I must to write a algorithm that exponentiates a base (integer or float) in a integer or float argument. I wrote this algorithm for Deluge (zoho.com), but it can only use integer exponents:
float math.potencia(float base, int expoente) { if(expoente>0) { base = base * thisapp.math.potencia(base, (input.expoente - 1)); } else if (expoente == 0) { base = 1; } return base; } (Deluge doesn't have a potentiation operator or function). Thanks!
Suppose that you can use sqrt, you can use the following algorithm:
double EPS = 0.0001; double exponentiation(double base, double exp) { if(exp >= 1) { double temp = exponentiation(base, exp / 2); return temp * temp; } else { double low = 0; double high = 1.0; double sqr = sqrt(base); double acc = sqr; double mid = high / 2; while(fabs(mid - exp) > EPS) { sqr = sqrt(sqr); if (mid <= exp) { low = mid; acc *= sqr; } else { high = mid; acc *= (1 / sqr); } mid = (low + high) / 2; } return acc; } } Well, more than 17 hours without a reply, finally I've found an answer to my own question:
In the simplest way, we can solve the problem using the value of "e" exponentiating to the logarithm of the number divided by the index:
e^(Log(number)/index)
where number is the radicand and index is the desired root.
Eg: The 10th root of the number 1024: e^(Log(1024)/10) = 2.
PS: the base of the Log function is also "e". the rounded value for "e" is: 2.718281828459045
I hope this technique may be usefull for you.
