A very interesting differential equation. 

Chalk-lifting is hard work.

lmao😂

Lol hhahahahah

one big one

bruh

As always with his aphorisms, he's right.

Antony Stark excellent point

Well, the easiest way, certainly

And you would be amazed at how useful that statement is, despite seeming so trivial.

Absolutely correct!!

Emphasis on "educated guess"

Like "PhD educated" guesses

And I'm sure they wouldn't deny it. You're not doing research in the classroom, you're learning basic, well-established mathematics - there is little need for guesswork

Accessible RU-vid video: "guess and check". Professional research paper: "ansatz"

It is actually a very deep statement. In the 1900's, there was a great debate in the matetatics community. There were 2 main sides, one represented by Pointcare and the other by Hilbert. The former defended that matematics required some kind of intuition by the matematicians side, while the latter defended that with methods and enough time spent they could map out the entire matematics. The side of Hilbert was winning, until Gödel came out and proved that not everything in matematics could be proven. So we are now left with the Pointcare side of the story.

Nah it makes sense that it'd be in there. Right when you see 1/r from inverse and r-1 from derivative it's familiar if you know how phi's like. (What number is itself plus the inverse of itself?)

Because e: :( busy with complex girlfriends Pi : sorry stuck in circle Phi : perfect for me 👍

@@branthebrave What number would that be? How could it be itaelf plus its inverse? I thought adding its inverse would always make a completely different number

Phi is algebraic. It pulls it's weight pretty damn good for being a generic number, it is not as talented as pi or e but very hard working ;)

These are not polynomials but yeah

@@rubberduck2078 but why wouldnt they be polynomial? They have const coefficients and x to the power of a const, isnt that the definition of a polynomial? Or does the exp have to be a whole number in order for it to be considered a polynomial?

@@hamburgerin6593 Yes, the exponent need to be non-negative integers for it to be considered a polynomial.

yeah but thats the same argument as the one used in the video. The extra step is just a matter of taste...

Exactly my thinking !

Contrary to what many women believe, it's fairly easy to develop along-term, stable, intimate, and mutually fulfilling relationship with a guy. Of course this guy has to be a Labrador retriever. With human guys, it's extremely difficult. This is because guys don't really grasp what women mean by the term relationship. Let's say a guy named Roger is attracted to a woman named Elaine. He asks her out to a movie; she accepts; they have a pretty good time. A few nights later he asks her out to dinner, and again they enjoy themselves. They continue to see each other regularly, and after a while neither one of them is seeing anybody else. And then, one evening when they're driving home, a thought occurs to Elaine, and, without really thinking, she says it aloud: "Do you realize that, as of tonight, we've been seeing each other for exactly six months?" And then there is silence in the car. To Elaine, it seems like a very loud silence. She thinks to herself: Geez, I wonder if it bothers him that I said that. Maybe he's been feeling confined by our relationship; maybe he thinks I'm trying to push him into some kind of obligation that he doesn't want, or isn't sure of. And Roger is thinking: Gosh. Six months. And Elaine is thinking: But, hey, I'm not so sure I want this kind of relationship, either. Sometimes I wish I had a little more space, so I'd have time to think about whether I really want us to keep going the way we are, moving steadily toward... I mean, where are we going? Are we just going to keep seeing each other at this level of intimacy? Are we heading toward marriage? Toward children? Toward a lifetime together? Am I ready for that level of commitment? Do I really even know this person? And Roger is thinking:... so that means it was... let's see...February when we started going out, which was right after I had the car at the dealer's, which means... lemme check the odometer... Whoa! I am way over due for an oil change here. Dave Barry

@@ernestmoney7252 Why did you write all of this ?! 😂

@@srijanbhowmick9570 Hey Srijan Here is an IQ test for you: why does the name "Dave Barry" (not my name) appear at the bottom of my post?

@@ernestmoney7252 Right after thinking, "I am way over due for an oil change here", he thought, "Dave Barry". Then he thought, "Why do I always think 'Dave Barry' after these internal monologues? Maybe I should talk to a professional".

@@Isitshiyagalombili Quite creative, but there is a more parsimonious explanation.

😂

MmmVomit well I’m not surprised he was a child doctor

Clearly you haven't met his friend, Ted.

I came to the comment section only for that reason

Doogie Howser, PhD

I was going to comment check out 3b1b loll

What about mathologer?

> Or perhaps the problems were first solved not with mathematical rigor, but with some sort of intuition This is true in a large majority of cases. There's also lucky accidents, proof by exhaustion, and so many other ways. Also, explaining intuition to someone else with enough brevity and clarity is no easy task.

Dr. Peyam also does some stuff like this.

Al Rats no, I have not.

The goal of a teacher is not just to enable a student to think on their own, but to make them unable to stop thinking. And then, that's exactly the right place to stop teaching.

Lol 🤣

I’m incredibly curious about f’(x) = f^n(x) now.

@@mateussouza3979 you made me curious too😂

Left as an exercise to the viewer :(

Numberphile needs to get on it.

This is a nonlinear DE so finding one solution doesn’t mean you’ve found all of them, I suspect there are probably more solutions are not of this general form.

Agreed. As stated in the video by Michael. I have just found one set of solutions by supposing f(x)=Ax^r but there might be others

Can u explain how did you solve for the general case? Thanks.

​@@rakshithsajjan3639 Suppose f(x) is of the Ax^r form, then f'(x) = rAx^(r-1) f''(x) = r(r-1)Ax^(r-2) f'''(x) = r(r-1)(r-2)Ax^(r-3) Until the nth derivative, which is r(r-1)(r-2)....(r-n+2)(r-n+1)Ax^(r-n). This can then be simplified to the following expression: (r! / (r-n)!) * Ax^(r-n) We know f^-1(x) = (x/A)^(1/r) You then use the same method as in the video, equating the 2 expressions and noticing that each of them must equal 1 for the equation to be true.

My calculator doesn't have a phi button but I calculated the phith root of one over phi to be 0.7427429446. Can anyone verify?

@@CandidDate, yup, that's correct: www.wolframalpha.com/input/?i=%281+%2F+golden+ratio%29%5E%281+%2F+golden+ratio%29

feef

one two three four phith

'To catch a phith'.

Do you use android? If so how do you type the dot (multiplication sign)? I only have • in my keyboard

...but then you'd miss the phi-squared-th root :O

To bed ? No!!! Nightmare!!!!

I thought the same, however, these are not polynomial functions, as the power doesn't have to be an integer. Yet i think it does still apply.

This is exactly what I came to say. Cuts out a couple steps.

1. You probably wanna say f(f'(x)) = x rather than f(f'(x)) = 1 2. Multiply a solution by (-1) doesn't necessarily yield another solution, cuz -f^(-1)(x) isn't necessarily the inverse of -f(x). ( -f'(x) is the derivative of -f(x) for sure. ) 3. The conclusion is partially correct: strictly monotonically increasing part can be defined only on positive numbers. However, we can find a decreasing solution on the negative axis. An ansatz f(x)=A(-x)^r works with A, r being undetermined negative coefficient, and it turns out r=bar(phi)=1-phi~ -0.618, A~ - (-bar(phi))^(-1/bar(phi))=-(phi-1)^phi. Putting thing together, a piecewise function defined on the whole real axis except {x=0}: f(x)= 0.618^0.618 * x^1.618 for x>0, -0.618^1.618 * x^(-0.618) for x

SW addresses the key issue with your point but I think it is worth extending both your and his remarks a bit further. 1. In fact, multiplying by -1 will never give you another solution. Since f must be either non-decreasing or non-increasing (the derivative can tend to 0 at a closed endpoint of the domain), the pre-image of its domain must be contained entirely in either the non-negative or non-positive real numbers. It follows that any non-decreasing solution must have domain bounded below, and likewise any non-increasing solution must have domain bounded above. Moreover, if f is non-decreasing then the left endpoint of the domain is bounded below by the value of f(0) (presuming this is defined, otherwise we need not worry about this), and vice-versa for a non-increasing function. Thus, we can guarantee that f^{-1)(x) >= x everywhere, and so f must be unbounded. It follows immediately that -f is not a solution. 2. Given this, one might ask if the part of a non-decreasing (resp. non-increasing) solution which lies on the "wrong" side of 0 (that is to say, the negative part of the domain of a non-decreasing solution) can be reflected in this manner and remain a solution. This too is impossible, since the values of f on the negative reals must be negative (resp. positive reals must be positive), and so negating the function results in an ill-defined inverse and thus does not solve our equation. 3. In fact, all that I've said above regarding non-increasing solutions can be safely disregarded, since there cannot be solutions defined on the negative real line. To see this, observe that any such solution must take on every negative value, be non-increasing, and have domain bounded above. It is not hard to see that this is a contradiction, since these last two imply f is bounded below. 4. We can do even better than this: There are no non-increasing solutions. Observe that any non-increasing solution must have domain bounded above by 0, and in fact must have value 0 at 0. But then it must take on a negative value somewhere, and be non-increasing. This is a contradiction. Thus we need only really worry about finding solutions with domain bounded below which are non-decreasing! (Note that SW has gone wrong in their third point. You cannot hope to take (-x)^r and get a real number, this is generically complex (and in fact will be for r = phi, so there really are no solutions on the negative reals)). 5. There are some comments on initial data that the video omits by restricting to the case f(0) = 0. One sees from my arguments above that if we pose data for the value of f at 0, then we must have that f(0)

@@erickilgore4869 Thanks for your long comment. I agree with your reasoning based on the presumption that f(0) is defined. Based on the presumption we can argue f(0)=0 and no such f(x) for x being negative. However, if we don't require the point x=0 to satisfy the original eq, my solution defined in the domain (-inf, 0) still works depite a typo (I was missing a minus sign before x). So my solution is: f(x)= 0.618^0.618 * x^1.618 for x>0 -0.618^1.618 * (-x)^(-0.618) for x

Drop down and GIVE ME TEN REASONS YOU WANT THIS PHD!

I'm watching this 3 years removed from upper division university math courses. I've been working and not really using everything I learned since then. I look back at something like this and think, "wow, I used to be able to do that!". Not so much now. To share your sentiment, this and a few other videos have made me want to get back into it just for the hell of it!

Doesnt change anything when you are solving via guess and check

V K phi

Well actually the greek letter φ is pronounced "fee". But english speaking folks keep mispronouncing it "fye" just as some other very common pronounciation mistakes. As like "Youler" is actually pronounced "Oiler"

V K Actually, he’s saying “phi”, but what he writes looks like a psi to me. It’s a cursive variant of phi that is very unfamiliar to me.

*quentin tarantino entered the chat*

@@simonstockinger9293 They also pronounce Pi (π) as Pie, not as Pee. I think "pee" would be too funny for English speakers.

The problem I have pronouncing phi as 'fee' is just consistency. Your pi letter is pronounced the same as our p. Because of that we pronounce pi as 'pie.' So just for consistency of sound, I pronounce phi as 'fie'

In French, phi gets pronounced as fee, but pi also gets pronounced as pee. ^_^

Sadly many words are mispronounced or wrongly written by non-native speakers. It is particularly a problem in the USA. It used to be in the UK but it seems less of a problem nowadays. An example is Krakatau, which Americans demand is written and spoken as Krakatoa and get extremely abusive when one points out the correct spelling and pronunciation. But then they still use bbls as a measure of volume.

@@boffeycn UK still using miles

The brits among you yell at me, for how I say the letter "phi". But ask a Greek, they won't deny, there's something odd in saying "phi"

Capitalization

It's really good, but I know someone who does even better. Then again, you have to give credits for bringing it to youtube, for free, so...

And the most ripped too, damn...

He's not British so he is a math professor.

I also like invoking phi-bar = 1 - phi = -1/phi as needed.

It is not a differential equation.

Adding that word to my vocabulary

You should have published your results at least on research gate. I got the same conclusion that there is a unique solution of type f:[a, +inf[ -> [a, +inf[. I have done this with inductive definition and convergence theorems. This is a longer proof, but can be applied to a more general case like f'(x) = F(f^-1(x)).

I don't think that function even has an inverse.

@@behzat8489 I think the inverse function of y=0 would be x=0, but that's not a function

@@ViniciusTeixeira1 yes you are right. For one moment i thought f(x)=0 as a point (0,0)

For a function to be invertible, you need perfect one-one correspondence in the domain and range; f(x) = 0 would only be invertible for x is a single constant (say a) and f^-1(0) = a. So technically yes it is invertible but not very interesting.

my exact thought process lmao

Almost generally, actually; I supposed analicity at 0. For the more general case the method would be nonetheless the same.

Differential equations is kinda of like algebra, but instead of values, your variables are functions.

Differential equations is kinda of like algebra, but instead of values, you lie to your parents about how well grad school is going

@@jiffylou98 that too.

Logically such thing exists, it's up to your genius to find it.

C*e^(x+t) = (C*e^t)*e^x = D*e^x, so the addition of t just changes the value of C. Similarly, Asin(x+c) + Bcos(x+t) = A(sinx*cosc + cosx*sinc) + B(cosx*cost - sinx*sint) = (Acosc-Bsint)sinx + (Asinc*Bcost)cosx = Csinx + Dcosx, so the addition of c and t here just change the values of A and B. The given solution is general and includes these cases.

Or every time he says "great".

@@watsisname one or the other. i need my liver

LOL. Don't drink and derive. :P

@@DancingRain Comment of the century

@@watsisname or "go ahead and.."

Which raises an interesting point. You might also find solutions if you can find a class of functions whose derivatives and inverse functions jump into the same class as each other, but a different one from the original class. Seems harder to think of such a case, though... Fred

@@pnneeth weirdchamp

@@pnneeth u realise what u did is take offense on someone elses behalf on what i thought was a compliment?

Why is this comment section so horny today?

Somebody I used to know No idea.

Think you need to check Penn’s personal website. Looks like he takes publish or perish seriously.

@@vladimirkhazinski3725 yes!

I had the same idea at the same moment bro, we simply have two functions of x equal to each other, by unicity the coefficient and the exponent just have to be the same.

So... In the end of the video Michael is giving us some really nice follow up questions and I actually went on looking at the third follow up question about other differential equations. If anyone is interested in taking a look at my solution, I'll leave a link here to an overleaf document. Hopefully I am not making a complete fool of myself. Note: I'm not a professional mathematician in any way. I just like doing these kinds of things recreationally. www.overleaf.com/read/mtywsgrzsbwh

I would argue for pedagogical reasons his explanation is worth the effort ;).

It's pretty east to pop these guys into Wolfram Alpha. Since it's a power series, the plot is not especially interesting. www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+%281%2F4%29%5E%281%2F1.61803%29+*+x%5E1.61803

@@vaxjoaberg naa, really not, but not showing them does not fit the format he is presenting in. Plotting the inverse/derivative alongside would be nice (sorry, i don't know if wolfram can do that)

It's x^2 and sqrt(x), scaled by some phi (ish) factor. All parabolas look the same, anyway :-)

Marvellous! Your intuition seems to work like mine.

Either that or he came up with that after already seeing the answer, and he wanted to seem smart.

@@gamerdio2503 That´s possible but I am over 30 and I am out of old doing something silly like that.

What is metallic ratio

@@NahinAndroid A generalization of the golden ratio.

How did you show that?

Francesco Cristantiello Wolframalpha :^D

Do you know the solution?

Hope ur not trolling with a non integratable function :D

The rigours proof is on my channel. But still working on kinks here and there. Still haven't perfected it.