root functions?
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I was sitting in math class and while the instructor was going over logarithmic functions, I was thinking about how you would go about making a root function (square root, cube root, etc.)?
I''m sure there are some in the "math.h" file or "cmath.h" file (in C++) but has anyone created one on their own?
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This forum is endlessly full of people screaming about how to make a wicked awesome tubular crazy fast square root function. Go back a few days, I think there just was one a day or two ago.
//Universal root function for whole numbers; returns the rootdouble root(double exp,int base){//Avoid computing known rootsif(base==0)return 0;else if(base==1)return 1;else if(base<0)return -1;//Compute unknown rootsint temp=base; //Value of base; used in calculationfor(double i=0.0001;i<=base;i+=0.0001){temp=base; //Restore valueif(int(pow(i,exp))==base)if(int(pow(int(i),exp)==base)) //Return single precision roots appropriatelyreturn int(i);else //Return double-precision roots appropriatelyreturn i;elsecontinue; //Try another number}return -1; //Error}it''s been a while since i''ve worked on it and it probably can be optomized.
Returning NaN or throwing a std::domain_error would be much more appropriate than returning -1
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The classical method of calculating sqrt() is to use Newton's method. There is a good concise explanation here.
For calculating the square root we use the equation x^2 - n, with the derivative 2x in Newton's method.
The following code is a working sqrt() implemenation using Newton's method:
Edit: Correction in algorithm for n = 0, fixed another bug with df being very close to zero.
[edited by - premandrake on October 16, 2003 1:06:42 PM]
For calculating the square root we use the equation x^2 - n, with the derivative 2x in Newton's method.
The following code is a working sqrt() implemenation using Newton's method:
#include <iostream>#include <limits>using namespace std;double mysqrt(double n) { const double TOL = 1e-12; const double N = 100; if (n < 0) return std::numeric_limits<double>::quiet_NaN(); double guess = n / 2.0; for (int i=0; i<N; i++) { double f = guess * guess - n; double df = 2 * guess; if (fabs(f) < TOL) return guess; double tmp = guess - f/df; if (fabs(guess - tmp) < TOL) return guess; guess = tmp; } return std::numeric_limits<double>::quiet_NaN();}int main() { cout << mysqrt(0) << endl;}Edit: Correction in algorithm for n = 0, fixed another bug with df being very close to zero.
[edited by - premandrake on October 16, 2003 1:06:42 PM]
It works when you put it into a function of its own as any sane person plus me would do:
inline double root(double num, double exp)
{
return pow(num, 1/exp);
}
inline double root(double num, double exp)
{
return pow(num, 1/exp);
}
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