# Thread: Convergence of fixed point iteration question

1. ## Convergence of fixed point iteration question

Let p>0 and $\displaystyle x = \sqrt{p+\sqrt{p+\sqrt{p+ \cdots }}}$ , where all the square roots are positive. Design a fixed point iteration $\displaystyle x_{n+1} = F (x_{n})$ with some F which has x as a fixed point. We prove that the fixed point iteration converges for all choices of initial guesses greater than -p+1/4.

Let $\displaystyle x_{n+1}=F(x_{n})= \sqrt{p+x_{n}}$ so x is a fixed point for F since F(x)=x.
Now let $\displaystyle g(x)=\sqrt{p+x}$. We have $\displaystyle g'(x)=\frac{1}{2 \sqrt{p+x}}$ .

I can see that for $\displaystyle x > -p + 1/4$, we have that g'(x) <1.

From there I am not sure how to proceed to obtain convergence for $\displaystyle x_{0} > -p +\frac{1}{4}$ .

2. Such a problem recently has been 'attacked' in...

http://www.mathhelpforum.com/math-he...nvergence.html

Here the sequence is defined as...

$\displaystyle x_{n+1} = \sqrt {p+x_{n}} \rightarrow \Delta_{n}= x_{n+1}-x_{n} = \sqrt{p + x_{n}} - x_{n} = f(x_{n})$ (1)

The fixed point is the $\displaystyle x_{0}$ for which is $\displaystyle f(x_{0})=0$ so that is...

$\displaystyle x_{0} = \frac{1 + \sqrt{1 + 4 p}}{2}$ (2)

... and that means that it must be $\displaystyle p > - \frac{1}{4}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$