# Math Help - Radius of Convergence

1. ## Radius of Convergence

Around x = 0 for ln(1-x).

Can someone explain what I'm looking for here? I don't want the problem worked, just a guide through the process. I tried looking up the definition for "radius of convergence" but I don't really understand it.

Thank you.

2. ## Re: Radius of Convergence

You need to find the Maclaurin series, then find radius of convergence.

3. ## Re: Radius of Convergence

Originally Posted by OneMileCrash
Around x = 0 for ln(1-x).

Can someone explain what I'm looking for here? I don't want the problem worked, just a guide through the process. I tried looking up the definition for "radius of convergence" but I don't really understand it.

Thank you.
It is possible to demonstrate that any power series $\sum_{n=0}^{\infty} a_{n}\ z^{n}$ converges in any point inside a circle centered in z=0 and with radius r. That's the defintion of 'circle of convergence'...

Kind regards

$\chi$ $\sigma$

4. ## Re: Radius of Convergence

No need to actually construct the series if that is not specifically asked for. The distance from the center point, 0, to the first point at which ln(1- x) is not smooth, x= 1, is 1. The radius of convergence is 1.

5. ## Re: Radius of Convergence

Is there a relationship between interval of convergence and radius of convergence?

HOI, what do you mean by smooth?

6. ## Re: Radius of Convergence

Originally Posted by OneMileCrash
Is there a relationship between interval of convergence and radius of convergence?...
You use the term 'interval of convergence' if the power series is in the real x variable, 'radius of convergence' if the power series in in the complex z variable. If the circle of convergence is centered in z=0, then the interval of convergence is twice the radius of convergence...

Kind regards

$\chi$ $\sigma$

7. ## Re: Radius of Convergence

It can be shown, by using the ratio test, perhaps, that a power series of the form $\sum a_n(x- a)^n$ must converge on some "interval of convergence" [a- r, a+ r] centered on a. r is the "radius of convergence. In the complex numbers, which is geometrically represented as a plane that "interval of convergence" is a disk and the "radius of convergence really is a radius.
By "smooth" I basically mean "differentiable".

On a qualifying test in graduate school, one of the problems was to find the radius of convergence for the MacLaurin series of $\frac{1}{1+ z^2}$ about the point z= 1 (there were a lot harder questions on the test!). One of my friends actually calculated the MacLaurin series and then the radius of convergence. I just argued that the function is differentiable (in fact, infinitely differentiable and even analytic) for all z except z= i and z= -i. The distance from 1 to i or -i is $\sqrt{2}$ so the radius of convergence is [itex]\sqrt{2}[/tex]. Much faster!