I don't think you need to use the factorial definition of the Gamma, since the factorial per sè isn't defined for non-natural numbers. You could just use the property that x•Gamma(x) = Gamma(x+1) and that (1/phi) = (phi-1) at 3:42 for a more "formal" solution. Great video tho! Loved the reupload fixing that mistake at the start, shows a great care and love for the craft.
@@uggupuggu objectively, no. n! is defined as n*(n-1)*...*2*1. Using the property that n! = n*(n-1)!, we can use the Gamma function to non-natural numbers, but by definition the factorial is defined only for positive integers. Saying that (1/2)! = √π/2 is an abuse of notation.
Dr Peyam also did a video on this integral, taking a completely different route. He actually guessed an antiderivative of the function instead. It wasn't an intuitive solution, but I still found it pretty surprising that a complicated function with irrational powers like this one has an elementary antiderivative. It shows just how special phi is!
It took a while, but I figured out how to do this with only u-substitution! Starting at the beginning, I set u=1+x^phi, and after following similar steps from the video, I rearranged the function to get: 1/u^phi * 1/(u-1) * x du all over phi. Solving for x in terms of u gets (u-1)^(1/phi), and plugging back in and dealing with exponents gets us: 1/u^phi * 1/[(u-1)^(1-1/phi)] du all over phi. Plugging in 1/phi = phi - 1 for the second term gets us: 1/u^phi * 1/[(u-1)^(2-phi)] du all over phi. Bringing the second term to the numerator and splitting up the exponent, we get: 1/u^phi * (u-1)^phi * 1/(u-1)^2 du all over phi. Because the first two terms have the same exponents, we can combine the terms inside of them: (1 - 1/u)^phi * 1/(u-1)^2 du all over phi. We want the second term in terms of 1 - 1/u, so taking a 1/u^2 out of it and flipping the signs (no change because -1 is squared), and combining with the first term, we get: (1 - 1/u)^(phi-2) * u^-2 du all over phi. Lastly, doing w-substitution where w = 1 - 1/u and dw = u^-2 will yield an integrand of: w^(phi-2) dw all over phi. And an anti derivative of: [w^(phi-1)]/(phi-1) all over phi, integrated from 0 to 1 - 1/(inf^phi + 1) if we did our u-subs correctly. Doing our strategy of taking the limit as a variable (b) goes to infinity for indefinite integrals and plugging in, we get: [(1-1/[b^phi + 1])^(phi-1)] / (phi-1) all over phi where b goes to infinity. Because b goes to infinity, the numerator just turns to 1, leaving us with: 1/(phi-1) * 1/phi And finally, plugging in phi - 1 = phi^-1 for the first term nets us our magical answer… 1 Very cool!
Another great video from Bri. This is a great solution and a really nice use of the Beta and Gamma functions, however, I evaluated the integral without using any special functions: Let u=1+x^phi, then, x=(u-1)^(1/phi), so dx=(1/phi)*(u-1)^(1/phi -1). Since, phi^2 = phi + 1, dividing by phi and taking 2 from both sides gives, 1/phi - 1 = phi - 2. Using this to simplify the expression for dx gives, dx = (1/phi)*(u-1)^(phi-2) du. At our bounds we have, x=0 -> u=1, x=inf -> u=inf. Re-writing the integral in terms of u gives: I = int( (1/phi)*(1/u^phi)*((u-1)^(phi-2)) du ) from 1 to infinity. By pulling the constant out the front and multiplying the integrand by (u^-2)/(u^-2) gives, I = (1/phi)*int( (u^-2)*(1/u^(phi-2))*(u-1)^(phi-2) du ) from 1 to infinity Rearranging the integrand, by noticing that the powers of u are now equal, and simplifying gives: I = (1/phi)*int( (u^-2)*(1 - 1/u)^(phi-2) du ) from 1 to infinity, Now let v = 1 - 1/u, so dv = u^-2 du. When u=1 -> v=0, u=inf -> v=1. This then gives: I = (1/phi)*int( v^(phi-2) dv) from 0 to 1. Integrating using the power rule and plugging in the bounds gives: I = (1/phi)*(1/(phi-1)) = 1/(phi^2 - phi) Since phi^2 = phi + 1, then subtracting phi from both sides gives, phi^2 - phi = 1, which reduced our integral to: I = 1 Hope that's clear enough for people to follow. I honestly didn't think I was gonna be able to solve it but then I noticed the beautiful substitution of v = 1 - 1/u.
Hi Robert. I basically found the same thing but after following Bri's first stage of u=x^phi rather than your u=1+x^phi. You then get the integral of 1/E where E is u^(2-phi)*(1+u)^phi = u^2*(1+(1/u))^phi and then a v=1/u substitution seems maybe a bit more obvious, followed by v -> v-1. OK, I'm doing v=(1/u)+1 rather than your final v=1-(1/u) [and getting a final integral from 1 to infty rather than 0 to 1], but maybe it is easier to spot with the positives.
This seems like a pretty brutal integral, don't know why I'd; *Evaluates to one, and uses the Beta and Gamma function* I could base a religion out of this
no need for gamma or beta function at all....it's much more simpler. @2:30 just let z=u/(1+u) then u=z/(1-z) and it becomes a simple algebraic integral.
I did by substituing for u = 1/(1+x^phi), after that there are some cancelations and the integral becomes very simple, without The need to know special functions. I would say however it is certainly more stylish to recognise the beta/gamma functions in there
Bro it always amazes me how you solve these abstract type integrals. Whenever I run into integrals like these I just jave no idea what to do but you make it look so simple. Good job
Instead of using the beta function it is possible to just transform u^(φ-2)du into u^φ*d(-1/u), substitute t = -1/u and then it simplifies and we get what we wanted
My way for calculating this integral Indefinite integral Int(1/(1+x^φ)^φ,x) can be quite easily integrated by parts Int(1/(1+x^φ)^φ,x)=Int((1+x^φ)/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Now we can integrate Int(1/(1+x^φ)^(φ-1),x) by parts with u = 1/(1+x^φ)^(φ-1) and dv = dx After integration by parts we can add integrals we will get Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)-Int(x*(-(φ-1)(1+x^φ)^(-(φ-1)-1))*φ*x^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ(φ-1)x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ*1/φ*x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+C If we want to calculate this integral using antiderivative we have to calculate the limit Now we need to calculate limit limit(x/(1+x^φ)^(φ-1),x=infinity) To calculate this limit all we need to do is some algebraic manipulations limit(x/(1+x^φ)^(φ-1),x=infinity)=limit(1/((1+x^φ)^(φ-1)/x),x=infinity) =1/limit(((1+x^φ)^(φ-1)/(x^(1/(φ-1))^(φ-1))),x=infinity) =1/limit((1+x^φ)^(φ-1)/x^(φ)^(φ-1),x=infinity) =1/limit(((1+x^φ)/x^φ)^(φ-1),x=infinity) =1/limit((1+1/x^φ)^(φ-1),x=infinity)
My math is very rough (what I remember from my C.S. degree), but integration has always seemed so ad hoc to me, like a collection of disjointed heuristics. Is it basically a tradeoff between the complexity of how we represent formulas vs. the complexity of the algorithm we use to integrate them?
I am amazed to see that mathematicians have come up with shortcuts for integration as well to save our time. Although I did not understand the technical stuff, I loved how this abstraction and clever substitution of beta function made this problem so beautiful. I remember studying gamma functions in my college. These are some college level concepts
This looked so scary but if you think about the definition of the golden ratio in terms of the sides of a rectangle it’s no wonder why integrating it gives such a nice number
is it ok to undo the analytic continuation of factorials (namely Γ(x)) when your argument isn't a natural number? it'll usually work, but you'll need to stipulate caveats regarding the evaluation of factorials, enforcing their arrangement as quotients of successive or integer-modulo arguments to the factorial function. Γ(x) also has the property that Γ(x + n) = x^n * Γ(x). so you could have just evaluated that directly. very pretty problem. i hardly ever see people talk about β(x, y)
Great video! But a Gamma function is equivalent to its argument factorial only when the argument is an integer, which is not the case for Phi. Is this correct?
What if b is larger? It just makes b=0. And if b is 0, a should be a negative number? And also, a+b/b will tend to infinity which means a/b=infinity which just means a=0 x infinity But a is negative, right? The golden ratio formula has a contradiction
ah what so the answer of one seem pretty reasonable because the integral of 1/x is ln x and the limit of (1+1/x)^x as x goes to infinity is e. The answer to ln e is 1. As an aside, 0! is not equal to 1 despite popular opinion. The gamma is not an exact answer for generalized factorials. The definition of factorial requires that the domain is the set of natural numbers. The gamma function is generalized in that it is an analytic continuation, but it loses meaning when things are more generalized.
Nice video, but notice you should have employed the properties of Gamma rather than the factorial at the end of the calculations. Gamma(phi-1)= (phi-1)Gamma(phi-2) and the result follows. Observe that factorial emerges only for natural numbers which is not the case for phi.
One represents unity and wholeness. The golden ratio is the geometric representation of multiple dimensions of "oneness." By this I mean it represents the reciprocating property of nature (the 1/) and the fact that nature builds upon itself to create new life beings the (1+). Therefore nature follows the Noble Path of the Golden code indefinitely 1+(1/(1+(1/(1+(1/...) The integral represents the touch of infinity and zero. It is the sum of infinitesimal small quantities of "dx" so small that humans can't even comprehend it. Combining the universal code, oneness, and the touch of the small into a mathematical expression is the act of seeking quantitative understanding of the Origin.
Instead of using some properties of the beta function which feels quite.. obscure I guess, you could've really proven that result by deriving the prof of that property in that specific case or something.
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I don't think you need to use the factorial definition of the Gamma, since the factorial per sè isn't defined for non-natural numbers. You could just use the property that x•Gamma(x) = Gamma(x+1) and that (1/phi) = (phi-1) at 3:42 for a more "formal" solution.
Great video tho! Loved the reupload fixing that mistake at the start, shows a great care and love for the craft.
I try :) Thanks so much for watching!
I did not see that one coming. 😀
factorial is defined for non-natural numbers, you are wrong
@@uggupuggu objectively, no. n! is defined as n*(n-1)*...*2*1. Using the property that n! = n*(n-1)!, we can use the Gamma function to non-natural numbers, but by definition the factorial is defined only for positive integers. Saying that (1/2)! = √π/2 is an abuse of notation.
"To solve this we have to know a few things about the golden ratio." Yeah the lack of information about phi isn't the problem here :D
😂
i loves when he still explain basic algebra and uses beta and gamma function
YouFeeAreIrr
When in doubt, the answer is probably 1 or 0. This helped me more times in my math undergrad wayyy more than it really should’ve lmao
Facts.
Who can't relate?
Math is too hard then, i guess, because physics is just the study of things that are zero
What about e and pi ?
@@muskyoxes what?
you really have one of the most entertaining and underrated math-related channel, keep up the work cause we all love it!!
Wow, thank you!
wow. you re-uploaded just to fix the a/b = (a + b)/b. wow.
Dr Peyam also did a video on this integral, taking a completely different route. He actually guessed an antiderivative of the function instead. It wasn't an intuitive solution, but I still found it pretty surprising that a complicated function with irrational powers like this one has an elementary antiderivative. It shows just how special phi is!
For sure!
I don't think the antiderivative itself is elementary, just that it has a nice value for these particular endpoints.
@@muskyoxes no, it's elementary lol. Go watch his video if you don't believe me!
@@muskyoxes
x(1+x^φ)^(-1/φ)+c
The fact that the integral is equal to 1 make my head explode, nice video
Thanks a ton!
It took a while, but I figured out how to do this with only u-substitution!
Starting at the beginning, I set u=1+x^phi, and after following similar steps from the video, I rearranged the function to get:
1/u^phi * 1/(u-1) * x du all over phi.
Solving for x in terms of u gets (u-1)^(1/phi), and plugging back in and dealing with exponents gets us:
1/u^phi * 1/[(u-1)^(1-1/phi)] du all over phi.
Plugging in 1/phi = phi - 1 for the second term gets us:
1/u^phi * 1/[(u-1)^(2-phi)] du all over phi.
Bringing the second term to the numerator and splitting up the exponent, we get:
1/u^phi * (u-1)^phi * 1/(u-1)^2 du all over phi.
Because the first two terms have the same exponents, we can combine the terms inside of them:
(1 - 1/u)^phi * 1/(u-1)^2 du all over phi.
We want the second term in terms of 1 - 1/u, so taking a 1/u^2 out of it and flipping the signs (no change because -1 is squared), and combining with the first term, we get:
(1 - 1/u)^(phi-2) * u^-2 du all over phi.
Lastly, doing w-substitution where w = 1 - 1/u and dw = u^-2 will yield an integrand of:
w^(phi-2) dw all over phi.
And an anti derivative of:
[w^(phi-1)]/(phi-1) all over phi, integrated from 0 to 1 - 1/(inf^phi + 1) if we did our u-subs correctly.
Doing our strategy of taking the limit as a variable (b) goes to infinity for indefinite integrals and plugging in, we get:
[(1-1/[b^phi + 1])^(phi-1)] / (phi-1) all over phi where b goes to infinity.
Because b goes to infinity, the numerator just turns to 1, leaving us with:
1/(phi-1) * 1/phi
And finally, plugging in phi - 1 = phi^-1 for the first term nets us our magical answer…
1
Very cool!
why cant to tkae x = 1??
Put 1+ 1/x^phi = u. Solves in second
Another great video from Bri. This is a great solution and a really nice use of the Beta and Gamma functions, however, I evaluated the integral without using any special functions:
Let u=1+x^phi, then, x=(u-1)^(1/phi), so dx=(1/phi)*(u-1)^(1/phi -1).
Since, phi^2 = phi + 1, dividing by phi and taking 2 from both sides gives, 1/phi - 1 = phi - 2.
Using this to simplify the expression for dx gives, dx = (1/phi)*(u-1)^(phi-2) du.
At our bounds we have, x=0 -> u=1, x=inf -> u=inf.
Re-writing the integral in terms of u gives:
I = int( (1/phi)*(1/u^phi)*((u-1)^(phi-2)) du ) from 1 to infinity.
By pulling the constant out the front and multiplying the integrand by (u^-2)/(u^-2) gives,
I = (1/phi)*int( (u^-2)*(1/u^(phi-2))*(u-1)^(phi-2) du ) from 1 to infinity
Rearranging the integrand, by noticing that the powers of u are now equal, and simplifying gives:
I = (1/phi)*int( (u^-2)*(1 - 1/u)^(phi-2) du ) from 1 to infinity,
Now let v = 1 - 1/u, so dv = u^-2 du. When u=1 -> v=0, u=inf -> v=1. This then gives:
I = (1/phi)*int( v^(phi-2) dv) from 0 to 1.
Integrating using the power rule and plugging in the bounds gives:
I = (1/phi)*(1/(phi-1)) = 1/(phi^2 - phi)
Since phi^2 = phi + 1, then subtracting phi from both sides gives, phi^2 - phi = 1, which reduced our integral to:
I = 1
Hope that's clear enough for people to follow. I honestly didn't think I was gonna be able to solve it but then I noticed the beautiful substitution of v = 1 - 1/u.
Hi Robert. I basically found the same thing but after following Bri's first stage of u=x^phi rather than your u=1+x^phi. You then get the integral of 1/E where E is
u^(2-phi)*(1+u)^phi = u^2*(1+(1/u))^phi and then a v=1/u substitution seems maybe a bit more obvious, followed by v -> v-1. OK, I'm doing v=(1/u)+1 rather than
your final v=1-(1/u) [and getting a final integral from 1 to infty rather than 0 to 1], but maybe it is easier to spot with the positives.
This seems like a pretty brutal integral, don't know why I'd;
*Evaluates to one, and uses the Beta and Gamma function*
I could base a religion out of this
Yayyy thank you so much Bri !
I wanted this!
You're so welcome!
Absolutely incredible
Glad you thought so!
no need for gamma or beta function at all....it's much more simpler. @2:30 just let z=u/(1+u) then u=z/(1-z) and it becomes a simple algebraic integral.
Anyone else get the shivers whenever he says: ‘pheee’ instead of phi
I still do, even though I am used to the fact that in Britain its pronounced phy and in the US its pronounced phee.
"fee" is closer to greek pronounciation:D
Yeah, just sounds wrong to me. I mean, we say "pie" instead of "pee" for "π", don't we?
01:20 oh wow thanks nobody has ever told me that I tend towards infinity 😊
Oh-
Aesthetically elegant. Thanks 4 sharing.
Glad you like it!
Wow!
Loved this.
So glad!
I did by substituing for u = 1/(1+x^phi), after that there are some cancelations and the integral becomes very simple, without The need to know special functions. I would say however it is certainly more stylish to recognise the beta/gamma functions in there
Bro it always amazes me how you solve these abstract type integrals. Whenever I run into integrals like these I just jave no idea what to do but you make it look so simple. Good job
An amazing video man. Keep it up!
Thanks a ton!
Dejà-vu, I've just been in this place before...
Instead of using the beta function it is possible to just transform u^(φ-2)du into u^φ*d(-1/u), substitute t = -1/u and then it simplifies and we get what we wanted
This is absolutely amazing
glad you thought so!
@@BriTheMathGuy thanks, I subscribed
grande spiegazione..bravo
Frickin awesome indeed… 👍👍
Glad you thought so! Thanks for watching :)
OMG! I'm a lover of integrals and your videos make me live them... You are amazing! 😃
Glad you like them!
This vid needs to be watched over and over
Glad you think so!
Wow... New info... Thank you for sharing...sir
So nice of you!
My way for calculating this integral
Indefinite integral Int(1/(1+x^φ)^φ,x) can be quite easily integrated by parts
Int(1/(1+x^φ)^φ,x)=Int((1+x^φ)/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x)
Now we can integrate Int(1/(1+x^φ)^(φ-1),x) by parts with
u = 1/(1+x^φ)^(φ-1) and dv = dx
After integration by parts we can add integrals we will get
Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)-Int(x*(-(φ-1)(1+x^φ)^(-(φ-1)-1))*φ*x^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ(φ-1)x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ*1/φ*x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x)
Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+C
If we want to calculate this integral using antiderivative we have to
calculate the limit
Now we need to calculate limit
limit(x/(1+x^φ)^(φ-1),x=infinity)
To calculate this limit all we need to do is some algebraic manipulations
limit(x/(1+x^φ)^(φ-1),x=infinity)=limit(1/((1+x^φ)^(φ-1)/x),x=infinity)
=1/limit(((1+x^φ)^(φ-1)/(x^(1/(φ-1))^(φ-1))),x=infinity)
=1/limit((1+x^φ)^(φ-1)/x^(φ)^(φ-1),x=infinity)
=1/limit(((1+x^φ)/x^φ)^(φ-1),x=infinity)
=1/limit((1+1/x^φ)^(φ-1),x=infinity)
Great videos! So well made and fun to watch.
Glad you like them!
at 1:52 x is just u^1/phi or a radical with index of phi
1:39 with respect to me? Never thought I’d get any respect at all 🥺
My math is very rough (what I remember from my C.S. degree), but integration has always seemed so ad hoc to me, like a collection of disjointed heuristics. Is it basically a tradeoff between the complexity of how we represent formulas vs. the complexity of the algorithm we use to integrate them?
Like in proving theorems, there is probably no general algorithm in solving integrals analytically
indeed, that is now my favourite integral, will save in my favorites omg
4:00 but how do you know that properties of factorials apply to non integer inputs?
I am amazed to see that mathematicians have come up with shortcuts for integration as well to save our time. Although I did not understand the technical stuff, I loved how this abstraction and clever substitution of beta function made this problem so beautiful. I remember studying gamma functions in my college. These are some college level concepts
Please do not confuse people! The beta function is written by B(x,y). Capital Greek beta.
beta is ß not B
sum intergral
When he said "pheee", I felt that
Mind= so blown even the quarks got split
🤯
I substituted x^(phi)=tan^2(theta). Nice problem.
Cool!
WHAT THE ACTUAL FUCK. THAT WAS SO FKIN SATISFYING
Dude, this year has been like math explosion festinghouse. Nice.
Mind blown, but more in a makes me want to cry way. I need to brush up on my integrals!
Please please keep your hands still. Thanks so much🙏
I swear every time a problem turns out like that I get a mathgasm, there, I said it
This is a very cool integral, but my favorite is probably the integral from -infinity to +infinity e^-x^2 dx, which comes out as √pi
THIS IS SO COOL
😎😎😎😎
That was very cool :)
Glad you thought so :)
Make some videos about zeta function , please
th-cam.com/video/siznb9u5xhI/w-d-xo.html
I literally just randomly googled this video lol. But I like it :)
Great to hear!
though I wont be able to remember in 2 minutes, it is indeed my favorite integral at the moment.
This is beautiful
What a brilliant integral
This looked so scary but if you think about the definition of the golden ratio in terms of the sides of a rectangle it’s no wonder why integrating it gives such a nice number
awesome video as always! only thing wrong with them is that they end :P
Thanks! 😄
Fascinating. I'm fascinated.
Glad to hear it :)
it's now my favourite integral lol
Nice!
Thank you! Cheers!
(You should pay a fine for calling Phi, Fee.)
👮♂️
I mean, that's how every language but english calls it
this is not maths but literature
🧐
Yo the algorithm finally recommending some good channels!
is it ok to undo the analytic continuation of factorials (namely Γ(x)) when your argument isn't a natural number? it'll usually work, but you'll need to stipulate caveats regarding the evaluation of factorials, enforcing their arrangement as quotients of successive or integer-modulo arguments to the factorial function.
Γ(x) also has the property that Γ(x + n) = x^n * Γ(x). so you could have just evaluated that directly.
very pretty problem. i hardly ever see people talk about β(x, y)
Yes, now I can integrate any expression, just by introducing a new function. 😀😀
Favoritegral.
😎
Great video! But a Gamma function is equivalent to its argument factorial only when the argument is an integer, which is not the case for Phi. Is this correct?
What if b is larger? It just makes b=0. And if b is 0, a should be a negative number? And also, a+b/b will tend to infinity which means a/b=infinity which just means a=0 x infinity
But a is negative, right? The golden ratio formula has a contradiction
How do you get that black background?...do you literally record it in a black planted rookm with a light on you?
Green screen :)
Hello! :) Could you tell me, please, the name of the software that you're using in the presentation? :) TY!
An interesting fact that I realized from playing around with this integral is that Γ(φ+1)=Γ(1/φ) even though φ+1≠1/φ
feynman’s technique for a parameter k in place of phi, so differentiating under the integral?
ah what so the answer of one seem pretty reasonable because the integral of 1/x is ln x and the limit of (1+1/x)^x as x goes to infinity is e. The answer to ln e is 1. As an aside, 0! is not equal to 1 despite popular opinion. The gamma is not an exact answer for generalized factorials. The definition of factorial requires that the domain is the set of natural numbers. The gamma function is generalized in that it is an analytic continuation, but it loses meaning when things are more generalized.
Γ(1+n) = n! for all natural n
Γ(1+0) = integral from 0 to inf of e^(-t) dt = 1
implies 0! = 1
If you love the Gamma Function, prove that gamma(z)*gamma(1-z)=pi/sin(pi*z) for all non integer z
I wonder the golden ratio has so many identities.
This was on cantor dust level 1…
some people pronounce φ as phai and others as pheeh. which one should i use? i’m not an english speaker and i’m confused
I am watching this and pretend I understand everything after the word "integral."
I hit the like button as soon as I heard you were pronouncing phi correctly LOL
I guess that the golden ratio has infinite possibilities ( If you can change the value of the variables such as x)
you might want to reconsider showing yourself in the picture.
What makes you say that? (genuinely asking)
....i swear ive seen this....
Loved this video! But you didn't have to do our boy Gamma like that! :P
I clickbaited because the differential variable wasn’t specified in the thumbnail :(
🥲
Nice video, but notice you should have employed the properties of Gamma rather than the factorial at the end of the calculations. Gamma(phi-1)= (phi-1)Gamma(phi-2) and the result follows. Observe that factorial emerges only for natural numbers which is not the case for phi.
At the beginning i thought this was an ex joke.
Damn it. I was hoping this didn't require a function I don't know of
There are other ways to do it!
Golden integral
New integral idea
Int from -infinity to infinity (e^((-(x-m)^2)/(2(n)^2))dx
very nice
Glad you thought so!
Where's the plot? I need a plot of the distro! Where is my plot!!
Is it just me or is there a discord ping sound at 0:10?
I think it's just u
One represents unity and wholeness. The golden ratio is the geometric representation of multiple dimensions of "oneness." By this I mean it represents the reciprocating property of nature (the 1/) and the fact that nature builds upon itself to create new life beings the (1+). Therefore nature follows the Noble Path of the Golden code indefinitely 1+(1/(1+(1/(1+(1/...)
The integral represents the touch of infinity and zero. It is the sum of infinitesimal small quantities of "dx" so small that humans can't even comprehend it.
Combining the universal code, oneness, and the touch of the small into a mathematical expression is the act of seeking quantitative understanding of the Origin.
Lol 🤣
Uughhh is it washable
Instead of using some properties of the beta function which feels quite.. obscure I guess, you could've really proven that result by deriving the prof of that property in that specific case or something.
I don’t think you can actually use the factorial for non-natural numbers. Why not use the properties of the gamma function instead?
Interesting video! Just a minor peeve, I think "phi" is actually pronounced fahy instead of fee. Great video!
Golden triangle
Nah Rendering Equation is my favorite
Does this integral have a practical use?
Here's a better way of writing the β function definition:
xβy=((x-1)!(y-1)!)/((x+y-1)!)