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 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.
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.
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.
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.
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.