Numerical method for finding the square-root. I found a picture of Evan O'Dorney's winning project that gained him first place in the Intel Science talent search. He proposed a numerical method to find the square root, that gained him $100,000 USD.
Below are some links of pictures of the poster displaying the method.


*

*Link 1 

*Link 2 

*Link 3
How does this numerical method work and what is the proof?
His method makes use of Moebius Transformation.

The iteration to find $\sqrt k$ is
$f(x) = \frac{d x+k}{x+d}$
where $d = \lfloor \sqrt k \rfloor$.
The iterations start with $x = d$.
If $x$ is a fixed point of this,
$x = \frac{d x+k}{x+d}$,
or $x(x+d) = dx + k$
or $x^2 = k$,
so any fixed point must be the square root.
Now wee see if the iteration increases or decreases.
If $y = \frac{d x+k}{x+d}$,
$$y - x = \frac{d x+k}{x+d} - x
= \frac{d x+k - x(x+d)}{x+d}
= \frac{k - x^2}{x+d}
$$
so if $x^2 < k$, $y > x$
and if $x^2 > k$, $y < x$.
Also,
proceeding like analyses of Newton's method,
$y^2-k = \frac{(d x+k)^2}{(x+d)^2} - k
= \frac{d^2 x^2 +2 d x k + k^2 - k(x+d)^2}{(x+d)^2}
= \frac{d^2 x^2 +2 d x k + k^2 - k(x^2 + 2dx +d^2)}{(x+d)^2}
= \frac{d^2 x^2 +2 d x k + k^2 - kx^2 - 2dkx -kd^2)}{(x+d)^2}
= \frac{d^2 x^2 + k^2 - kx^2 -kd^2)}{(x+d)^2}
= \frac{d^2 (x^2-k) + k^2 - kx^2)}{(x+d)^2}
= \frac{d^2 (x^2-k) - k(x^2-k))}{(x+d)^2}
= \frac{(d^2-k) (x^2-k)}{(x+d)^2}
= (x^2-k)\frac{d^2-k}{(x+d)^2}
$.
Since $d = \lfloor \sqrt k \rfloor$,
$d < \sqrt k < d+1$
or $d^2 < k < d^2 + 2d +1$
or $-2d - 1 < d^2 - k < 0$,
so $|d^2-k| < 2d+1$.
Using this,
$|y^2-k|
< |x^2-k|\frac{2d+1}{(x+d)^2}|
= |x^2-k|\frac{2d+1}{x^2+2dx+d^2}
$,
so $|y^2-k|< |x^2-k|$,
and each iteration gets closer to the square root.
Since the starting iterate is $d$,
all following iterates exceed $d$
so 
$|y^2-k|
< |x^2-k|\frac{2d+1}{(d+d)^2}|
< |x^2-k|\frac{2d+1}{4d^2}|
< |x^2-k|\frac{1+1/(2d)}{2d}|
\le 3|x^2-k|/4$
since $d \ge 1$.
This show that the iteration converges.
However, this does not show that it converges quadratically like Newton's,
only that it converges linearly.
