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How to get number power in J2ME

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Implementation of the mathematical function J2ME (double, double)

I am developing a simple j2me application. There I need to get the power of the number, as in pow(double num1, double num2)in java. But, as I found out, j2me does not support this pow () method. Any useful option is welcome.

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public double pow(double num1, double num2) {
  double result = 1;
  for (int i = 0; i < num2; i++)
    result *= num1;
  return result;
}
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When developing applications for mobile devices using Java, you may need mathematical methods that are not available on your specific Java virtual machine. You can use this code. This will help 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;
    }
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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);
}

Java ME Math.pow()

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