1. ## maclaurin series

find the maclaurin series for f(x)=ln(1-x) and its radius of convergence. how do i do this? is it similar to that of ln(1+x)?

2. ## Re: maclaurin series

Taking the derivative, you find $f'(x) = -\dfrac{1}{1-x} = -\sum_{n\ge 0}x^n$. Integrating both sides, we get $f(x) = -\sum_{n\ge 0}\dfrac{x^{n+1}}{n+1}+C$. Since $f(0) = 0 = C$, this becomes $\ln(1-x) = -\sum_{n\ge 1}\dfrac{x^n}{n}$ (after you reindex to start at one instead of zero).

3. ## Re: maclaurin series

Let $g(x)=\ln(1+x)$

$f(x)=g(-x)$

We know (or can easily find out) that

$g(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{x^k}{k}$

$f(x)=g(-x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{(-x)^k}{k}$

$f(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}(-1)^k\dfrac{x^k}{k}$

$f(x)=\displaystyle{\sum_{k=1}^\infty}-\dfrac{x^k}{k}$

4. ## Re: maclaurin series

thank you! and how do i find R?

5. ## Re: maclaurin series

Originally Posted by tinspire
thank you! and how do i find R?
The radius of convergence is the same as that of ln(1+x) (or if you use the method I gave you, it is the radius of convergence of the geometric sum, which is 1).