Mittag-Leffler and Runge's Theorem — Examples

Suppose polynomials $p_n$ converge uniformly to $1\mathrm{/}z$ on ${\gamma }^{\ast }$. Then contour integration along the fixed rectifiable contour $\gamma$ preserves the limit, so ${\int }_{\gamma }{p}_n(z)dz\to {\int }_{\gamma }dz\mathrm{/}z$.

By [L1], every ${\int }_{\gamma }{p}_n(z)dz$ is $0$. But [L2] gives ${\int }_{\gamma }dz\mathrm{/}z={\int }_0^{2\pi }idt=2\pi i$. This contradiction shows that no such polynomial sequence exists.

Each closed disc ${\bar{D}}_j$ is disjoint from $K$, and the pairs $(2,5\mathrm{/}2)$, $(5\mathrm{/}2,3)$, and $(3,7\mathrm{/}2)$ lie in $D_1$, $D_2$, and $D_3$ respectively. Thus the displayed data form a pole-pushing chain from $2$ to $7\mathrm{/}2$.

Apply clause 1 of [L1] to that chain to obtain, for any prescribed $\epsilon >0$, a rational function with only pole $7\mathrm{/}2$ that approximates $(z-2{)}^{-1}$ uniformly on $K$. For the polynomial conclusion, every $z\in K$ satisfies $\mathrm{\mid }z\mathrm{\mid }<3\mathrm{/}2<7\mathrm{/}2$, so clause 2 of [L1], with $R=3\mathrm{/}2$, gives a polynomial approximating $(z-2{)}^{-1}$ uniformly on $K$ to within $\epsilon$.

Substitute $z=ia$ into [L1]. Since $\frac{1}{ia-n}=\frac{-ia-n}{n^2+a^2},$ the terms for $n$ and $-n$ combine to $-2ia\mathrm{/}(n^2+a^2)$, while the $n=0$ term is $1\mathrm{/}(ia)$.

Thus $\pi \cot (\pi ia)=\frac{1}{ia}-2ia{\sum }_{n\ge 1}\frac{1}{n^2+a^2}=-ia{\sum }_{n\in \mathbb{Z}}\frac{1}{n^2+a^2}\mathrm{.}$ Using $\cot (iw)=-i\coth (w)$ and dividing by $-ia$ gives the displayed closed form.

The set $\mathbb{Z}$ is discrete in $\mathbb{C}$, and each $p_n(z)=(z-n{)}^{-2}$ is a finite negative Laurent polynomial at $n$.

Applying [L1] to this data yields a meromorphic function $f$ whose principal part at every integer is exactly $(z-n{)}^{-2}$. Because that principal part is nonzero and contains only the degree $-2$ term, each pole has exact order $2$.

The unit circle is compact, and $1\mathrm{/}z$ is holomorphic on a neighbourhood of it.

If the displayed claim were true, then $1\mathrm{/}z$ would be uniformly approximable on that compact set by polynomials. This contradicts [L1].

Let $r$ be any rational function whose poles all lie in the unbounded complementary component of $K$. Then $r$ is holomorphic on the closed unit disc, so [L1] gives ${\int }_{\mathrm{\mid }z\mathrm{\mid }=1}r(z)dz=0$.

But ${\int }_{\mathrm{\mid }z\mathrm{\mid }=1}dz\mathrm{/}z=2\pi i$. Therefore no such rational function can approximate $1\mathrm{/}z$ uniformly on $K$, because the contour integral on the inner circle would preserve the limit. So the inner complementary component also needs a pole representative.

Fix any noninteger positive real $x$. Then $p_n(x)=-n\mathrm{/}(x-n)=1+x\mathrm{/}(n-x)\to 1$ as $n\to \mathrm{\infty }$.

Therefore the naive series ${\sum }_{n\ge 1}{p}_n(x)$ has terms that do not tend to $0$, so it diverges. This is exactly why the correction terms from [L1] are not optional in Mittag-Leffler's theorem.