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I Why are there no other gamma functions?

  1. May 5, 2017 #1
    hi, i was thinking that every function that satisfies the conditions
    $$f(0)=1$$
    $$f(n+1)=(n+1)f(n)$$
    could be a generalization of the factorial function, and why the gamma function is the only function that complies with this conditions?

    I mean why dont exist other functions, or functions based of integrals that also complies with the 2 requirements?

    Even I can obviate the first property and we get a "dephased" factorial function (like gamma function) so every function that complies with the property $$f(n+1)=(n+1)f(n)$$ could be an extension of the factorial, why the UNIQUE function that make that is the gamma?

    Thanks
     
  2. jcsd
  3. May 5, 2017 #2

    jedishrfu

    Staff: Mentor

  4. May 5, 2017 #3

    jedishrfu

    Staff: Mentor

  5. May 6, 2017 #4

    FactChecker

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    It's really not the only function, but it is the only analytic function. If you require that the function be analytic (a very good property to have and very restrictive) and that f(0)=1 and f(x+1)=x⋅f(x) for all x>0, then it will be unique.
    Two analytic functions that agree that much will be identical (see https://www.encyclopediaofmath.org/index.php/Uniqueness_properties_of_analytic_functions). ##\Gamma##(z) is analytic except for simple poles at the negative integers. (In fact 1/##\Gamma##(z) is an entire function. So it is the unique entire function with the correct values on the real line.)
     
  6. May 6, 2017 #5
    thanks, I Look at the links that you give me.
    @FactChecker
    why is so important that a function is analytic and why is so restrictive?.
    i heard that one can extend the domain of a function (like riemann zeta) in one unique way and maintaining the function analytic property.

    if i correct, if a function complies with the cauchy-riemann equations in a subset of ##\mathbb{C}## then is analytic in that region,
    but what consequences it implies that the function complies with this equations?

    its has a laurent representation for every ##z_0## in that subset?
    its Infinitely differentiable?
    why this is so important?
     
  7. May 6, 2017 #6
    i have to view the proof , but the link to the hadamard function , is not just functions based on gamma?
    like if we define the function
    $$P(x)=\Gamma (x)\, sin\left ( \frac{\pi }{2}+2\pi x \right )$$
    then is a valid extension for factorial but is obvius trivial, why the hadamard function is different?
     
  8. May 6, 2017 #7

    FactChecker

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    Those are good questions that deserve a lot of thought. Analytic functions have so many profound properties. Being infinitely differentiable implies that the function is completely smooth to an infinite level. The existence of a convergent Taylor series means that we have at least one known way to represent and approximate the function. And representing it with powers of z, the simplest thing, is such a natural extension of polynomials. The derivatives are also easy to determine from the Taylor series. The values of an analytic function in a disk are completely determined by its values on the circumference. The line integrals of analytic functions have special properties. For instance, a closed line integral of an function that is analytic in a simply connected region containing the line is always zero. The list of special properties of analytic functions goes on and on. You may be interested in looking at some complex analysis books, if only to become aware of what types of things they address.
     
  9. May 18, 2017 #8
    The Gamma function is not the only analytic function with those properties; you need to add the condition that the positive real part of the function is log convex, i.e. that log f(x) is a convex function. Note that two analytic functions that satisfy those two properties will not have to agree that much, only on the nonnegative integers is agreement required.
     
  10. May 19, 2017 #9

    FactChecker

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    Oh, thanks. You are correct (I think). I was thinking that f(0)=1 and f(x+1)=x⋅f(x) for x>0 was stronger than it is. But it only determines the positive integers. Having to use convexity seems to make the proof of uniqueness much more difficult. Sorry.
     
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