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Is the a function $f(x)$ other then the Gamma function with said properties.

  1. $f(x)=(x-1)!$ when x is a non-negative integer.
  2. $f(x)$ is smooth (infinitely differentiable.)
  3. $f(x)$ is convex.
  4. $f(x)=xf(x-1)$. for x>1

I know that the Gamma function is the only solution if 3. is strengthened to being Logarithmically convex.

blademan9999
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  • A related question: https://math.stackexchange.com/questions/26680/the-uniqueness-of-the-gamma-function – Kavi Rama Murthy Jun 09 '20 at 07:27
  • I'm requiring the function to be Smooth. – blademan9999 Jun 09 '20 at 07:55
  • You may want to know that if you strengthen the conditions to $\log \circ f$ being convex and $f$ being analytic there is only one such function. See Bohr-Mallerup Theorem – Dunnò000 Jun 09 '20 at 14:24
  • Aren't you nailing the function you need pretty much to the Gamma function by setting those properties? And could you not differentiate the Gamma function (in the positive Real) by multiplying the digamma by the gamma? (https://www.youtube.com/watch?v=kjK9WfmLElo) - What do you plan on using the function you are looking for for? – bukwyrm Jun 10 '20 at 05:21
  • If I weaken 2 to simply be Continuous differentiable I can come up with other functions. $y=0.5((xmod(1)-0.5)^2+1.75)∏max(x-i,1)$ from i=0 to i=infinity – blademan9999 Jun 10 '20 at 05:26

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You might try Hadamard's gamma function? $$ H(x) = \frac{1}{\Gamma (1-x)}\,\dfrac{d}{dx} \left \{ \ln \left ( \frac{\Gamma ( \frac{1}{2}-\frac{x}{2})}{\Gamma (1-\frac{x}{2})}\right ) \right \} $$ This looks reasonably smooth and convex (not sure about the bump and have no proof of this). But.. it only satisfies your functional relationship for positive integers $x$ and is otherwise $$ H(x+1) = xH(x) + \frac{1}{\Gamma(1-x)} $$ all credit to Wikipedia.

Benedict W. J. Irwin
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