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Proof by induction - I'm stuck!

  1. Oct 14, 2004 #1

    quasar987

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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

    [tex]x_{1+1}=x_2=a^1x_1+b\frac{1-a^1}{1-a}[/tex]

    [tex]x_2=b+ax_1[/tex],

    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

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

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

    [tex]x_{n+1+1}=x_{n+2}=a^{n+1}x_1+b\frac{1-a^{n+1}}{1-a}[/tex]

    but after "simplifing" to

    [tex]x_{n+2}=aa^nx_1+b\frac{1-aa^n}{1-a}[/itex],

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

    Fredrik

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

    quasar987

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    Ah yes! Very clever! Thanks Fredrik!
     
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