## Slide 2 Determining Radius of Convergence for a power series. The Ratio Test

In earlier lectures we were told for a power series $\Sigma_{k=0}^{\infty}a_k(z - z_0)^k$there is a value of R such that if distance of any z from the “centre” $z_0$ is less than R, i.e. | z – $z_0$| < R then the power series converges whereas if |z – $z_0$| > R the series diverges.

How do we find R.?

Theorem: if $|\frac{a_k}{a_{k+1}}|$ reaches a limit R as n tends to infinity, then this limit is the radius of convergence.

## Slide 3 Examples

$\Sigma_{k=0}^{\infty}(z)^k$, so $z_0$ = 0, a_k = 1, so $|\frac{a_k}{a_{k+1}}| = \frac{1}{1} \to 1$ then R = 1 (about the point z = 0).

$\Sigma_{k=0}^{\infty}k(z )^k$, so $z_0$ = 0, a_k = k, $|\frac{a_k}{a_{k+1}}| = \frac{k}{k+1} \to 1$, so R = 1

$\Sigma_{k=0}^{\infty}\frac{(z)^k}{k!}$, so $z_0$ = 0, a_k = 1/k!, $|\frac{a_k}{a_{k+1}}| = \frac{(k+1)!}{k\;!} \to k + 1$ therefore R = $\infty$

## Slide 7 Relationship Analytic Functions and Power series

If f: U -> C is analytic

and ${|z - z_0| < r } \subset U$ i.e. considering the z values in a disk of radius r surrounding the point $z_0$ << diagram >>, then in this disk f HAS a power series representation and that representation is $\Sigma_{k=0}^{\infty}a_k(z - z_0)^k \textrm{ where } a_k=\frac{f^{(k)}z_0}{k!}$ and its radius of convergence is at least the radius of the disk surrounding $z_0$, i.e. $|z \;-z_0|$.

## Slide 8 Examples of Taylor series of exp(z) about different points in complex plane

Summary of Taylor series expansion

$\Sigma_{k=0}^{\infty}a_k(z - z_0)^k, |z \;-z_0|. Note carefully the expression for the coefficient.

consider $f(z) = e^z$, then all derivatives $f^{(k)}(z) = e^z$. I fin the Taylor expansion we set z_0 = 0, then  $f^{(k)}(0) = e^0 = 1$ for every derivative, thus $a_k = \frac{1}{k!}$ for all k so

$e^z = \Sigma_{k=0}^{\infty}\frac{z ^k}{k!}\; ,z \in C$

Similarly, about $z_0 = 1$

$e^z = \Sigma_{k=0}^{\infty}\frac{e}{k!}z ^{k-1}\; ,z \in C$

## Slide 9 Series sin z about 0

Sin (z) is analytic in C. Then about 0

 f(z) sin(z) f(0) sin(0)=0 f’(z) cos(z) f’(0) cos(0)=1 f’’(z) -sin(z) f’’(0) -sin(0)=0 f’’’(z) -cos(z) f’’’(0) -cos(0)= -1 f(4)(z) sin(z) f(4)(0) sin(0)=0

so $sin (z) = 0 + \frac{1}{1!}(z) + \frac{0}{2!}(z)^2 + \frac{-1}{3!}(z)^3 + \frac{0}{4!}z^4 + \frac{1}{5!}(z)^5 + ...$

$z - \frac{1}{3!}(z)^3 + \frac{1}{5!}(z)^5 - \frac{1}{7!}(z)^7 + \frac{1}{9!}(z)^9 + ...$

## Slide 10 Series cos z about 0

Differentiating term by term

Cos z = $1 - \frac{1}{2!}(z)^2 + \frac{1}{4!}(z)^4 - \frac{1}{6!}(z)^6 + \frac{1}{8!}(z)^8 + ...$

## Slide 11 Analytic function

An analytic function is determined by all its derivatives $f^{(k)}(z_0)$ at the centre $z_0$ of the disc.