Monday Math 19: The Gamma Function (Part 2/?)

The Gamma Function Part 2: Weierstrass form and Stirling’s formulas
(Part 1 is here)

Let us now define a sequence of functions . We note first that has simple poles at 0, -1, -2, …, -n. Next,
We also look at the logarithm of and the derivatives thereof:


So is convex.

Now, let us consider the limit of this sequence, which can be shown to exist via complex analysis:

We thus see from the above that has poles at 0, -1, -2, -3,…; and the properties of above tell us:
And is convex. As we noted at the end of the previous part, the only logarithmically convex function , x>0, for which and is the gamma function. Thus

Next, I will remind readers of the Euler-Mascheroni constant, γ, which is defined by the limit:

giving γ=0.57721566…
We note:

Now, suppose we multiply each term in the product by , multiplying out front with the reciprocal terms:
Now, we take the limit as , and see that the limit of the bracketed terms in the first exponent is –γ, obtaining:
This in known as the Weierstrass form of the gamma function.

Also useful are Stirling’s approximations, which allow one to approximate for large x, and n! for large n:

A derivation of the second approximation using Laplace’s method can be found here.


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8 Responses to “Monday Math 19: The Gamma Function (Part 2/?)”

  1. Physics Friday 21 « Twisted One 151’s Weblog Says:

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  5. Monday Math 54 « Twisted One 151’s Weblog Says:

    […] Monday Math 54 By twistedone151 Let us consider an integral of the form , with parameter λ>0, and where f(x) has a single local minimum in the interval (a,b). The method of steepest decent, also known as saddle-point integration, is a useful method for approximating such an integral in the small λ limit. Namely, if the local minimum occurs at x=x0, then we can expand f(x) in the Taylor series about this point; as it is a local minimum, f‘(x0)=0 and f”(x0)≥0. Here, we assume that this second derivative is nonzero. Thus, our Taylor series is . Now, we note that as λ→0+, the term in the exponent becomes ever more negative, and the integrand approaches zero; the local minimum at x0 is the “slowest” to approach, and thus the function near that point comes to dominate the rest of the integral; and so and Now, we see that the last integral is part of a gaussian integral; the peak of our gaussian is in the region of integration, and as λ→0+, the gaussian’s width goes to zero as well, so that the tails become negligible in the limit, and . Now, recall that the integral of the gaussian over all real numbers is given by . Thus , and so . Let us perform an example: . This does not fit the form as it is, but can be transformed to do so. First, let us make the substitution . Then we have: . Second, we note that for positive u, . Thus, we have , and the integral above fits our form with and ; thus our approximation will be for n→∞. Now, , which is zero for u=1. , so , and . Thus, our first saddle-point approximation says and so which you may recognize as Stirling’s approximation (see here). […]

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  7. Sumire Saita Says:

    How to prove “Euler integral is convergence ?

  8. Monday Math 124 « Twisted One 151's Weblog Says:

    […] for which the approximation is valid? From our binomial distribution, we have and . Now, recall Stirling’s approximation for large n. Then and so Now, to approximate the and terms, we use . Now, we expand to […]

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