Sequence of converging polynomials

morganfor · #1 $Old$ June 6th, 2009, 09:33 AM

I need help proving the following true or false: there is a sequence of polynomials {p_n} which converge uniformly to f(z)=1/z on the unit circle |z|=1

putnam120 · #2 $Old$ June 6th, 2009, 11:36 AM

Use the Stone-Wierstrass theorem which states that the set of polynomials viewed as a subset of $C(X)$ , form a dense subset. (where $X$ is a compact set).

Note that on the unit circle $f(x)=\frac{1}{x}$ is continuous.

Gamma · #3 $Old$ June 6th, 2009, 11:12 PM

polynomials have residue 0, while 1/z has residue 1.

It is false.

Gamma · #4 $Old$ June 6th, 2009, 11:22 PM

Oh another nice proof is just to use cauchy's formula. polynomials are entire, so take the usual path $\gamma(t)=e^{i t}$ for $t\in [0,2\pi ]$ so their integral is 0, while $\int_{\gamma}z^{-1}=2\pi i$ , but since $p(z) \rightarrow z^{-1}$ uniformly,

$o=\int_{\gamma}p_i(x)\rightarrow \int_{\gamma}z^{-1}=2\pi i$ a contradiction.

chisigma · #5 $Old$ June 8th, 2009, 01:07 AM

On the unit circle is $z=e^{i\cdot \omega}$ so that for $f(z)=\frac{1}{z}$ is...

$f(e^{i\cdot \omega})= e^{-i\cdot \omega} = \sum_{k=0}^{\infty} (-1)^{k} \cdot \frac{(i\cdot \omega)^{k}}{k!}$ (1)

From (1) it derives that the polynomial sequence...

$p_{n} (z) = \sum_{k=0}^{n} (-1)^{k} \cdot \frac{z^{k}}{k!}$ (2)

... converges uniformly to $\frac{1}{z}$ for $z=e^{i\cdot \omega}$ ...

Kind regards

$\chi$ $\sigma$