Power series representation????

**Blackberry410** · April 15th 2008, 03:36 PM

Can anyone please help me???

It asks to find the power series representation and radius of convergence for

f(x) = ln(a - x) for a>0
As an answer I got the series 1 to infinity of
__x^n___
a^n+1
I don't know if it is right and the radius of convergence I got was a but I don't know if that is right

f(x) = __x^3___
(1-4x)^2
I have no idea how to do this one...

Please help me

**TheEmptySet** · April 15th 2008, 04:19 PM

Originally Posted by Blackberry410

Can anyone please help me???

It asks to find the power series representation and radius of convergence for

f(x) = ln(a - x) for a>0
As an answer I got the series 1 to infinity of
__x^n___
a^n+1
I don't know if it is right and the radius of convergence I got was a but I don't know if that is right

f(x) = __x^3___
(1-4x)^2
I have no idea how to do this one...

Please help me

Let

$f(x)=\ln(a-x)$ then
n
$f'(x)=\frac{-1}{a-x}=-\frac{1}{a} \cdot \frac{1}{1-\frac{x}{a}}$

using the defintion of the geometric series we get

$f'(x)=-\frac{1}{a} \sum_{n=0}^{\infty} \left( \frac{x}{a}\right)^n= -\sum_{n=0}^{\infty} \frac{x^n}{a^{n+1}}$

integrating both sides we get

$f(x)= \ln(a-x)= \left( - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n+1)} \right) +C$

To find the constant C we evaluate

$f(0)= \ln(a)= \left( - \sum_{n=0}^{\infty} \frac{0^{n+1}}{a^{n+1}(n+1)} \right) +C \iff \ln(a) = C$

$f(x)= \ln(a-x)= \ln(a) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n+1)}$

Since this based on the geometric series it will converge when r < 1 so

$|\frac{x}{a}| < 1 \iff -1 < \frac{x}{a} < 1 \iff -a < x < a$