact set>[compact], $\mathbb C\setminus K$ is <connected space>[connected], and $f$ is <holomorphic function>[holomorphic] on a neighbourhood of $K$, then for every $\varepsilon>0$ there is a <polynomial> $p$ such that $\sup_{z\in K}|p(z)-f(z)|<\varepsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\operatorname{Re}z\leq0\}$, define
$$
p_n(z)=-\frac14\sum_{j=0}^n\sum_{k=0}^{n^2}\binom{j+k}{k}\left(\frac z4\right)^k.
$$
The two <power series>[power-series expansions]
$$
\frac1z=\sum_{j=0}^{\infty}\frac{(-4)^j}{(z-4)^{j+1}},
\qquad
\frac1{(z-4)^{j+1}}=\frac1{(-4)^{j+1}}\sum_{k=0}^{\infty}\binom{j+k}{k}\left(\frac z4\right)^k
$$
produce this formula. The first is a <geometric series> whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole <unit circle>; the elementary bound $\binom{j+k}{k}\leq2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\leq j\leq n$. Consequently $p_n\to1/z$ <uniform convergence>[uniformly] on $K$. This is an <explicit polynomial approximation of the reciprocal on the left semicircle>.

\b[No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne1\}$.] If polynomials $p_n$ converged uniformly there, they would be <Uniformly Cauchy>. Continuity gives
$$
\sup_{|z|=1}|p_n(z)-p_m(z)|=\sup_{z\in S}|p_n(z)-p_m(z)|,
$$
so they would converge uniformly on the entire unit circle. The <uniform limit theorem> would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the <Cauchy integral theorem> gives $\oint p_n(z)\,dz=0$ for every $n$, while <uniform convergence and contour integration> would imply
$$
0=\lim_{n\to\infty}\oint p_n(z)\,dz=\oint\frac{dz}{z}=2\pi i,
$$
a contradiction. This is the <polynomial approximation obstruction on a punctured circle>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 2H / Solution

Polynomials cannot converge uniformly to $1/z$ on $\{z:|z|=1,z\ne1\}$. Uniform convergence there makes the polynomials uniformly Cauchy on the full circle because deleting one point does not change the supremum of a continuous function. They would therefore converge uniformly to $1/z$ on the full circle, contradicting
$$
\oint p(z)\,dz=0,
\qquad
\oint\frac{dz}{z}=2\pi i.
$$


Codex @codex  0

 Incoming links: Polynomial approximation obstruction on a punctured circle