This Will Be Your Favorite Integral
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- Опубликовано: 29 сен 2024
- 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
indeed, that is now my favourite integral, will save in my favorites omg
I hit the like button as soon as I heard you were pronouncing phi correctly LOL
Your thumbnail is missing dx. But you know what? Maybe don't fix it, the only reason I clicked the video was to see what was going on with the missing dx.
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.
wow. you re-uploaded just to fix the a/b = (a + b)/b. wow.
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!
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
Dejà-vu, I've just been in this place before...
The fact that the integral is equal to 1 make my head explode, nice video
Thanks a ton!
01:20 oh wow thanks nobody has ever told me that I tend towards infinity 😊
Oh-
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
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?
(You should pay a fine for calling Phi, Fee.)
👮♂️
I mean, that's how every language but english calls it
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
Absolutely incredible
Glad you thought so!
Yayyy thank you so much Bri !
I wanted this!
You're so welcome!
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?
Like in proving theorems, there is probably no general algorithm in solving integrals analytically
Nice!
Thank you! Cheers!
1:39 with respect to me? Never thought I’d get any respect at all 🥺
Frickin awesome indeed… 👍👍
Glad you thought so! Thanks for watching :)
THIS IS SO COOL
😎😎😎😎
Interesting video! Just a minor peeve, I think "phi" is actually pronounced fahy instead of fee. Great video!
I literally just randomly googled this video lol. But I like it :)
Great to hear!
Nah Rendering Equation is my favorite
X^(fi)=Me
Thank you 🙂
I clickbaited because the differential variable wasn’t specified in the thumbnail :(
🥲
....i swear ive seen this....
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
Aesthetically elegant. Thanks 4 sharing.
Glad you like it!
Nice
Thanks!
OMG! I'm a lover of integrals and your videos make me live them... You are amazing! 😃
Glad you like them!
I substituted x^(phi)=tan^2(theta). Nice problem.
Cool!
you might want to reconsider showing yourself in the picture.
What makes you say that? (genuinely asking)
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.
Make some videos about zeta function , please
ruclips.net/video/siznb9u5xhI/видео.html
grande spiegazione..bravo
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
sum intergral
Yes, now I can integrate any expression, just by introducing a new function. 😀😀
This is the most beautiful integration problem I've seen on RUclips.
Wow!
Loved this.
So glad!
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
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?
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
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.
Hello! :) Could you tell me, please, the name of the software that you're using in the presentation? :) TY!
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)
I don’t think you can actually use the factorial for non-natural numbers. Why not use the properties of the gamma function instead?
very nice
Glad you thought so!
It's fi not fee. Your videos are great but like a true yank you're ruining it with pronunciation (it has a u in it not an o). You're on youtube so I'm sure you can find something on the Greek alphabet to learn it, don't you have $100k a year to get told these things at college? :D
I love you
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?
New integral idea
Int from -infinity to infinity (e^((-(x-m)^2)/(2(n)^2))dx
Finally, someone that pronounces φ fi and not phai(I'm Greek btw)
That was very cool :)
Glad you thought so :)
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
It's all wonky Greek alphabets all I can see. If I'm about to write the solution, and come up with 1, I'd doubt my work too
Mind= so blown even the quarks got split
🤯
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?
An amazing video man. Keep it up!
Thanks a ton!
You're an academic. The academic pronunciation is either /faay/ or /pHee/.
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
That was super entertaining but I still can't understand most of the things :)
Damn it. I was hoping this didn't require a function I don't know of
There are other ways to do it!
though I wont be able to remember in 2 minutes, it is indeed my favorite integral at the moment.
this is not maths but literature
🧐
awesome video as always! only thing wrong with them is that they end :P
Thanks! 😄
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
I swear every time a problem turns out like that I get a mathgasm, there, I said it
How do you get that black background?...do you literally record it in a black planted rookm with a light on you?
Green screen :)
This is absolutely amazing
glad you thought so!
@@BriTheMathGuy thanks, I subscribed
This vid needs to be watched over and over
Glad you think so!
Please please keep your hands still. Thanks so much🙏
its mildly infuriating how he says phee instead of phi
Where's the plot? I need a plot of the distro! Where is my plot!!
Favoritegral.
😎
Great videos! So well made and fun to watch.
Glad you like them!
am i the only one who didn't understand a thing but still watched till the end! 🤣
Shout out for pronouncing the Greek correctly!
by any chance can someone start Scitube so I can get off the you tube platform
great video
👨🔬
I wonder the golden ratio has so many identities.
WHAT THE ACTUAL FUCK. THAT WAS SO FKIN SATISFYING
This was on cantor dust level 1…
ew. you didnt. that was not my favorite
it's now my favourite integral lol
Is it just me or is there a discord ping sound at 0:10?
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
At the beginning i thought this was an ex joke.
Please do not confuse people! The beta function is written by B(x,y). Capital Greek beta.
beta is ß not B
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
When he said "pheee", I felt that
Pretty interesting, although I don't understand 99% of the content
gonna be back in 12 months with a full understanding hopefully
Here's a better way of writing the β function definition:
xβy=((x-1)!(y-1)!)/((x+y-1)!)
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.