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
Use the Stone-Wierstrass theorem which states that the set of polynomials viewed as a subset of $\displaystyle C(X)$, form a dense subset. (where $\displaystyle X$ is a compact set).
Note that on the unit circle $\displaystyle f(x)=\frac{1}{x}$ is continuous.
Oh another nice proof is just to use cauchy's formula. polynomials are entire, so take the usual path $\displaystyle \gamma(t)=e^{i t}$ for $\displaystyle t\in [0,2\pi ]$ so their integral is 0, while $\displaystyle \int_{\gamma}z^{-1}=2\pi i$, but since $\displaystyle p(z) \rightarrow z^{-1}$ uniformly,
$\displaystyle o=\int_{\gamma}p_i(x)\rightarrow \int_{\gamma}z^{-1}=2\pi i$ a contradiction.
On the unit circle is $\displaystyle z=e^{i\cdot \omega}$ so that for $\displaystyle f(z)=\frac{1}{z}$ is...
$\displaystyle 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...
$\displaystyle p_{n} (z) = \sum_{k=0}^{n} (-1)^{k} \cdot \frac{z^{k}}{k!} $ (2)
... converges uniformly to $\displaystyle \frac{1}{z}$ for $\displaystyle z=e^{i\cdot \omega}$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$