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Math Help - induction proof

  1. #1
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    induction proof

    I've been working on this for the past hour, but haven't gone anywhere with it. If anyone can help to complete it, it would be highly appreciated. Thanks

    Let 0< a1< b1 and define

    an+1= √anbn

    bn+1=(an+bn)/2


    a) Use induction to show that
    an<an+1<bn+1<bn

    Thus prove that an and bn converge.
    b) Prove that they have the same limit.
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  2. #2
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    How far along the 'path' have you gotten?
    Have you been able to start with the basic step?
    Please tell us where you are in the proof.
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  3. #3
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    Greatly appreciate it if someone took me through the entire problem, don't know where to start. Thank you
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  4. #4
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    This entire problem revolves about the following facts:
    \begin{array}{l}<br />
 x^2  + y^2  \ge 2xy \\ <br />
 x = \sqrt a \quad \& \quad y = \sqrt b \quad  \Rightarrow \quad \frac{{a + b}}{2} \ge \sqrt {ab} \\ <br />
 0 < a_N  < b_N \quad  \Rightarrow \quad a_N  < \frac{{a_N  + b_N }}{2} < b_N  \\ <br />
 \end{array}.

    Begin with this, if 0 < a_N  < b_N we can use the above and derive:
    \begin{array}{l}<br />
 a_{N + 1}  = \sqrt {a_N b_N }  > \sqrt {a_N a_N }  = a_N  \\ <br />
 a_N  < \frac{{a_N  + b_N }}{2} < b_N \quad  \Rightarrow \quad a_N  < b_{N + 1}  < b_N  \\ <br />
 b_{N + 1}  = \frac{{a_N  + b_N }}{2} \ge \sqrt {a_N b_N }  = a_{N + 1}  \\ <br />
 \end{array}.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Swamifez View Post
    Can someone just show me a full blown proof of this problem, I really would have posted everything I had, but its really hard for me to type everything I had. Thanks

    Let 0< a1< b1 and define

    an+1= √anbn

    bn+1=(an+bn)/2


    a) Use induction to show that
    an<an+1<bn+1<bn
    Suppose that for some k \in \mathbb{N} that:

    0<a_{k-1}<a_k<b_k<b_{k-1}

    then a_{k+1}=\sqrt{a_kb_k}>a_k as b_k>a_k,

    and b_{k+1}=(a_k+b_k)/2<b_k as a_k<b_k.

    Also by the geometric-arithmetic mean inequality:

    a_{k+1}<b_{k+1} (equality is not a possibility here as a_k \ne b_k),

    so we have shown that:

    0<a_{k}<a_{k+1}<b_{k+1}<b_{k}.

    The base case is demonstrated by showing that:

    0<a_1<a_2<b_2<b_1

    using essentialy the same methods, which proves the required result by
    mathematical induction.

    RonL
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Swamifez View Post
    Let 0< a1< b1 and define

    an+1= √anbn

    bn+1=(an+bn)/2


    a) Use induction to show that
    an<an+1<bn+1<bn

    Thus prove that an and bn converge.
    b) Prove that they have the same limit.
    Part a) shows that a_n is an increasing bounded sequence, hence converges,
    also b_n is a decreasing sequence bounded below and hence converges.

    The limits satisfy the equations:

    a=\sqrt{a\,b}

    b=(a+b)/2.

    The solution to these is a=b.

    RonL
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