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Math Help - induction help

  1. #1
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    induction help

    I keep confusing myself when I try and do this.
    Let a_{1} and a_{0} be distinct real numbers. Define a_{n} = \frac{a_{n-1} + a_{n-2}}{2} for positive integers n \geq 2 . Show a_{n+1} - a_{n } = (-\frac{1}{2})^{n}(a_{1}- a_{0})

    I show when n =1 is true a_{2} - a_{1 } =  (-\frac{1}{2})^{1}(a_{1}- a_{0})
    => a_{2} - a_{1 } = (-\frac{1}{2})}(a_{1}- a_{0})

    Then I must assume n = k is true, and show for n = k+1 is true.
    However I keep wanting to reduce into terms of a_{1} and a_{0} and I get mixed up and forget what Im trying to do.
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  2. #2
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    Re: induction help

    Quote Originally Posted by delgeezee View Post
    I keep confusing myself when I try and do this.
    Let a_{1} and a_{0} be distinct real numbers. Define a_{n} = \frac{a_{n-1} + a_{n-2}}{2} for positive integers n \geq 2 . Show a_{n+1} - a_{n } = (-\frac{1}{2})^{n}(a_{1}- a_{0})

    I show when n =1 is true a_{2} - a_{1 } =  (-\frac{1}{2})^{1}(a_{1}- a_{0})
    => a_{2} - a_{1 } = (-\frac{1}{2})}(a_{1}- a_{0})

    Then I must assume n = k is true, and show for n = k+1 is true.
    However I keep wanting to reduce into terms of a_{1} and a_{0} and I get mixed up and forget what Im trying to do.
    Assume true for n, i.e.

    $\begin{align*}&a_{n+1}-a_n=\left(-\dfrac 1 2\right)^n (a_1-a_0)\\ \\

    &\text{Now show true for }n+1 \\ \\

    &a_{n+2}-a_{n+1} = \dfrac 1 2 (a_{n+1} + a_n) - a_{n+1}= \\ \\

    &-\dfrac 1 2 (a_{n+1}-a_n)=\\ \\

    &-\dfrac 1 2 \left(-\dfrac 1 2\right)^n (a_1-a_0)= \\ \\

    &\left(-\dfrac 1 2\right)^{n+1} (a_1-a_0)\end{align*}$

    and as you showed it true for $n=2$ this completes the proof.
    Thanks from delgeezee
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  3. #3
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    Re: induction help

    You need to show that if n=1, the statement that a_{1+1}-a_1 = \left(-\dfrac{1}{2}\right)^1(a_1-a_0) is true. You are not supposed to show that n=1 is true ( n=1 is true when you let n=1, so there is nothing to show).

    Based on the definition:

    a_2 = \dfrac{a_1+a_0}{2}

    Subtracting a_1 from both sides gives:

    a_2-a_1 = -\dfrac{1}{2}a_1 + \dfrac{1}{2}a_0 = -\dfrac{1}{2}(a_1-a_0)

    That shows the statement is true when n=1. Assume for all positive integers less than or equal to n, a_{n+1}-a_n = \left(-\dfrac{1}{2}\right)^n(a_1-a_0).

    \begin{align*}a_{(n+1)+1} - a_{n+1} & = \dfrac{a_{n+1}+a_n}{2} - a_{n+1} \\ & = -\dfrac{1}{2}\left(a_{n+1} - a_n\right)\end{align*}

    By the induction hypothesis, a_{n+1}-a_n = \left(-\dfrac{1}{2}\right)^n(a_1-a_0). Plugging that in:

    a_{n+2}-a_{n+1} = \left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)^n(a_1-a_0) = \left(-\dfrac{1}{2}\right)^{n+1}(a_1-a_0)

    This shows the induction statement is true for n+1, as well. (Feel free to add in k and k+1 if you like). This is the general idea of the proof.
    Thanks from delgeezee
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  4. #4
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    Re: induction help

    Hi,
    The following is an inductive proof, a little tricky, I admit. If you know about linear recurrence relations, the whole problem is easier to solve without induction.
    Edit: Slipeternal's standard inductive proof is much easier, but maybe you want to read this anyway.

    Last edited by johng; April 21st 2014 at 09:19 PM.
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  5. #5
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    Re: induction help

    Thank you for your elaboration. I can see where I was going wrong.
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