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?

Books I like:

Abstract Algebra:

Number Theory:

Calculus:

[antonystark]

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

[carlosdanielvelazquezflore]

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

[Darkstar]

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.

[aienbalosaienbalos]

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.

[diegoparodi]

Completely random equation: exists
Phi: let us introduce ourselves

[Celastrous]

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

[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

[ButiLao]

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

[titusng]

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

[SoonRaccoon]

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

[stevenpurtee]

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-riachi]

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....

[Beniguitar]

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

[AlexTrusk]

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

[mina]

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.

[karangupta]

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

[ПавелЩичко-еж]

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.

[intergalakti]

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.

[KarstenJohansson]

I need more differentials in my workout.

[oliverinspace]

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.