4 ms·

Probably not. gamma(N) is the single mostly-analytic function for which gamma(x+1) = x*gamma(x). This is a requirement.

• deleted 11y ago
  [deleted]
• tanderson92 11y ago
  > single mostly-analytic function I think by "mostly-analytic" here you mean meromorphic. One really cool thing about the gamma function is it is not the only meromorphic function which satisfies that recurrence relation on the integers! E.g.: gamma(x) + sin(x*pi) is meromorphic and agrees with the factorial.
  • Chinjut 11y ago
    gamma(x) + sin(x * pi) agrees with the factorial [shifted by one, because the gamma function has that stupid shift in its definition for no good reason] on integer arguments, but does not satisfy the recurrence relation on non-integer arguments. That having been said, one can devise infinitely many analytic functions which do satisfy the recurrence relation in general. See my other comment.
    • tanderson92 11y ago
      Yes indeed, that's why I only said they agreed "on the integers". Your other example is deviously clever, nicely done.
• Chinjut 11y ago
  There are lots of analytic functions with this property (and agreeing with the ordinary gamma function on integer arguments). For example, gamma(x) * cos(2πx) [or, in complete generality, gamma(x) * f(x) for any analytic function f of period 1, normalized to take value 1 at integers; thus, there are infinitely many examples]. Rather, the gamma function is the unique function satisfying this recurrence, taking the usual value at integers, with the asymptotics that gamma(n + x)/(gamma(n) * n^x) approaches 1 as x is held fixed and n grows large. [See http://www.quora.com/How-exactly-does-the-gamma-function-extend-the-factorial-function-to-the-reals/answer/Sridhar-Ramesh http://www.quora.com/How-exactly-does-the-gamma-function-ext... for a detailed exposition]