Accord.Math.Normal.Inverse C# (CSharp) Method

        public static double Inverse(double y0)
        {
            if (y0 <= 0.0)
            {
                if (y0 == 0)
                    return Double.NegativeInfinity;
                throw new ArgumentOutOfRangeException("y0");
            }

            if (y0 >= 1.0)
            {
                if (y0 == 1)
                    return Double.PositiveInfinity;
                throw new ArgumentOutOfRangeException("y0");
            }


            double s2pi = Math.Sqrt(2.0 * Math.PI);
            int code = 1;
            double y = y0;
            double x;



            if (y > 0.8646647167633873)
            {
                y = 1.0 - y;
                code = 0;
            }

            if (y > 0.1353352832366127)
            {
                y -= 0.5;
                double y2 = y * y;
                x = y + y * ((y2 * Special.Polevl(y2, inverse_P0, 4)) / Special.P1evl(y2, inverse_Q0, 8));
                x *= s2pi;
                return x;
            }

            x = Math.Sqrt(-2.0 * Math.Log(y));
            double x0 = x - Math.Log(x) / x;
            double z = 1.0 / x;
            double x1;

            if (x < 8.0)
            {
                x1 = (z * Special.Polevl(z, inverse_P1, 8)) / Special.P1evl(z, inverse_Q1, 8);
            }
            else
            {
                x1 = (z * Special.Polevl(z, inverse_P2, 8)) / Special.P1evl(z, inverse_Q2, 8);
            }

            x = x0 - x1;

            if (code != 0)
                x = -x;

            return x;
        }

        /// <summary>
        ///   Inverse complemented error function (<see cref="Erfc(double)"/>.
        /// </summary>
        ///
        public static double Ierfc(double y)
        {
            double s = Normal.Inverse(-0.5 * y + 1);
            double r = s * Math.Sqrt(2) / 2.0;

            return(r);
        }