// The point of this method is to compute // e G_{e}(x) = \frac{1}{\Gamma(x)} - \frac{1}{\Gamma(x + e)} // To do this, we will re-use machinery that we developed to accurately compute the Pochhammer symbol // (x)_e = \frac{\Gamma(x + e)}{\Gamma(x)} // To do this, we use the reduced log Pochhammer function L_{e}(x). // \ln((x)_e) = e L_{e}(x) // To see why we developed this function, see the Pochhamer code. The Lanczos apparatus allows us to compute // L_{e}(x) accurately, even in the small-e limit. To see how look at the Pochhammer code. // To connect G_{e}(x) to L_{e}(x), write // e G_{e}(x) = \frac{(x)_e - 1}{\Gamma(x + e)} // = \frac{\exp(\ln((x)_e)) - 1}{\Gamma(x + e)} // = \frac{\exp(e L_{e}(x)) - 1}{\Gamma(x + e)} // G_{e}(x) = \frac{E_{e}(L_{e}(x))}{\Gamma(x + e)} // where e E_{e}(x) = \exp(e x) - 1, which we also know how to compute accurately even in the small-e limit. // This deals with G_{e}(x) for positive x. But L_{e}(x) and \Gamma(x + e) still blow up for x or x + e // near a non-positive integer, and our Lanczos machinery for L_{e}(x) assumes positive x. To deal with // the left half-plane, use the reflection formula // \Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin(\pi z)} // on both Gamma functions in the definition of G_{e}(x) to get // e G_{e}(x) = \frac{\sin(\pi x)}{\pi} \Gamma(1 - x) - \frac{\sin(\pi x + \pi e)}{\pi} \Gamma(1 - x - e) // Use the angle addition formula on the second \sin and the definition of the Pochhammer symbol // to get all terms proportional to one Gamma function with a guaranteed positive argument. // \frac{e G_{e}(x)}{\Gamma(1 - x - e)} = // \frac{\sin(\pi x)}{\pi} \left[ (1 - x - e)_{e} - \cos(\pi e) \right] - \frac{\cos(\pi x) \sin(\pi e)}{\pi} // We need the RHS ~e to for small e. That's manifestly true for the second term because of the factor \sin(\pi e). // It's true for the second term because (1 - x - e)_{e} and \cos(\pi e) are both 1 + O(e), but to avoid cancelation // we need to make it manifest. Write // (y)_{e} = \exp(e L_{e}(y)) - 1 + 1 = e E_{e}(L_{e}(y)) + 1 // and // 1 - \cos(\pi e) = 2 \sin^2(\half \pi /e) // Now we can divide through by e. // \frac{G_{e}(x)}{\Gamma(y)} = // \sin(\pi x) \left[ \frac{E_{e}(L_{e}(y))}{\pi} + \frac{\sin^2(\half \pi e)}{\half \pi e} \right] - // \cos(\pi x) \frac{\sin(\pi e)}{\pi e} // and everything can be safely computed. // This is a different approach than the one in the Michel & Stoitsov paper. Their approach also used // the Lanczos evaluation of the Pochhammer symbol, but had some deficiencies. // For example, for e <~ 1.0E-15 and x near a negative integer, it gives totally wrong // answers, and the answers loose accuracy for even larger e. This is because the // computation relies on a ratio of h to Gamma, both of which blow up in this region. private static double NewG(double x, double e) { Debug.Assert(Math.Abs(e) <= 0.5); // It would be better to compute G outright from Lanczos, rather than via h. Can we do this? // Also, we should probably pick larger of 1 - x and 1 - x - e to use as argument of // Gamma function factor. double y = x + e; if ((x < 0.5) || (y < 0.5)) { double h = MoreMath.ReducedExpMinusOne(Lanczos.ReducedLogPochhammer(1.0 - y, e), e); if (e == 0.0) { double t = MoreMath.SinPi(x) * h / Math.PI - MoreMath.CosPi(x); return(AdvancedMath.Gamma(1.0 - y) * t); } else { double s = MoreMath.SinPi(e) / (Math.PI * e); double s2 = MoreMath.Sqr(MoreMath.SinPi(e / 2.0)) / (Math.PI * e / 2.0); double t = MoreMath.SinPi(x) * (h / Math.PI + s2) - MoreMath.CosPi(x) * s; return(AdvancedMath.Gamma(1.0 - y) * t); } } return(MoreMath.ReducedExpMinusOne(Lanczos.ReducedLogPochhammer(x, e), e) / AdvancedMath.Gamma(x + e)); }
/// <summary> /// Computes the Pochammer symbol (x)<sub>y</sub>. /// </summary> /// <param name="x">The first argument.</param> /// <param name="y">The second argument.</param> /// <returns>The value of (x)<sub>y</sub>.</returns> /// <remarks> /// <para>The Pochhammer symbol is defined as a ratio of Gamma functions.</para> /// <para>For positive integer y, this is equal to a rising factorial product.</para> /// <para>Note that while the Pochhammer symbol notation is very common, combinatorialists sometimes /// use the same notation for a falling factorial.</para> /// <para>If you need to compute a ratio of Gamma functions, in most cases you should compute it by calling this /// function instead of taking the quotient of two calls to the <see cref="Gamma(double)"/> function, /// because there is a large fraction of the parameter space where Gamma(x+y) and Gamma(x) overflow, /// but their quotient does not, and is correctly computed by this method.</para> /// <para>This function accepts both positive and negative values for both x and y.</para> /// </remarks> /// <seealso href="http://mathworld.wolfram.com/PochhammerSymbol.html"/> public static double Pochhammer(double x, double y) { // Handle simplest cases quickly if (y == 0.0) { return(1.0); } else if (y == 1.0) { return(x); } // Should we also handle small integer y explicitly? // The key case of x and y positive is pretty simple. Both the Lanczos and Stirling forms of \Gamma // admit forms that preserve accuracy in \Gamma(x+y) / \Gamma(x) - 1 as y -> 0. It turns // out, however, to be ridiculously complicated to handle all the corner cases that appear in // other regions of the x-y plane. double z = x + y; if (x < 0.25) { bool xNonPositiveInteger = (Math.Round(x) == x); bool zNonPositiveInteger = (z <= 0.0) && (Math.Round(z) == z); if (xNonPositiveInteger) { if (zNonPositiveInteger) { long m; double p; if (z >= x) { m = (long)(z - x); p = Pochhammer(-z + 1, m); } else { m = (long)(x - z); p = 1.0 / Pochhammer(-x + 1, m); } if (m % 2 != 0) { p = -p; } return(p); } else { return(0.0); } } if (y < 0.0) { // x negative, y negative; we can transform to make x and y positive return(1.0 / Pochhammer(1.0 - x, -y) * MoreMath.SinPi(x) / MoreMath.SinPi(z)); } else if (z <= 0.25) { // x negative, y positive, but not positive enough, so z still negative // we can transform to make x and y positive return(Pochhammer(1.0 - z, y) * MoreMath.SinPi(x) / MoreMath.SinPi(z)); } else { // x negative, y positive enough to make z positive return(Gamma(1.0 - x) * Gamma(z) * MoreMath.SinPi(x) / Math.PI); } } else { if (z >= 0.25) { // This is the standard case: x positive, y may be negative, but not negative enough to make z negative double P; if ((x > 16.0) && (z > 16.0)) { P = Stirling.ReducedLogPochhammer(x, y); } else { P = Lanczos.ReducedLogPochhammer(x, y); } return(Math.Exp(y * P)); } else { // if y is very negative, z will also be negative, so we can transform one gamma if (Math.Round(z) == z) { return(Double.PositiveInfinity); } else { return(Math.PI / (Gamma(x) * Gamma(1.0 - z) * MoreMath.SinPi(z))); } } } }