MATH3401

Let $f$ be analytic on $B_R(z_0)$. Then $f$ has a power series representation on $B_R(z_0)$: $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{(z-z_0)}^n\text{ for }|z-z_0|<R,$$ where $a_n={\displaystyle \frac{f^{(n)}(z_0)}{n!}}$, $n\in {\mathbb{N}}_0$.

For $z_0=0$: Maclaurin series

Take $z_0=0$, otherwise translate.

Choose $z\in B_R$, $|z|=r$. Fix some $r_0\in (r,R)$ and set $C=C_{r_0}$ positively oriented.

Cauchy $⟹$ $$f(z)=\frac{1}{2\pi i}{\int }_C\frac{f(\zeta )d\zeta }{\zeta -z}(1)$$

Note

${\displaystyle \frac{1}{\zeta -z}}={\displaystyle \frac{1}{\zeta }}\left({\displaystyle \frac{1}{1-\frac{z}{\zeta }}}\right)$ $={\displaystyle \frac{1}{\zeta }}\left({\displaystyle \sum _{n=0}^{N-1}{\left(\frac{z}{\zeta }\right)}^n+\frac{{\left(\frac{z}{\zeta }\right)}^N}{1-\frac{z}{\zeta }}}\right)$

$={\displaystyle \sum _{n=0}^{N-1}\frac{1}{{\zeta }^{n+1}}{z}^n+\frac{z^N}{(\zeta -z){\zeta }^N}}$

Multiply by ${\displaystyle \frac{f\left(\zeta \right)}{2\pi i}}$ and then integrate over $C$.

$⟹{\displaystyle \frac{1}{2\pi i}{\int }_C\frac{f(\zeta )}{\zeta -z}d\zeta }$

$={\displaystyle \sum _{n=0}^{N-1}\frac{1}{2\pi i}{\int }_C\frac{f(\zeta )z^n}{{\zeta }^{n+1}}d\zeta +\underset{\text{Call this }{\rho }_{N-1}(z)}{\underbrace{\frac{z^N}{2\pi i}{\int }_C\frac{f(\zeta )}{(\zeta -z){\zeta }^N}d\zeta }}}$

$(1)$ and the extended Cauchy integral formula $⟹$ $$f(z)=\sum _{n=0}^{N-1}\frac{f^{(n)}(0)z^n}{n!}+{\rho }_{N-1}(z).$$

If we can show that ${\displaystyle \underset{N\to \mathrm{\infty }}{lim}{\rho }_{N-1}(z)=0}$ we are done.

Take $\zeta \in C⟹|\zeta |=r_0$. Suppose there exists $M_N$ such that $$(H)\left|\frac{f(\zeta )}{(\zeta -z){\zeta }^N}\right|\le M_N\text{ on }C.$$

Then   $|{\rho }_{N-1}(z)|\le {\displaystyle \frac{r^N}{2\pi }}{M}_N\cdot \ell (C)$ $=r_0{r}^N{M}_N$. $(\ast \ast )$

To get $M_N$: $f$ is analytic $⟹|f|$ is continuous, $C$ is closed and bounded, so by the extreme value theorem there exists $\mu$ such that $$|f|\le \mu \text{ on }C.$$

Further $|\zeta {|}^N=r_0^N$ and

Hence $(\ast \ast )$ $⟹$

In $\mathbb{R}$, Taylor series may converge, but fail to converge to the function: c.f. Lecture 16.

Now consider the series expansion $\sum a_n(z-z_0{)}^n$.

We can calculate the Radius of convergence $R$:

Set $\mathrm{\Lambda }={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|}$, then set $R={\displaystyle \frac{1}{\mathrm{\Lambda }}}$.

$\mathrm{\Lambda }=0⟺R=\mathrm{\infty }$ and $\mathrm{\Lambda }=\mathrm{\infty }⟺R=0$.

Example 1: For $f(z)=e^z$, we have $e^z={\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{z}^n}$

Thus $R=\mathrm{\infty }$ and $$e^z=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{z}^n\text{ on }\mathbb{C}.(\ast )$$

Example 2: Find Maclaurin series for $f(z)=z^2{e}^{3z}$.

Answer: Since $f$ is entire, $(\ast )$ implies $$e^{3z}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{3^n{z}^n}{n!}.}$$

Therefore $z^2{e}^{3z}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{3^n{z}^{n+2}}{n!}}$ $={\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{3^{n-2}{z}^n}{(n-2)!}}$.