Convergence of fixed point iteration question

**math8** · April 19th 2010, 01:34 PM

Let p>0 and $x = \sqrt{p+\sqrt{p+\sqrt{p+ \cdots }}}$ , where all the square roots are positive. Design a fixed point iteration $x_{n+1} = F (x_{n})$ 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 $x_{n+1}=F(x_{n})= \sqrt{p+x_{n}}$ so x is a fixed point for F since F(x)=x.
Now let $g(x)=\sqrt{p+x}$ . We have $g'(x)=\frac{1}{2 \sqrt{p+x}}$ .

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

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

**chisigma** · April 20th 2010, 04:35 AM

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

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

Here the sequence is defined as...

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

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

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

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

Kind regards

$\chi$ $\sigma$

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