the craziest of differential equations!
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- Опубликовано: 23 дек 2023
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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?
The aim seems not to be to find all solutions, just some solutions.
"Trust me, Bro."
"I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." - words of greatest mathematicians
This ODE is highly nonlinear. It is likely impossible to analytically determine how many solutions it has.😊
@@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!
Yes, audio volume is very low.
16:40 I can’t say identity function this time 😢
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!
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.
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!
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.
I graphed this in polar and I liked it. Spiral that moves around spiral while a portion stays on the origin
Interesting!
Note that r(r-1)...(r-n+1)=r!/(r-n)! which is easier to write.
only works if r is a positive integer
@@Xeroxias Euler would like to have a word with you ...
Thanks!
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.
@ 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.
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.
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?
Hi. I'd like to know what subject studies if some solution set of functional equations like that is the complete set or not.
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.
@@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)
What do you mean a=0? A^0 = 1 right?
Forget the exponent, look at the coeficient. It gives you 0, so the RHS has to be 0 too
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.
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?