Floating exponent exponentiation algorithm

Question

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!

Answer 1

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.

Answer 2

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(abs(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;
  }
}