Complex analysis: Gamma function
ฝัง
- เผยแพร่เมื่อ 6 ก.ย. 2024
- This lecture is part of an online undergraduate course on complex analysis.
We describe the basic properties of the gamma function, including its functional equations and the duplication formula, and give a characterization of it in terms of its functional equation and growth rate.
Correction: On sheet 5 and 8 (at 19:20) I forgot to add the assumption that the functions are holomorphic except possibly at integers.
For the other lectures in the course see • Complex analysis
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Prof. Borcherds, I am really enjoying these videos. If there is time in the course I would love to see a discussion of Riemann surfaces.
Very nice video as always! I was expecting the famous theorem that says that Gamma is the only log convex function satisfying f(s+1) = s f(s). Does anyone know an application of this theorem?
(Don't forget f(1) = 1) That's sort of the analogue of 16:49 for real numbers. You can prove all the identities in this video without complex analysis by showing log convexity instead of boundedness in strips. The drawbacks are that the results are less general (you only prove the real case) and the proofs are harder. In Artin's 'The Gamma Function' there's a nice proof of the reflection formula using log convexity but it requires a diff EQ argument that I wouldn't think of in a million years.
In the Artin's literature Gamma function, Gamma is characterized similarly where the second condition is replaced with the log convexity. And the rest of arguments go with log convexity on the real line without using complex analysis. Is there any connection between growth rates of function on the imaginary direction and log convexity?
Professor, what is the name of the book you showed at the beginning of the video? With the description of the gamma function?
Hi Professor, when proving the characterization of Gamma function at 19:21, do we also need to require the function to have no poles for Re(s)>0? Otherwise why can't the gamma_0(s) have other poles in the region 0
Yes, I accidentally omitted this condition.
At 10:55, shouldn’t it be g(s+1) = -g(s) instead?
I think he misses a minus sign.
Hi Dr. Borcherds. When you say the poles and the growth rate are the key to understanding a meromorphic function this hints at something I've been wondering about. You can write sin(x) as a product of linear terms just by factorizing as though it were a polynomial, using the zeros. You can get a series for csc(x) just by add terms of the form 1/(x - b) to get poles where you need them. Is there a general rule for when you can express a meromorphic function in a simplistic way like this i.e. just using the zeros, poles, and normalization?
Look up Weierstrass factorization and Mittag-Leffler's theorem.
@@shapeoperator Thanks!
It depends on how fast the function grows. If a holomorphic function grows less than exponentially fast then it is determined (up to a constant) by its zeros, but the faster it grows the more ambiguity there is. For a meromorphic function something similar is true but it is trickier to define the growth rate.
@@richarde.borcherds7998 I see, thank you
What is the book that had the graph of the gamma function?
Mentioned in the previous video
Que libro es?
18:42 I feel like I'm missing something: How do we see that Gamma=Gamma_0 from this? The best I can see is that Gamma/Gamma_0 must satisfy a bunch of conditions that presumably force it to be =1. But I think some argument is needed?
For instance Gamma_0(s) = 2^(s-1/2) Gamma(s) seems to also satisfy Gamma_0(s)Gamma_0(1-s)=pi/sin(pi s) but it doesn't satisfy Gamma_0(1)=1. I think perhaps the best we can get from this functional equation is that Gamma_0(s) = a^(s-1/2) Gamma(s) (and then the normalization at s=1 gives us that Gamma_0=Gamma)..but I haven't tried to work out the details, in case I'm missing something simple!
Are you good with Γ_0(s) Γ(1-s) = π/sin(πs)? Because then it equals Γ(s) Γ(1-s) and you can divide both sides by Γ(1-s).
@@martinepstein9826 I think that is a typo in the video and should be Gamma_0(s) Gamma_0(1-s) = pi/sin(pi s), i.e., Gamma_0 satisfies the same functional equation that Gamma satisfies.
Actually upon rewatching it, he mentions that it’s ok to change the first Gamma to Gamma_0, so maybe it wasn’t a typo after all... hmm.
If you create f(s) = gamma(s)gamma0(1-s) where gamma0(s) has the conditions laid out there, you can see that f(s) still follows the three properties at 12:15 and so the identity works.
@@calvindang7291 Yes. It's sneaky! I like it.
i divide people into those who can convincingly sneak 'black magic' into a mathematics lecture; and the rest.
no one ever expects the change of variables trick. lol
yeeee