Power Series.

We consider a power series $ S(z)=\underset{k \geq 0}{\sum} a_k (z-z_0)^k$ . Denote by $ R$ its radius of convergence.

Theorem 8.0.1  
  1. The sum of a power series is an analytic function on the open set $ \{ z \in \mathbb{C} ; \; \vert z-z_0\vert < R \}$ .
  2. The successive derivatives of $ S(z)$ are obtained by term-by-term differentiation.

This theorem is a consequence of Corollary 3.6 and Thm 3.7.

Try this with the examples in 1.2.

Example 8.0.2   Let $ S(z)= \underset{k \geq 0}{\sum} z^k = 1+z+z^2+z^3+ \dots$ . This is a geometric series whose ratio is equal to $ z$ . It is convergent for $ \vert z\vert<1$ and its sum is equal to $ S(z)= \frac {1}{1-z}$ .

By term-by-term differentiation, we get the successive series, all convergent for $ \vert z\vert<1$ :

$\displaystyle \frac {1}{1-z}$ $\displaystyle = \underset{n \geq 0}{\sum} z^n$    
$\displaystyle \frac {1}{(1-z)^2}$ $\displaystyle = \underset{n \geq 0}{\sum} nz^{n-1} = \underset{n \geq 1}{\sum} nz^{n-1}$    
$\displaystyle \frac {1}{(1-z)^3}$ $\displaystyle = \frac 12 \left( \underset{n \geq 0}{\sum} n(n-1)z^{n-2} = \underset{n \geq 2}{\sum} n(n-1)z^{n-2} \right)$    
$\displaystyle \frac {1}{(1-z)^4}$ $\displaystyle = \frac 16 \left( \underset{n \geq 0}{\sum} n(n-1)(n-2)z^{n-3} = \underset{n \geq 3}{\sum} n(n-1)(n-2)z^{n-3} \right)$    



Subsections
Noah Dana-Picard 2007-12-24