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Prove that Rn(x) is zero

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data

    How do I prove that Rn(x) is zero as n approaches infinity for the nth taylor polynomial of 1/(1-x)
    when x is between -1<x<1

    2. Relevant equations


    1/(1-x)
    Rn(x)=M*(x-a)^(n+1)/(n+1)!
    nth derivative of 1/(1-x) is n!/(1-x)^n+1
    3. The attempt at a solution

    the nth derivative of 1/(1-x) is n!/(1-x)^(n+1) so the max of the n+1 derivative between 0 and d if abs(d)<1 is (n+1)!(d)^(n+1)/(n+1)!(1-d)^(n+2). I simplify it and get d^(n+1)/(1-d)^(n+2) But there seems to be values of d between -1 and 1 such that when n approaches infinity, Rn approaches infinity
     
  2. jcsd
  3. Nov 10, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You have to evaluate your derivatives at x=0 to get the individual terms of the taylor expansion.
     
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