Results 1 to 5 of 5
Like Tree2Thanks
  • 2 Post By Opalg

Thread: limit with composite function

  1. #1
    Member
    Joined
    Dec 2010
    Posts
    106
    Thanks
    2

    limit with composite function

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

    My Solution::

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

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

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

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

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

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

    $\displaystyle \underbrace{f(f(f(x)))}_{n-times} = $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: limit with composite function

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

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

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

    taking y=g(x)

    $\displaystyle \implies y(1+y)=1$

    $\displaystyle \implies y^2+y-1=0$

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

    $\displaystyle \implies g(x)=\frac{\sqrt{5}-1}{2}$
    Last edited by sbhatnagar; Jan 14th 2012 at 12:10 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2010
    Posts
    106
    Thanks
    2

    Re: limit with composite function

    But Sir Here why we discard other root
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: limit with composite function

    I discarded the other root because , g(x) is clearly positive...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10

    Re: limit with composite function

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

    $\displaystyle f^{(n)}(x) \approx\frac{\lambda_+^nx+\lambda_+^{n+1}}{\lambda _+^{n+1}x+\lambda_+^{n+2}} = \frac1\lambda_+ = \frac2{1+\sqrt5} = \tfrac12(\sqrt5-1).$
    Thanks from jacks and General
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Composite function
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: Apr 7th 2011, 09:42 PM
  2. Composite Function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Apr 18th 2010, 04:53 AM
  3. Show the limit of two composite functions aren't =
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 20th 2009, 05:42 PM
  4. Composite function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 11th 2007, 11:21 PM
  5. Composite Function
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Oct 21st 2006, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum