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Is the gamma function $\Gamma(n+1)$ the only continuous function and defined derivative with the same recursive definition as of $n!$ for $n>-1$ ? (When using real numbers.)

The recursive definition of $n!$ I'm referring to:

$0!=1$
$n!=n\cdot(n-1)!\quad$ for $\quad n\in\mathbb Z^+$

Clarification: let $y$ be a function with the restrictions:

$y(0)=1$
$\forall x\in\mathbb R^+:y(x)=xy(x-1)$

Find a continuous function $y(x)$ in the intervall $x>-1$ with a defined $y'(x)$ in the intervall $x>-1$. Is $y(x)$ always the function $\Gamma(x+1)$ ?



Notice: I edited $\Gamma(x-1)$ to $\Gamma(x+1)$.

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  • $\begingroup$ Welcome to Math.SE! Please add some context to your question for better understanding and better answers. Also, it helps to show what have you done (and how you did it) to add richness to the question. $\endgroup$ Commented Sep 25, 2014 at 16:20
  • $\begingroup$ Take any function defined on $[-1,0]$, equal to $0$ and with a null derivative at $-1$ and $0$. Use it as the basis of the recursion. These conditions can be relaxed. $\endgroup$ Commented Sep 25, 2014 at 16:28

2 Answers 2

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Unless the function is logarithmically convex, it won't be unique (the Gamma function). See the Bohr-Mollerup theorem.

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Also for a function/sequence $f(n) = a_n$, consider $f(n)\cos(2\pi\,n) = a_n$ for integer values of $n$.

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