Results 1 to 8 of 8

Math Help - Induction on n square roots

  1. #1
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1

    Induction on n square roots

    Hey,

    I posted this before and didn't get any replies, so I'm posting again in case anyone can help. I've going round the houses with this problem and getting nowhere . The problem is as follows

    Consider the equation

    x=\sqrt{x+\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}

    where there are n square roots. Show that the real roots of this expression are independent of n and find them.

    The real roots are x=3 and x=0. My proof for the fact that these are independent of n goes:

    If we define F_1=x^2 and F_n=\left(\frac{F_{n-1}-x}{2}\right)^2 then we show by induction that F_n=3x has real roots x=3 and x=0 for all n\geqslant 1,\,n\in\mathbb{N}.

    Clearly $F_1=3x\Rightarrow x^2=3x\Rightarrow x(x-3)=0$ has these real roots, so assuming this holds for n-1 the case for n is

    F_n=3x\Rightarrow \left(\frac{F_{n-1}-x}{2}\right)^2=3x\Rightarrow \left(\frac{3x-x}{2}\right)^2=3x because F_n=3x for all n. This reduces again down to x(x-3)=0, so the real roots are independent of n.

    My problem is that in all cases, F_{n-1} = x(x-3)p(x) for some polynomial p which has no real roots, and I don't think my proof as it stands is valid.

    I'm desperate for some help with this problem it's been driving me mad, please help.

    Thanks.
    Last edited by hairymclairy; April 16th 2011 at 12:34 PM. Reason: for some reason latex isn't working...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by hairymclairy View Post
    Hey,

    I posted this before and didn't get any replies, so I'm posting again in case anyone can help. I've going round the houses with this problem and getting nowhere . The problem is as follows

    Consider the equation

    x=\sqrt{x+\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}

    where there are n square roots. Show that the real roots of this expression are independent of n and find them.

    The real roots are x=3 and x=0. My proof for the fact that these are independent of n goes:

    If we define F_1=x^2 and F_n=\left(\frac{F_{n-1}-x}{2}\right)^2 then we show by induction that F_n=3x has real roots x=3 and x=0 for all n\geqslant 1,\,n\in\mathbb{N}.

    Clearly $F_1=3x\Rightarrow x^2=3x\Rightarrow x(x-3)=0$ has these real roots, so assuming this holds for n-1 the case for n is

    F_n=3x\Rightarrow \left(\frac{F_{n-1}-x}{2}\right)^2=3x\Rightarrow \left(\frac{3x-x}{2}\right)^2=3x because F_n=3x for all n. This reduces again down to x(x-3)=0, so the real roots are independent of n.

    My problem is that in all cases, F_{n-1} = x(x-3)p(x) for some polynomial p which has no real roots, and I don't think my proof as it stands is valid.

    I'm desperate for some help with this problem it's been driving me mad, please help.

    Thanks.

    It seems to be the LaTeX compiler doesn't work and then trying to work this kind

    of questions without seeing clearly what's going on is very hard.

    At any rate, it is beyond my capabilities.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by hairymclairy View Post
    Consider the equation

    x=\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}} . . . . . (I'm assuming there is a missing 2 there.)
    where there are n square roots. Show that the real roots of this expression are independent of n and find them.
    I think that the best way to attack this problem is by considering it as a dynamical system. For fixed x>0 define a function f(t) for t>0 by f(t) = \sqrt{x + 2t}. This function has a unique fixed point, at t_0 = 1 + \sqrt{1+x}, and its derivative satisfies 0 < f'(t_0) < 1. It follows that the fixed point t_0 is attractive, and |f(t) - t_0| < |t - t_0| for all t. This means that f(t) cannot have any periodic points other than the fixed point.

    Therefore t_0 is the only point such that f^{(n)}(t) = t (where f^{(n)} means f composed with itself n times). This applies in particular if t_0 = x, in which case an easy calculation shows that x=3.

    Since x has to be non-negative in order for the square roots to exist, the only possibility left open by the above analysis is when x=0, where again we get a fixed point.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by hairymclairy View Post
    Hey,

    I posted this before and didn't get any replies, so I'm posting again in case anyone can help. I've going round the houses with this problem and getting nowhere . The problem is as follows

    Consider the equation

    x=\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}

    where there are n square roots. Show that the real roots of this expression are independent of n and find them.

    The real roots are x=3 and x=0. My proof for the fact that these are independent of n goes:

    If we define F_1=x^2 and F_n=\left(\frac{F_{n-1}-x}{2}\right)^2 then we show by induction that F_n=3x has real roots x=3 and x=0 for all n\geqslant 1,\,n\in\mathbb{N}.

    Clearly $F_1=3x\Rightarrow x^2=3x\Rightarrow x(x-3)=0$ has these real roots, so assuming this holds for n-1 the case for n is

    F_n=3x\Rightarrow \left(\frac{F_{n-1}-x}{2}\right)^2=3x\Rightarrow \left(\frac{3x-x}{2}\right)^2=3x because F_n=3x for all n. This reduces again down to x(x-3)=0, so the real roots are independent of n.

    My problem is that in all cases, F_{n-1} = x(x-3)p(x) for some polynomial p which has no real roots, and I don't think my proof as it stands is valid.

    I'm desperate for some help with this problem it's been driving me mad, please help.

    Thanks.
    You can use an induction logic.

    P(k) has "k" square roots and P(k+1) has "k+1" square roots.

    Then if you square your P(k+1) proposition equation, you get

    x^2=x+2P(k)

    x^2=x+2x=3x

    x^2-3x=0

    x(x-3)=0

    (you are missing the first "2" in your expression inside the external square root).

    The base case is x=\sqrt{x+2x}

    Here is a portable document to compensate for the lack of Latex.
    Attached Files Attached Files
    Last edited by Archie Meade; April 18th 2011 at 04:14 AM. Reason: added a file
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1
    Quote Originally Posted by Opalg View Post
    ...This function has a unique fixed point, at t_0 = 1 + \sqrt{1+x}, and its derivative satisfies 0 < f'(t_0) < 1....
    Thank you so much, I hadn't even considered it from that angle. I don't follow how you find t_0. Setting f(t)=t, I can't seem to derive the same point, please could you clarify this?

    Thanks again.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1
    Sorry I've worked that out, I'm just struggling with the expression |f(t)-t_0|<|t-t_0|. how is this justified?

    Thanks
    Last edited by hairymclairy; April 18th 2011 at 02:43 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by hairymclairy View Post
    I'm just struggling with the expression |f(t)-t_0|<|t-t_0|. how is this justified?
    With TeX not available at present, I'm not inclined to attempt an analytic explanation for that. But it's easy to see geometrically. Draw the graphs of y = f(t) = \sqrt{x+2t} (for some notional value of x) and y=t. You will see that the parabola is above the line when t<t_0, and below the line when t>t_0. That means that f(t) is always closer to t_0 than t is.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1
    Sure, I understand why you'd feel inclined not to do that. Does anyone know when it will be fixed? I can see geometrically why it is true, but could really do with a nudge in the right direction to show it analytically. Thanks again.
    Last edited by hairymclairy; April 18th 2011 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square roots help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 15th 2009, 03:10 PM
  2. Sum of two square roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 2nd 2009, 04:16 AM
  3. square roots
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 19th 2008, 09:15 AM
  4. square roots and i
    Posted in the Math Topics Forum
    Replies: 12
    Last Post: December 7th 2006, 01:25 PM
  5. Square Roots
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: February 28th 2006, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum