sqrt

I would like to look at the way Python does computes square roots, so I tried to find the definition for math.sqrt() , but I can't find it anywhere. I have looked in _math.c , mathmodule.c , and elsewhere. I know that python uses C's math functions, but are these somewhere in the Python distribution, or are they linked to code elsewhere? I am using Mac OS X. Where is the algorithm in math.sqrt() ? Alexei Sholik It depends on the implementation. CPython is using math functions from the standard C library. Jython is most likely using Java's math methods. And so on. In fact, Python has nothing to

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 was finding out the algorithm for finding out the square root without using sqrt function and then tried to put into programming. I end up with this working code in C++ #include <iostream> using namespace std; double SqrtNumber(double num) { double lower_bound=0; double upper_bound=num; double temp=0; /* ek edited this line */ int nCount = 50; while(nCount != 0) { temp=(lower_bound+upper_bound)/2; if(temp*temp==num) { return temp; } else if(temp*temp > num) { upper_bound = temp; } else { lower_bound = temp; } nCount--; } return temp; } int main() { double num; cout<<"Enter the number\n"; cin

This question already has an answer here: Why am I getting “undefined reference to sqrt” error even though I include math.h header? [duplicate] 6 answers I don't know if I'm missing something obvious, but it appears that I'm unable to compute square roots of a variable in C; the sqrt() function only seems to work on constants. This is my code: #include <math.h> #include <stdio.h> int main() { double a = 2.0; double b = sqrt(a); printf("%f", b); return 0; } When I run this program, I get the following error: gcc -Wall -o "test2" "test2.c" (in directory: /home/eddy/Code/euler) /tmp/ccVfxkNh.o: