## Slide 2 Definition Taylor series

Series of form, Centred at z_0 \in C.

$\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$

Examples

$\Sigma_{k=0}^{\infty} (z )^k \textrm{ where } a_k = 1 \textrm{ and } z_0 = 0$, converges for |z| < 1

$\Sigma_{k=0}^{\infty} \frac{(-1)^k}{2^k}z^{2k} \;= \;1 - \; \frac{z^2}{2}\; + \; \frac{z^4}{4}\; - \; \frac{z^6}{8}= \Sigma_{k=0}^{\infty} (\frac{-z^2}{2})^k = \Sigma_{k=0}^{\infty} (w)^k$

where w = $\frac{-z^2}{2}$

this converges when $|w| < 1$

converges when $|z| < \sqrt{2}$ and diverges when $|z| \geq \sqrt{2}$

## Slide 3 Radius Convergence Theorem *important *

Let $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ be a power series.

There exists a real number R, $0 \leq R \leq \infty$ , such that the series

converges absolutely in $|z \; - \; z_0|$ < R

and diverges in $|z \; - \; z_0|$ > R

convergence is uniform for each r < R

## Slide 4 Examples Radius Convergence

$\Sigma_{k=0}^{\infty} k^kz^k$ Pick an arbitrary $z \in C \backslash{0}$

Note: $|k^kz^k| = (k|z|)^k \geq 2^k \textrm{ as soon as } k \geq \frac{2}{|z|}$

But eventually this will happen, no matter what the value of |z| is, k will cause the terms  of the series to be increasing in size and the series does not converge.

In contrast: $\Sigma_{k=0}^{\infty} \frac{z^k}{k^k}$ . Similarly to above pick an arbitrary z, then $| \frac{z^k}{k^k}| = \frac{|z^k|}{k^k} \textrm { which will eventually } \leq (\frac{1}{2})^k$ when k reaches the value $k \geq 2|z|$ and beyond this value of k all subsequent terms are less than powers of 1/2 so the series will converge. Since this is always true the series has infinite radius of convergence.

## Slide 5 Analycity of Power Series

Theorem. For the power series $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ with radius of convergence R > 0, the claim is:

f(z) = $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ is analytic in $|z \; - \; z_0| < R$

and since it is analytic it can be differentiated (infinitely) as follows (note change of limits):

f ‘ (z) = $\Sigma_{k=1}^{\infty} a_k .k(z \; - \; z_0)^{k-1}$

f ” (z) = $\Sigma_{k=2}^{\infty} a_k .k.(k-1)(z \; - \; z_0)^{k-2}$

eventually

$f^{(k)}(z_0) = a_k .k!$ (since it makes sense to define $0^0= 1$).

Then by rearranging we get an expression for every coefficient in the original power series

$a_k = \frac{f^{(k)}(z_0)}{k!} \textrm{ for } k \geq 0$

## Slide 6 Differentiating term-by-term

Because $\Sigma_{k=0}^{\infty} z^k \; = \; \frac{1}{1-z}$ with radius of convergence 1, the sum is analytic when |z| < 1

So we can differentiate term by term and end up with

$\Sigma_{k=0}^{\infty} (k+1)z^k \; = \; \frac{1}{(1-z)^2}$

## Slide 7 Integrating term-by-term

Similarly , with certain certain conditions (to be added), can integrate infinite series term by term.

## Slide 8 Integration example to get series for Log(z)

Justification that $\int_0^w\;\Sigma_{k=0}^{\infty} z^k dz\; = \;-Log(1-z) \; = \;\Sigma_{k=1}^{\infty} \frac{w^k}{k}$ when |z|< 1.

By fiddling with variable we can get: $\Sigma_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(z-1)^k = Log(z)$ for |z – 1| < 1, we have a power series for Log.