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Math Help - Infinite series (7)

  1. #1
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    Infinite series (7)

    Define the sequence \{x_n \} by x_1=3 and x_{n+1}=x_n^2 - 2, \ \ n \geq 1. Evaluate \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

     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  2 \cosh(2^{n-1}y)

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

    so the product of first  n terms is

     x_1 x_2 x_3 .. x_n =

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

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

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


     = \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 :


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


     \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  \frac{1}{a^2 - 1}  - \frac{1}{a^2 - 1} next to the series .

    I got

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

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

    for  x_1 = 3  , m \to \infty

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

    or an elegant-appearance

     \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  10^{-6} , wow
    Last edited by simplependulum; December 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

     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  2 \cosh(2^{n-1}y)

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

    so the product of first  n terms is

     x_1 x_2 x_3 .. x_n =

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

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

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


     = \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 :


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


     \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  \frac{1}{a^2 - 1} - \frac{1}{a^2 - 1} next to the series .

    I got

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

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

    for  x_1 = 3 , m \to \infty

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

    or an elegant-appearance

     \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  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 \sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n} is a "telescoping" series because \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 \sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n} is a "telescoping" series because \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 :

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


    my method is to multiply  \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 ...


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


    we have



     x_1 f(x_1) = 1 + f(x_2) = 1 + f( x_1^2 - 2 )

    so i obtain a functional equation

     x f(x) = 1 + f(x^2 - 2 )

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

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

    then  x g(x) = g(x^2 - 2)

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

    but the problem is here :

     k \sqrt{ x^2 - 4 } is also the solution but how do we get the exact value of  k ? (  k = -\frac{1}{2} )
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  7. #7
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    here's the trick: x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4 and thus x_{n+1}^2 - 4 = x_n^2(x_n^2-4). therefore 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: x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4 and thus x_{n+1}^2 - 4 = x_n^2(x_n^2-4). therefore 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 ,  \sqrt{5} is the limit .
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