A very interesting differential equation: derivative equals composition

Here's a nice intro to functional differential equations which are sort of a niche in the study of DEs
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[@manstuckinabox3679]

for this differential equation we apply my favorite method in all of mathematics, assumptions. Great to have you back boss.

[@user-ky5dy5hl4d]

Yes, here I can actually assume anything for this function. So, why did he use his assumption when for dreivative of the function I can use the third derivative of this function.

[@wabc2336]

​@@user-ky5dy5hl4d The third derivative is f^4 f^3 f^2 f^2 + f^4 f^3 f^3, how does that help

[@user-ky5dy5hl4d]

@@wabc2336 It does not help. What I was tuning to was that the video shows arbitrary any substiturion that can be applied to any function even in third derivative.

[@insouciantFox]

It's not a ln application of the quad formula in REAL life. It's an application in complex life.

[@chaitanyasinghal1098]

Nice 😂

[@maz0000]

I don't think there is a real life opposed to a complex one. Life is life and nature is what it is. Real numbers are not more connected to nature than complex ones, they are just much easier to connect to what surrounds us. Both are just abstract tools invented by humans to describe and to tame nature. Extremely powerful ones, but an endless quest all the same. At least that's how I see it 😊

[@helwasyaharinramadhan9371]

I see this problem and i think f(x) = 0 is solution automatically

[@spiderwings1421]

solution was lost when we divided by zero at 2:17

[@gabitheancient7664]

that's the first thing I try everytime I see an equation with functions lmao

[@datguiser]

I’d assume it is an answer, but not the exhaustive answer

[@peachesaupear8455]

You're not wrong!

[@Player-pj9kt]

I lot of differential solutions have solution f(x)=0 or a constant but it usually doesn't work when there's a initial condition

[@martijnvankuppeveld1997]

I think this is not just two solutions, but a countable infinity of them with the standard trick of (e^a)^b = e^(a b + 2 pi i n b) for arbitrary n and non-integer b.

[@MarcoMate87]

I don't understand why b has to be non-integer. The formula is correct even with an integer b. Instead, n has to be integer and not simply arbitrary.

[@jobliar937]

Nah those solutions are all identical I wouldn’t say infinitely many

[@duncanhw]

That doesn't give more solutions, just more equivalent ways to write them

[@codinghub3759]

For non-integral values of b, they are genieunly new solutions. However, for countable infinite solutions, I believe b must be non-rational.

[@bobh6728]

Did he say, “Express these in polar form, because they look cooler.”? Or did he say “Express these in polar form, because they look Euler.”?!

[@apm77]

Here's something I played with a while back. We know that the function f(x)=i*x has the property that f(f(x)) yields the opposite of whatever x we started with, if by "opposite" we mean "negative". Likewise f(x)=x^i has the property that f(f(x)) yields the opposite of whatever x we started with, if by "opposite" we mean "reciprocal". In both cases f(f(f(f(x)))) = x. So I wondered about the function such that f(f(x)) equals the negative reciprocal of x, and with caveats, this turned out to be f(x) = i*e^(pi/2)*x^i - the caveat is that there are multiple solutions and only one of them is the one we want. If I ask Wolfram Alpha to "plot f(f(f(f(x)))) where f(x) = i*e^(pi/2)*x^i", the graph is the same as for f(x)=x when x is large, but it plots a different solution when x is small. I am sure you could tell me something interesting about this equation.

[@charlie3k]

I challenge you to solve the following differential equation: d^i/dx^i (d^i/dx^i(f)) = d^(-1)/dx^(-1) f In other words, find a function f that has the property such that when you apply two imaginary-order derivatives it results in the anti-derivative of f.

[@icodestuff6241]

how do you take an imaginary derivatve

[@charlie3k]

​@@icodestuff6241 Consider the function sin(x). The n'th order derivative of sin(x) is: d^n/dx^n sin(x) = sin(x + n*pi/2) The intuition behind this is that derivatives of sin(x) are just a phase-shifted versions of sin(x). Letting n=i, we then have the imaginary-order derivative of sin(x): d^i/dx^i sin(x) = sin(x + i*pi/2) Using the sum of angles trigonometric identity, we can expand the RHS: = sin(x)cos(i*pi/2) + sin(i*pi/2)cos(x) You evaluate the complex sine and cosine functions by rewriting them as complex exponentials: = sin(x)*(e^i(i*pi/2) + e^-i(i*pi/2))/2 + cos(x)*(e^i(i*pi/2) - e^-i(i*pi/2))/2i Which after some simplication results in: = sin(x)*(e^(-pi/2) + e^(pi/2))/2 + icos(x)*(e^(pi/2) - e^(-pi/2))/2 This is the imaginary order derivative of sin(x)! :)

[@denyraw]

d^a/dx^a e^(b x) = b^a e^(b x) Works for all positive integers a. It can be defined (generalized) to work for all complex numbers a. Using the Laplace transform generalizes this to all functions.

[@LittleWhole]

@@denyraw Or you could also generalize the Cauchy formula for repeated integration to negative and non-real values by swapping the factorial for a gamma function (called a _differintegral_ - very creative name, I know).

[@jonathanaarhus224]

My only constructive criticism is that when writing complex numbers in polar form don't forget the include the phase factor of e^(2pi*i*n) for all integers n.

[@pacolibre5411]

There are ordinary differential equations, then there are extraordinary differential equations.

[@emilioturbay667]

Brilliant math. Very fun to watch.

[@FLCoeur]

I have another point: let's assume f(a)=a, and expand the equation near a. So f(x)=f(a)+f'(a)(x-a)+... and I can solve for the coefficients as : f'(a)=a,f''(a)=a^2,f^{(3)}(a)=a^3 (a+1),f^{(4)}(a)=a^4 \left(a^3+a^2+4 a+1 ight),f^{(5)}(a)=a^5 \left(a^6+a^5+4 a^4+8 a^3+11 a^2+11 a+1 ight),f^{(6)}(a)=a^6 \left(a^{10}+a^9+4 a^8+8 a^7+22 a^6+22 a^5+60 a^4+58 a^3+66 a^2+26 a+1 ight)...

[@CamiKite]

It's way cooler when you write f(x) in rectangular form, with some strange sqrt(x)*cos(sqrt(3)*ln(x)/2) 🙂 By the way, is there other solutions? (Other than f(x)=0)

[@maths_505]

I haven't found any yet

[@dukenukem9770]

Fun stuff! I’ll do this little problem with my son today…

[@illumexhisoka6181]

Is there a way to know how many solutions there is ? (I have never studied deferential equations)

[@mokhtarmougai5088]

I think that's obvious from the was you wrote it 💀 Sorry if you felt bad, it's okay to make mistekes :)

[@maths_505]

Well no not exactly. You just integrate DEs to uncover solutions.

[@mathunt1130]

Let u=f(x), then f'(x)=u'(x), and the equation becomes u'(x)=f(u).

[@g4_61]

I’m new to this channel. Really surprised at how doable this problem was! Is there a way to do it on the assumption that f(x) has more terms tho?

[@ahsgdf1]

1) Expanding the complex expression for f(x) we find {Re[f(x)],Im[f(x)]} = {1/2 E^(\[Pi]/(2 Sqrt[3])) Sqrt[x] Sqrt[3] Cos[1/2 Sqrt[3] Log[x]] - Sin[1/2 Sqrt[3] Log[x]], 1/2 E^(\[Pi]/(2 Sqrt[3])) Sqrt[x] (Sqrt[3] Sin[1/2 Sqrt[3] Log[x]] + Cos[1/2 Sqrt[3] Log[x]])} 2) For an ODE of first order there must be one arbitrary constant in the solution. But I can't see it. For instance f'(x) = f(f(x)) and f(1)=1, also I we can request the solution to be real.

[@charlied.4683]

Perhaps the constant comes from the equality of the complex value for beta to the exponential form, because you can add a + 2*n*pi into the exponent for more solutions? I haven't taken ODE classes yet, but is this arbitrary constant valid as it can only be an integer?

[@sylvainm-v7919]

The solution space isnt a vectorial space, the equation is linear, there doesnt have to be a constant

[@sylvainm-v7919]

Like for the equation - (f'-1)^2 = (f-1)^2 the only sol is f=1 and there is no other arbitrary constant involved

[@duncanhw]

This is not an ODE

[@wolfganghintze732]

​@@duncanhwYou are right. So drop the O and you are done. But my point remains in force.

[@joku237]

Can you prove uniqueness to the solution? F.e. using some other kind of ansatz function than a polynomial for the solution. Or is this just A solution to the problem?

[@Sir_Isaac_Newton_]

It's just A solution. Try f(x) = 0x and it becomes apparent that it's a solution too.

[@user-fv2oo3kw6u]

Great Vid! You deserve more recognition!

[@haku1082]

Why are there no solutions of forms other than alpha * x^beta?

[@MrWael1970]

Good idea, wish you all of success. Thank you

[@yoav613]

Functional de is very interesting please solve more like this. I saw michael penn has also nice de f'(x)=f(1/x).

[@moregirl4585]

Other solutions?

[@maths_505]

Left as an exercise to the viewer?

[@sylvainm-v7919]

@@maths_505 u trully funny

[@ashutoshganguly1805]

hey really loved the video. However it would be great if you could speak a bit about what does it really mean for the rate of change of a function be f(f(x)). Or maybe show some visual so that noobs like me could get a feel of what is it.

[@strangeWaters]

I wonder what categories this construction works in.

[@poly_hexamethyl]

6:27 There must be more solutions than that. For example, the function that's 0 everywhere is an obvious solution.

[@Sir_Isaac_Newton_]

f(x) = 0x works too

[@DrLukeWendt]

Can you solve this? e^x = f( f(x) ) I believe it is referred to as a functional square root.

[@hqTheToaster]

2^(ax)^(ax)+2^(bx)+2^c+d = 0

[@soyoltoi]

I have to ask. What the hell do you mean by structure? Normally that's an algebra term, not analysis unless it's more abstract analysis.

[@susanafaciolince7755]

f(x)=0 for all x is another solution.

[@Fasalytch]

Yes, he missed it when divided by alpha

[@BedrockBlocker]

what about uniqueness?

[@Anmol_Sinha]

Why did we assume that y is in the form of ax^b in the beginning?

[@TheOiseau]

Because he knew in advance that was the answer.

[@fullfungo]

Intuition

[@maths_505]

Derivatives of power functions are power functions Compositions of power functions are power functions That's why

[@Anmol_Sinha]

@maths_505 thanks!

[@dispatch1347]

Isn’t it 2b not b^2?

[@gianluca4893]

How can you prove these are the only functions?

[@Speed001]

We'll miss the days when humans did math like this one day. That might've been a day years ago but the point stands.

[@kaous5690]

Ok so you made y'= &ßx^(ß-1) but what happens when we log in things like sin, cos, e^x for x... Doesn't it break since what you did is true for polinomial equations only?

[@olivergreer3690]

The entire solution process is predicated on the assumption that the solution is a polynomial. X is the dependent variable here what do you mean by loging in other functions for x?

[@nicolascamargo8339]

Wow genial

[@unixux]

What about nonmonomial solutions ?

[@maths_505]

Haven't tried those. Besides the final solution is more trigonometry than polynomial.

[@TheOneMaddin]

The problem is interesting. But then you assume the form of the solution and it becomes instantly boring because solving it is now just doing some basic computations. Also, the solutions are dubious. A complex number to the power of a complex number is not a well defined expression. It can result in several values depending on the branch cut of the complex logarithm. So you have to figure out which of these infinitely many solutions actually constitutes a solution. Before you did this, there might as well be NO solution.

[@willpudupakkam]

How come we can guess that it's of the form Ax^B, and not some other function like sin, cos, tan, ln, etc.?

[@codinghub3759]

He didn't find the general solution here, he just showed *a* possible solution.

[@muddle.]

Can anyone explain how we just assume that f(x) can be represented as ax^b

[@fullfungo]

It’s really simple. He said “let’s assume”, so we assumed.

[@muddle.]

@@fullfungo ye i know he said that but im asking how can we be so sure

[@fullfungo]

@@muddle. we are not. We are just assuming.

[@fullfungo]

@@muddle. Let’s say I want to solve f’’(x)=1. Assume that f(x)=a•x^2+5•x+7. Then, f’’(x)=2•a, so f(x)=0.5•x^2+5•x+7 is a solution. This does not guarantee that 0.5•x^2+5•x+7 is the only solution. But we *did* find a solution.

[@muddle.]

@@fullfungo got it, thanks

[@squdioodellover2589]

But is this solution complete or just a special chase?

[@Stimulus7]

What was the intuition behind starting with a polynomial ansatz?

[@maths_505]

Well an exponential function was out of question so I tried fitting a polynomial. Little did I know that the polynomial needed complex exponentials.

[@manumagal3894]

brooo you gotta drop tensors for babies ep 2 🙏🏽🙏🏽 great vid btw

[@oringimz2848]

Fantastic for sur but what's the concept for using ãlfa,beta and the relationship between theme and x ??? Plz i need an explanation 😊

[@maths_505]

I just tried fitting a solution

[@Yon215]

Here he is trying to find a polynomial function st solves the DE. The polynomials are useful because of their properties

[@lambda653]

Adobe enhance audio moment

[@yoav613]

Nice.

[@shrirammaiya9867]

Isn't alpha equal to e^(pi^2/9)? This is just another place where e and pi randomly show up unexpectedly

[@dylanwhite2949]

u can't do that when u have something to the power of a complex number

[@vishnuhu9086]

Whats a function that’s equal to the derivative of itself? That’s the question I get asked when looking at the equations. It’s not really showing up randomly. There’s also two other trial solutions that the video didn’t go over, sin and cos. I forgot when the rules when you have to trial those solutions.

[@eylonshachmon6500]

Is there a proof that these are the only solutions (there aren’t any solutions that aren’t polynomial)?

[@maths_505]

Left as an exercise to the viewer?

[@eylonshachmon6500]

@@maths_505 { sin(1/x) | x is rational { 0 | x is irrational I believe it satisfies the condition (on the basis that {} = {} ) No idea how to find / disprove the existence of a solution where f’(x) must be an actual function

[@miloradtomic]

Here is another proof, that doing Theoretical mathematics, means be a poet.

[@gabitheancient7664]

the 0 function works, bye

[@mokouf3]

Missing solution: zero function f(x) = 0

[@sirnewton77]

Bro ,why dont you solve Putnam problems?❤❤❤

[@anupamamehra6068]

hi maths 505 which country do you live in?

[@maths_505]

Pakistan

[@jobliar937]

Yes this is TOTALLY a real life application of the quad formula you DEFINITELY ARENT COPING

[@sabriath]

the answer is 0

[@PracticeMakePerfectMuslim93]

this solution is antishortcut 😂

[@Anonymous-Indian..2003]

Nois