A fascinating differential equation: when does the derivative equal the composition?

[@GeraldPreston1]

Insane maths gameplay

[@User-SSHVMBEY]

Cranking derivatives

[@gdmathguy]

360 no-scoping integrals

[@imadeddinefethallah3662]

9:44 "i don't think there's other solutions to this differential equation " The omnipresent 0 function cries in the corner as it mourns on its acknowledgement

[@manateepink9100]

alpha = 0

[@tomkerruish2982]

1:18 He rules it out since it's boring.

[@ACertainMan]

​@@tomkerruish2982 That's the funny part.

[@opensocietyenjoyer]

i cannot think of a non-trivial differential equation that has only 0 as a solution

[@imadeddinefethallah3662]

@@opensocietyenjoyer non homogenous ODEs, differenial equations with y in the denominator

[@nikolaoszervos96]

Another way to look at the problem is putting f(x) in the place of x. Then you have f'(f(x))=f(f(f(x))) with the RHS being f(f'(x)). Then f'=f^-1 which is a problem I believe you have already solved on the channel.

[@maxgoldman8903]

I think f’(x)=f^-1(x) => f’(f(x))=f(f’(x))=x, but it may not be the other way around, i.e., f’(f(x))=f(f’(x))x unless we can show f’(x)=f^-1(x) f’(f(x))=f(f’(x)).

[@nikolaoszervos96]

@@maxgoldman8903 you are correct, in general fog=gof doesn't mean g=f^-1, e.g. f=x, g=x+1 then gof=fog. This means that f'=f^-1 is A solution to the problem but not the only one. I don't know if f'of=fof'=>f'=f^-1 can be shown.

[@phyllipkjellberg7160]

Cool that the preposition "of" adds an extra "of" when you read fof(x) like fofofex

[@Calcprof]

How about something like f'(x) = f*f, where * is some sort of convolution. Solve by integral transforms!

[@maths_505]

Excellent idea

[@emanuellandeholm5657]

This is a cool problem! You have proved that if f(x) is of the form alpha x^beta, these, plus the trivial solution alpha=0, are the solutions. But are we sure there aren't any other solutions?

[@maths_505]

@@emanuellandeholm5657 someone commented a couple of cool solutions. If you scroll down you should find them.

[@roberttelarket4934]

@@maths_505: You are "maths" which tells me you’re not from the U.S. but probably from Britain. Yet you don’t have an accent!

[@maths_505]

@@roberttelarket4934 I'm actually from Pakistan and we call it both maths and math here but I grew up calling it maths.

[@redroach401]

Parental advisory

[@MrWael1970]

Nice Analysis. Thanks.

[@terdragontra8900]

This is very interesting, my first thought was “there must be a one parameter family of solutions, though probably without a closed form, just set f(0) to some value, and newtons method approximates a solution”, but because of the composition that doesn’t work because you need to know the behavior of f near f(0), too (when f(0) = 0 you do, which just gives you f(x) = 0)

[@Winium]

Question for you (from someone uneducated in functional analysis) about "polynomial" functional equations, but the "exponent" is actually iteration. e.g. f(f(f(x))) + f(f(x)) + x = 0, and where '+' is not necessary integer addition -- what are these things and are they "interesting"? I guess, another way to phrase it might be: are there ring-like objects where instead of 'multiplication', we have function application?

[@モハメドイブラヒム-k8f]

0x is also a valid solution too ,it's the first thing came to my mind it's nice and simple but But it kind of has no fun.😅

[@sandem4592]

What a neat diff eqⁿ and all, but could you attempt the integral from 0 to 1 of x^(1/x) dx? A solution has evaded me for a bit too long. Maybe you know a special function I don't, would be cool to see mr integral man.

[@maths_505]

Aight

[@XxAspect23xX]

There will definately be more functions, just not in the polynomial form.. let the degree of f(x) by a then derivation makes the degree a-1 and f(f(x)) always has degree 2a so a=-1. thus the polynomial CAN BE IN THE FORM OF- a/x + b/x^2 + c/x^3........ till whatever term. this can be tough to find but by case bashing (say its of the form a/x, then a/x + b/x2 and so on) we can find functions and if one is lucky there could by a possibility for a parameter which would imply there are finitely such functions... nevertheless when talking about functions there are endless possibilities.. and the possibility of there being a function with maybe factorials and other stuff is def possible

[@maths_505]

Now this is the type of comment I was hoping for

[@deinauge7894]

hm... if f(x) = 1/x then f(f(x)) = x somewhere your logic is flawed, since the degree of f(f(x)) is not -2 😉

[@DGQQ78]

One can try with power series...

[@TheAzwxecrv]

Then what is wrong with the following argument? f (prime) = f of f mens, (d/dx) f(x) = f(f(x)). Put y = f(x). Then, dy / dx = f(y). So, we are free to choose for f(y). For example, if we choose f(y) = 1 / y, then y dy = dx, so y^2 / 2. = x + c, or y = f(x) = sqr rt ( 2x + c1).

[@pokemantrainer]

By choosing f(y), you also choose f(x) (in your case f(x) = 1/x). Yet, in your final solutiin you get a different f(x) which is is a contradiction

[@maths_505]

Exactly

[@mileroqueiro]

If f is invertible, with inverse g, then f' = f○f implies 1/g' = f'○g = f. That problem is equivalent to the fg' = 1 problem (derivative times the inverse equals 1).

[@TheDhdk]

Alpha_1 can be simplified to e^(π(√3+i)/6) alpha_2 can also be similarly simplified

[@evanwilliams7376]

I will memorize these equations so I can have a shortcut that is simpler to calculate than the power rule for this specific case.

[@edilon619]

Já enjoei essas questões.

[@brunojani7968]

f(x)=0 is a solution

[@05degrees]

Multi-valued functions are continuous, you just don’t need to look at continuity in weird ways not compatible with Riemann surfaces! 😶 Well at least those multi-valued functions that one can get in complex analysis.

[@frimnpi7473]

thumbnail goes hard

[@XxAspect23xX]

Here'sanother way to look at this, by integrating both sides we get-> f(x)= int( f(f(x)) dx ....... (1) so in (1) put x--> f(x) so we get f(f(x))= int (f(f(f(x)))dx ....... (2) now substituting f(f(x)) from (2) in (1) we get f(x)= int(int(f(f(f(x))) dx) dx now the pattern maybe a little tough but continuing the same process we see that f(x)= in(int(int.......( f(f(f(......(x)))))) )dx) dx) dx..... where the number of integrals is n-1 and the number of f's in f(f(f....(x)))) is n so by algebrec anaylitic continuation we can put n=0 so when the number of integral becomes -1 and f(f(...x)))) becomes 0, thus we get f(x)= d/dx(1)= 0 this is one such function then putting n=-1 we get f(x)= d2/d2x ( f-(x)) this may look a little tough buut we get the same function mentioned in the video. Both ideas work its just the method shown in the video is a little incomplete whereas the method used here is basically impossible to solve....

[@maths_505]

So it's a comparison between the incomplete and the impossible to solve😂😂😂

[@deinauge7894]

you cannot substitute the x in the integral by f(x). because in reality you have (1): f(x) = int (from somewhere to x) f(f(t)) dt then substituting x->f(x) changes the integration bounds and not the integrand

[@isaswa1602]

just curious how can we prove that power functions are the only solutions to this problem?

[@tomashernandez8711]

What program do u use?

[@mikelevels1]

LET HIM COOK 🗣️🗣️🗣️🔥🔥🔥!!!

[@decaf3821]

I like to think you record your videos first try with no notes or prep

[@VincentKok458]

Did you change the thumbnail?

[@pjb.1775]

i honestly thought f(x) = 0 would be a solution that someone would care to put in a video

[@Loyis]

Thanks for the explicit content warning lmao

[@-jason1912]

This is why i quit calculus, too many sweats

[@viktor-kolyadenko]

"(complex number)^(complex number)" - very bad idea.

[@charlievane]

isn't surreal^surreal worse ?

[@roberttelarket4934]

You didn’t have enough room in the margin? Is your name Fermat?!

[@DerMathematicker]

f(x) = 0

[@maths_505]

@@DerMathematicker indeed

[@toony966]

Could you make a discord server?

[@maths_505]

Would that be useful to the community?

[@toony966]

@@maths_505 I think it would because many people who are interested in mathematics and are viewers of your channel can communicate and discuss about mathematics and get help for mathematics.

[@TheArtOfBeingANerd]

Yes! I want a math discord server that isnt swarming with users

[@stefanalecu9532]

I'm sure you've actually solved this exactly differential equation 10 months ago... I think the video was called "A very interesting differential equation: derivative equals composition"

[@maths_505]

Now that I've released 400 videos it's become hard to keep track of the problems I've solved😭

[@Mario_Altare]

@@maths_505 kzbin.info/www/bejne/eWfJi4mtps2VpKc&ab_channel=Maths505 😀

[@premdeepkhatri1441]

Very Fast thinking and solving problem. One question is make simple when we put there any appropriate constant?

[@KaliFissure]

Graph it

[@davidalejandrolopeztorres1083]

Awesome content as always. I would like to share that I was precisely thinking about this kind of Differential Equations yesterday after KZbin recommended me this video: kzbin.info/www/bejne/qH-4l5yHnM9pkKs from Michael Penn (pretty similar approach, not sure if that was the source of inspiration of this video as well). However, I was thinking more in the equation f'(x) = f_n(x) (being f_n the nth composited function f). Doing the same trick you can lead to interesting conclusions about the parameters of the power functions that make the job. I haven't reach any marvelous result yet, but probably you could take a look if interested. It was nice to share results (and thoughts) with you today. Always a pleasure to see your videos, definitely they had become a recurrent highlight of the day. Kind regards from Mexico!

[@maxvangulik1988]

f^(n-1)(x)=f°^n(x) a•ß!/(ß+1-n)!•x^(ß+1-n)=a^sum[k=0,n-1](ß^k)•x^ß^n ß^n=ß+1-n a•ß!/(ß+1-n)!=a^((1-ß^n)/(1-ß)) ß!/(ß+1-n)!=a^(n/(1-ß)) a=(ß!/(ß+1-n)!)^((1-ß)/n) let n=2 ß^2=ß-1 ß^2-ß+1=0 ß=(1+-isqrt(3))/2 ß=e^(+-ipi/3) a=ß^((1-ß)/2) a=e^+-(ipi/6•e^(-+ipi/3)) if n=3, f"(x)=f(f(f(x))) aß(ß-1)x^(ß-2)=a^(1+ß+ß^2)•x^ß^3 ß^3=ß-2 aß(ß-1)=a^((3-ß)/(1-ß)) ß(ß-1)=a^(2/(1-ß)) a=(ß^2-ß)^((1-ß)/2) this is solvable with the cubic formula, but I'd rather not

[@maths_505]

Fascinating

[@varunnadgir4513]

f(x) = 0 😂

[@mau9639]

allo :))

[@quantum_psi]

Math was fun and interesting up until differential equations. I then realized how most of it is fake and arbitrary. Sad

[@alexterra2626]

balls

[@orionspur]

1sties

[@diogeneslaertius3365]

Please stop saying "OK cool". It's extremely annoying.

[@Javy_Chand]

"Ok cool!" Became the catchphrase of the channel. Stopping would be like changing π for some random letter

[@Xnoob545]

OK cool

[@fullfungo]

Nuh uh

[@maths_505]

The ohhkayy coool is the reason why people have subscribed in the first place!

[@diogeneslaertius3365]

@@maths_505 Interesting. Could you add time codes for OK cool so I can press / unpress m on time? :)

[@guruone]

AsymptoticSum[(-Log[1 - t x]/t)/z/. t -> n/z, {n, 1, z}, z -> Infinity]

[@deinauge7894]

this looks as if it gives 0 for all x. but maybe not 😅 without the z in the denominator the sum converges to the integral of log(1-t x)/t for t from 0 to 1. Ok this is a divergent integral, and the 1/z seems to counteract this divergence. Have you plotted the result?