square-root

I am using Anthony Williams' fixed point library described in the Dr Dobb's article " Optimizing Math-Intensive Applications with Fixed-Point Arithmetic " to calculate the distance between two geographical points using the Rhumb Line method . This works well enough when the distance between the points is significant (greater than a few kilometers), but is very poor at smaller distances. The worst case being when the two points are equal or near equal, the result is a distance of 194 meters, while I need precision of at least 1 metre at distances >= 1 metre. By comparison with a double

I've heard of the "fast inverse square root", discussed here , and I wanted to put it in my Java program (just for research purposes, so ignore anything about the native libraries being faster). I was looking at the code, and the C code directly converts the float into an int with some C pointer magic. If you try to do this in Java with casts, it doesn't work: java truncates the float (as you would expect), and you can't get the pointer of a primitive (as you can in C). So how do you do this? Riking Remember to benchmark your code before using this. If it turns out you don't need it, or it's

This question already has an answer here: C program to convert Fahrenheit to Celsius always prints zero 8 answers I have this little program in C that calculates the square root x of a positive integer N using a recursive function (implemented using a while loop). If I calculate x using this: x = (1/2)*(x + N/x) //x0 = 1.0 Then x keeps growing to inf and then nan. However if I use this: x = (x + N/x)/2 //x0 = 1.0 It works fine, why? Thanks. 1/2 does integer division, its result is 0 , change either or both operand to double , e.g: 1.0/2 来源： https://stackoverflow.com/questions/26643536/trouble

Is there a library that will find the square root of a BigInteger? I want it computed offline - only once, and not inside any loop. So even computationally expensive solution is okay. I don't want to find some algorithm and implement. A readily available solution will be perfect. Just for fun: public static BigInteger sqrt(BigInteger x) { BigInteger div = BigInteger.ZERO.setBit(x.bitLength()/2); BigInteger div2 = div; // Loop until we hit the same value twice in a row, or wind // up alternating. for(;;) { BigInteger y = div.add(x.divide(div)).shiftRight(1); if (y.equals(div) || y.equals(div2))

How is the square root function implemented? Srujan Kumar Gulla Source here (Babylonian method) . Problem statement: Given x>0, find y such that y^2=x => y=x/y (this is the key step). Guess some value g for y and test it. Compute x / g. If x / g is close enough to g, return g. Otherwise, try a better guess. double test(double x, double g) { if closeEnough(x/g, g) return g; else return test(x, betterGuess(x, g)); } boolean closeEnough(double a, double b) { return (Math.abs(a - b) < .001); } double betterGuess(double x, double g) { return ((g + x/g) / 2); } sqrt(2) | Guess g x / g | New guess,