1. ## Direct Analytical Continuations

Hi any help with this would be greatly appreciated:

Many thanks and "Mathshelpforum" rocks

2. Originally Posted by moolimanj
Hi any help with this would be greatly appreciated:

Many thanks and "Mathshelpforum" rocks
$\displaystyle f(z)=\sum_{n=0}^{\infty}(2+z)^n, \ \ (|z+2|<1)$

$\displaystyle g(z)=-2 \sum_{n=1}^{\infty}\left( \frac{1}{3+2x}\right)^n, \ \ (|z+3/2|>1/2)$

The series defining function $\displaystyle f(z)$ is a geometric series which sums to:

$\displaystyle f(z)=\frac{-1}{1+z}, \ \ (|z+2|<1)$

Similarly the series defining $\displaystyle g(z)$ may also be summed with the
aid of the geometric series formula to:

$\displaystyle g(z)=2-2 \sum_{n=0}^{\infty}\left( \frac{1}{3+2x}\right)^n$$\displaystyle =2-\frac{2}{1-\frac{1}{3+2z}}=\frac{-1}{1+z} , \ \ (|z+3/2|>1/2)$

Thus $\displaystyle f(z)$ and $\displaystyle g(z)$ are identical on the overlap of their regions of convergence, and so are analytic continuations of one another.

RonL