Coursera Complex Analysis 6.3 Radius of convergence of a series

Aug 11, 2016

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)^kthere 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 4 Root Test << to be done >>

Slide 5 Root Test examples << to be done >>

Slide 6 Cauchy Hadamard

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|<r, \textrm{ where } a_k=\frac{f^{(k)}z_0}{k!}. 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.

 

 

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