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.

You need to find the Maclaurin series, then find 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

$\chi$ $\sigma$

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.

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

HOI, what do you mean by smooth?

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

$\chi$ $\sigma$

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.
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!