I looked through the site but still am unable to solve my problem. Please help!
$x_{k+1} = \frac{\frac{x_{k}+1}{x_{k}}}{2}$ assume $x_{0}$ > $0$ and that the iteration converges.
(a) what number does the iteration converge to? (I took the sequence to the limit as n approaches infinity and then substituted in n and got 1. However, this is a fixed point iteration and it converges to p where g(p) = p. 1 does satisfy this but did I solve for it correctly?
(b) what is the order of convergence? I know this also involves taking the limit to infinity of $\frac{abs(x_{k+1}) - L}{abs(x_{k} - L)^{q}} =C$: q will give the order of convergence and L is the limit (which comes from part (a) = 1) but I am not sure how to actually compute for q to get a value.
$2x_{k+1}x_k=1+x_k.$ If $L=\lim_{k\to \infty}$ exists then $L=\lim_{k\to \infty}x_{k+1}$ so $$2L^2=\lim_{k\to \infty}x_{k+1}x_k=\lim_{k\to \infty}(1+x_k)=1+L.$$ The only non-negative solution to $2L^2=1+L$ is $L=1.$
If $x_k>0$ then $x_{k+1}=(1+x_k)/(2x_k)>0$. And we have $x_0>0$. So by induction on $k$ we have $x_k>0$ for all $k.$ So $L\geq 0.$ Therefore $L=1.$
Let $x_k=1+d_k$.
(i). If $x_0=1$ for all $k$ we have $x_k=1$ and $d_k=0.$
(ii). $d_k\ne 0 \implies d_{k+1}=-d_k/(2+2d_k)\ne 0.$ So if $d_0\ne 0$ then $d_k\ne 0$ for all $k$ by induction on $k.$ And $d_k\to 0$.
So if $x_0\ne 1$ then $d_{k+1}/d_k=-1/(2+d_k)$ converges to $-1/2. $