Results 1 to 4 of 4

Math Help - induction

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    23

    induction

    The questions asks ...

    Suppose that R > 0, x_{0}>0, and
    x_{n+1}= \frac{1}{2}(\frac{R}{x_{n}}+x_{n}), n\geq0

    Prove: For n\geq1, x_{n}>x_{n+1}>\sqrt{R} and
    x_{n}-\sqrt{R}\leq\frac{1}{2^n}\frac{(x_{o}-\sqrt{R})^2}{x_{o}}

    MY attempt at a solution:

    I really don't know how to go about this... at first i tried to show that if x_{n}>x_{n+1}>\sqrt{R} is true then x_{n}-x_{n+1} > 0.

    So that means
    x_{0}- \frac{1}{2}(\frac{R}{x_{0}}+x_{0})>0
    \frac{x_{0}}{2}-\frac{R}{2x_{0}}>0
    x_{0}-\frac{R}{x_{0}}>0

    My problem is that i don't know how big R in comparison to x_{0}, so unless i restrict R to be less than x_{0} i can't prove that x_{n}>x_{n+1}>\sqrt{R}


    Basically, i need help with this question... can anyone help ?

    I'm doing a self study by myself so i would appreciate a detailed solution if possible.. please and thank u.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    I believe you are right, in fact if x_0<\sqrt{2} (for
    instance x_0=1 and R=4 then the sequence increases in the first terms x_1>x_0, but x_1>x_2>2 and then the hypothesis seems to hold (for
    n\geq 1 as stated).

    However, if x_0=\sqrt{R} we have the constant sequence (\sqrt{R})

    I conjecture without doing any more calculations that this is the unique problem with the hypothesis, we have to require x_0\neq \sqrt{R}.

    I suggest
    a) Try to prove the inequality if x_0>\sqrt{R}, it seems an easy induction at least for the first one.
    b) Try to show using differential calculus that f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right)
    maps [0,\sqrt{R}) into (\sqrt{R},\infty). This would solve the case fos "small x_0's"
    because it would allow to apply a)
    I haven't made the calculations but it seems to me the way of attacking it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    23
    Sorry i'm not very sure about what you're asking me to do.

    My point was that x_{n}>x_{n+1} is not true for all n if R>  x_{0}. So do i restrict R so that this case is satistfied ?

    If i do this then it would seem that the sequence is decreasing at first but then it would blow up to infinity .

    The question is puzzuling to me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    113
    My idea is to divide the problem in the following cases:

    a) If x_0>\sqrt{R} then x_0>x_1>\cdots x_n >\sqrt{R} and the sequence tends to \sqrt{R}. I believe it follows from the inequality you obtained in the first post, by an easy induction. Observe that the inequality you got is x_0-x_1=\frac{x_0^2-R}{2x_0}. If x_0>\sqrt{R} then you have clearly x_0>x_1>\sqrt{R} and you can apply induction. For showing x_1>\sqrt{R} you have to use that f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right) is increasing in [\sqrt{R},\infty) and f(\sqrt{R})=\sqrt{R}.

    b) If x_0=\sqrt{R} then x_n=\sqrt{R} for all n.

    c) If x_0<\sqrt{R} then x_1>\sqrt{R}>x_0 and then x_1>x_2>\cdots x_n >\sqrt{R} by the same argument of a). The key for obtaining x_1>\sqrt{R} is to observe that
    f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right) is decreasing in (0,\sqrt{R}] and f(\sqrt{R})=\sqrt{R}.

    I mean that f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right), x\in \mathbb{R}^+ has a unique attracting point , the minimum, and that any iteration goes to this point. The case b)
    is an exception to your statement since it involves strict inequalities. c) is NOT an exception since you have to show x_n>x_{n+1}>\sqrt{R} for n\geq 1,

    If it is still not clear I will try to write any step detailedly, but I don't have now the necessary time.
    Last edited by Enrique2; August 25th 2009 at 01:47 PM. Reason: Adding and correcting some details
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strong induction vs. structural induction?
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: April 21st 2011, 12:36 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. induction help
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: April 19th 2010, 05:39 AM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  5. Induction!
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 7th 2008, 04:10 PM

Search Tags


/mathhelpforum @mathhelpforum