I prefer the version of the equation that features the Pi function as opposed to the Gamma function. Π(z) = Γ(z - 1), thus Γ(1 - z) = Π(-z), and Γ(z) = Π(z - 1). Multiply the equation by z, so that the formula reads Π(z)Π(-z) = πz/sin(πz). The normalized cardinal sine function is defined by the equation sinc(z) = lim sin(πt)/(πt) (t -> z), so for nonzero z, πz/sin(πz) = 1/sinc(z). Meanwhile, for z = 0, Π(z) = Π(-z) = 1 = lim sin(πt)/(πt) (t -> 0) = sinc(0) = 1/sinc(0). Therefore, for all z, Π(z)Π(-z) = 1/sinc(z). I find this formula far more elegant than the popular version. And presumably, the proof would be simpler since you would not need to deal with those annoying factors of z that appear in the formulas.
You said it was hard to derive if you just substitute into the formula. This took me a few hours, but you can simplify this expression and factorise the part in the infinite product. Then multiply the inside of the infinite product by ((k^2-z^2)/k^2) *(k^2/(k^2-z^2)). The first of these terms should cancel with the pre-existing stuff (+ the 1/(1-z)) and should only be left with the second half. This will yield the reflection formula.
Ha- Bouz It does not have the same linear factors as sine. Remember that sin(t) has no singularities, while tan(t) does, so some of the linear factors are located in the denominator. He did a video on this, actually, expressing the six basic trigonometric functions as an infinite product of linear factors.
I'd guess it's called the reflection formula because 1-z is the complement in some sense. I'd think of it like in probability, the probability of an event A's complement happening is 1 - P(A). Probably not on the right track, but my first thought given the prevalence of the gamma function in probability.
As soon as I saw the video I wrote out the proof that I thought of, I'm curious to see if you did it the same way. Edit: I used the Euler product formula for the gamma function and I did go down the route of plugging in 1-z
Girlfriend of famable maths: What are you thinking about? You look so concentrating... Famable maths: Oh its nothing darling... i just think about our futura as a pair... Flamable maths mind: Euler macaroni waifu
Take the x^2 out of the root, then substitute x^2 for t, multiply and divide by the constant. √10-x^2 has a semi standard result or so I believe, as √a^2-x^2.
i don't understand how maths RUclipsrs know so much maths, like I'm in the second year of a maths degree and i can barely remember all the stuff from first year.... do you guys read and revise everything again before each video?
Papa Flammey, you should play me in chess. If you reply with an affirmative response, I will send you a link. You may choose the time control format and the color of pieces that you play with.
Flammable Maths lichess.org/flyOOn17 Time format is 10 minute with 10 second increments (Each move grants you 10 seconds). The pieces are going to be random. If you don’t see it or have changed your mind then don’t worry
"I am not going to spoiler what the Solution ist" said the person who put the Solution in the Thumbnail :D
greetings from austria (europe)
Gotta make sure people dont confuse it with Australia huh
I was going to write the exact same thing, but I'm from Australia (Australia). coincidence???
Such a gorgeous identity. Nearly as breathtaking as you, Papa.
Interestly enough, gamma((1-x)/2)*gamma((1+x)/2) changes the sine into cosine
That's because the 1/2 introduces a pi/2 into the sine. Trivial yo
@@nontrivialdog Yep, specially on x/2 tho
I awaiting the integral video. Papa promised it would be long and hard one! Sounds fun ;)
The most elegant identity in mathematics also along with the complex version of this that involves hyperbolic Sine function!
@@PapaFlammy69 Make one for that too just as a skit or something...
> Says I don’t want to spoiler the solution
> Puts it in the Thumbnail anyway
mfw
What a flammable surprise!
The fourier series of cos(ax) can also give the result
Wait what
Gotta try it
Sin(x pi)=x pi
Take it or leave it
Isn't it a reflection because you are reflecting around the 0 of the factorial?
Any formula which relates f(a - x) with f(x) for any a, is a reflection formula.
Source: Wikipedia
this channel is spewing good stuffs!
thanks! =D
Captain Hook and gamma function, a flammable combination. Thank you for nice derivation.
Thanks Mr. Anthony Fantano, Nice Flannel. I give this video a 10.
I fucking love your videos. You are saving my life for my project
I prefer the version of the equation that features the Pi function as opposed to the Gamma function. Π(z) = Γ(z - 1), thus Γ(1 - z) = Π(-z), and Γ(z) = Π(z - 1). Multiply the equation by z, so that the formula reads Π(z)Π(-z) = πz/sin(πz). The normalized cardinal sine function is defined by the equation sinc(z) = lim sin(πt)/(πt) (t -> z), so for nonzero z, πz/sin(πz) = 1/sinc(z). Meanwhile, for z = 0, Π(z) = Π(-z) = 1 = lim sin(πt)/(πt) (t -> 0) = sinc(0) = 1/sinc(0). Therefore, for all z, Π(z)Π(-z) = 1/sinc(z). I find this formula far more elegant than the popular version. And presumably, the proof would be simpler since you would not need to deal with those annoying factors of z that appear in the formulas.
Flammable maths does product k >/1 mean the same as proudct from 1 to inf, ive not seen this notation before
Yes, it's just shorthand
You said it was hard to derive if you just substitute into the formula. This took me a few hours, but you can simplify this expression and factorise the part in the infinite product. Then multiply the inside of the infinite product by
((k^2-z^2)/k^2) *(k^2/(k^2-z^2)). The first of these terms should cancel with the pre-existing stuff (+ the 1/(1-z)) and should only be left with the second half. This will yield the reflection formula.
Amazing. Exactly the same problem appeared on my Übungsblatt today :D
@@PapaFlammy69 truuu
5:55 : You lied to us, you are secretly presenting a physics video !
@@PapaFlammy69 You are again copying shamelessly this poor andrew dotson x)
What is the application of this function in reality?
Can't understand the linear factors thing. Following the same sense, doesnt tan(x) have the same linear factors as sin(x), making sin(x) = tan(x)?
Ha- Bouz It does not have the same linear factors as sine. Remember that sin(t) has no singularities, while tan(t) does, so some of the linear factors are located in the denominator. He did a video on this, actually, expressing the six basic trigonometric functions as an infinite product of linear factors.
God this was beautiful to watch
This is extremely well explained. The only thing I don't understand is the Newton joke :)
Hehe :D Glad you liked the video :3
Thats cool, so we found that the minimal value of Γ(z) Γ(1-z) is π, with z=π/2+2πn
Are you going to start using this analytic stuff to prove number-theoretic results about the integers?
HE CALLED ME FELLOW MATHMATICIAN...MOM I DID IT IM A MATHMATICIAN NOW
I'd guess it's called the reflection formula because 1-z is the complement in some sense. I'd think of it like in probability, the probability of an event A's complement happening is 1 - P(A). Probably not on the right track, but my first thought given the prevalence of the gamma function in probability.
What is an equation of motion when weight hanging and sliding down on oblique elastic rope?
Finally the reflect formula :'D
Another way to write this is
G[1+z]G[1-z]=PI[1-(z/k)^2]^-1
Can we write x! =x•(x-1)! even for non integral x?
Didn't you forget a minus sign from the - z?
It's called reflection formular because it is a reflection at the axis with real part 1/2.
Can you suggest some reading online on this?
I love when you say “meaning” papa, that turns me on
You should do a video working with the Barnes G-Function
Is papa flammy doing a homage to the melon with the yellow flannel?
I must rate this proof “not good” due to a lack of the fundamental theorem of engineering
Maybe meetup next time? I’ll be in München at Oktoberfest week 40. You wanna join?
As soon as I saw the video I wrote out the proof that I thought of, I'm curious to see if you did it the same way.
Edit: I used the Euler product formula for the gamma function and I did go down the route of plugging in 1-z
Me: I'm Chilean and I'm going to watch the eclipse.
*Papa Flammy videos pop up now"
Me: fuck off sun.
I can prove that using the riemann functional zeta equation
Σ Infinity boi + Π infinity cirl best pair ever
grill
Girlfriend of famable maths:
What are you thinking about? You look so concentrating...
Famable maths:
Oh its nothing darling... i just think about our futura as a pair...
Flamable maths mind:
Euler macaroni waifu
Hello
I
Can we evaluate divergent integrals ?
@@PapaFlammy69 but we can use cauchy's principal value 🔥
Flammable Maths I think he meant it in the Ramanujan sense or something similar to that.
@@angelmendez-rivera351 i am talking about using cauchy's principal value 🔥
My boi flammy can i get in those credits
@@PapaFlammy69 yayy
someone plz help me with the integral of (10x^2-x^4)^1/2 dx
Wolfram Alpha???
Maybe you can factor out the x^2 and then do a trig sub.
@@morganmitchell4017 If you dragged out the x² of the radical sign then you'd have a direct u-sub situation in your hands
Take out the x^2 term out of the root and then parametrize it using sqrt of 10 sint
Take the x^2 out of the root, then substitute x^2 for t, multiply and divide by the constant. √10-x^2 has a semi standard result or so I believe, as √a^2-x^2.
Used a similar legendre duplication formula a while back,for proving the orthogonality relation of legendre polynomials... It was a thing of beauty..😋
i don't understand how maths RUclipsrs know so much maths, like I'm in the second year of a maths degree and i can barely remember all the stuff from first year.... do you guys read and revise everything again before each video?
Legendre duplication next pls pls pls :)
Maybe next time explain this problem: f(x)=x(f(x)) ! How about that?!
What the fuck
I have a math trick for you! Do you believe that when multiplying is like subtraction?!
Here it is: 2nx=2n-x (where n € R)
Papa Flammey, you should play me in chess. If you reply with an affirmative response, I will send you a link. You may choose the time control format and the color of pieces that you play with.
Flammable Maths lichess.org/flyOOn17
Time format is 10 minute with 10 second increments (Each move grants you 10 seconds). The pieces are going to be random. If you don’t see it or have changed your mind then don’t worry
Flammable Maths hey the link expired so maybe another time uwu
nah video
First?
yeah
hey that's cute owo
Do you know speak something in portuguese?😅
@@PapaFlammy69 😊
:)
Wow. Nice accurate, true profile picture ;)
@@morganmitchell4017 :)
Papa, what's engrrrrrriiiiissssshhhhhhhh