Question

I am just playing around with approximating functions, and was curious if anyone can find a better or different way to approximate sqrt(x). Hopefully we'll find something interesting or clever.

Answer 1

Here's my code written in Java, but any language is fine.
```
public static double sqrtApprox(int n) {
double i = 1;
while (i * i < n)
i++;
i--;
return .5 * (n + i * i) / i;
}
```
It's based on the idea that we can sort of use the limit definition, so it's better the closer your number is to a perfect square. \[\Large f'(a) = \frac{f(a+h)-f(a)}{h} \implies f(a)+h f'(a) \approx f(a+h)\]

Answer 2

Well, it can always bounce around...

Answer 3

if the first estimate is close enough, you can iterate using Newton's method:
x1=x-f(x)/f'(x)
so
x1=x-(x^2-N)/(2x)
where N is number whose square-root is to be found.
Accuracy (in number of significant figures) doubles with every iteration (if it converges).

Answer 4

Using your proposed algorithm, you can improve accuracy by multiplying by 10^(2n), and dividing the result by 10^n.
However, in that case, you're better off using the bisection algorithm or else it will run for a long time.
Example:
sqrt(2)=sqrt(20000)/100=141/100=1.41