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Math Help - How to continue? Induction.

  1. #1
    Junior Member simplysparklers's Avatar
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    How to continue? Induction.

    Hi everyone!
    I'm not sure what to do with this induction question, can anyone help me please?

    Let a_1=0, a_2=3, and, for all n>=3 let
    a_n= 1/2(a_(n-1) + a_(n-2))
    By induction on n, show that for all n>=2,
    a_n= 2 + 4(-1/2)^n.

    What I did so far was work with the 2nd equation, and said:
    let n=1
    0= 2 + 4(-1/2)^1
    = 2 - 2
    =0
    therefore it is true for n=1

    Assume true for n=k
    a_k= 2 + 4(-1/2)^k

    ...what's the next step for the proof?Or have I gone completely wrong from the beginning?? Thanks in advance for your help guys!
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    First step: a_3  = \frac{{a_1  + a_2 }}{2} = \frac{3}{2} = 2 + 4\left( { - \frac{1}{2}} \right)^3.

    Now assume that a_K  = 2 + 4\left( { - \frac{1}{2}} \right)^K is true.
    Look at the next step.
    a_{K + 1}  = \frac{{a_K  + a_{K - 1} }}{2} = \frac{{2 + 4\left( { - \frac{1}{2}} \right)^K  + 2 + 4\left( { - \frac{1}{2}} \right)^{K - 1} }}{2}
    \frac{{2 + 4\left( { - \frac{1}{2}} \right)^K  + 2 + 4\left( { - \frac{1}{2}} \right)^{K - 1} }}{2} = 2 + 2\left( { - \frac{1}{2}} \right)^{K + 1} \left[ {\left( { - \frac{1}{2}} \right)^{ - 1}  + \left( { - \frac{1}{2}} \right)^{ - 2} } \right]
    Can you finish?
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  3. #3
    Super Member PaulRS's Avatar
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    In this case you'll have to assume that it's true for a_{n-1} and a_{n} and then show it's true for a_{n+1}

    (this is a stronger form of induction)
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    Quote Originally Posted by PaulRS View Post
    In this case you'll have to assume that it's true for a_{n-1} and a_{n} and then show it's true for a_{n+1}

    (this is a stronger form of induction)
    "Strong induction" is the name

    This means that the property is true for any k<n, and then show that it's true for n ^^
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by simplysparklers View Post
    Hi everyone!
    I'm not sure what to do with this induction question, can anyone help me please?

    Let a_1=0, a_2=3, and, for all n>=3 let
    a_n= 1/2(a_(n-1) + a_(n-2))
    By induction on n, show that for all n>=2,
    a_n= 2 + 4(-1/2)^n.

    What I did so far was work with the 2nd equation, and said:
    let n=1
    0= 2 + 4(-1/2)^1
    = 2 - 2
    =0
    therefore it is true for n=1

    Assume true for n=k
    a_k= 2 + 4(-1/2)^k

    ...what's the next step for the proof?Or have I gone completely wrong from the beginning?? Thanks in advance for your help guys!
    Our base case is n=3

    a_3=\frac{1}{2}(0+3)=\frac{3}{2}
    also
    a_3=2+4\left( -\frac{1}{2}\right)^3=2-\frac{1}{2}=\frac{3}{2}

    So the base case checks

    We need to use "strong" (you may know it by another name)mathematical induction

    assume true of all k < n use this to prove n is true

    We want to show \frac{1}{2}(a_{n-1}+a_{n-2})=a_n

    since n-1,n-2 < n they are true by the induction hypothesis

    \frac{1}{2}(a_{n-1}+a_{n-2})=\frac{1}{2}(2+4\left( -\frac{1}{2}\right)^{n-2}+2+4\left( -\frac{1}{2}\right)^{n-1})=

    2+2\left( -\frac{1}{2}\right)^{n-2}+2\left( -\frac{1}{2}\right)^{n-1}=2+2\left( -\frac{1}{2}\right)^{n-2}\left( 1+\left(-\frac{1}{2}\right) \right)=

    2+2\left( -\frac{1}{2}\right)^{n-2}\left( \frac{1}{2}\right)=2+\left( -\frac{1}{2}\right)^{n-2}=

    2+\underbrace{(4)\left( -\frac{1}{2}\right)^2}_{=1}\left( -\frac{1}{2}\right)^{n-2}=2+4\left( -\frac{1}{2}\right)^n=a_n

    QED
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  6. #6
    Junior Member simplysparklers's Avatar
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    Follow up

    Thank you so much all you guys for your help!

    One more question , and then I think I'm done with this lot of questions , the end of this question after the induction is:
    deduce that (a_n) \rightarrow 2.

    What I did so far was:
    |a_n- \alpha | = |2+4(-1/2)^n-2|
    = |4(-1/2)^n|

    ....what do I do from here please? How do I prove 2 is \alpha ??
    Last edited by simplysparklers; May 12th 2008 at 08:52 AM. Reason: Couldn't use Latex properly! :P
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  7. #7
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    Quote Originally Posted by simplysparklers View Post
    One more question ,
    deduce that (a_n) \rightarrow 2.
    Well, \left( {\frac{{ - 1}}{2}} \right)^n  \to 0.
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  8. #8
    Junior Member simplysparklers's Avatar
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    But what if n is negative, then it doesn't approach 0??
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    Quote Originally Posted by simplysparklers View Post
    But what if n is negative, then it doesn't approach 0??
    How can n be negative?
    It is an index of a sequence.
    To find the limit of the sequence, n approaches infinity and is therefore positive.
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  10. #10
    Junior Member simplysparklers's Avatar
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    Oh of course!Duh me! Thanks Plato!!& sorry for all the silly questions, I'm just not getting the whole concept of sequences at all!!
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