# Getting stuck on series - can't develop Taylor series.

• Oct 4th 2010, 03:55 AM
liquidFuzz
Getting stuck on series - can't develop Taylor series.
$\displaystyle \frac{Log z}{sin(\pi z)}$ Find convergences radius at z = 1 and develop a Taylor series.

Just by looking at the function I see that deviser and numerator has zeros in z = 1, so theres a limit in z = 1. The numerator isn't defined in z = 0. So the convergence radius is |z-1| < 1. (See deviser also, zeros in z = 2, 3, 4 ...)

First the log function, I think I've got it right. The Taylor expansion is also pretty straight forward.
$\displaystyle \log z$ = $\displaystyle \log (z - 1 + 1)$ = $\displaystyle \log 1 + (z - 1)$ =
$\displaystyle (z-1) - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4}+O((z-1)^5)$

In this part I'm starting to get shaky.
$\displaystyle \sin \pi z$ = $\displaystyle - \sin \pi (z - 1)$

$\displaystyle \frac {1}{\sin \pi z}$ = $\displaystyle -\frac{1}{\pi(z-1)} \frac{1}{1 - \frac{\pi^2}{6}(z-1)^2 - O((z-1)^4)}$ - This is where I need some pointers. I can't simplify this enough to multiply it with the log z part to get the whole series.
• Oct 4th 2010, 08:56 AM
Bruno J.
Are you just trying to find the first few terms of the expansion, or also the general coefficient? Because this would be much more complicated. The cosecant has a mean-looking series expansion involving the Bernoulli numbers.
• Oct 4th 2010, 12:44 PM
liquidFuzz
Oh sorry, I forgot to mention that. In the exercise it's stated that the expantion should be in the power of 3.
• Oct 5th 2010, 12:49 AM
liquidFuzz
Update
I asked a teacher about this exercise and was shown a solution. However, the part where I got stuck still puzzles me. It wasn't even in the solution... (Wondering)

$\displaystyle \frac{1}{\sin \pi z}$ = $\displaystyle -\frac{1}{\pi (z-1)} \frac{1}{1-\left(\frac{\pi^2}{6}(z-1)^2 + O((z-1)^4)\right)}$ = ??? = $\displaystyle -\frac{1}{\pi (z-1)}\left(1+ \frac{\pi^2}{6}(z-1)^2 + O((z-1)^4)\right)$

From here on I'm ok, but if someone could explain the trick above...
• Oct 5th 2010, 07:49 AM
Bruno J.
Well it's just the geometric series $\displaystyle \frac{1}{1-x}=1+x+x^2+\dots$... right?
• Oct 5th 2010, 08:11 AM
liquidFuzz
Quote:

Originally Posted by Bruno J.
Well it's just the geometric series $\displaystyle \frac{1}{1-x}=1+x+x^2+\dots$... right?

It depends on what you need, but you must always make sure that whatever $\displaystyle x$ is, it is less than 1 in absolute value.