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Thread: Infinite series (7)

  1. #1
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    Infinite series (7)

    Define the sequence $\displaystyle \{x_n \}$ by $\displaystyle x_1=3$ and $\displaystyle x_{n+1}=x_n^2 - 2, \ \ n \geq 1.$ Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}.$
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  2. #2
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    wow !


    How can people ( is that you ?) find out this this expression/series for irrational numbers ?


    we have

    $\displaystyle x_1 = 3 , x_2 = 7 , x_3 = 47 , x_4 = 2207 , x_5 = 4870847 ... $


    we can see that the series converges so quickly !


    for this problem ,

    I use a quite complicated method , is there any method better than mine ?

    Go through it !


    the general term of the seqence is $\displaystyle 2 \cosh(2^{n-1}y) $

    where $\displaystyle y = \cosh^{-1}(x_1 /2 ) $

    so the product of first $\displaystyle n $ terms is

    $\displaystyle x_1 x_2 x_3 .. x_n = $

    $\displaystyle 2^n \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $

    $\displaystyle = \frac{2^n}{\sinh(y)} \cdot \sinh(y) \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $

    $\displaystyle = \frac{2^{n-1}}{\sinh(y)} \cdot \sinh(2y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$


    $\displaystyle = \frac{2^{n-2}}{\sinh(y)} \cdot \sinh(4y) \cosh(4y) ... \cosh(2^{n-1}y) = \frac{\sinh(2^n y)}{ \sinh(y) } $

    the series can be written in this form :


    $\displaystyle 2\sinh(y) \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$


    $\displaystyle \sqrt{x_1^2 - 4 } \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

    To evluate this series , we just need to create $\displaystyle \frac{1}{a^2 - 1} - \frac{1}{a^2 - 1} $ next to the series .

    I got

    $\displaystyle \sqrt{ x_1^2 - 4 } \left( \frac{1}{ e^{2y} - 1} - \frac{1}{ e^{ 2^{m+1} y } - 1 } \right ) $

    $\displaystyle e^y = \frac{ x_1 + \sqrt{ x_1^2 - 4 }}{2} $

    for $\displaystyle x_1 = 3 , m \to \infty $

    I got $\displaystyle \frac{ 3 - \sqrt{5} }{2} $

    or an elegant-appearance

    $\displaystyle \sqrt{5} = 3 - \frac{2}{3} - \frac{2}{3\cdot7} - \frac{2}{3 \cdot 7 \cdot 47 } - \frac{2}{3\cdot7\cdot47\cdot2207} - .. $

    the fifth term is about $\displaystyle 10^{-6} $ , wow
    Last edited by simplependulum; Dec 8th 2009 at 02:30 AM.
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    wow !


    How can people ( is that you ?) find out this this expression/series for irrational numbers ?


    we have

    $\displaystyle x_1 = 3 , x_2 = 7 , x_3 = 47 , x_4 = 2207 , x_5 = 4870847 ... $


    we can see that the series converges so quickly !


    for this problem ,

    I use a quite complicated method , is there any method better than mine ?

    Go through it !


    the general term of the seqence is $\displaystyle 2 \cosh(2^{n-1}y) $

    where $\displaystyle y = \cosh^{-1}(x_1 /2 ) $

    so the product of first $\displaystyle n $ terms is

    $\displaystyle x_1 x_2 x_3 .. x_n = $

    $\displaystyle 2^n \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $

    $\displaystyle = \frac{2^n}{\sinh(y)} \cdot \sinh(y) \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $

    $\displaystyle = \frac{2^{n-1}}{\sinh(y)} \cdot \sinh(2y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$


    $\displaystyle = \frac{2^{n-2}}{\sinh(y)} \cdot \sinh(4y) \cosh(4y) ... \cosh(2^{n-1}y) = \frac{\sinh(2^n y)}{ \sinh(y) } $

    the series can be written in this form :


    $\displaystyle 2\sinh(y) \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$


    $\displaystyle \sqrt{x_1^2 - 4 } \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

    To evluate this series , we just need to create $\displaystyle \frac{1}{a^2 - 1} - \frac{1}{a^2 - 1} $ next to the series .

    I got

    $\displaystyle \sqrt{ x_1^2 - 4 } \left( \frac{1}{ e^{2y} - 1} - \frac{1}{ e^{ 2^{m+1} y } - 1 } \right ) $

    $\displaystyle e^y = \frac{ x_1 + \sqrt{ x_1^2 - 4 }}{2} $

    for $\displaystyle x_1 = 3 , m \to \infty $

    I got $\displaystyle \frac{ 3 - \sqrt{5} }{2} $

    or an elegant-appearance

    $\displaystyle \sqrt{5} = 3 - \frac{2}{3} - \frac{2}{3\cdot7} - \frac{2}{3 \cdot 7 \cdot 47 } - \frac{2}{3\cdot7\cdot47\cdot2207} - .. $

    the fifth term is about $\displaystyle 10^{-6} $ , wow
    your solution looks a little too complicated but it seems ok and your final answer is correct. well-done! i'll post my solution later. let's see if anybody else would like to try it.
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  4. #4
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    here's another approach: the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}$ is a "telescoping" series because $\displaystyle \frac{1}{x_1x_2 \cdots x_n} = \frac{1}{2} \left( \frac{x_n}{x_1 x_2 \cdots x_{n-1}} - \frac{x_{n+1}}{x_1x_2 \cdots x_n} \right).$
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    here's another approach: the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}$ is a "telescoping" series because $\displaystyle \frac{1}{x_1x_2 \cdots x_n} = \frac{1}{2} \left( \frac{x_n}{x_1 x_2 \cdots x_{n-1}} - \frac{x_{n+1}}{x_1x_2 \cdots x_n} \right).$

    I wanna know how you evaluate the limit :

    $\displaystyle \lim_{n\to\infty} \frac{ x_{n+1}}{ x_1 x_2 x_3 ... x_n } $


    my method is to multiply $\displaystyle \sqrt{ x_1^2 - 4 } $ both the numerator and the denominator ...
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  6. #6
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    I have consider another method but got stuck somewhere ...


    $\displaystyle f(x_1) = \frac{1}{x_1} + \frac{1}{x_1 x_2} + \frac{1}{x_1 x_2 x_3} + .... $


    we have



    $\displaystyle x_1 f(x_1) = 1 + f(x_2) = 1 + f( x_1^2 - 2 ) $

    so i obtain a functional equation

    $\displaystyle x f(x) = 1 + f(x^2 - 2 ) $

    and i know $\displaystyle \frac{x}{2}$ is a one of the solution

    so i let $\displaystyle f(x) = \frac{x}{2} + g(x) $

    then $\displaystyle x g(x) = g(x^2 - 2) $

    i found that $\displaystyle g(x) = \sqrt{x^2 - 4 } $

    but the problem is here :

    $\displaystyle k \sqrt{ x^2 - 4 } $ is also the solution but how do we get the exact value of $\displaystyle k $ ? ( $\displaystyle k = -\frac{1}{2} $ )
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  7. #7
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    here's the trick: $\displaystyle x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4$ and thus $\displaystyle x_{n+1}^2 - 4 = x_n^2(x_n^2-4).$ therefore $\displaystyle x_{n+1}^2-4=x_n^2x_{n-1}^2(x_{n-1}^2 - 4)=x_n^2 x_{n-1}^2x_{n-2}^2(x_{n-2}^2 - 4) = \cdots$ got it?
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  8. #8
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    Quote Originally Posted by NonCommAlg View Post
    here's the trick: $\displaystyle x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4$ and thus $\displaystyle x_{n+1}^2 - 4 = x_n^2(x_n^2-4).$ therefore $\displaystyle x_{n+1}^2-4=x_n^2x_{n-1}^2(x_{n-1}^2 - 4)=x_n^2 x_{n-1}^2x_{n-2}^2(x_{n-2}^2 - 4) = \cdots$ got it?

    oh , i see .

    this method is similar to my method , $\displaystyle \sqrt{5} $ is the limit .
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