If the gamma function didn't have that pesky -1, the reflection formula would look like this: Γ(z)Γ(-z) = πz/sin(πz) You can easily see there is a reflection with the minus sign, and also the RHS looks like the reciprocal of sin(x)/x, an important function used in calculus to find trig derivatives.
@@fartoxedm5638 Γ(z) = (z-1)! with the usual definition (for positive integers, but let's extend it) The reflection formula is Γ(z)Γ(1-z) = π/sin(πz) Replace gammas with factorial: (z-1)!(-z)! = π/sin(πz) Multiply with z: (z)!(-z)! = πz/sin(πz) Change notation to make gamma match with factorial: Γ(z)Γ(-z) = πz/sin(πz)
@@f5673-t1h You have literally wrote the mapping from factorial world to the gamma world in your first line. You can't simply go with "Nah, let's take Г(x) = x!" In the end
This is great! I took a complex analysis class 41 years ago, but I was always weak on contour integration and applying the concept to real integrals (I was weak, or the course was weak). Inspired me to look at your complex analysis videos on MathMajor, and I'll probably go through those. Thank you for doing all of this! BTW, if you make or already have a video series on multi-variable calculus, I'll be reviewing that as well!
It seems like we're doing something shady with the contour integral. In particular, the replacement u -> exp(i2pi) u is baffling, since there should be no change. I figure Michael is leaving out some t -> 0+ from some of these definitions and for C3 the replacement is actually u -> exp(i2pi - i2t) or something of that nature.
12:00 We have to restrict the value of z much earlier: The integral definitions of the Gamma function which Michael uses right from the start are only valid for Re(z) > 0 and Re(z) < 1, respectively. 15:00 Here it's not regardless of what z is, this only works for Re(z) > -1.
I used contour integration to derive this identity as well, but started with this representation of the beta function, the integral of t^(a-1)/(1+t)^(a+b) from 0 to infinity for t.
I would be very interested in a discussion of convergence on this integral. Normally, it’s not something I care about, but because the integral defining the gamma function is only defined for z>0, meaning this integral should diverge for z outside (0,1), meaning you actually sneakily did some analytic continuation here.
The restriction is necessary to evaluate the product Γ(z)Γ(1-z), because the integral representations of both Γ(z) and Γ(1-z) need to simultaneously converge and this only happens in the "critical strip" 0 < Re(z) < 1. Once this expression is evaluated, it turns out to simplify to π/sin(πz), giving us a valid /equation/ that holds in the critical strip. But once the equation is proved, it may be re-interpreted as a /formula/ for computing values of Γ in places where the integral representation does not converge (i.e. thereby "getting rid of the restriction"). In fact, treating the equation as a formula is the /unique/ way to extend Γ to the rest of the complex plane while maintaining its nature as an analytic function.
I make a Proof of this identity using contour integral on my notes. Later I made a Proof of the Riemann and Hurwitz Zeta Functional equation using complex contour integral (which have an infinite number of poles...)
As for moi - whenever the presentation goes off at an extreme tangent covering some gross new things they seem to be eminently forgettable. But I reserve the right to be wrong on this :-) Basis of my conjecture: math is not frightening, math is eminently doable. Nonetheless - great video, great swooping intro to some gigantic new things (I feel like calling them monsters and that is okay)
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If the gamma function didn't have that pesky -1, the reflection formula would look like this: Γ(z)Γ(-z) = πz/sin(πz)
You can easily see there is a reflection with the minus sign, and also the RHS looks like the reciprocal of sin(x)/x, an important function used in calculus to find trig derivatives.
Actually, It would be -pi / (z * sin(pi * z)). The thing you are reffering to is Г(z + 1) * Г(1 - z)
@@fartoxedm5638 Γ(z) = (z-1)! with the usual definition (for positive integers, but let's extend it)
The reflection formula is Γ(z)Γ(1-z) = π/sin(πz)
Replace gammas with factorial: (z-1)!(-z)! = π/sin(πz)
Multiply with z: (z)!(-z)! = πz/sin(πz)
Change notation to make gamma match with factorial: Γ(z)Γ(-z) = πz/sin(πz)
@@f5673-t1h You have literally wrote the mapping from factorial world to the gamma world in your first line. You can't simply go with "Nah, let's take Г(x) = x!" In the end
@@fartoxedm5638 that's the point I'm trying to make when I said "if gamma didn't the -1"
Please read
@@f5673-t1h ah, got it. Sorry for misunderstanding
This is great! I took a complex analysis class 41 years ago, but I was always weak on contour integration and applying the concept to real integrals (I was weak, or the course was weak). Inspired me to look at your complex analysis videos on MathMajor, and I'll probably go through those. Thank you for doing all of this!
BTW, if you make or already have a video series on multi-variable calculus, I'll be reviewing that as well!
It seems like we're doing something shady with the contour integral. In particular, the replacement u -> exp(i2pi) u is baffling, since there should be no change. I figure Michael is leaving out some t -> 0+ from some of these definitions and for C3 the replacement is actually u -> exp(i2pi - i2t) or something of that nature.
12:00 We have to restrict the value of z much earlier: The integral definitions of the Gamma function which Michael uses right from the start are only valid for Re(z) > 0 and Re(z) < 1, respectively.
15:00 Here it's not regardless of what z is, this only works for Re(z) > -1.
I'm fairly certain the integral is well defined for all Re(z) > 0 why wouldn't it be for re(z)>1?
nvm I get what you're saying because of the z and the 1-z ignore me I'm an idiot 😂
@@brendanmiralles3415 No problem. I think it was my fault, I didn't explain very well what I meant.
20:02
I think, with a couple of appropriate hints, this derivation would make a very nice final exam question for a complex analysis class
Complex analysis is by far my favorite field of mathematics. So elegant and powerful!
I used contour integration to derive this identity as well, but started with this representation of the beta function, the integral of t^(a-1)/(1+t)^(a+b) from 0 to infinity for t.
book recommendation for complex analysis?
I would be very interested in a discussion of convergence on this integral.
Normally, it’s not something I care about, but because the integral defining the gamma function is only defined for z>0, meaning this integral should diverge for z outside (0,1), meaning you actually sneakily did some analytic continuation here.
That was gorgeous. props
This really makes me want to learn complex analysis! I just need to find the energy and make the time.
Very interesting, I often see this done using Euler or Weierstrass product
This is an excellent video, a ton of dirty details without getting bogged down in the algebra.
Eulers reflection formula is one of my favourite identities in math! Thank you for the video.
I can already guess that the integral is 1/(1+x^n) or its counterparts.
Edit: after integration by parts it's a simple substitution for my integral.
Been wanting to see this for a while! Got stuck midway and didn’t know how to proceed
14.32
He forgot to put "i" in front of the integral but that is not a problem because the integral goes to zero
Excellent video
How do you show that this formula is valid for Re(1+z) > 2?
Had a heart stroke at 6:56
This proof assumes Re(z+1)
We have that f(z) := Γ(z)Γ(1-z)sin(πz) satisfies f(z) = π on Re(z+1)
The restriction is necessary to evaluate the product Γ(z)Γ(1-z), because the integral representations of both Γ(z) and Γ(1-z) need to simultaneously converge and this only happens in the "critical strip" 0 < Re(z) < 1. Once this expression is evaluated, it turns out to simplify to π/sin(πz), giving us a valid /equation/ that holds in the critical strip.
But once the equation is proved, it may be re-interpreted as a /formula/ for computing values of Γ in places where the integral representation does not converge (i.e. thereby "getting rid of the restriction"). In fact, treating the equation as a formula is the /unique/ way to extend Γ to the rest of the complex plane while maintaining its nature as an analytic function.
Thanks!
Glorious!
Interesting, because this approaches 1/z as pi approaches 0.
رائع جدا كالعادة
What about Jackson integral?
you don't need to use a contour integral, just use the beta function
@16:23 how is this not dividing by 0?
e^(2pi*iz)=1^z=1 so 1-e^(2pi*iz)=0 and we are dividing by 0
isn't the branch cut supposed to be in the negative real axis?
I make a Proof of this identity using contour integral on my notes. Later I made a Proof of the Riemann and Hurwitz Zeta Functional equation using complex contour integral (which have an infinite number of poles...)
We had to assume that the real part of z is less than 1?
Nice one
This crazy
Hi. Please make videos on another math's ares's like abstract algebra, differencial geometry, algebric geometry and etc....
With regards
Elegant
hi teacher how can i contact you
As for moi - whenever the presentation goes off at an extreme tangent covering some gross new things they seem to be eminently forgettable. But I reserve the right to be wrong on this :-)
Basis of my conjecture: math is not frightening, math is eminently doable. Nonetheless - great video, great swooping intro to some gigantic new things (I feel like calling them monsters and that is okay)
I'm glad it wasn't my method 😅.
????????????
But we've only proven this for the case of Re(z+1) < 2?
Use the identity principle from complex analysis to extend the Identity to all of C (minus multiples of π)
And Re(z+1) > 0, otherwise the integral with the epsilon wouldn't vanish.