1. Apr 19, 2016

### shedrick94

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

How would you find the radius of convergence for the taylor expansion of:

$$f(z)=\frac{e^z}{(z-1)(z+1)(z-3)(z-2)}$$

I was thinking that you would just differentiate to find the taylor expansion and then use the ratio test but this seems far too tedious to be the right way to do it! Any help?

Last edited by a moderator: Apr 19, 2016
2. Apr 19, 2016

### Svein

The Taylor expansion of ez converges for all z. The denominator introduces poles at -1, 1, 2 and 3, so you need to be sufficiently far away from those values. Now determine "sufficiently far away"...

3. Apr 20, 2016

### Ray Vickson

Should we assume you want to expand around $z = 0$?

4. Apr 20, 2016

### Ray Vickson

If (as I asked in #3 but did not receive an answer!) you are expanding about $z = 0$, you can just expand each factor separately, then multiply the series together. The resulting series will be messy and not easy to write explicitly, but at least you can say something about the radius of convergence, since a lower bound on the radius of convergence of a product of series is known in terms of the individual radii of convergence. Google 'product of power series' or something similar.

5. Apr 20, 2016

### shedrick94

Sorry I understand this now. The expansion was about z=i, but I understand you would just the distance between the place you are expanding around and the closest singularity.