# Thread: Fixed Points

1. ## Fixed Points

Hi,

I am having trouble working out this. I have got this equation, x^2 -x - 0.5 = 0. Then the answer is 0.5√3, which is +/-0.366. How can you find that value?

Any help needed.

2. Originally Posted by mark090480
Hi,

I am having trouble working out this. I have got this equation, x^2 -x - 0.5 = 0. Then the answer is 0.5√3, which is +/-0.366. How can you find that value?

Any help needed.
Quadratic formula is fastest:

For $ax^2+bx+c=0$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\frac{1 \pm \sqrt{3}}{2} = \frac{1}{2} \pm \frac{\sqrt3}{2}$

3. What, exactly, is the question? You title this "fixed point" but then just give a quadratic equation. The quadratic equation can, of course, be solved using the quadratic formula as e^(i*pi) said.

4. The question is about finding the limit using the fixed point rule. You can do that using the quadratic equation, not sure if that is what I should do. The question is this:

x0 = xn+1 = xn^2 -1/2 (n= 0,1,2...)

Generated by this function f(x) = x^2 - 1/2 appears to converge to a limit between -0.5 and 0. Assuming that this convergence does occur, find the value of the limit.

This question came after learning the fixed point rule, which I don't understand. If you have a link or knowledge about that; that would be useful.

Thanks.