TITLE: Approximating $1/z$ by polynomials
QUESTION [18 upvotes]: Let $C=\{\mathrm e^{\mathrm it}, 0\le t\le 3\pi/2\}$ and $f(z)=1/z$. By Runge's theorem, there is a sequence of polynomials $p_n(z)$ such that $$\lim_n p_n(z)=f(z)$$ uniformly on $C$. 
Does anyone know such a sequence?

REPLY [8 votes]: This solution is specialized to the particular problem. As in my other solution, I am working with the arc $C = \{ e^{i \theta} : \pi/4 < \theta < 7 \pi/4 \}$. Let $T_n(z)$ be the $n$-th Chebyshev polynomial, so it is the polynomial with leading term $2^{n-1} z^n$ which has $|T_n(z)|\leq 1$ for $-1 \leq z \leq 1$. 
Set
$$g_n(z) = z^{-1} - \frac{(1 + 1/\sqrt{2})^n  z^{n - 1}}{2^{n-1}}
 T_n\left(\frac{z + z^{-1} +1-1/\sqrt{2}}{1 + 1/\sqrt{2}} \right)$$
Okay, what's going on here? The map $z \mapsto z + z^{-1}$ takes $C$ to $[-2, \sqrt{2}]$. The linear function inside the Chebyshev polynomial takes $[-2, \sqrt{2}]$ to $[-1,1]$, so the $T_n$ term is $\leq 1$. So the whole second term has absolute value at most $(1+1/\sqrt{2})^n/2^{n-1} \approx 2 \cdot 0.853^n$. This will be much less than $z^{-1}$, for $n$ large, so $g_n(z) \approx z^{-1}$. 
On the other hand, $T_n\left( \mbox{stuff} \right)$ will be a Laurent polynomial with most negative term $\frac{2^{n-1} z^{-n} }{(1+1/\sqrt{2})^n}$. 
After multiplying by $\frac{(1 + 1/\sqrt{2})^n  z^{n - 1}}{2^{n-1}}$, the second term will be of the form $z^{-1} + \mbox{polynomial}$. So $g_n(z)$ is a polynomial. 

In this picture, the black arc is $e^{i \theta}$ for $\pi/4 \leq \theta \leq \pi$. The red, blue and green arcs are $g_{10}$, $g_{20}$ and $g_{30}$ evaluated on the black arc. 
How did I come up with this? Roughly: I want $z^{-1} \approx p(z)$ for a polynomial $p$. I want $z^{-1} - p(z) \approx 0$. I want $z^{-n} + \cdots + z^n \approx 0$. I want Laurent polynomials with leading term $1$ and very small values on $C$. I know a family of polynomials with leading term $1$ and very small values on $[-1, 1]$: Namely, $2^{-n+1} T_n(z)$. How can I turn one into the other?
Note that these have much smaller degree than the polynomials in my other solution. The thing I called $f_N$ in my other solution has degree $N^3$, this $g_n$ has degree $2n-1$.