A very interesting differential equation.

We present a solution to a very interesting differential equation. In particular, we find a solution to the differential equation f'(x)=f^(-1)(x). That is, we answer the question: When is the derivative of a function the same as its inverse?
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[@antonystark9240]

"The only good way to solve a differential equation is to know the answer already." --- Richard Feynman

[@AndyZach]

As always with his aphorisms, he's right.

[@jayyoung9766]

Antony Stark excellent point

[@Hecatonicosachoron]

Well, the easiest way, certainly

[@joda7697]

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

[@dheerajsharma355]

Absolutely correct!!

[@carlosdanielvelazquezflore6512]

I wonder how many integrals I need to do until my arms get that big

[@md2perpe]

Chalk-lifting is hard work.

[@user-lw5oc1tt8k]

lmao😂

[@MohaMMaDiN55]

Lol hhahahahah

[@_kopcsi_]

one big one

[@andrefreitas9936]

bruh

[@Eichro]

"Educated guesses and checking them is how a lot of pure research-level mathematics is done" I wish my teachers were in this room to hear this

[@ultraollie]

Emphasis on "educated guess"

[@p-aluneau5136]

Like "PhD educated" guesses

[@turtlellamacow]

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

[@donporter1247]

Accessible KZbin video: "guess and check". Professional research paper: "ansatz"

[@oguzcanoguz5977]

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.

[@Celastrous]

*Asks some incredibly hard questions* "Ok well, this is a good place to stop."

[@35571113]

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.

[@dananajj]

Lol 🤣

[@mateussouza3979]

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

[@dheerajsharma355]

@@mateussouza3979 you made me curious too😂

[@shmojelfed9664]

Left as an exercise to the viewer :(

[@titusng2483]

Her: He must be thinking about another woman Him: What function's derivative is same as its inverse?

[@ernestmoney7252]

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

[@srijanbhowmick9570]

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

[@ernestmoney7252]

@@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?

[@Isitshiyagalombili]

@@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".

[@ernestmoney7252]

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

[@Darkstar2342]

I would have done just a "coefficient comparison" at the step where you have rAx^(r-1) = (1/A)^(1/r)x^(1/r). Because for these two polynomials to be equal (for all x), the factors in front of the x as well as the power of x have to be equal at the same time. So you immediately get r-1 = 1/r and rA = (1/A)^(1/r), solve the first for r and use it in the second to get A. No need to shuffle constants and x's left/right.

[@rubberduck2078]

These are not polynomials but yeah

[@hamburgerin6593]

@@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?

[@TC159]

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

[@rodrigolind5665]

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

[@LachezarTsakov]

Exactly my thinking !

[@diegoparodi3854]

Completely random equation: exists Phi: let us introduce ourselves

[@branthebrave]

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?)

[@ourgoalisto6737]

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

[@SP-qi8ur]

@@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

[@hybmnzz2658]

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 ;)

[@aienbalosaienbalos4186]

I have only seen two videos. But it seems like these videos have a really, really good level of both rigor and just that general hunger/curiosity for finding knowledge, not because it’s useful or for the knowledge, but for the pursuit itself. Which makes this channel unique as far as I know. It has the same joyful take on problems as 3b1b, that is not “Here is problem. Here is solution. Here is proof”. The process of finding the solution is the main point of the video, not the solution itself. And while 3b1b is clearly superior in the production quality (in which it is the absolute best), they usually leave out the mathematical rigor. Other sources that do have the mathematical rigor, like differential equation textbooks, at least the few I’ve read, I may have been unlucky, tend to be so incredibly dull and uninterested in the topic, perhaps in favor of brevity and efficiency?, and pay little importance to how the problems were solved, focusing only on presenting solutions to equations and proofs. At least me personally, I find mathematics textbooks to care little for how much “sense” a proof makes, or how natural a reasoning it is, only that it is technically correct. Of course these might all be consequences of my non-mathematician brain. Or perhaps the problems were first solved not with mathematical rigor, but with some sort of intuition, that only later was rigorously proven with some technical ungodly proof. Furthermore, as an engineering student I am all too tired of having to learn “true” things, that are very powerful and useful, without having the time or expertise to know how we know they are true. I tend to give them a crack, but usually I find I am not prepared and don’t have the time to completely understand/prove, so just end up figuring out intuitions, for which 3b1b helps a lot. It’s very refreshing to just wonder about things, and not have to worry about their usefulness. For all these reasons, I am glad to have found this channel. Thanks for sharing these.

[@gregoryG540]

I was going to comment check out 3b1b loll

[@shoam2103]

What about mathologer?

[@shoam2103]

> 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.

[@badhbhchadh]

Dr. Peyam also does some stuff like this.

[@aienbalosaienbalos4186]

Al Rats no, I have not.

[@ButiLao44]

Me: "Oh wow I'm so glad I'm finally through with my math final and am rid of math forever" KZbin: hey wanna watch this video about MATH? Me: *slams play button*

[@SoonRaccoon]

I never knew Neil Patrick Harris was so good at math.

[@jocider5698]

😂

[@justinhill8170]

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

[@dananajj]

Clearly you haven't met his friend, Ted.

[@oliver_siegel]

I came to the comment section only for that reason

[@acobolew1]

Doogie Howser, PhD

[@christianherenz5072]

That was quality 15 minutes.

[@stevenpurtee5062]

This was a nice example. I'd argue that you made it more complicated than it needed to be after 7:45. The only way those two power functions can be equal for all values of x is if the coefficients are equal and if the powers are equal. That gives you the same two relationships.

[@fadiel-riachi6675]

Nice video! I immediately felt like solving for the general case. If you solve the following equation : nth derivative of f(x) equals inverse of f(x), you get a very nice closed-form epression for f(x)=Ax^r where r is solution to r^2-nr-1=0, the so-called ''metallic ratios'' and A=( (r-n)!/r! )^(r/(r+1)). Something interesting also happens when n is odd....

[@shaileshrana7165]

Numberphile needs to get on it.

[@possessedchair8144]

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.

[@fadiel-riachi6675]

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

[@rakshithsajjan3639]

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

[@fadiel-riachi6675]

​@@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.

[@Beniguitar94]

Very fun exercise! Thank you for sharing. Though I followed similar rationale, I solved it differently (and with way less algebra): f'(x) = f-1(x) f(f-1(x)) = x Hence: f(f'(x)) = x Assuming a shape of the form: f(x) = A · x^r f'(x) = A · r · x^(r-1) f(f'(x)) = A^(r+1) · r^r · x^(r^2 - r) = x = x^1 From here, it is easy to see that terms of x need to be equal in both sides, hence: r^2 - r = 1; r = (1±sqrt(5))/2 Then, given that r^r is a mess, you define A to cancel out the mess: A^(r+1) · r^r = 1; A = (1/r^r)^(1/(r+1)) Finally getting: f(x) = (1/r^r)^(1/(r+1)) · x^r, where r = (1±sqrt(5))/2

[@theuserings]

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

[@AlexTrusk91]

Thanks, the first 5 minutes helped me with a wildly different math problem. When you talked about classes of functions, that gave me the right hint what i had to look for

[@mina86]

7:33 - at this point you could have just said that by the power of polynomials, either rA = 0 (which is easy to show that would not give solutions) or rA=(1/A)^(1/r) and r-1=1/r.

[@vivien2184]

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.

[@alert.272]

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

[@jerrysstories711]

When you find yourself thinking about "phith" roots, it's time to go to bed.

[@dananskidolf]

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

[@dominiquebercot9539]

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

[@KarstenJohansson]

I need more differentials in my workout.

[@karangupta4978]

Love the frequent uploads, keep going!! Your videos are very informative for high school students preparing for competitive exams.

[@oliverinspace9252]

Great video! A nice intuition of why φ arises from this DE is that φ and its conjugate, -1/φ, is 1 more than their reciprocal (i.e. 1/φ and -φ). For the explanation below, we consider the case r=φ, as outlined in the video. As the function is in polynomial form, the derivative of the function would be φ-1, thus giving you 1/φ. Similarly, the inverse of f(x) would also be in polynomial form, with reciprocal power, as we take the root of the power to obtain the inverse. Considering the case where r=-1/φ, we would find that A=(-1)^φ=e^(iπφ), thus obtaining a complex function in polynomial form. Sadly, this does not give a solution to the DE because the derivative differs from the inverse by a factor of 1/φ.

[@abelpalmer552]

"The phith root" lol

[@CandidDate]

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?

[@JivanPal]

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

[@Fasteroid]

feef

[@PhilipGLee]

one two three four phith

[@AndyZach]

'To catch a phith'.

[@intergalakti176]

I don't think I would call this a differential equation, its more of a functional equation, as we want f(f'(x))=1. Sadly, the machinery of ODEs doesn't work in this case, i.e. Picard Linelöf isn't applicable; that would have maybe answered the uniqueness question. However, a view remarks are in order: - We can multiply any solution by (-1) to get another, different solution. - Any solution must be strictly monotonically increasing or decreasing if we want it defined on a connected subset of R, so that the inverse exists. By the above point, we may assume wlog. that it is strictly monotonically increasing. But then f'(x)>=0, so f^-1(x)>= 0 too, so f can not be defined on the whole of R, but only on nonnegative numbers!!! The function in the video is obviously only well-defined for nonnegative numbers, too! Therefore, no solution can be defined on all real numbers.

[@sw4379]

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

[@erickilgore4869]

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)

[@sw4379]

@@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

[@usptact]

When you solve enough differential equations, you get six pack.

[@JynxSp0ck]

He inhales so much I'm 4 minutes in and already exhausted.

[@SwapnilAnand1998]

This is one of the most beautiful math video involving one of the most beautiful mathematical equation I've ever come across.

[@digitig]

"We're almost at the end" - the video progress bar begs to differ.

[@ПавелЩичко-е3ж]

Why should one calculate the inverse of f, when one can just compose candidate f' with f to show that it is the inverse indeed.

[@oldnordy2665]

I transformed the question into f( f'(x) ) = x, and also immediately thought of (not necessarily integer) polynomials fitting the bill. Starting with x^n, you immediately get that you need to satisfy n(n-1) = 1 and the golden ratio solution and the one with the negative sign in front of the square root. Then you only have to worry about the constant in front, which can take on different forms (but same numerical value) because of the properties of "phi." Seems a bit faster - but then I did not have to write on a black / white bord ... I try to Americanize my Greek letters' pronunciation, because otherwise my students don't know what I am talking about. Anyone else say "vfee?" I admire the author for doing so - at best, I give both options (mjoo, mü ...).

[@Tiqerboy]

When I took a course on differential equations, and it has been a LONG time, I'm sure glad that question wasn't on the final. I needed a coffee to get through all that. But that guy is better than the math profs I did have, LOL

[@observ_2008]

When you already know some math but you're still intrigued into it. Nice content dude.

[@jimgoldie967]

What a delight to have such a lucid presentation of how to approach this curious differential equation. It made me feel as though I still can do it decades after Math 46 (Diff. Eqns.)!

[@alanxie5907]

I keep have this feeling that this dude would somehow suddenly start giving you random workout tutorials🤣

[@peter_godman]

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

[@samallen3327]

This video singly handlily made me want to get back into math. I may know an approach I could take to finding out if this is the unique solution but it sounds over complicated. I may need to do some research!!

[@chirrrs]

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!

[@ichigo_nyanko]

I used the educated guess Ax^B just like you, but then I used the fact that if f'(x)=f^-1(x), then f(f'(x))=x. That gives you A(ABx^(B-1))=x which then gives you the system of equations B^2-B-1=0 (instantly recognisable as B=phi) and A^(B+1)*B^B=1 which can be easily rearranged to get A=phi^-1/phi

[@klausg1843]

Very nice, as always.🤸‍♂️But I could propose to start by getting rid of the inverse function by replacing x by f(x) to get f’(f(x) = x, which is easier to manipulate

[@wynautvideos4263]

Doesnt change anything when you are solving via guess and check

[@dcterr1]

Wow, I never thought about this type of differential equation before. Very interesting, and great solution!

[@Someone-cr8cj]

I am greek and I have to thank you for pronouncing φ the right way

[@PaulMartin-qt9ux]

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'

[@TheOiseau]

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

[@boffeycn]

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.

[@leswhynin913]

@@boffeycn UK still using miles

[@daviskeene363]

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"

[@Serghey_83]

1. Справочник по дифференциальным уравнениям с частными производными первого порядка - Зайцев В.Ф., Полянин А.Д. 2. Справочник по обыкновенным дифференциальным уравнениям - Зайцев В.Ф., Полянин А.Д. 3. Справочник по интегральным уравнениям. Методы решения - Манжиров А.В., Полянин А.Д. 4. Справочник по обыкновенным дифференциальным уравнениям - Э. Камке

[@lawrence18uk]

The warmup question is quite relevant for today, where we're considering exponential growth, where the rate of growth = the growth. Now that really is scary.

[@lukedavis6711]

Capitalization

[@おき-m6x]

初めて外国人が数学の授業をしている動画を見ましたが、英語の勉強にもなって素晴らしいです。 I watched the video that foreigns taught mathematics for the first time I was interested in this video because I can learn both math and English

[@pianoclassico718]

for the third question , i believe that taking the inverse function on both sides results in an interesting equation that is not so different from the current one f^-1(f'(x))=f(x) as for the direct solution we can get to f(f'(x))=x we can reduce it so f(f(x)=1/f'(x) ( very similar to question 3) which may yield to some hint towards the analytical solution without a guess, though I am not sure.

[@tayyihcheung7450]

There is a much simpler way to solve it: rewrite the equation as f(x)df(x)=dx; integrating both sides yields (1/2)f(x)_squared=x+c. Hence, f(x)=(+/-)square_root(2x+C), where c, C is an arbitrary constant. Because the problem asks for only one solution, take the positive one and set C=0.

[@CotopaxiAH1968]

That was quite enjoyable! The right combination between creativity (educated guesswork) and technique.

[@sams6454]

Seriously, keep up the great work. I could see you contributing like 3blue1brown. Just interesting problems and a chalkboard. You don't need anything too fancy. People have been doing amazing math on the chalkboard for centuries

[@VK-sp4gv]

The feeth root of one over fee. I love it.

[@dat2125]

V K phi

[@simonstockinger9293]

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"

[@StuartSimon]

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.

[@nicktosti7487]

*quentin tarantino entered the chat*

[@p-kotov]

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

[@Chewy427]

This is such a fascinating equation. After staring at y=x for a while it seems obvious that the solution would be a power function whose exponent is between 1 and 2. If the exponent of f was greater than 2, then both it and its derivative would be superlinear but the inverse would be sublinear. If the exponent of f was less than 1, then f and its derivative would be sublinear but the inverse would be superlinear.

[@sujitbhattacharyya3705]

That's probably the best differential equation I've ever seen.

[@frenchimp]

It is not a differential equation.

[@txikitofandango]

I did it. Yay! I got to the same equation as you: (1/b)^(1/a)*x^(1/a) = abx^(a-1) but then I set the exponents equal to each other AND THEN SEPARATELY set the coefficients equal to each other. Setting the exponents equal gives you the quadratic equation that leads to the Golden Ratio for a. And then it's easy to solve for b in the coefficients. It's always handy to know that 1/phi = phi - 1, it saves you from some gnarly expressions involving phi.

[@txikitofandango]

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

[@pacolibre5411]

It feels weird to think of this as a differential equation since you can’t make a BVP or IVP out of it. Its very neat

[@xinfuchen5355]

The general solution is f(x)= a^{1-a} [ x-A]^a +A, x in [A,\infty) where A is any real number and a is the root of a=1+1/a.

[@joshbarthelmess4796]

This man is the swollest math teacher I’ve ever seen in my life

[@JoaoVictor-gy3bk]

Can you elaborate on that?

[@khemirimoez8661]

Suppose there exists a math teacher more swole but that's just wrong. QED ■

[@gentlemandude1]

Have you never seen Prof. Leonard? kzbin.info/www/bejne/rpeQZHSKr6usgaM

[@joshbarthelmess4796]

gentlemandude1 I had not. Hot damn.

[@foreachepsilon]

Pietro Bouselli

[@nacl-sn7156]

AAAA, Why do I only get this video now?! I've studied this problem 6~7 years ago when my maths teacher couldn't answer... From what I remember, I proved there are an infinite amount of solutions: one for each positive real number "a" (the solution is then defined on "[a, +inf[") plus an other one. The one you show would be for "a=0". There's an other one with the same exact shape I call "a=0-", where you solve the same steps staring with "f(x) = A*(-x)^r" and the solution is defined on "]-inf, 0[". I only ever found the Taylor series of the other ones, no simple way to write them out... Still have the Python scripts I used to plot them! ^^ You can approach a solution to the DE with function series: "f_0(x) = x" and "f_{n+1} = \int_{a}^{x} f_{n}^{-1}(t) dt + f_{n}(a)". You get very interesting results, like how for each solution "a" in "[0, 1]" there is an other solution "b" in "[1, phi]" so that both functions overlap for "x >= b". Also tried a lot in the complexes, with binary functions, matrices... Gosh, I have pages and figures upon scripts about this simple equation that obsessed me for a solid year and a half, I don't know how I feel about this now!! >< (Great video by the way! Never thought it would take a maths video for me to post a KZbin comment...)

[@joshuabirkhoff9704]

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)).

[@gagsgsiso2378]

Just saying that f(x)=0 is a solution for f'(x)=f(x) Nevermind, just realized it's included in f(x)=ce^x... oops

[@amaarquadri]

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

[@ViniciusTeixeira1]

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

[@behzat8489]

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

[@ravinchowdhury5215]

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.

[@agginswaggin]

my exact thought process lmao

[@mathurinrouan3971]

A pretty difficult exam from France is based on showing that this equation has a solution, and that it is unique, but without expliciting it.

[@adenpower249]

The world's best maths professor strikes again.

[@cmilkau]

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...

[@hziebicki]

And the most ripped too, damn...

[@mikehodge5251]

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

[@grobertoac2430]

2:02 I think that it works to add C, but we have to add it to x, which makes e(x+C) = e(x)*e(C) = C'*e(x), where C' = constant. It is the same, but I see it like this in my mind, because: When you have f'(x)=f(x), then f'(x)/f(x) = 1. And f'/f is the derivate of ln(f), which means ln(f(x)) = x + C, where C is a constant. So f(x) = e(x+C) = C'*e(x). So, in a way, the constant is still an additive constant when we primitive

[@leswhynin913]

This seemingly simple differential equation really yields an interesting result. Tomorrow I'll have to tackle the follow up questions

[@AndyZach]

How'd you do on the followup questions?

[@leswhynin913]

​@@AndyZach 1) I'm not sure how it can be solved without guess and test. I looked at Laplace transforms but I don't think it's possible because of f(-1)(x). 2) It is difficult to prove uniqueness without a direct solution method, plus what about non elementary functions or yet to be discovered functions? 3) f'(x)=f composed of itself, I think has solutions like y= (0.5*(1+i*sqrt(3))^(2/(1+i*sqrt(3))*x^(0.5*(1+i*sqrt(3)) and the same but with the complex conjugate. It's done using the same method as the video, but I am too lazy to back check this right now.

[@leswhynin913]

@@AndyZach If you are interested, I saw today that Dr. Peyam solved the number 3 follow-up question here: kzbin.info/www/bejne/mYnTd3eifq6WsMk. His answer looks equivalent to mine from yesterday. What timing!

[@luck3949]

From the moment where you have rA x^(r-1) = blabla* x^(1/r) you can say that since it holds for all x, then it also holds for case when x=1, and therefore rA = blabla. Therefore you can cancel out rA and blabla, and get x^(r-1) = x^(1/r) immediately. I really love this trick of assignin x=1 or x=0 in equations that hold for all x, it allows to quickly get rid of many letters at once.

[@BobCliffe]

Nice video! I played with something similar before: (f^{-1})’ = (f’)^{-1}

[@bentupper4614]

Similar challenges from Douglas Hofstadter's Metamagical Themas: 1) Find a real-valued function f: R -> R where f(f(x)) = 1/x. 2) Find a real-valued function g: R -> R where g(g(x)) = -x

[@harrystuart7455]

Does anyone know if there are any other solutions? Would be really interesting to see if there are any or if this is the most general solution

[@juliendodet2513]

If you derive the DE you get f''(x)=(f-1)'(x)=1/f'(f(x)) But since f'(x)=f-1(x), f'(f(x))=f-1(f(x))=x, hence f''(x)=1/x Integrating twice you get f(x)=x ln(x) +Ax + B (A and B are constants)

[@yonil256]

Great video as usual! Love all of them so far. But what about f(x) = SQR(2x)? f'(x) = 1/ SQR(2x)

[@KolasName]

I also couldn't get it until I've understood it is inversed function, means if Y = f (x) then x = f ^ (- 1) (Y) not (f (x)) ^ (- 1). I wasn't taught so and I don't like this notation. it's misleading, think, finv is better.

[@belemusic]

I solved this in another way without putting x on one side and constants on the other side. I just simply compared the coefficient one step before which for me seemed a little easier and resulted in the same outcome. Great video btw, really interesting stuff!

[@JureGrg]

Hello! I'm excited about your channel. Your lessons are great. I will take the way you use the board as an example, as I am a math teacher myself. I conclude that many lessons have also been influenced by quarantine and your sports energy has focused even more on math. Thank you for many lessons and best regards, Jure Grgurevic!

[@S0meWho]

I've read several comment below and noticed that nobody mentioned that for two problems of warm-up we can add several solutions. For the first one C*e^(x+t) (where t and C are constants) is the solution as well as A*sin(x+c)+B*(x+t) (where A, B, c and t are constants) is the solution for the second one.

[@maxchemtov3482]

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.

[@ev3rything491]

I tried solving the problem and when I reached the equation at 7:33 in the video I did it a bit differently than you. I am wondering if my step is legitimate or not. With regards to the equation r*A*x^(r-1) = (1/A)^(1/r)*x^(1/r) I simply made the assumption that in order for equality to hold for all x, the exponent of x on the left side has to be equal to the exponent of x on the right side. That is r-1=1/r. Furthermore the coefficients has to be equal. Is this an OK approach?

[@ev3rything491]

@@vladimirkhazinski3725 yes!

[@enzbenzz]

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.

[@ev3rything491]

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

[@konstantinos3184]

(have not checked thoroughly for similar answers, sorry if I missed one), Let g = f^-1. Then g' = -f^-1 f' f^-1 = -g^3. => dg/g^3 = -dt ... etc.

[@ichbinluis2343]

I have not even completed Pre-Calculus, I don’t know what I’m doing here.

[@itar10n]

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

[@jiffylou98]

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

[@itar10n]

@@jiffylou98 that too.

[@raphaelreichmannrolim25]

I was very pleased with your answer to the problem, mr. Penn. So beautiful... Through expansion of the respective functions into power series, I was able to find generally an equivalent system of equations that the Taylor coefficients of f(x) must satisfy to be a solution. They are complicated, however, and I won't try to go any further.

[@raphaelreichmannrolim25]

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

[@si48690]

It's beautiful that golden ratio pops like that serendipitously!! Awesome

[@theimmux3034]

Adding that word to my vocabulary

[@stone7327]

I understood little of this video (same with many of your other videos), but what I did understand I found fascinating.

[@ewoud2688]

sometimes a video like this one has to pop up to remind me how beautiful maths is :)

[@mitancheyflorian6570]

6:02 if y=A*x^^r it doesn't means that x=A*y^^r. It is x=(y/A)^^(1/r)

[@joshuaburlington2898]

Man, I kept getting lost messing around with the relation [f^-1]' = 1/(f'(f^-1))

[@thiagomilanezi1026]

The beauty of mathematics... simply amazing! 👏👏👏 Thank you for this great class! 🙏

[@lordthiccusiii]

This guy looks likes ur average white late 30s maths teacher if he lifted

[@98danielray]

@@pnneeth weirdchamp

[@lordthiccusiii]

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

[@somebodyiusedtoknow2012]

Why is this comment section so horny today?

[@NStripleseven]

Somebody I used to know No idea.

[@williamlyerly3114]

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

[@exoplanet11]

At 9:00 when the equation: r - 1 + 1/r = 0 Appeared, I solved it as: r-1 = 1/r This equation itself suggests the original problem in that r-1 is what the exponent becomes after a derivative is taken and 1/r suggests the inverse. How delightful to see it all solve for φ !!!

[@JustMaiyak]

I cannot concentrate on what this class is about. I wonder why. 💪

[@paulstclairterry6971]

I am a mathematician with a focus on set theory and metamathematics and differential equations are not really part of my world. That said, I have a deep love-affair with all mathematics and so enjoyed your exposition hugely. I have one material criticism - your pronunciation of the Greek phi. I am ancient and read Greek as well as Latin at school. Do you remember Jack and the Beanstalk? The giant would announce his evil intention with the prelude, "Fee, Fi, Fo, Fum…". You pronounced the symbol as "Fee". Admittedly, no ancient Greeks remain around today to check our pronunciation but, by convention, we pronounce the symbol as "Fi" to rhyme with eye. Yes, it is the sort of criticism that you would expect from someone like me! 🙂

[@molybd3num823]

oh you haven't seen Flammable Maths pronouncing phi 'fa'

[@Joe-nh9fy]

This is really cool. Does anyone know examples in nature that use this differential equation?

[@liviu445]

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

[@KW-12]

Woah!!! that the golden number is the solution to r is really interesting! I never imagined that property for this number. If you use the inverse function’s derivative theorem you can tell that this also satisfies (d^2f(f(x))/dx^2)*(df(x)/dx)^2=1 and df/dx(df(x)/dx)*d^2f(x)/dx^2=1

[@trbz_8745]

Take a shot every time he says "phi"

[@watsisname]

Or every time he says "great".

[@Metalhammer1993]

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

[@DancingRain]

LOL. Don't drink and derive. :P

[@jiffylou98]

@@DancingRain Comment of the century

[@andrewtaylor8899]

@@watsisname or "go ahead and.."

[@xinfuchen5355]

Similarly, for any positive constant A>=1, the following initial value problem has a unique solution: g'(y)= 1/ g(g(y)) for y in (A, infty) g(A)=A

[@Rundas69420]

Just seeing that differential equation made mw think: In the end, either e, pi or the golden ration must pop up from nowhere.

[@gregoryfenn1462]

Are there any areas in maths that this solution would be applicable? Like for example the exponential and trig functions are useful for describing growth and rotation, which have many applications. But I can't think of a reason you'd need f'(x) = f^-1(x) :/ . Still very interesting to see how the solution is found though.

[@BeattapeFactory]

this video kinda blew up. thats lots of clicks in one day for a small math channel

[@HotsumaOboro17]

I had long been searching for an answer to this question, suspected that the phi number was involved. Now I already know the answer. It is very beautiful.

[@yoavshati]

Doesn't A*exp(x+B) also work for the first question?

[@carlohu9745]

It's the same answer, cause you can combine A*exp(B) as exp(B) is a constant.

[@photonicsauce7729]

@@carlohu9745 yep ur right

[@txikitofandango]

Didn't he explain that this doesn't work? Because the inverse of exp is log, and the derivative of exp is exp, and so log cannot equal exp.

[@blugio]

A*exp(x+B) = A*exp(B)*exp(x) = C*exp(x), so it's the same solution in the end

[@vinlebo88]

@@txikitofandango OP refers to the first "warm up" question, not the main one.

[@GeorgeTsiros]

I still remember the first day of our introductory diff eq class. Prof was like "okay. What is the first method of finding a solution to a diff eq?" (silence in class). "Guess" (class is now mildly confused). "No, I mean _the first method is guessing the solution_ " (class is thoroughly confused as to how this could possibly work). "Let's go through some examples".

[@alexeyvlasenko6622]

Is anything known about the uniqueness of this solution? In particular, the solution doesn't have any adjustable parameters, so can it really be the general solution to the equation?

[@ktuluflux]

The second follow up question is what I was thinking about through the whole video. What if I didn't think of Ax^r as a possible class of solutions? Is there an explicit way to find all families of solutions to this?

[@Julschgan]

You can see that is must be the golden ratio way earlier from the exponents: 1/r = r-1 is solved by the golden ratio

[@active285]

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

[@TheBrutalDoomer]

Maybe that's better for the explanation, but a huge piece of calculating values of A and r can be simplified by saying: "Two power functions should be equal for any x. Thus, LHS x power = RHS x power and LHS coefficient = RHS coefficient, which results in a system". Yet that's a bit more implicit.

[@Killerspieler]

"this is a good place to stop" NOO!!!! plot the function dammit! yes, they look kind of boring, but this is a video, you can not just not show that ... they do still have a nice curvature and intersect at (phi;phi). kind of cool.

[@vaxjoaberg]

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

[@Killerspieler]

@@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)

[@WackyAmoebatrons]

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

[@cardelf6159]

1:45 f(x) = 0 is also a solution because f(x) = f'(x) = 0 Edit: I suppose that if C = 0, then Ce^x = 0 and therefore, 0 is accounted for.

[@NumdegasedUHC]

“The phifth root” lol

[@connormckelvey8986]

Im about 2 years late to the comments on this video, but c*e^x isnt the only family of functions whos derivative is itself. b*e^(x-a) is a function which is its own derivative, and passes through the point (a,b).

[@easymathematik]

For the third quetion my feeling tell me, that the so called "metallic ratios" are involved.

[@nahidhkurdi6740]

Marvellous! Your intuition seems to work like mine.

[@gamerdio2503]

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

[@easymathematik]

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

[@NahinAndroid]

What is metallic ratio

[@easymathematik]

@@NahinAndroid A generalization of the golden ratio.

[@neologicalgamer3437]

Y prime = Y inverse, inverse function of x = function of y, therefore Y prime = X, Y equals integral of X, therefore y=(x^2)/2+c, no?