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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

@BriTheMathGuy

2 жыл бұрын

😂

@FreeGroup22

2 жыл бұрын

i loves when he still explain basic algebra and uses beta and gamma function

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

@BriTheMathGuy

2 жыл бұрын

Facts.

@tsan_jey

2 жыл бұрын

Who can't relate?

@muskyoxes

2 жыл бұрын

Math is too hard then, i guess, because physics is just the study of things that are zero

@quentind1924

2 жыл бұрын

What about e and pi ?

@angelmendez-rivera351

2 жыл бұрын

@@muskyoxes what?

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.

@BriTheMathGuy

2 жыл бұрын

I try :) Thanks so much for watching!

@YossiSirote

2 жыл бұрын

I did not see that one coming. 😀

@uggupuggu

10 ай бұрын

factorial is defined for non-natural numbers, you are wrong

@vitorcurtarelli254

10 ай бұрын

@@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.

you really have one of the most entertaining and underrated math-related channel, keep up the work cause we all love it!!

@BriTheMathGuy

2 жыл бұрын

Wow, thank you!

The fact that the integral is equal to 1 make my head explode, nice video

@BriTheMathGuy

2 жыл бұрын

Thanks a ton!

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

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!

@BriTheMathGuy

2 жыл бұрын

For sure!

@muskyoxes

2 жыл бұрын

I don't think the antiderivative itself is elementary, just that it has a nice value for these particular endpoints.

@violintegral

2 жыл бұрын

@@muskyoxes no, it's elementary lol. Go watch his video if you don't believe me!

@xinpingdonohoe3978

3 ай бұрын

​@@muskyoxes x(1+x^φ)^(-1/φ)+c

01:20 oh wow thanks nobody has ever told me that I tend towards infinity 😊

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

Anyone else get the shivers whenever he says: ‘pheee’ instead of phi

@MichaelRothwell1

2 жыл бұрын

I still do, even though I am used to the fact that in Britain its pronounced phy and in the US its pronounced phee.

@zeozen

2 жыл бұрын

"fee" is closer to greek pronounciation:D

@nordicexile7378

2 жыл бұрын

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!

@BriTheMathGuy

2 жыл бұрын

You're so welcome!

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.

@mchmch6185

2 жыл бұрын

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.

Dejà-vu, I've just been in this place before...

OMG! I'm a lover of integrals and your videos make me live them... You are amazing! 😃

@BriTheMathGuy

2 жыл бұрын

Glad you like them!

Absolutely incredible

@BriTheMathGuy

2 жыл бұрын

Glad you thought so!

Wow! Loved this.

@BriTheMathGuy

2 жыл бұрын

So glad!

Aesthetically elegant. Thanks 4 sharing.

@BriTheMathGuy

2 жыл бұрын

Glad you like it!

Great videos! So well made and fun to watch.

@BriTheMathGuy

2 жыл бұрын

Glad you like them!

An amazing video man. Keep it up!

@BriTheMathGuy

2 жыл бұрын

Thanks a ton!

indeed, that is now my favourite integral, will save in my favorites omg

Please please keep your hands still. Thanks so much🙏

Frickin awesome indeed… 👍👍

@BriTheMathGuy

2 жыл бұрын

Glad you thought so! Thanks for watching :)

grande spiegazione..bravo

Wow... New info... Thank you for sharing...sir

@BriTheMathGuy

2 жыл бұрын

So nice of you!

Mind blown, but more in a makes me want to cry way. I need to brush up on my integrals!

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

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!

@ass123qdwqdw

6 ай бұрын

why cant to tkae x = 1??

@sahiljain1504

3 ай бұрын

Put 1+ 1/x^phi = u. Solves in second

This vid needs to be watched over and over

@BriTheMathGuy

2 жыл бұрын

Glad you think so!

This is absolutely amazing

@BriTheMathGuy

2 жыл бұрын

glad you thought so!

@simon39wang43

2 жыл бұрын

@@BriTheMathGuy thanks, I subscribed

Dude, this year has been like math explosion festinghouse. Nice.

1:39 with respect to me? Never thought I’d get any respect at all 🥺

This is beautiful

WHAT THE ACTUAL FUCK. THAT WAS SO FKIN SATISFYING

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

Yes, now I can integrate any expression, just by introducing a new function. 😀😀

Hello! :) Could you tell me, please, the name of the software that you're using in the presentation? :) TY!

When he said "pheee", I felt that

Nice!

@BriTheMathGuy

2 жыл бұрын

Thank you! Cheers!

awesome video as always! only thing wrong with them is that they end :P

@BriTheMathGuy

2 жыл бұрын

Thanks! 😄

I literally just randomly googled this video lol. But I like it :)

@BriTheMathGuy

2 жыл бұрын

Great to hear!

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?

@user-jx7cv2td4y

2 жыл бұрын

Like in proving theorems, there is probably no general algorithm in solving integrals analytically

Yo the algorithm finally recommending some good channels!

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

though I wont be able to remember in 2 minutes, it is indeed my favorite integral at the moment.

I swear every time a problem turns out like that I get a mathgasm, there, I said it

Make some videos about zeta function , please

@mathematicsmi

2 жыл бұрын

kzread.info/dash/bejne/pZ2u0MRypZjemKw.html

That was very cool :)

@BriTheMathGuy

2 жыл бұрын

Glad you thought so :)

What a brilliant integral

I hit the like button as soon as I heard you were pronouncing phi correctly LOL

(You should pay a fine for calling Phi, Fee.)

@BriTheMathGuy

2 жыл бұрын

👮‍♂️

@chanderule605

2 жыл бұрын

I mean, that's how every language but english calls it

THIS IS SO COOL

@BriTheMathGuy

2 жыл бұрын

😎😎😎😎

Mind= so blown even the quarks got split

@BriTheMathGuy

2 жыл бұрын

🤯

feynman’s technique for a parameter k in place of phi, so differentiating under the integral?

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

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

Fascinating. I'm fascinated.

@BriTheMathGuy

2 жыл бұрын

Glad to hear it :)

Loved this video! But you didn't have to do our boy Gamma like that! :P

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)

This was on cantor dust level 1…

very nice

@BriTheMathGuy

2 жыл бұрын

Glad you thought so!

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)

it's now my favourite integral lol

I substituted x^(phi)=tan^2(theta). Nice problem.

@BriTheMathGuy

2 жыл бұрын

Cool!

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

How do you get that black background?...do you literally record it in a black planted rookm with a light on you?

@BriTheMathGuy

2 жыл бұрын

Green screen :)

Does this integral have a practical use?

I guess that the golden ratio has infinite possibilities ( If you can change the value of the variables such as x)

Favoritegral.

@BriTheMathGuy

2 жыл бұрын

😎

I wonder the golden ratio has so many identities.

That was super entertaining but I still can't understand most of the things :)

Interesting video! Just a minor peeve, I think "phi" is actually pronounced fahy instead of fee. Great video!

Golden integral

sum intergral

Where's the plot? I need a plot of the distro! Where is my plot!!

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

@BriTheMathGuy

2 жыл бұрын

There are other ways to do it!

Nice

@BriTheMathGuy

2 жыл бұрын

Thanks!

Please do not confuse people! The beta function is written by B(x,y). Capital Greek beta.

@Nebulisuzer

Жыл бұрын

beta is ß not B

Golden triangle

New integral idea Int from -infinity to infinity (e^((-(x-m)^2)/(2(n)^2))dx

If you love the Gamma Function, prove that gamma(z)*gamma(1-z)=pi/sin(pi*z) for all non integer z

Shout out for pronouncing the Greek correctly!

I don’t think you can actually use the factorial for non-natural numbers. Why not use the properties of the gamma function instead?

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.

@orkedar8786

2 жыл бұрын

Lol 🤣

@lukandrate9866

9 ай бұрын

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

....i swear ive seen this....

Hehe, funny numbers.

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.

Ratio

@BriTheMathGuy

2 жыл бұрын

Ratio :)

This is the most beautiful integration problem I've seen on KZread.

I am watching this and pretend I understand everything after the word "integral."

Is it just me or is there a discord ping sound at 0:10?

Finally, someone that pronounces φ fi and not phai(I'm Greek btw)

Nah Rendering Equation is my favorite

Here's a better way of writing the β function definition: xβy=((x-1)!(y-1)!)/((x+y-1)!)

this is not maths but literature

@BriTheMathGuy

2 жыл бұрын

🧐

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.

@lukandrate9866

9 ай бұрын

Γ(1+n) = n! for all natural n Γ(1+0) = integral from 0 to inf of e^(-t) dt = 1 implies 0! = 1