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Binomial expansion of (1+(1/x))^(-1)

  1. Jul 12, 2014 #1
    Expand the following functions as a series of ascending powers of x up to and including the term x^3. In each case give the range of values of x for which the expansion is valid.


    3. The attempt at a solution
    1 + (-1)(1/x) + (-1)(-2)(1/x^2)/2 + (-1)(-2)(-3)(1/x^3)/3!
    = 1 - (1/x) + (1/x^2) - (1/x^3)
    -1< 1/x <1
    x<-1 and x>1

    Which is invalid, but I can't see why. The way I deduced the ranges of values of x wasn't very rigorous so I suspect that might have something to do with it.

  2. jcsd
  3. Jul 12, 2014 #2

    Ray Vickson

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    Homework Helper

    You were asked for ascending powers of ##x ##, that is, for the series in terms of ##x, x^2, x^3, \ldots##. Instead, you gave descending powers ##1/x =x^{-1}, 1/x^2 = x^{-2}, 1/x^3 = x^{-3}, \ldots##.
  4. Jul 12, 2014 #3
    Thank you, I get it now
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