How to get the power of a number in J2ME [duplicate]

Answer 1

public static double pow(final double a, final double b) 
{

final int x = (int) (Double.doubleToLongBits(a) >> 32);

final int y = (int) (b * (x - 1072632447) + 1072632447);

return Double.longBitsToDouble(((long) y) << 32);
}

you can also find information about it

Creating a Java ME Math.pow() Method

Answer 2

When developing applications for mobile devices using Java, you may require mathematical methods not available on your particular Java VM. You can use this code. It will helps you.

public double pow(double x, double y)
    {
        return powTaylor(x,y);
    }

    public double powSqrt(double x, double y)
    {
        int den = 1024, num = (int)(y*den), iterations = 10;
        double n = Double.MAX_VALUE;

        while( n >= Double.MAX_VALUE && iterations > 1)
        {
            n = x;

            for( int i=1; i < num; i++ )n*=x;

            if( n >= Double.MAX_VALUE ) 
            {
                iterations--;
                den = (int)(den / 2);
                num = (int)(y*den);
            }
        }   

        for( int i = 0; i <iterations; i++ )n = Math.sqrt(n);

        return n;
    }

    public double powDecay(double x, double y)
    {
        int num, den = 1001, s = 0;
        double n = x, z = Double.MAX_VALUE;

        for( int i = 1; i < s; i++)n *= x;

        while( z >= Double.MAX_VALUE )
        {
            den -=1;
            num = (int)(y*den);
            s = (num/den)+1;

            z = x; 
            for( int i = 1; i < num; i++ )z *= x;
        }

        while( n > 0 )
        {
            double a = n;

            for( int i = 1; i < den; i++ )a *= n;

            if( (a-z) < .00001 || (z-a) > .00001 ) return n;

            n *= .9999;                          
        }

        return -1.0;
    }

    double powTaylor(double a, double b)
    {
        boolean gt1 = (Math.sqrt((a-1)*(a-1)) <= 1)? false:true; 
        int oc = -1,iter = 30;
        double p = a, x, x2, sumX, sumY;

        if( (b-Math.floor(b)) == 0 )
        {
            for( int i = 1; i < b; i++ )p *= a;
            return p;
        }

        x = (gt1)?(a /(a-1)):(a-1);
        sumX = (gt1)?(1/x):x;

        for( int i = 2; i < iter; i++ )
        {
            p = x;
            for( int j = 1; j < i; j++)p *= x;

            double xTemp = (gt1)?(1/(i*p)):(p/i);

            sumX = (gt1)?(sumX+xTemp):(sumX+(xTemp*oc));

            oc *= -1;
        }

        x2 = b * sumX;
        sumY = 1+x2;

        for( int i = 2; i <= iter; i++ )
        {
            p = x2;
            for( int j = 1; j < i; j++)p *= x2;

            int yTemp = 2;
            for( int j = i; j > 2; j-- )yTemp *= j;

            sumY += p/yTemp;
        }

        return sumY;
    }

Answer 3

public double pow(double num1, double num2) {
  double result = 1;
  for (int i = 0; i < num2; i++)
    result *= num1;
  return result;
}