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

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

$\frac{Log z}{sin(\pi z)}$ 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.
$\log z$ = $\log (z - 1 + 1)$ = $\log 1 + (z - 1)$ =
$(z-1) - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4}+O((z-1)^5)$

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

$\frac {1}{\sin \pi z}$ = $-\frac{1}{\pi(z-1)} \frac{1}{1 - \frac{\pi^2}{6}(z-1)^2 - O((z-1)^4)}$ - 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.

2. 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.

3. Oh sorry, I forgot to mention that. In the exercise it's stated that the expantion should be in the power of 3.

4. 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...

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

From here on I'm ok, but if someone could explain the trick above...

5. Well it's just the geometric series $\frac{1}{1-x}=1+x+x^2+\dots$... right?

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

Thank you for pointing out the obvious. I'll never forget it! Is it always sufficient to use the two first factors of the geometric series while tinkering with series?

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