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585:. The claim that it converges everywhere is dubious. You might want to do some numerical exploration. I see two issues: binomial coefficients get huge before they get small, so you risk having a sum that has immense, giant terms in the middle that have to delicately cancel each other out. The other issue is that you have an infinite number of logarithms in sum. Each logarithm has a cut. You can put the cut wherever you want, but managing them all is a trick. Understanding the
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divergent ranges, but singular points which are singular, and so the function converges to the true value of the function in an infinite radius of convergence. The function is thus fully described in it's entirety, it is an or has an entire description. How do? Same man who added entire gamma formula to bottom of this page
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The nature of the entire description means the negative axis is correct, and the Euler graph is wrong. This is to do with the termination of the Euler definition of factorial ending as a count down to 1, and not continuing to minus infinity. So this ENTIRE gamma is based on an infinite product of the
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Well my markup foo is lacking, but ..... The series always locally converges. At a series of small points arranged around a singularity pole, the radius converges upto the singularity, so the singularities are surrounded by convergence from all sides. Thye singular points of divergence are then not
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that (you claim) converges everywhere on the complex plane (except at the poles, of course). But that does not mean that the gamma function is "entire" -- its the same as it always was. The word "entire" does not mean what you think it means. It means
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The Euler gamma makes a mistake of including an absolute magnitude operation in the algorithm by carrying out an inverse plus one iteration on negative parameter values. This prevents entirity, and also using the Risch algorithm on the Euler gamma.
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I think this proves Gamma is entire. It requires the use of L'Hopitals rule when in the limit, and using the differential in the second argument need integrating. And stuff the whole lot into the 0th polygamma definition.
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Why do differential equations need like the Bessel, or Bessel-Clifford need alternate solution functions? The negative gamma values? Another note on the multiplicative inverse of entire functions. Think about it!
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The series, and other formulas, should include a discussion of the regions of absolute and conditional convergence. I slapped an expert tag on the section introducing this idea.
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where (q-1) is a natural number. It reduces to the simpler form on writing a=1/q (and recalling that Zeta(s,1) is the
Riemann zeta function).
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How does it generalize PolyGamma if the identity in fact holds only for integers? Usually PolyGamma is generalized in a different way.--
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is tricky. You should compare your formula to the very first formula given in the section called "the log-gamma function" here:
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There is a useful generalization of the multiplication theorem for the
Hurwitz zeta function. In Maple notation:
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pdf bottom of page, final page has formula. just need the constant calculating to fit positives.
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Sorry - I made an error. The lower bound of the sum (over p) should be zero, not unity! So -
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The section or sections that need attention may be noted in a message below.
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https://sites.google.com/site/thebohrmagneton/the-entire-gamma-function
407:{\displaystyle \sum _{p=0}^{q-1}\zeta (s,a+p/q)=q^{s}\,\zeta (s,qa)}
281:{\displaystyle \sum _{p=1}^{q-1}\zeta (s,a+p/q)=q^{s}\,\zeta (s,qa)}
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This topic is in need of attention from an expert on the subject
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The
Hurwitz zeta function generalizes the polygamma function:
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So put it into the article. The above equation in LaTeX is
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Gamma = Const Exp Log, {k, 0, n}], {n, 0, Infinity}]]
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You have discovered an identity for the (log of the)
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Assuming the binomial n!/((n-k)!k!) and z in
Complex
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