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Math Help - Sequence Convergence

  1. #1
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    Sequence Convergence

    Hello,

    Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

    Show that a^2n > 7

    So i've said a^2(n+1) > 7

    So then,

    a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

    And then I hit a wall, any thoughts?

    Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MattWT View Post
    Hello,

    Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

    Show that a^2n > 7

    So i've said a^2(n+1) > 7

    So then,

    a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

    And then I hit a wall, any thoughts?

    Thanks
    This makes no sense. Can you write this in LaTeX or use better typsetting? Are you prove that \displaystyle a_{n+1}=\frac{1}{2}\left(a_n+\frac{7}{a_n}\right)\  implies a_n^2>7?
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  3. #3
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    Yes, sorry for that, not to sure on how to use the Math tags, it dislikes my brackets.
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  4. #4
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    Quote Originally Posted by MattWT View Post
    Hello,

    Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

    Show that a^2n > 7

    So i've said a^2(n+1) > 7

    So then,

    a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

    And then I hit a wall, any thoughts?

    Thanks

    If by "a^2n" you meant a^n_2 then it is true only for n\geq 2 , and inductively:

    a^2_{n+1}=\frac{1}{4}\left(a^2_n+14+\frac{49}{a_n^  2}\right)>\frac{1}{4}\left(63+\frac{49}{a_n^2}\rig  ht) .

    Now this last expression is greater than 7, since \displaystyle{\frac{1}{4}\left(63+\frac{49}{a_n^2}  \right)>7\Longleftrightarrow \frac{49}{a_n^2}>28-63} , and since

    this last inequality is trivial we're done.

    Tonio
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by MattWT View Post
    Hello,

    Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

    Show that a^2n > 7

    So i've said a^2(n+1) > 7

    So then,

    a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

    And then I hit a wall, any thoughts?

    Thanks
    The difference equation defining the sequence is...

    \displaystyle \Delta_{n}= a_{n+1}-a_{n} = -\frac{a_{n}}{2} + \frac{7}{2 a_{n}} = \varphi(a_{n}) (1)

    The function \varphi(x) for x>0 [for x<0 is \varphi(-x)= - \varphi(x)...] is illustrated here...



    There is an 'attractive fixed point' at x_{0} = \sqrt{7} = 2.645751311.... Any a_{1}>0 will produce a sequence converging at x_{0} but, because is [see 'red line'...] |\varphi(x)|>|x_{0} - x| for x< x_{0} and |\varphi(x)|<|x_{0} - x| for x> x_{0} , any a_{1}< x_{0} will produce a sequence a_{n}> x_{0} for n>1 ...

    Kind regards

    \chi \sigma
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