Question

find a power series for f(x)=ln(1-x) centered at x=0. someone please help!

Answer 1

this is a famous one
\[-\sum\frac{x^n}{n}\]

Answer 2

can you show me some work please im lost on this one

Answer 3

\[\frac{d}{dx}\ln(1-x)=\frac{1}{x-1}=-\frac{1}{1-x}\] and the last one is a well known power series

Answer 4

\[\frac{1}{1-x}={1+x+x^2+...}=\sum_{k=1}^\infty x^n\] integrate term by term

Answer 5

so which one is it your first one or somply x^n like your second?

Answer 6

it is
\[\ln(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}\]

Answer 7

you can also take successive derivatives and see that it works

Answer 8

ok i get ya. whats the radius of convergance

Answer 9

\[f(x)=\ln(1-x)\]
\[f'(x)=\frac{1}{x-1}=(x-1)^{-1}\]
\[f''(x)=-(x-1)^{-2}\]
\[f'''(x)=2(x-1)^{-3}\]
\[f^{(n)}(x)=(-1)^{n-1}(n-1)!(1-x)^{-n}\]

Answer 10

plug in 0, divide by
\[n!\] and get \[-\frac{x^n}{n}\] for each term

Answer 11

think this is good on the domain,
\[x<1\]