Method pow

JSAT/src/jsat/math/FastMath.java:161–186

Computes a b . @param a the base @param b the power @return a b

(double a, double b)

     * @return a<sup>b</sup>
     */
    public static double pow(double a, double b)
    {

        /*
         * Wright out a^b as 2^(b log2(a)) and then replace a with 'm 2^e' to get
         * 2^(b * log2(m*2^e)) which simplifies to 
         * m^b 2^(b e) when m, e, and b are positive. 
         * 
         * m & e are by IEEE defintion positive 
         */

        if (b < 0)
            return 1 / pow(a, -b);//b is now made positive

        long rawBits_a = doubleToLongBits(a);
        long mantissa_a = getMantissa(rawBits_a);
        final int e_a = Math.getExponent(a);

        //compute m^b and exploit the fact that we know there is no need for the exponent
        double m = longBitsToDouble(1023L << 52 | mantissa_a);//m in [1, 2]
        
        final double log2m = 1.790711564253215 + 0.248597253161674 * m - 3.495545043418375 / (0.714309275671154 + 1.000000000000000 * m);

        //we end up with 2^(b * log_2(m)) * 2^(b * e), which we can reduce to a single pow2 call
        return pow2(b * log2m + b * e_a);//fun fact, double*int is faster than casting an int to a double...
    }
    
    private static final double expPowConst = 1.0/Math.log(2);