Це відео не доступне.
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This Will Be Your Favorite Integral
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- Опубліковано 20 жов 2021
- The golden ratio is a pretty famous math inequality. BUT have you heard of (what I call) the Golden Integral?
This Golden Ratio Integral integrates to a pretty remarkable result!
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #goldenratio
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"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
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.
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?
wow. you re-uploaded just to fix the a/b = (a + b)/b. wow.
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!
The fact that the integral is equal to 1 make my head explode, nice video
Thanks a ton!
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
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
01:20 oh wow thanks nobody has ever told me that I tend towards infinity 😊
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.
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?
Yayyy thank you so much Bri !
I wanted this!
You're so welcome!
Absolutely incredible
Glad you thought so!
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
Dejà-vu, I've just been in this place before...
Aesthetically elegant. Thanks 4 sharing.
Glad you like it!
OMG! I'm a lover of integrals and your videos make me live them... You are amazing! 😃
Glad you like them!
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
Frickin awesome indeed… 👍👍
Glad you thought so! Thanks for watching :)
When he said "pheee", I felt that
1:39 with respect to me? Never thought I’d get any respect at all 🥺
WHAT THE ACTUAL FUCK. THAT WAS SO FKIN SATISFYING
An amazing video man. Keep it up!
Thanks a ton!
This vid needs to be watched over and over
Glad you think so!
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
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
Please please keep your hands still. Thanks so much🙏
Great videos! So well made and fun to watch.
Glad you like them!
I swear every time a problem turns out like that I get a mathgasm, there, I said it
grande spiegazione..bravo
indeed, that is now my favourite integral, will save in my favorites omg
(You should pay a fine for calling Phi, Fee.)
👮♂️
I mean, that's how every language but english calls it
though I wont be able to remember in 2 minutes, it is indeed my favorite integral at the moment.
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
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)
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.
Mind= so blown even the quarks got split
🤯
Dude, this year has been like math explosion festinghouse. Nice.
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
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
Please do not confuse people! The beta function is written by B(x,y). Capital Greek beta.
beta is ß not B
I substituted x^(phi)=tan^2(theta). Nice problem.
Cool!
Wow... New info... Thank you for sharing...sir
So nice of you!
THIS IS SO COOL
😎😎😎😎
Nice!
Thank you! Cheers!
This is beautiful
Mind blown, but more in a makes me want to cry way. I need to brush up on my integrals!
I literally just randomly googled this video lol. But I like it :)
Great to hear!
Make some videos about zeta function , please
ua-cam.com/video/siznb9u5xhI/v-deo.html
Interesting video! Just a minor peeve, I think "phi" is actually pronounced fahy instead of fee. Great video!
this is not maths but literature
🧐
awesome video as always! only thing wrong with them is that they end :P
Thanks! 😄
Yes, now I can integrate any expression, just by introducing a new function. 😀😀
Fascinating. I'm fascinated.
Glad to hear it :)
That was very cool :)
Glad you thought so :)
it's now my favourite integral lol
I clickbaited because the differential variable wasn’t specified in the thumbnail :(
🥲
Favoritegral.
😎
I hit the like button as soon as I heard you were pronouncing phi correctly LOL
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
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
What a brilliant integral
sum intergral
I don’t think you can actually use the factorial for non-natural numbers. Why not use the properties of the gamma function instead?
At the beginning i thought this was an ex joke.
This was on cantor dust level 1…
I wonder the golden ratio has so many identities.
Yo the algorithm finally recommending some good channels!
I guess that the golden ratio has infinite possibilities ( If you can change the value of the variables such as x)
feynman’s technique for a parameter k in place of phi, so differentiating under the integral?
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)
Where's the plot? I need a plot of the distro! Where is my plot!!
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
Hello! :) Could you tell me, please, the name of the software that you're using in the presentation? :) TY!
Damn it. I was hoping this didn't require a function I don't know of
There are other ways to do it!
This is the most beautiful integration problem I've seen on UA-cam.
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
New integral idea
Int from -infinity to infinity (e^((-(x-m)^2)/(2(n)^2))dx
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.
Golden integral
Isn't Γ(x)=x! iff x belongs to the set of natural numbers? Here, φ is not a natural number, so shouldn't we not be using notion of factorial?
Nah Rendering Equation is my favorite
you might want to reconsider showing yourself in the picture.
What makes you say that? (genuinely asking)
very nice
Glad you thought so!
Shout out for pronouncing the Greek correctly!
How do you get that black background?...do you literally record it in a black planted rookm with a light on you?
Green screen :)
....i swear ive seen this....
If you love the Gamma Function, prove that gamma(z)*gamma(1-z)=pi/sin(pi*z) for all non integer z
Loved this video! But you didn't have to do our boy Gamma like that! :P
Finally, someone that pronounces φ fi and not phai(I'm Greek btw)
I am watching this and pretend I understand everything after the word "integral."
There is a famous saying, if there is a nasty integral the answer is a gamma or beta function.
Does this integral have a practical use?
If you think about this, such play of formulae just shows us how we can use mathematical formulae as tools to play with concepts, also in schools we are taught formulae. We can learn to play with abstract concepts using them but you can't really understand deeply what is going on. It is just like learning to drive a car without really understanding how the car is made. So, everyone should get a chance to do this without thinking what are all these formulae and where did they come from? Just take them for granted, play a little and if the question persists, maybe you have that mathematical inclination in you. I don't think this channel is good for that.
I have recently completed my undergraduate on and I discovered this channel and plan to use his videos to have fun with these other worldly objects 😂 (as Ptolemy would say)
Is it just me or is there a discord ping sound at 0:10?