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Proof by induction - I'm stuck!

  1. Oct 14, 2004 #1


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    We want to show that the nth + 1 term of a sequence that is defined as [itex]x_{n+1} = b + ax_n, n\geq1 \ a,b \in \mathbb{R} [/itex] is given by

    [tex] x_{n+1} = a^nx_1 + b\frac{1-a^n}{1-a} \ \mbox{if} \ a\neq 1 [/tex]

    The manual suggests that we proceed by induction. Let us proceed by induction. :smile:

    P(1) is that



    which is in agreement with the expression of [itex]x_2[/itex] implied by the definition of [itex]x_{n+1}[/itex]

    Let us suppose P(n) to be true, i.e. that [itex]x_{n+1}[/itex] is in fact


    And let's see if P(n)being true implies that P(n+1) is true. P(n+1) is that


    but after "simplifing" to


    I don't see how to go any further than that!
    Last edited: Oct 14, 2004
  2. jcsd
  3. Oct 14, 2004 #2


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    The word you're looking for is "sequence" (but I see you figured that out while I was writing this), and I believe that this is the piece of the calculation you're missing:

    [tex]x_{(n+1)+1}=b+ax_{n+1}=b+a\bigg(a^n x_1+b\frac{1-a^n}{1-a}\bigg)=b+a^{n+1}x_1+ab\frac{1-a^n}{1-a}=a^{n+1}x_1+b\bigg(1+a\frac{1-a^n}{1-a}\bigg)=[/tex]
  4. Oct 14, 2004 #3


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    Ah yes! Very clever! Thanks Fredrik!
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