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Taylor Expansion where the derivatives are undefined?

  1. May 7, 2013 #1
    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. jcsd
  3. May 7, 2013 #2
    Are you sure you don't want the Laurent expansion instead of the Taylor expansion?
  4. May 7, 2013 #3
    Sorry about that, yes I believe that is the case.
  5. May 7, 2013 #4


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    Are you confused about how (x-1)-1 + 1 is a Laurent series?
  6. May 7, 2013 #5
    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.
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