This first guessing method requires to check that it produces all the solutions. How do we check that there are no other types of functions?

@aqeel6842

7 ай бұрын

The aim seems not to be to find all solutions, just some solutions.

@tomkerruish2982

7 ай бұрын

"Trust me, Bro."

@user-nt7cg6ok6f

7 ай бұрын

"I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." - words of greatest mathematicians

@spaghetti1383

7 ай бұрын

This ODE is highly nonlinear. It is likely impossible to analytically determine how many solutions it has.😊

@kingplunger6033

7 ай бұрын

​@@spaghetti1383hail wise spaghetti

CRAZY DIFFERENTIAL EQUATION! Michael: its a monomial.

Michael, just a heads up: your audio is getting lower at each video. Anyway, thanks for the videos!

@alexeykamenev1348

7 ай бұрын

Yes, audio volume is very low.

16:40 I can’t say identity function this time 😢

I graphed this in polar and I liked it. Spiral that moves around spiral while a portion stays on the origin

@Alan-zf2tt

7 ай бұрын

Interesting!

I think replacing "x" with "x + c" for an arbitrary constant c works too. Wondering if we can add additional constants for the higher order cases.

@einszwei6847

6 ай бұрын

How do you arrive at that? The +c vanishes in the derivative but it doesn't vanish in the composition because you would have something like f(f(x))=a(ax^r+c)^r+c which is not a pure power law anymore. So I think just adding a constant does not lead to a valid solution, but maybe I overlooked something, I would like to be proven wrong! I'm pretty much a layperson my self, but I would guess that an attempt at a general solution would probably consist of two (rather difficult) steps. Step 1: Classify all solutions. Maybe they can all be written as a power series or something like that. Step 2: Use this as an ansatz. For example in the case of a power series there are rules for coefficients of compositions, so one might be able to find some kind of recursion there. I you want, you can try a power series in one of the easier cases. Maybe write a program that can calculate the coefficients up any given order (if you like programming, that is)? But I can't guarantee that it will lead somewhere or even work at all.

Happy holidays to all of you!

Thanks!

n=1 gives r=1/φ, a=φ^φ => f(x)=φ^φ x^(1/φ) as a solution to f'(x) = f(1/f(x)) ... which seems like a "nice" result to me!

Third post on this one - it does intrigue me! The DE suggests acceleration on f increases the larger f is and conversely acceleration on f decreases the smaller f is. But even more intriguingly as a relation is based on multiplicative inverse of f at x this makes a strange combination of happenings. There are cases of behaviors to look at in that relation. I cannot help but feel that the relation makes asymptotic partitions in R2 plane and that f(x)= zero is equally interesting.

Note that r(r-1)...(r-n+1)=r!/(r-n)! which is easier to write.

@Xeroxias

7 ай бұрын

only works if r is a positive integer

@einszwei6847

6 ай бұрын

@@Xeroxias Euler would like to have a word with you ...

Hello, As someone who wants to try this out, could you please maybe put in the solutions for what you have shown at the 10:20 mark. I just want to be able to check my work if I chose to try out the differentiation.

I'm at 7:30, just beginning the power function solution for the generalized problem, but I'm wondering: can we know for sure there aren't other solutions not of the form a*x^b?

@ 14:32 The reason why r=1 does not give a solution is because that makes the coefficient of the power of x on the rhs a^-2, which cannot be 0. Furthermore, if r=-2, we get f(f(x))=0 for all x, making our original equation 0=1/0, which is a contradiction.

Hi. I'd like to know what subject studies if some solution set of functional equations like that is the complete set or not.

@einszwei6847

6 ай бұрын

I think you can just look up "functional differential equations" or something like that. There are probably books or other texts about that. I don't think (or at least don't know) that this has a specific name. Probably just something like "the theory of functional (differential) equations" or something similar.

@kobtron

6 ай бұрын

@@einszwei6847 Thanks.

@ 1:00 this property seems to look something like translational symmetry,? With f(x) related to reflective symmetry ~ inverse of f(x) Scaling symmetry seems a given by constraints on placeholders called "a" and "x" Stuck there but there may be more as google/wiki tells me that there are 8 symmetries in math and these are: 1 Euclidean symmetries in general. 2 Reflectional symmetry. 3 Point reflection and other involutive isometries. 4 Rotational symmetry. 5 Translational symmetry. 6 Glide reflection symmetry. 6.1 Rotoreflection symmetry. 7 Helical symmetry. 8 Double rotation symmetry. I tried to figure out helical symmetries but it gave me a headache drawn on by looking at and thinking about progression on index of power on placeholder "x" Interesting to see a power solution assumption leads almost fractal like to potential nesting of power solutions. It must have been a powerful brain that contrived this thing I'm going for a cup of tea now ps EDIT: I started by looking at Reflexive Symmetric and Transitive symmetries but could not quite remember the third - it is a gnarlie beastie for sure 🙂 pps EDIT: be interesting to know what complex power series might show and/or even quaternion power series might show (if they exist)

I have a challenging functional equation for you to solve, it is featured in a question in Madhava Mathematics Competition 2019. The question is:- Let f(x) = a_0 x^(n)+a_1 x^(n-1)+· · ·+a_n be a non-constant polynomial with real coefficients satisfying f(x)f(2x^2) = f(2x^3 + x) for all real numbers x. Find a polynomial f satisfying f(x)f(2x^2) = f(2x^3 + x) for all real numbers x.

What do you mean a=0? A^0 = 1 right?

@OuroborosVengeance

7 ай бұрын

Forget the exponent, look at the coeficient. It gives you 0, so the RHS has to be 0 too

@krisbrandenberger544

7 ай бұрын

Yes, a^(r-1)=1 on the rhs when r=1. Thus, equating the coefficients of the powers of x on both sides gives 0=1, which is a contradiction. Thus, there is no solution for the r=1 case.

Hi, 3:10 : "like a place to start".

What twisted soul created that nightmare? 🙂

That is not a differential equation, for being ordinary it would look like F(x, f, f’, f”) = 0, with F a known function. At best we can call it a functional differential equation.

@sucroseboy4940

2 ай бұрын

While that is true, you would be able to write this equation in the general form: f(x)’’-f(1/f(x))=0. For that reason, isn’t it still a differential equation, just one not written in the general form?