# Convergence of Taylor Series

1. Nov 14, 2013

### Hertz

1. The problem statement, all variables and given/known data

Where does the Taylor series converge? [You do not need to find the Taylor Series itself]
$f(x)=...$

I have a few of these, so I'm mainly curious about how to do this in general.

3. The attempt at a solution

I haven't really made an attempt yet. If I were to make an attempt, it would be to determine the taylor series and then test its convergence, but I'm assuming you're not supposed to do that considering the problem says that you don't have to.

I've done a bunch of research on convergence of taylor series (because calc 2 was so long ago >.<) and all I can really find is stuff about the ratio test. I don't remember learning how to check convergence without actually finding the series

2. Nov 14, 2013

### Staff: Mentor

As you say, the ratio test is commonly used to find the interval of convergence for a given Taylor series. Since we can represent a Taylor series this way -
$$\sum_{n = 0}^{\infty}\frac{a_n (x - a)^n}{n!}$$

then your question boils down to determining the interval for which this inequality holds.
$$\lim_{n \to \infty} \frac{a_{n + 1}|x - a|}{a_n(n + 1)} < 1$$
a and the an's will depend on whatever series you're looking at.

3. Nov 14, 2013

### Hertz

The question explicitly states that you don't have to compute the taylor series. Power series are unique right? So theoretically you should be able to determine radius of convergence just from the function and the center of the taylor series, right? (Yes, it gives me the center of the taylor series. I should have mentioned that originally.)

I'm thinking complex numbers here. In the complex plane, the radius of convergence is simply the shortest distance from the center of the series to a singularity. So, couldn't I make the same argument for real values? The radius of convergence being the shortest distance to a singularity IN THE COMPLEX PLANE?

E.g.
The taylor series for the function $f(x)=\frac{1}{1+x^2}$ centered at zero will have a radius of convergence of 1, because the distance from zero to the closest singularity, i, is 1.

I don't see any reason why this shouldn't work now that I think about it, but of course, I'm not confident enough to use it on my test in a few hours! Can anybody confirm this?

4. Nov 15, 2013

### vanhees71

Sure, that's a correct argument. The question is, whether you are allowed to use complex function theory if the problem is asked in a real-analysis course ;-).