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Use induction to show that

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Define an = (an-1 + an-2)/2 for each positive integer ≥ 2. Use induction to show that: an+1 - an = (-1/2)n(a1 -a0)


    2. Relevant equations
    First show it is true for base case. Assume if it is true for (k), then show it is true for (k + 1).


    3. The attempt at a solution
    Base case n=1. Then a2 - a1 = (-1/2)1(a1 - a0) = (a0 - a1)/2
    Check: a2 - a1 = (a1 + a0)/2 - a1 = (a1 + a0 - 2a1)/2 = (a0 - a1)/2.

    So then do I assume that ak+1 - ak = (-1/2)k(a1 - a0)? Then what?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 1, 2012 #2

    lanedance

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    Homework Helper

    so assume it is true for k-1,k as below (note you had the power wrong above)
    [tex] a_k - a_{k-1} = (-1/2)^k(a_1 - a_0)[/tex]

    Now using that can you show
    [tex] a_{k+1} - a_{k} = (-1/2)^{k+1}(a_1 - a_0)[/tex]
     
  4. Feb 7, 2012 #3
    No, I did not have the power wrong. This is from Introduction to Analysis, Gaughan, Prob. 23.

    It says you may want to use induction to show that:
    an+1 - an = (-1/2)^n(a1 - a0).
     
  5. Feb 7, 2012 #4
    Got the induction part:

    Assume ak+1 - ak = (-1/2)k(a1-a0), then

    ak+2 - ak+1 = (ak+1 + ak)/2 -[(-1/2)k(a1 - a0) + ak]

    = (ak+1 + ak - 2ak)/2 - [(-1/2)k(a1 - a0)]
    = (ak+1 - ak)/2 - [(-1/2)k(a1 - a0)]
    = (1/2)(-1/2)k(a1 - a0) - [(-1/2)k(a1 - a0)]
    = -(-1/2)k+1(a1 - a0) - [(-1/2)k(a1 - a0)]
    = (-1/2)k(a1 - a0)[1/2 - 1] = (-1/2)k+1(a1 - a0) as desired.

    The rest of the proof uses Triangle Inequality Theorem.
     
    Last edited: Feb 7, 2012
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