root of the function $f(x)=\sqrt{2}-x$ by using fixed point iteration How can I find the approximate value of $\sqrt{2}$ by using the fixed point iteration? I have tried
$x-\sqrt{2}=0$,
$x^2=2$,
$x^2-2+x=x$,
$g(x)=x^2-2+x$,
$g\prime(x)=2x+1$
And i choose $x_0=-\frac{1}{2}$. 
But i cant find the approximate value correctly. Where am i wrong?

In general, you can produce a function $g(x)$ with $\sqrt2$ as a fixed point by letting
$$g(x)=x+(x^2-2)h(x)$$
with pretty much any function $h(x)$.  However, as André Nicolas pointed out, if you want $\sqrt2$ to be an attracting fixed point for $g$, which is what you need if you want to approximate $\sqrt2$ by iterating the function $g$, then you need $|g'(\sqrt2)|\lt1$.  Moreover, as André pointed out, you're really best off if $g'(\sqrt2)=0$.  Since
$$g'(x)=1+2xh(x)+(x^2-2)h'(x)$$
we have
$$g'(\sqrt2)=1+2\sqrt2h(\sqrt2)$$
It's convenient at this point to take $h$ to have the form $h(x)=cxk(x^2)$, because then we have
$$g'(\sqrt2)=1+4ck(2)$$
in which case we can get $g'(\sqrt2)=0$ by letting $c=-1/(4k(2))$ and all we need is to choose a function $k(x)$ such that $k(2)\not=0$.  The simplest such function is $k(x)\equiv1$, which gives $c=-1/4$ and thus
$$g(x)=x-{1\over4}x(x^2-2)={6x-x^3\over4}$$
Alternatively, $k(x)=1/x$ leads to $h(x)=-1/(2x)$ which gives
$$g(x)=x-{1\over2x}(x^2-2)={x\over2}+{1\over x}$$
as in André's answer.  But as I said, you can really let $k$ be anything you like (as long as $k(2)\not=0$) and get a function $g$ that has $\sqrt2$ as an attracting fixed point.
Finally, as André also pointed out, whatever $g$ you use, you need to start close enough to $\sqrt2$ so that you're within the "basin of attraction" of the fixed point.  I'll leave that for someone else to discuss in detail, but merely note that $x_0=1$ works as a starting point for either $(6x-x^3)/4$ or $x/2+1/x$ as the function to be iterated.
