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Math Help - limit with composite function

  1. #1
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    limit with composite function

    If \displaystyle f(x) = \frac{1}{1+x} and g(x) = \underbrace{f(f(f(x)))}_{n-times}. Then \displaystyle \lim_{n\rightarrow \infty}g(x) =

    My Solution::

    f(x) = \frac{1}{x+1} Then

    f(f(x)) = \frac{x+1}{x+2}

    f(f(f(x))) = \frac{x+2}{2x+3}

    f(f(f(f(x)))) = \frac{2x+3}{3x+5}

    .................................................. .......

    .................................................. ...

    \underbrace{f(f(f(x)))}_{n-times} =
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  2. #2
    Member sbhatnagar's Avatar
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    Re: limit with composite function

    f(x)=\frac{1}{1+x} and g(x)=\underbrace{f(f(f(\cdots f(x) \cdots )))}_{\text{Infinite Times}}

    g(x)= \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots  }}}}

    \implies g(x)=\frac{1}{1+g(x)}

    taking y=g(x)

    \implies y(1+y)=1

    \implies y^2+y-1=0

    \implies y=\frac{\sqrt{5}-1}{2}

    \implies g(x)=\frac{\sqrt{5}-1}{2}
    Last edited by sbhatnagar; January 14th 2012 at 12:10 AM.
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  3. #3
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    Re: limit with composite function

    But Sir Here why we discard other root
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  4. #4
    Member sbhatnagar's Avatar
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    Re: limit with composite function

    I discarded the other root because , g(x) is clearly positive...
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  5. #5
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    Re: limit with composite function

    For a slightly different way of looking at it, you might notice that the n'th iterate f^{(n)}(x) is given by f^{(n)}(x) = \frac{F_nx+F_{n+1}}{F_{n+1}x + F_{n+2}}, where F_n is the n'th Fibonacci number. This is given by the formula F_n = \bigl(\lambda_+^n-\lambda_-^n\bigr)/\sqrt5, where \lambda_+ = \tfrac12(1+\sqrt5) and \lambda_- = \tfrac12(1-\sqrt5). But |\lambda_-|<1, and so \lambda_-^n\to0 as n\to\infty. Thus for large values of n, F_n\approx \lambda_+^n/\sqrt5 and therefore

    f^{(n)}(x) \approx\frac{\lambda_+^nx+\lambda_+^{n+1}}{\lambda  _+^{n+1}x+\lambda_+^{n+2}} = \frac1\lambda_+ = \frac2{1+\sqrt5} = \tfrac12(\sqrt5-1).
    Thanks from jacks and General
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