# Taylor Expansion where the derivatives are undefined?

1. May 7, 2013

### pandaBee

1. The problem statement, all variables and given/known data
Expand x/(x-1) at a=1
The book already gives the expansion but it doesn't explain the process. The expansion it gives is:
x/(x-1) = (1+x-1)/(x-1) = (x-1)^(-1) + 1

2. Relevant equations

3. The attempt at a solution
I've already solved for the Mclaurin expansion for the same polynomial which gives me -x - x^2 - x^3 - x^4 - .....
The derivatives all have a power of (x-1) as its denominator therefore all of its derivatives are undefined. I don't know any other way to expand polynomials other than term by term, I've been staring at

x/(x-1) = (1+x-1)/(x-1) = (x-1)^(-1) + 1

But it doesn't make sense. All they did was just split the fraction up into two parts.

2. May 7, 2013

### micromass

Staff Emeritus
Are you sure you don't want the Laurent expansion instead of the Taylor expansion?

3. May 7, 2013

### pandaBee

Sorry about that, yes I believe that is the case.

4. May 7, 2013

### Office_Shredder

Staff Emeritus
Are you confused about how (x-1)-1 + 1 is a Laurent series?

5. May 7, 2013

### pandaBee

I'm confused about the method used to derive it. As in I don't understand the method of the reasoning behind it.

Also I don't see how they can come up with that answer when the function x/(x-1) is undefined around x=1.