A very interesting differential equation.
Вставка
- Опубліковано 12 тра 2020
- 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?
Please Subscribe: ua-cam.com/users/michaelpennma...
Personal Website: www.michael-penn.net
Randolph College Math: www.randolphcollege.edu/mathem...
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
Books I like:
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Number Theory:
Crisman(online): math.gordon.edu/ntic/
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Calculus:
OpenStax(online): openstax.org/subjects/math
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
"The only good way to solve a differential equation is to know the answer already." --- Richard Feynman
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!!
I wonder how many integrals I need to do until my arms get that big
Chalk-lifting is hard work.
lmao😂
Lol hhahahahah
one big one
bruh
"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
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 UA-cam 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.
Her: He must be thinking about another woman
Him: What function's derivative is same as its inverse?
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.
I never knew Neil Patrick Harris was so good at math.
😂
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
Completely random equation: exists
Phi: let us introduce ourselves
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 ;)
*Asks some incredibly hard questions*
"Ok well, this is a good place to stop."
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 :(
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.
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 !
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.
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.
Me: "Oh wow I'm so glad I'm finally through with my math final and am rid of math forever"
UA-cam: hey wanna watch this video about MATH?
Me: *slams play button*
That was quality 15 minutes.
"The phith root" lol
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'.
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....
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.
Love the frequent uploads, keep going!!
Your videos are very informative for high school students preparing for competitive exams.
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
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.
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
Do you use android? If so how do you type the dot (multiplication sign)? I only have • in my keyboard
I need more differentials in my workout.
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.)!
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.
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
This is one of the most beautiful math video involving one of the most beautiful mathematical equation I've ever come across.
Wow, I never thought about this type of differential equation before. Very interesting, and great solution!
Really cool video! The pace was perfect and I liked your reasoning for what you were doing.
Very nice topic, I would love to see a follow up video showing off any other functions which are solutions, and maybe a more concrete way of finding such functions. It was easy to follow along and you managed to explain everything very clearly, thank you!
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.
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!
That was quite enjoyable! The right combination between creativity (educated guesswork) and technique.
Revised a lot of concepts so fast through that problem!
Thank you!
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.
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
I understood little of this video (same with many of your other videos), but what I did understand I found fascinating.
The beauty of mathematics... simply amazing! 👏👏👏
Thank you for this great class! 🙏
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.
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.
When you find yourself thinking about "phith" roots, it's time to go to bed.
...but then you'd miss the phi-squared-th root :O
To bed ? No!!! Nightmare!!!!
This was very good. Always nice to see some basics again
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!
When you already know some math but you're still intrigued into it.
Nice content dude.
When you solve enough differential equations, you get six pack.
Great video! very unique way of teaching, it makes it intuitive and simple. Thank you
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
"We're almost at the end" - the video progress bar begs to differ.
He inhales so much I'm 4 minutes in and already exhausted.
great presentation! I enjoyed every second of that! ❤️
I clicked this video out of curiosity. I'm taking calc 3 this summer and both linear algebra and differential equations this fall. I'm so psyched!
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 ...).
I am greek and I have to thank you for pronouncing φ the right way
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"
分かりやすくてすごい
I love this channel! Tried my hand at the fof case left as a problem in the end, one solution I got is
Y = Ax^r
A = ln(w)/w , r = w
w is the solution of x^2-x+1=0 so r = omega basically.
The feeth root of one over fee. I love it.
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 Or Wheeler: ua-cam.com/video/4Xp4F1h0YZM/v-deo.html
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!!
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!
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
This was outstanding. Thank you for posting.
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
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
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
Thank you such much man!!!,U make Math quite interesting and easy💯💯
This is beautiful! Thanks for sharing.
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.
Capitalization
Nice video! I played with something similar before: (f^{-1})’ = (f’)^{-1}
Yes, it is really a very interesting differential equation. I like the warm up and guess part as exploration method.
Not a differential equation.
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.
I keep have this feeling that this dude would somehow suddenly start giving you random workout tutorials🤣
Drop down and GIVE ME TEN REASONS YOU WANT THIS PHD!
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
Doesnt change anything when you are solving via guess and check
Great video. Could you please make a video about functions which satisfy f(x)=f^(-1)(x) i.e. symmetric functions with respect to line y=x?
The teaching is so clear and clear that it is understood
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
sometimes a video like this one has to pop up to remind me how beautiful maths is :)
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.
Almost generally, actually; I supposed analicity at 0. For the more general case the method would be nonetheless the same.
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?
This is really cool. Does anyone know examples in nature that use this differential equation?
Logically such thing exists, it's up to your genius to find it.
Great video as usual! Love all of them so far.
But what about f(x) = SQR(2x)? f'(x) = 1/ SQR(2x)
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.
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.
Thanks for the digestible knowledge. Good stuff.
This guy looks likes ur average white late 30s maths teacher if he lifted
@@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.
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.
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
Appreciate if you could discuss more about why such an problem is interesting? Or is it from some fundamental problem or powerful applications?
This seemingly simple differential equation really yields an interesting result. Tomorrow I'll have to tackle the follow up questions
How'd you do on the followup questions?
@@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.
@@AndyZach If you are interested, I saw today that Dr. Peyam solved the number 3 follow-up question here: ua-cam.com/video/cXpFDlIIczg/v-deo.html. His answer looks equivalent to mine from yesterday. What timing!
This man is the swollest math teacher I’ve ever seen in my life
Can you elaborate on that?
Suppose there exists a math teacher more swole but that's just wrong. QED ■
Have you never seen Prof. Leonard? ua-cam.com/video/xf-3ATzFyKA/v-deo.html
gentlemandude1 I had not. Hot damn.
Pietro Bouselli
What a fantastic way of solving this. Great video
It seems that you really like to teach. I enjoyed it, thank you!
It's beautiful that golden ratio pops like that serendipitously!! Awesome
Adding that word to my vocabulary
The world's best maths professor strikes again.
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.
5:13 if exp and sin/cos have inverse functions from a different class, a power function has a inverse function from the same class so r*x^(r-1)=x^(1/r) can be written. but as r appears as factor and power, it means more trouble
THANK YOU FOR PRONOUNCING Φ CORRECTLY
I cannot concentrate on what this class is about. I wonder why. 💪
That's probably the best differential equation I've ever seen.
It is not a differential equation.
You are amazing. So many nice solutions.
Having not thought about math in a year, this was a very fun review. Amazing, how fast you lose basic skills, even basic language.
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?
Man, I kept getting lost messing around with the relation [f^-1]' = 1/(f'(f^-1))
I could never imagine Math would sound this fun to me some day!
Good video, but I think, you could simplify calculations by using more powers instead of roots and inverse functions. Phi-th root of 1/phi could be phi ^ (-1/phi) and in my opinion its easier to work with this representation
Take a shot every time he says "phi"
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.."
I have not even completed Pre-Calculus, I don’t know what I’m doing here.
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.
Amazing problem solving technique. Never thought of this idea thinking in terms of classes that go to themselves after some transformation
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
¡¡Genial loco!! Buena demostración....
Doesn't A*exp(x+B) also work for the first question?
It's the same answer, cause you can combine A*exp(B) as exp(B) is a constant.
@@carlohu9745 yep ur right
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.
A*exp(x+B) = A*exp(B)*exp(x) = C*exp(x), so it's the same solution in the end
@@txikitofandango OP refers to the first "warm up" question, not the main one.
this video kinda blew up. thats lots of clicks in one day for a small math channel
Super interesting, never saw this one before!
Thank you for this Mr. Michael 💙