Problem related to fixed point iteration How can I use fixed point iteration to solve $x^2 = 3$ using $g(x) = x^2 + x - 3$ to find the numerical value of the solution $x = +\sqrt{3}$. What happens? Then I use $g(x) = (x + 3/x)/2$. For which values of $x_0$ is this guaranteed to converge to the solution $x = +\sqrt{3}$?
 A: First question: . Let $g(x)=x^2+x-3$, and consider solving $g(x)=x$ by Fixed Point Iteration. 
If we are lucky enough to choose as our first "guess" $x_0$ the value $x_0=\sqrt{3}$, there will be trivial convergence.  There are other values of $x_0$ such that the $n$-th iterate $x_n$ just happens to be exactly $\sqrt{3}$ for some $n$. But we show that there cannot be approach to $\sqrt{3}$.
Note that $\sqrt{3}$ is between $1$ and $2$. Suppose that the $n$-th iterate $x_n$ is between $1$ and $2$, and not exactly equal to $\sqrt{3}$. By the Mean Value Theorem, 
$$\frac{|x_{n+1}-\sqrt{3}|}{|x_n-\sqrt{3}|}= \frac{|g(x_n)-\sqrt{3}|}{|x_n-\sqrt{3}|}=|f'(c)|$$
for some $c$ between $1$ and $2$. Since $g'(x)=2x+1$, we have $|g'(c)|\gt 3$ in the interval $(1,2)$.  
This says that $x_{n+1}$ is at least $3$ times further from $\sqrt{3}$ than $x_n$ is. 
Remark: The point $x=\sqrt{3}$ is a fixed point of $g(x)=x$, but because the absolute value of $g'(x)$ is "big" near $x=\sqrt{3}$, the point $x=\sqrt{3}$ is a repelling fixed point. 
Second question: We will be brief, leaving a number of details to you. Obviously a negative $x_0$ is no good, for then $x_n$ is always negative.  Let
$$g(x)=\frac{x+\frac{3}{x}}{2}.$$
Then 
$$g'(x)=-\frac{3}{2x^2}.$$
If $x\gt \sqrt{3}$, then $|g'(x)|\lt \frac{1}{2}$.
It follows by a Mean Value Theorem argument that if $x_0\gt \sqrt{3}$, then the fixed point iteration will converge to $\sqrt{3}$. 
But actually any positive number $x_0$ will do the job. For if $0\lt x_0\le \sqrt{3}$, then $x_1\gt \sqrt{3}$, so we get convergence. 
The rate of convergence is fairly quick: when we are near $\sqrt{3}$, the error gets multiplied by a factor $\lt 1/2$ with each iteration. 
