Coursera Complex Analysis 6.2 Taylor Series

Aug 9, 2016

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.

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