Member Avatar for tubby123

Hey, i was trying to implement the sqrt function in c, and this was what i came across, and it works abs fine.
But i dont get this , is this a standard logic based on Taylor, Maclaurin or something ?

Is this more of a programming question or a Math formula(which has proof) and has just been put into code.

g2=n/2;
 do
 {
	g1=g2;
	g2=(g1+n/g1)/2;
 }
 while((g1-g2)!=0);

 printf("The root of %f is ",n);
 printf("%f",g2);

}
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Nice experiment — ’s loop shows the approach that identified. A few practical points will make it safe and useful beyond toy tests.

Don’t use equality like (g1-g2)!=0 as your stop condition; floating‑point iterates rarely hit exact equality. Use either a residual test (|xx - n| small) or a relative step test such as `|x_next - x| <= eps max(1, |x_next|). Also handle edge cases up front:n == 0,n < 0(choose NaN or complex), and an iteration cap to avoid infinite loops. For production, prefermath.h’ssqrt()`; implement your own only for learning, special precision needs, or embedded constraints.

A compact, robust pattern in C looks like this:

#include <math.h>
#include <float.h>

double newton_sqrt(double n) {
    if (n < 0.0) return NAN;
    if (n == 0.0) return 0.0;

    double x = (n > 1.0) ? n/2.0 : 1.0;
    const double tol = 2.0 * DBL_EPSILON;
    for (int i = 0; i < 50; ++i) {
        double next = 0.5 * (x + n / x);
        if (fabs(next - x) <= tol * fabs(next)) return next;
        x = next;
    }
    return x;
}

A few more notes: Newton has quadratic convergence near a simple root, so iterates gain digits very fast — typically 3–6 iterations for double precision from a decent guess. It can slow or fail when the derivative is zero or the root has multiplicity >1 (convergence drops to linear), or when an iterate lands where f' is tiny. Common defenses: bracket the root and use bisection until close, or use a hybrid (bisection → Newton) or secant method if computing derivatives is expensive. For natural logs, a far more common strategy is range reduction (n = m * 2^e, so ln n = e·ln2 + ln m) and then approximate ln m with a small polynomial — using Newton to solve exp(x)=n is workable but usually costlier because each step needs exp().

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Member Avatar for nezachem

It is a standard logic based on a Newton-Raphson method. Here is how it works:
We are trying to solve an equation
x^2 - n = 0
Given an iteration X0, the next iteration X1 is an x-intercept of a tangential at X0. In a general form the tangential is
(y - Y0)/(x - X0) = f'(X0)
In our case Y0 is X0^2 - n and f' is 2*X0, that is
(y - X0^2 + n)/(x - X0) = 2*X0
For x-intercept, y is 0:
(-X0^2 + n)/(x - X0) = 2 * X0
or
n - X0^2 = 2*x*X0 - 2*X0^2
or
x = (X0 + n/X0)/2
which is your formula.

commented: Very clear +11
Member Avatar for tubby123

hey nezachem thanks a lot for that.
I did find about the Newton-Raphsson method after posting, but none of those explanations were as precise and clear as urs. Thanks a lot for that.
Is there a method to find the log of a number like this ?
I mean, i want to be able to write code to find the log of a number.
On a lighter note , there is a reference to the Newton-Raphsson method in the movie 21

Member Avatar for nezachem

Yes and no. Of course there is a N-R formula for a logarithm, but it involves calculating an exponent at each iteration; that makes it way less attractive than a square root one.

A log n is a solution of exp(x) = n. Applying the same logic as above, you'd arrive to X1 = X0 + n/exp(X0) - 1 If you have a good exponentiator, you're all right. Otherwise, it may become tricky.

Member Avatar for tubby123

hey nezachem, thanks for that. Yes , i followed ur advice and was able to reach ur answer along the same lines.
I also tried for cube root of x and got the solution as
x=(n/x0^2 + 2x0)/3; :)
I tried for y=sinx and got the sol as
x=-tanx0 + x0

I dont get it,
1)does this method work for every function in the world ?
2)What are the cons of this method ?
Is it that, you have to keep on doing iteratively to get a better answer all the time ?
3)Can u probably give me a couple of examples where this doesnt work ?
I mean, i am able to find the answer to literally everything using this that, i really cant understand why you would need anoter method to approximate functions

Member Avatar for nezachem

N-R is a very powerful method. It is easy to implement, it converges fast, it doesn't even need a closed form: it is possible to work directly from the initial equation
X1 = X0 - Y0/f'(X0)
and approximate the derivative numerically.

The problem with N-R is that sometimes it doesn't converge at all. This chapter of a Wiki gives some good examples. A proof of convergence in a given case could be very hard and painful.

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