A fun functional equation!!

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We solve a nice functional equation.
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14 дек 2020

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Комментарии : 302

@lucasdepetris5896 3 года назад

Isn't the uniqueness of the function already garanted since you didn't make any choice in the process of isolating f?

@timurpryadilin8830 3 года назад

Yes, that's what I thought! Probably a superfluous part of the argument. I believe, though, that instead, Michal should have checked that the found f(x) actually satisfies *. We only proved the forward direction: * => f = f(x), but not backwards: f = f(x) => *

@lucasdepetris5896 3 года назад

@@timurpryadilin8830 Right! And for the sake of the truth, that would also have got the video to pass 10min

@peraperic4633 3 года назад

Precisely. But what he did in fact does not guarantee that it is a solution (for instance, the solution for other two linear system entries might not agree). So he got it completely flipped.

@aa-lr1jk 3 года назад

@@peraperic4633 But this already were surpefluous, since the system was well definied from the beginning and all of the operations that he made (and here i consider the substitution x->1/(1-x) as such) were biunivoc (for example, taking the square root is not biunivoc). He really did something that he did'n needed to do.

@aa-lr1jk 3 года назад

@@angelmendez-rivera351 It is, dude. There was no assumption about domains, and if y = 1/(1-x), then x = 1-1/y, and this shows that the substitution is biunivoc.

@stewartzayat7526 3 года назад

You put the fun in functional, Michael!

@goodplacetostart4606 Год назад

Without him, there would only be ctional equations.

@BrunoVisnadi1 3 года назад

What you showed in the first half of the video was that if a function satisfies your functional equation, if also satisfies f(x) = [x^3 - x + 1]/2x(x-1). The next step should be to verify that this function indeed satisfies the equation. I don't understand why it's necessary to verify uniqueness

@BrunoVisnadi1 3 года назад

Actually the second half of the proof only 'works' because you assume fh is unique. Could have assumed f is unique in first place

@user-jc2lz6jb2e 3 года назад

Useful facts: The function g is a Mobius transformation, which are functions that have the form (ax+b)/(cx+d), not to be confused with the Mobius function. You can compose Mobius transformations easily by putting the entries in a square matrix, e.g. a,b,c,d from above become [a, b] [c, d] And the composition of two functions like this is just the entries of their matrix product. So g(x) = (0x+1)/(-x+1) becomes [0, 1] [-1, 1] Multiply it with itself to get g(g(x)) [-1, 1] [-1, 0] But you can tidy up things by multiplying both rows by -1 (you are allowed to multiply both rows at the same time by anything, since it's actually a ratio and not a matrix), and then you get g(g(x)) = (x-1)/x, same thing as what Michael got.

@hybmnzz2658 3 года назад

@@angelmendez-rivera351 agreed, it is more of a fun fact.

@user-jc2lz6jb2e 3 года назад

@@angelmendez-rivera351 if you had to compose it many times, it's easier to think of it as a matrix and diagonalize then raise it to a power.

@morgengabe1 11 месяцев назад

My friend, you deserve a prize for that comment!

@DS-xh9fd 3 года назад

The solution is only unique if you assume that 0 and 1 are not in the domain. Otherwise you can set f(0) = k for some arbitrary k and set f(1) = -k, and the rest as you did, to create an infinite family of solutions.

@DS-xh9fd 3 года назад

@@angelmendez-rivera351 The given functional equation is not defined for x=1, and that is precisely why you can set f(0) arbitrarily. For my answer to be wrong, you must be able to provide a value of x that I can plug into the given functional equation and produce an incorrect result. But there is no such value.

@DS-xh9fd 3 года назад

@@angelmendez-rivera351 The problem statement does not say that the functional equation holds for all x in dom(f). It is not implied either.

@shaiavraham2910 3 года назад

There is an error in 10:45: It should be f_h((x - 1) / x) according to g(g(x)) and not f_h(x / (x - 1)). Luckily, it doesn't matter because these cancel when you solve the system anyways.

@jamirimaj6880 3 года назад

there's always an error in almost each of his videos uggghhh

@Zack-xz1ph 3 года назад

@@jamirimaj6880 to err is human, to forgive is divine :)

@mcwulf25 3 года назад

Michael makes mistakes in most of his videos. However unlike the rest of us, his don't seem to affect the results!!!

@mcwulf25 3 года назад

@@jamirimaj6880 But he still gets to the answer. Why doesn't that happen for me?

@DylanNelsonSA 3 года назад

Actually at the point where you claimed that we have shown that f is a solution, but not that it is the only solution, what you had actually shown was exactly the opposite: You had shown that if f is a solution then it has to be the function in question, but not that the function in question is actually a solution. It is true that it is a solution to the system of linear equations that we derive. By solving the system, we'd also get a value for f(1/(1-x)). There's probably a good reason that the value that you get for f(1/(1-x)) is just the expression that you get for f(x) but with 1/(1 - x) substituted for x, but a priori I don't think that it's automatically true. You could probably make an argument using symmetry: the system of equations doesn't change if you replace every x with a 1/(1 - x), and so if we do the same thing in every line of reasoning that we used to deduce that f(x) = ..., then we'd obtain f(1/(1-x)) = (same expression with 1/(1-x) in the place of x) But here the symmetry of the situation is crucial. If we instead had somehow obtained the system of functional equations f(x) + f(x + 1) = 2x + 1 f(x + 1) + f(x + 2) = 2x + 3 f(x + 2) + f(x) = 2x we'd get the "solution" f(x) = x - 1, and f(x + 1) = x + 2, and f(x + 2) = x + 1. We could then correctly draw the conclusion that if there is a solution then it is given by f(x) = x - 1, (i.e. f(x) = x - 1 if the only possible solution) but we can't draw the conclusion that f(x) = x - 1 is a solution to the system of equations because in fact the equations aren't compatible with each other, and there are no functions f that satisfy the system of functional equations.

@DanielWalvin 3 года назад

7:45 "We've got *a* function [...] Now we want to show that it's the only function [...]" Perhaps you technically do still need to show this, to be rigorous, but you've already done all the mathematics for it at this point*, it could be completed with just a basic logical implication. All of your steps up to that point have been implications of previous steps or of facts, so, referring to the statement "f(x) + f(1/(1 - x)) = x (for arbitrary x ≠ 1)" as "A", and the statement "f(x) = (x³ - x + 1)/(2x(x - 1)) (for arbitrary x ≠ 0 or 1)" as "B", you've done this: A => ... => ... ... ... => B. I.e. you have A => B. Then the contrapositive of that is "!B => !A", where "!X" means "not X", and this is directly logically implied by "A => B". That logical implication could be demonstrated by, e.g., translating "A => B" into plain English, something like "Whenever A is true, B is true", then "So you can't have that A is true while simultaneously B is not true", then "So whenever B is not true, A is not true", and then you can convert that back into symbolic logic: "!B => !A".) !B => !A, i.e. if f(x) ≠ (x³ - x + 1)/(2x(x - 1)) (for arbitrary x ≠ 0 or 1), then f(x) + f(1/(1 - x)) ≠ x (for arbitrary x ≠ 1). I.e. f(x) = (x³ - x + 1)/(2x(x - 1)) is the unique solution to the original functional equation. I think that's a tighter argument, going back to a fundamental logical implication rather than having to do more mathematical working. Although, the way presented in the video allowed for the introduction of the ideas of "homogeneous solution" and "particular solution", which is worthwhile in its own way. *Up to what the guy with the Haruhi picture, Seth Harwood, pointed out, that x = 0 and x = 1 are separate cases from the rest, and any solution with f(0) = c, f(1) = -c, works. That doesn't affect the rest of the argument though, since the original functional equation only pairs them to each other.

@goodplacetostop2973 3 года назад

11:11 Someone asked few days ago for a combinatorics problem. So here’s an exercise for the 1994 IMO shorlist. 1994 girls are seated at a round table. Initially one girl holds n tokens. Each turn a girl who is holding more than one token passes one token to each of her neighbours. Show that if n < 1994, the game must terminate. Show that if n = 1994 it cannot terminate. As a variant, take 1991 girls and show that the game cannot terminate for n

@goodplacetostop2973 3 года назад

Solution for the problem... prase.cz/kalva/short/soln/sh94c5.html

@kshitijsharma3170 3 года назад

Thank you, That's a good place to stop.

@ofekn 3 года назад

you didnt need to prove that f is unique until 7:52 you proved that: if f(x)+f(1/(1-x))=x so f(x)=(x^3-x+1)/(2x(x-1)) and because of that f(x) is unique

@alphapolimeris 3 года назад

Only true if domain is R\{0;1} because of the nature of the variable change.

@justintroyka8855 3 года назад

The solution is not unique: for any number c, the given equation is satisfied by f(x) = { c if x = 0; -c if x = 1; (x^3-x+1)/(2x(x-1)) else }. You can see that this works by evaluating the equation at x=0, which yields f(0) + f(1) = c-c = 0. The flaw in the proof of uniqueness is that it assumes that f is undefined at 0 and 1.

@RobertBlair 3 года назад

I was looking for a similar tricky answer, looks like you found one.

@spacescopex 2 года назад

MY SOLUTION: ：ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2p8mWGxIqvY.html

@biinghannchia3442 3 года назад

There are a few errors/inefficiencies. 1) g^3(x) is not the same as x. Of course g^3 meaning g(g(g())). Try g^3(0). The problem is, g is bijective only in R - {0,1}. So you need to restrict your solution only to that domain. Turns out, you can pretty much choose any value for f(1) and f(0), as the only thing constraining them are f(0) + f(1) = 0. 2) You dont need to show uniqueness. In your entire solution, each line is necessary to the line preceding it. In other words, if there exists f s.t. the original constraint is met, then f necessarily must have the form you obtained. What you MIGHT need to check is to plug your answer back into the original constraint (I'm not too sure about this).

@panyachunnanonda6274 2 года назад

Thank you very much, I love this pretty method.

@user-mz4jz5pg2m 3 года назад

I have read a lot of comments stating that 1 or 0 or both are NOT in the domain of f. Since the given functional equation does not specify which set x should be in, we have to assume that set is R\{1}, namely the 'biggest' subset of R in which the equation makes sense. (I am not talking about the domain of f just yet.) So the original question should be : Find all functions f, such that f(x)+f(1/(1-x))=x for all x in R\{1}. That said, 0 must obviously be in the domain of the requested function f. But how about 1 ? The Answer is also YES, because the requested function f must satisfy the equation for x=0, that is f(0)+f(1)=0. So the domain of the requested function f must be R. More generally, if the exercise goes like : Find all functions f, such that . . . . f(g(x)) . . . = . . . . . for all x in A. (That is, f(g(x)) is part of the functional equation where g is a given function and A, a given set.) Then obviously g(A) must be a subset of the domain of the requested function f. Usually some g is just the identity function x, hence A ( =g(A) ) is also a subset of the domain of the requested function f, but that does not mean that dom(f) = A, because, like the particular example we are discussing , there exists an f(1/(1-x)) in the functional equation making { 1/(1-x) : x in R\{1} } a subset of dom(f) too. Hence dom(f) = R. I apologize for my English and I hope my contribution will be helpful.

@jeromepatoux9719 3 года назад

There are things that would have required some clarification. In the first part, you multiply by X and X-1. So we need to assume X not equal to 0 or 1. So the function you found needs to be defined on a definition domain that excludes 0 and 1 (note that the original function f is also not defined on any particular definition domain). Then you would need to study the case for 0 (can we find a function f that satisfies f(0) + f(1) = 0?). By default you would assume x different than 1 to be able to write f(1/(X-1)) in the subject. This being said, the video is still pretty cool. :-)

@jamirimaj6880 3 года назад

I don't think the domain matters here. It will only matter in a math exam, when the examiner also wants you to find f(0) or f(1) or etc. That aside, this is just finding families of functions that satisfy the given conditions.

@aliasgharheidaritabar9128 3 года назад

It was so great sir.thank you

@2false637 3 года назад

Great video!

@constantinogeorgiou7801 3 года назад

Very good!!! From Greece have a nice 2021!!!!

@rongliao9255 3 года назад

This is amazing 🤩 Very impressive!!

@spacescopex 2 года назад

MY SOLUTION歡迎指教：ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2p8mWGxIqvY.html

@starsmasher568 3 года назад

Wow, the solution was so elegant 😃 Also, on a side note, the sounds of chalk hitting the white board were strangely satisfying 😌

@nournote 3 года назад

The domain of definition of f does not include 1, that's clear. Now suppose f is defined at 0, we would necessarily have : *f(0) = -f(1)* But that's not possible, so f is not defined at x=0. We keep all the rest as is and we don't need to prove unicity.

@carcaperu4041 3 года назад

A QUESTION This problem was easy once I realized that (it took me hours) 1/(1-x) is the solution of : T(T(T(x)))=x Does any one knows the solution for the case of 4? T(T(T(T(x))))=x and how one gets it for any N? Thanks Nobody answered my question but I found the answer, it is 1/(a-a**2/2) for any "a", for example 2/(2-x). I got itby trying functions of the form 1/(a+bx), but I still do not have an easy form for a general case of N.

@danielmilyutin9914 3 года назад

One can prove uniqueness by relying on fact that solution of system of general linear equations with nonzero determinant is unique. Actually you did prove this fact but for this one system for f.

@ArgumentumAdHominem 3 года назад

Yes, but his system of equations is not linear. It is only linear with respect to the function itself, but not with respect to the parameters of the function. Consider the system {x^2 + y^2=2, x^2 - y^2=0}. The linear system has only 1 solution (x^2=1, y^2=1), but in total there are 4 solutions because x and y can be negative

@danielmilyutin9914 3 года назад

@@ArgumentumAdHominem I did mean linear in respect to function itself. I don't see any contradiction.

@ArgumentumAdHominem 3 года назад

@@danielmilyutin9914 On second thought, I think you are right. For every value of x, the linear system prescribes a single exact value to the function. So, considering the system for all values of x should fix the values of the function everywhere, resulting in a unique solution. My original intuition seems to have been wrong, thanks for noting

@dmitrystarostin2814 3 года назад

Great job. 1. lim of this function at + inf is x/2 - 1/x, which is essentially a quasi-diagonal straight line in asymptote. Looks like a radio lamp or a laser charge-up. :) 2. There is a method of teaching when it is told who and when proposed and solved this equation first. But maybe it was just the specific of the physics department. It looks like the 18th-century problem. If it is cubic, it may have originated in the works of Omar Hayam, Al Tusi and Nicolo Tartaglia.

@bobzarnke1706 Год назад

The above procedure solves the more general functional equation: f(x) + f(1/(1-x)) = h(x), giving 2f(x) = h(x)-h(1/(1-x))+ h((x-1)/x). Then 2f(x) + 2f(1/(1-x)) = (h(x)-h(1/(1-x))+h((x-1)/x)) + (h(1/(1-x))-h((x-1)/x)+h(x)) = 2h(x). h(x) = x above.

@maimounchokri1135 3 года назад

Thanks alot

@valovanonym 3 года назад

Problem suggestion: Find the value (after demonstrating the it converges) of: sum for n=0 to infinity of (-1)^n/(1+3n) Shouldn't be to hard

@jabunapg1387 3 года назад

That's really nice

@sil1235 3 года назад

There are hidden assumptions on domain of f, and domain of functional equation variables, which I think should be made explicit in functional equations problems... Namely the whole solution assumes that we are interested in f with domain R - {0,1}, and that x in the equation is from R - {1}. However in fact f could be from R to R, and only the domain of x in the functional equation could be R - {1}... then we would have to additionally say how f behaves on x=0 and x=1. From the equation we see (from x=0) that f(0)+f(1)=0, so we can choose any real number c and then put f(0)=c, f(1)= -c, which would complete the solution (and so by the choice of c, we have in fact infinitely many solutions).

@HagenvonEitzen 3 года назад

7:53 Up to now you showed *uniqeness*. What's missing is that it actually fulfills the functional equation!

@yoav613 3 года назад

the proof that f is the only function at the end was great!!

@CglravgHRjsksgS 2 года назад

Which is the domain of f that is given in the beginning? This function f is the only one that will satisfy this functional equation if x€R-{0,1}. The fact that x≠1 is obvious because we have an 1/(1-x) term in the first functional equation. But the x=0 case is not excluded. The operations that you did though cannot be done without taking the restriction x≠0...

@ojasdeshpande7296 2 года назад

Plug in x=0 in the equation you end up with f(0)=-f(1) which is undefined...

@adrianamor8472 3 года назад

6:00 you take 3rd equation - 2nd equation gets you f(x) - f(1/(1-x)) = (x-1)/x - 1/(1-x) then from 1st eq. replace f(1/(1-x)) with x - f(x) and then you get 2f(x) = x + (x-1)/x - 1/(1-x) Then show that this function satisfies the condition. I think that's it, no more arguments are needed.

@pwmiles56 3 года назад

Investigating the identity g(g(g(x)))=x, we can write g as a Moebius transformation with real coefficients: [g1;g2] = M [x1;x2]. where M=[0 1; -1 1] and g=g1/g2, x=x1/x2. The desired property is M^3 = kI with k non-zero, I the identity matrix. Other solutions are possible e.g. M=[2 -4; 3 2]. g(x)=2*(x-2)/(3*x+2)

@romanburczymorda4313 3 года назад

Hmm, maybe from Polish IMO SELECTION TEST : Find functions f such that: f( (x+y)/(x-y) ) = ( f(X) + f(y) )/( f(X) - f(y) )

@brunoparra7204 3 года назад

I'm attempting now, have you figured it out?

@OuroborosVengeance 3 года назад

Whats the difference between Capital X and little x?

@williamadams137 3 года назад

@@OuroborosVengeance I think it’s a typo. The functional equation is probably f( (x+y)/(x-y) ) = ( f(x) + f(y) )/( f(x) - f(y) ) .

@OuroborosVengeance 3 года назад

@@williamadams137 yeap. Thanks

@williamadams137 3 года назад

@@angelmendez-rivera351 Maybe the domain is all (x,y) in R² , and I think the question asks us to find SOME function”s” satisfying that equation, so I guess we just need to find at least 2 functions. 🤷🏻‍♂️

@shivansh668 3 года назад

My favorite 😍 topic of MATH 😊😊

@konstantinsarychev9305 3 года назад

I'm quite far from mathematics, but it's curious, why do we need to prove that the function we found is the only one, if the equation is solved in generic form and there were no assumptions?

@bulzangeorge1054 3 года назад

You read my mind with that question.

@KuroboshiHadar 3 года назад

I might be wrong, but there were probably some assumptions that were implicit along the way, it's usually the case with this kind of things..

@WillCummingsvideos 3 года назад

Lots of possible zero denominators kicking about, could be to do with that

@aa-lr1jk 3 года назад

@@WillCummingsvideos That would be the case, but there was no assumptions about the domain and all he did was algebraic manipulations over "formal" variables.

@JM-us3fr 3 года назад

Well it wasn't solved in generic form. It solved to a very specific form. When doing algebraic steps, it is common to sweep assumptions under the rug, which may have unnecessarily restricted the kind of solutions at which we arrive. So doing a uniqueness proof is a way of catching anything we might have missed. For example, the functional equation makes the assumption that x is not 1, but Michael's algebraic steps made the additional assumption that x is not 0. This is why Michael needs to prove uniqueness. However, his uniqueness proof again only proved it for x not 1, so there's still a slight gap by the end of the video (though it's easily fixed).

@vijayakrishna07 2 года назад

Finally I am fulfilling my thirst for functional equations. Thanks

@polyhistorphilomath 11 месяцев назад

The easy way (maybe?): (1 + exp(x^2 D) exp(D))f=x. F x = -i sqrt(π/2) δ'(ω) or so, depending on your conventions. Then the inverse of -i sqrt(π/2) δ'(ω) + sqrt(π/2) δ(ω) + 1/2 i sqrt(π/2) (-1 + e^(i ω)) sgn(ω) is f(x) as defined around 7:45

@zakichan85 2 года назад

In the lasr part you've just demonstrated that a solution if it exists it's unique, still you need to verify your solution since your reasoning wasn't by successive equivalences

@typha 3 года назад

The solution is actually not unique. It is only unique up to changes at finitely many points. The general solution for any real number c is: f(x) = (x^3-x+1)/2x(x-1) when x is not 0 or 1, and f(0) = c f(1) = -c

@MrRyanroberson1 3 года назад

I wonder what is the likit as x goes to 0 or 1? At least one of them must be removable

@typha 3 года назад

@@MrRyanroberson1 Hi Ryan, as it happens neither of them are, they both look a little like the sort of discontinuity 1/x has. Or if you know what 1/(1-x^2) looks like off the top of your head (hey, some people do), the whole thing kind of looks like that. So there's no particularly 'natural' choice of f(0) or f(1), but once you decide what you want f(0) to be, the functional equation tells us that f(0)+f(1/(1-0))=0, so f(1)= -f(0). And it's important to note that our function equation doesn't tell us anything else about either f(0) or f(1).

@canalMatUem 3 года назад

É legal essa, basta fazer composições até chegar na identidade, caiu no ITA uma parecida aqui no Brasil, tempos atrás

@spacescopex 2 года назад

MY SOLUTION：ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2p8mWGxIqvY.html

@user-bu2uk7mx9m 3 года назад

this was in STEP 3 2016

@balazsgyekiczki1140 3 года назад

What about the Domain of the function?

@goodplacetostart9099 3 года назад

It goes without saying that it's your damn homework

@KingStuartPhD 3 года назад

Exactly. As stated the problem has no solutions

@RMGiroux 3 года назад

Interesting! What's missing is the "why"... Why was repeated composition chosen as the tool for solving this, and when is it applicable?

@miguelangelsancheznina2986 3 года назад

experience i guess

@tiripoulain 3 года назад

Like Miguel said, it comes down to just having solved these types of equations before. The cyclical nature of 1/(1-x) really jumps at you for this problem.

@henk7747 3 года назад

@@tiripoulain slight objection: I think what pops up more than the cyclic nature of 1/1-x is that the LHS is f(x) + f(1/1-x) and it seems substituting x->1/1-x may help us isolate f(x). The fact that it is cyclic works nice but even if it didn't it would be the first step I would try.

@firstfirst809 3 года назад

I believe this question was from a STEP (English maths) paper. The first part of the question was a simpler variant which heavily hinted that this method would work on the second part of the question

@alexsmart2612 2 года назад

Try to solve a simpler equation: 2f(x) + f(1/x) = x. You can think of f(x) and f(1/1-x) as the two "unknown" functional forms in the original equation. Problem: there are two unknowns and one equation. In general, any new substitution will give you one new "unknown" one new equation. The hope in trying these substitutions is that eventually one of these terms will cycle back so that in the end you have n equations and n unknowns, so that you can eliminate all but one of them, giving you the form of f(x).

@peraperic4633 3 года назад

Nope, by solving for f(x) you've already shown that the function has to be equal to that, i.e. that the solution is unique if it exists. In the second part you need to check that it is a solution, NOT that it is unique. Or, you could just start with the formula (which you may guess) and check it works (which is straightforward), and then prove uniqueness. In any case, you seem to have misunderstood what needs to be done by using your method.

@peraperic4633 3 года назад

Also you need perhaps to consider values of x equal 1 separately, as this value does not work with your method.

@peraperic4633 3 года назад

In fact, solution is neither unique, nor complete as you have described it. You have that f(0)+f(1)=0, and the equality for x equal to q does not make sense. For all other x, you have your solution, but values for 0 and 1 you pick as two arbitrary real numbers that are of oposite sign. That is if your domain is R (or indeed C or Q) and you interpret the equality as whatever makes sense. If you would include infinity in your domain, you get f(inf)+f(0)=inf, f(1)+f(inf)=1 f(0)+f(1)=0, which cant be consistently interpreted, so you have to exclude infinity from domain - only if you also exclude 1 and 0 from domain, you get your unique solution...

@pepefrogic3034 3 года назад

@@angelmendez-rivera351 You are such a mor on. The comment is spot on, it precisely describes the issues - the man just mixed up proving that this is solution with uniqueness, plus the domain issue. The conditiins that you mention are neither mentioned, needed or used.

@jkid1134 3 года назад

As interesting as it is, is the proof of uniqueness really necessary? If there is some step in the beginning which could obscure a solution, I'm not sure which part, or how.

@thornadotrigger3073 2 года назад

I want to try something like that with functions . Equations for me only were linear and quadratic until coming across other equations exponential with the letter E and trigonometric with waves Sinx and cosx . This is a new with functions .

@ojasdeshpande7296 2 года назад

It's studied specifically like trigonometry is done

@ArgumentumAdHominem 3 года назад

Really cool method. I'm not 100% sure about the x=0 case though. It ends up in the denominator at some point.

@want-diversecontent3887 3 года назад

So far just by plugging in numbers I've figured out that f(-1) has to be 0.25, f(0.5) has to be -1.25, and f(2) has to be 1.75.

@chrisbattey 3 года назад

If that sequence of steps wasn't sufficient to show that f(x) = (x^3 - x + 1)/(2x(x-1)) was a unique solution to (*), why was it sufficient to show that f_h(x) = 0 was a unique solution to f_h(x) + f_h(1/(1-x)) = 0?

@josemath6828 2 года назад

Interesante ejercicio.

@CM63_France 3 года назад

Hi Michael, The first part of your proof leads you to a necessary condition, so the second part (the uniqueness) is not usefull. For fun: 1:12 : "the denominator and the denominator", 1 "let's may be go ahead and do that", 1 "let's may be go ahead and",

@bryceherdt2363 2 года назад

I think you missed that the "unique" function you gave cannot be evaluated at 0 or 1. Now, since f(1/(1/1)) has no meaning, the only other constraint seems to be that f(0)+f(1/(1-0))=f(0)+f(1)=0.

@Igorious92 2 года назад

Yes, it's a family of functions: f(x) = ... for x != 0, x != 1; f(0) = c and f(1) = -c.

@AmrishKelkar 2 года назад

Coming up with 3 simultaneous equations looks amazing.. But what is the intuition of how one decides to investigate the direction of g(g(x)) and g(g(g(x)))?

@shashankkatiha9439 3 года назад

How about putting x=cos theta

@tomholroyd7519 5 месяцев назад

Wasn't expecting a cubic

@sawyerw5715 3 года назад

In the proof at the point that you had the fh equation, you could note that the original had the same arguments but resulted in x on the rhs while this equation results in 0 on rhs, thus being incompatible or am I seeing something wrong?

@tokajileo5928 3 года назад

if x=1 how you calculate g(g(g(x))) ? you get 1/0 on the road

@romanburczymorda4313 3 года назад

Find all continuous functions f: R-> R such that for any x, y there is equality f(x) +f(y) + 5xy(x+y) =f(x+y)

@md2perpe 3 года назад

f(x) = (5/3)x^3 + cx for any constant c in R. Note that f(tx) + f(ty) - f(tx+ty) = 5(tx)(ty)(tx+ty) = t^3 5xy(x+y) = t^3 (f(x) + f(y) - f(x+y)). This suggests that f(x) = ax^3 for some constant a. Inserting this solution into the equation and solving for a gives a = 5/3. Now assume that g is another solution. Then (f-g)(x) + (f-g)(y) - (f-g)(x+y) = (f(x) + f(y) - f(x+y)) - (g(x) + g(y) - g(x+y)) = 5xy(x+y) - 5xy(x+y) = 0, i.e. (f-g)(x+y) = (f-g)(x) + (f-g)(y). This has continuous solutions (f-g)(x) = cx, where c is a constant. Thus, all solutions are given by f(x) = (5/3)x^3 + cx.

@md2perpe 3 года назад

@@angelmendez-rivera351 You are right, I made a sign mistake. But it doesn't matter in the end. My argument for concluding f(x) = ax^3 is a bit handwavy, but then I show that any other solution differs with a solution to the Cauchy functional equation, so I can conclude that all solutions are given by f(x) = ax^3 + cx, where a=5/3 and c is arbitrary.

@aleks0id Год назад

One moment.. when we begin our judgment about f1 it means that we don’t rely on the “g(g(x))+g(g(g(x)))” game in that it guarantees us a singleness of f() we found. But since then we cannot rely that fh=0 for all x found by the same game is the only one function as well. Isn’t it a little bit contradictory?

@IsmeGenius 3 года назад

I wonder how can we redefine x -> 1/(1 - x) and them sum resulting equations, in each of which x has a different meaning. How can we rigorously show this is correct to do that?

@Alians0108 3 года назад

x is a dummy variable. Might as well change it to something else if you want.

@backyard282 3 года назад

x does not have a different meaning. For the same given x, we are evaluating the function at x in one instance and in the other at 1/(1-x).

@VideoFusco 3 года назад

to demonstrate the uniqueness of the solution was it not enough to say that it was guaranteed by the uniqueness of the solution of linear systems with equal number of unknowns and equations?

@lucagagliano5118 2 года назад

Can anyone suggest a good reference for functional equations?

@nathanisbored 3 года назад

is there classifications of functional equations similar to differential equations, and are there strategies for solving different types in general? i never learned anything about pure functional equations like this, but i can imagine there being a broad theory of them just like ODEs

@bourhinorc1421 3 года назад

I imagine that you can numerate some properties that has to appear like distributivity or commutativity from these types of functions, but I think that you can't deduce with that the fonctions that can satisfy the thing because you can always invent a new fonction that satisfy these properties

@nathanisbored 3 года назад

@@bourhinorc1421 what if you add some condition like the function has to be "analytic" or something

@julienliottier2094 3 года назад

According to the equation we should have f(0) + f(1) = 0 but there our solution isn't defined on 0 and 1, isn't it an issue? Should we define a random value c = f(0) = -f(1) and then lose the uniqueness and the continuity ?

@julienliottier2094 3 года назад

@@angelmendez-rivera351 I agree that the initial equation is not defined for x=1 but it is for x=0 and it gives us f(0) + f(1/1-0) = 0 i.e f(0) + f(1) = 0 so you can (must ?) define both f(1) and f(0). On the opposite saying that f(0) or f(1) doesn't exit don't make any sens (for me) because then the initial equation makes no sens for x=0 Moreover, because of the multiple compositions both the demonstration of the general form and (implicitly) the demonstration of the uniqueness are for x =/= 0 and x =/= 1 so even if we chose a random value for f(0) and f(1) (s.t f(0) = -f(1)) they are still valid.

@julienliottier2094 3 года назад

@@angelmendez-rivera351 the more you explain the more I disagree. ^^" The domain of the solutions and the domain of the equation are not the same things. An equation which is not defined for a value of x doesn't imply that its solutions are not defined on this value. The equation is not defined for x=1 because 1/1-x is not defined for this value but the original equation IS defined for x=0. You do it in the wrong way, you see the solutions in the video and then conclude that the function can't be defined for x = 0 or x = 1 but the functionnal equation doesn't imply that. It is true for every x which gives a sens to this equation, in particular for x = 0 which gives f(0) + f(1) = 0. The fact that the solution in the video is not defined on 0 and 1 is due to the fact that since the beginning the equation (and not the function) is not defined for x=1 and then later 1-x/x is introduced in the demonstration so the reasoning is only for x =/= 1 and x =/= 0 BUT it doesn't mean that the function is not defined on 1 or 0, it just means that its form is the one in the video for every x distinct of 0 or 1 and then everything can happen on these values if you disagree can you explain me why f defined by : * f(0) = 2 * f(1) = -2 * f(x) = the f of the video for every other x is not a solution even if it fully satisfies the original equation (even for x = 0)? What I want to show is that a functional equation doesn't imply anything on the domain of the solution. If there is a value on which the equation is not defined it doesn't mean that the solutions are not defined for this value. However a solution of this equation must fully satisfies this equation and there the solution of the video doesn't satisfy f(0) + f(1) = 0 so it's not a complete solution.

@julienliottier2094 3 года назад

@@angelmendez-rivera351 "For example, the functional equation f(x + 1) = 2·f(x) is solved by 2^x because it is satisfied everywhere. It is also satisfies exactly by g(x) = x if x = -1, but g(x) is definitely not a solution to the equation, and I would boldly assume that you agree." I totally agree with that and that's the problem I have with Michael's solution, it doesn't satisfy the equation everywhere, so, according with your example you should agree that it's not a solution because it just works for chosen points and not for others. "No, it is not, because the functional equation implies both f(1) + f(1/0) = 1 and f(0) + f(1) = 0, and the former implies f(1) is undefined, and f(1) is undefined implies by the latter that f(0) is undefined." I don't understand this argument since the beginning. It makes no sens for me. You can't write f(1) + f(1/0) = 1 because the equation is not defined for x=0, that's all but it doesn't imply that f(1) doesn't exist. 1 + 2 + 3 + 4 + 1/0 is not a number but 1, 2, 3 and 4 are numbers, it's not because they are written in an undefined form that they are undefined. But even if I disagree with your idea that f can't be defined on 1. You may have right on your definition of a solution (I find definitions of functional equations but no definition of a solution of a functional equation on the web), especially thanks to your log example. "functions which satisfy the equation everywhere in their domain" may be the true definition (and I begin to believe that's it^^) so Michael's solution is a solution. But it implies that mine is a solution too or that "constructed" functions such that Michael's function but only defined on ]0,1[ are solutions too becaus both of them satisfy the equation on their domain?

@chato9327 3 года назад

Le sabe demasiado.

@Plomeg 3 года назад

Shouldn't we give values for f(1) and f(0)? The problem is defined for x =/= 1 but we need a value for f(0) and thus f(1) (the problem and the function set are different). Thus it gives f(0) = -f(1) and we can set f(0) to any constant leading to infinite solutions

@ricardomejias156 3 года назад

Can you recommend me some books for similar problems to your videos please??

@matthewcroy7704 3 года назад

what are the applications of these?

@patsplats 2 года назад

g(g(x)) happens to be inverse of g(x)?

@vsevolodyeroshenko1149 3 года назад

I do not fully understand why this three f (smth) are linearly independent? I mean I see arguments are, but there is some mapping to the space of f() which in general doesn't mean they should be...

@CarlosMarante 3 года назад

Does f(0)+f(1)=0 holds as it should be from the original equation, when computing the limit?

@mayamodraf 3 года назад

Unclear

@user-A168 3 года назад

Good

@gabrielmuhammad3361 3 года назад

f(x)=0 does also work as a solution to this functional equation

@brankoco 3 года назад

For the uniqueness part of the proof, wasn't there a mistake by writing fh(x/(x-1)) instead of fh((x-1)/x))? Thanks for the clarification

@omarhassan2124 3 года назад

It's a mistake but when we do his method the f_h(x/(x-1)) cancel itself so it doesn't affect

@hypergaming2051 2 года назад

How He find the value of g(1/1-x)????

@kratoss_sparta 3 года назад

Don't u need to find the Dg(g(x)) first before you actually find the function?

@vatsan2483 3 года назад

How do u know taht that's the tip we require as in problem??? Like how do u think about it???

@no_one6749 3 года назад

If it was obvious it wouldn't need to be said

@kevinmartin7760 3 года назад

I don't think the "uniqueness proof" part really does anything. It seems to me that it is going through the same procedure as the original derivation (with the deck chairs rearranged a bit) so I don't think we should feel either surprised or delighted that f1 must be the same as f.

@zealiskander 3 года назад

> It seems to me that it is going through the same procedure as the original derivation (with the deck chairs rearranged a bit) so I don't think we should feel either surprised or delighted that f1 must be the same as f. Well, yeah. It's fairly common though, => is trivial to prove, it just flows from the manipulations we've done with f, and there's technically no need to actually do the uniqueness proof, because you could simply prove that you also have . (and you know that you *can* prove because there is only one solution...) That said, proving is often hard to properly justify. You have to go over each step and justify how none of them break , and that can get messy pretty fast. (especially in that case, where there's soooo much you could argue because technically, the function isn't properly defined for x=1, or x=0.) So, most people don't bother with trying to prove , and while you don't need to prove the uniqueness and could go another way, it's gonna be shorter than the alternative, even if a bit repetitive.

@martinepstein9826 3 года назад

You're right. The first part of the video was already a proof that "if f satisfies the equation then f(x) = ..." What's left to show is that the function he found does in fact satisfy the equation.

@claudiosaccon1951 3 года назад

Also I think that you need to verify that the function found in the first step actually solves the functional equation

@kevinmartin7760 3 года назад

@@claudiosaccon1951 It all depends on whether each step involves equivalence ("if and only if") or just implication ("only if"). If all the steps are equivalences you have nothing extra to prove (other than checking your work for mistakes).

@claudiosaccon1951 3 года назад

@@kevinmartin7760 I agree. But I would say that what he does in the video is more or less "if the functional equation holds, THEN taking g(x) instead of x gives ...., then solving the linear system..." ending up with only one possible form of f(x) (this makes the second part of the video unnecessary). Now to do the reverse path you should SAY that g is invertible (which is simple since the inverse of g is the twofold iteration of g). I was not doubting that the results is true, just commenting that he didn't check that all the steps he did were reversibile.

@Mohamed.Soltan1991 3 года назад

Wonderful 💖💖💖

@MixMasterMohan 3 года назад

the math checks out and its a great solution but as I was watching this all I was thinking was how did he know to do that, especially in the beginning steps

@Hiltok 3 года назад

I have a bachelors degree in maths but am by no means as knowledgeable as many of the commenters here, but I can describe how I arrived at a similar solution to Michael's by firstly playing around with the equation by plugging in some real values for x to see if I could find some values of f. x=1 was out from the start for the obvious reason 1/(1-x). So, I tried x=0 --> f(0) + f(1) = 0 This equation just tells us that f(0)=-f(1), even though we already ruled out allowing x=1. It's a strong hint that x=0 will also be a problem. Then I looked at x=-1 --> f(-1)+f(1/2) = -1 Then what about x=1/2? x=1/2 --> f(1/2)+f(2)=1/2 So naturally I then looked at x=2 --> f(2)+f(-1)=2 Hmmm. Back to -1. Interesting. Now, that system with x=-1, 1/2, and 2 is solvable and you get values for f: f(-1)=1/4, f(1/2)=-5/4 and f(2)=7/4. I calculated these values because I wanted to get a feel for what the function might look like. I goes from positive to negative back to positive as x goes from -1 to 1/2 to 2. But that isn't much to look at, so I then plugged in x=-2 --> f(-2)+f(1/3)=-2 Then, following the process of getting more values, x=1/3 --> f(1/3)+f(3/2)=1/3 And then x=3/2 --> f(3/2)+f(-2)=3/2 I then calculated solutions for f: f(-2)=-5/12, f(1/3)=-19/12 and f(3/2)=23/12 because getting some values to look at was what I was really trying to do, but then the big 'Aha!' moment was realizing that the process cycled back to the original input again and in the same number of steps. Would it do this every time? I then used the same process but instead of using real numbers, I just left x as it was, and followed the algebra of cycling through inputs x, 1/(1-x), and then of course you are led to 1/(1-(1/(1-x))=(1-x)/(-x)=(x-1)/x which in turn cycles to 1/(1-((x-1)/x))=x. The system of 3 equations gives solutions by adding any two and subtracting the other. For f(x) you choose to add the two equations with f(x) and the subtraction removes the f(1/(1-x)) and f((x-1)/x) terms. Many of these problems are solved by first messing around with them in very messy ways before you recognize what approach will yield a neat solution.

@MixMasterMohan 3 года назад

@@Hiltok Great explanation! Logical, easy to understand rather than pulling out equations from a hat. This addressed my confusion. Thank you sir!

@Hiltok 3 года назад

@@MixMasterMohan Thank you. Enjoy your mathematics studies. Michael makes mistakes occasionally, but not as many as me. :)

@tammammohammed4442 2 года назад

From the question, shouldn't the domain be (x does not = 1)? Why the domain of the solution is (x does not = 0 and 1)?

@alphapolimeris 3 года назад

So the solution is unique if it is unique ? Yikes ! A careful and rigorous writing would demonstrate that on R \ {0;1} all solutions have to coincide with what you found. But there are an infinite number of different solutions : you can set any value 'a' to f(0) as long as you set f(1)=-a. (because f(0) + f(1)= 0 ) Cool problem though ! A bit tough without the hint. Edit: And I will gladly admit that the "rigorous" writing wouldn't make for a very fun part on video format. So that's easily forgivable. But that "proof" of uniqueness is a sin of the highest degree (circular reasoning).

@Idsa0 3 года назад

My approach was looking at the function g(x)=1/(1-x) and inverting it and it works the same way since g(g(x))=g^-1(x). awesome problem, I really enjoy your functional equation videos!

@user-og1xw6bl1g 3 года назад

This method is called : "Obviously to notice", see also "Follows from mathanalisys course", "By not clever transformation" methods)))

@user-jt1nx1se5v 3 года назад

"оставим как упражнение читателю" еще, куда без него

@MrGyulaBacsi 3 года назад

Very good derivation! Thumbs up. One question though: you knew that that helper lemma you derived at the beginning will help you find the solution. But how could someone figure out that lemma in the first place? Eg. I was given this problem to solve, but I'm not aware of this helper tool. How am I supposed to get there?

@tiripoulain 3 года назад

You really just have to notice the cyclical nature of 1/(1-x) under composition. Now you have, and if you every come across a similar-looking rational function in a functional equation, you might try composing it with itself a bunch of times

@Hiltok 3 года назад

I just wrote a bit of a long reply to someone with a similar question. I'll copy it below. The bottom line is you don't need the lemma. The cycle x --> 1/(1-x) --> (x-1)/x --> x comes naturally from the algebra as you work towards a solution. Michael just pulled that bit out to make the algebra a bit quicker and neater on the way through. +++ I have a bachelors degree in maths but am by no means as knowledgeable as many of the commenters here, but I can describe how I arrived at a similar solution to Michael's by firstly playing around with the equation by plugging in some real values for x to see if I could find some values of f. x=1 was out from the start for the obvious reason 1/(1-x). So, I tried x=0 --> f(0) + f(1) = 0 This equation just tells us that f(0)=-f(1), even though we already ruled out allowing x=1. It's a strong hint that x=0 will also be a problem. Then I looked at x=-1 --> f(-1)+f(1/2) = -1 Then what about x=1/2? x=1/2 --> f(1/2)+f(2)=1/2 So naturally I then looked at x=2 --> f(2)+f(-1)=2 Hmmm. Back to -1. Interesting. Now, that system with x=-1, 1/2, and 2 is solvable and you get values for f: f(-1)=1/4, f(1/2)=-5/4 and f(2)=7/4. I calculated these values because I wanted to get a feel for what the function might look like. I goes from positive to negative back to positive as x goes from -1 to 1/2 to 2. But that isn't much to look at, so I then plugged in x=-2 --> f(-2)+f(1/3)=-2 Then, following the process of getting more values, x=1/3 --> f(1/3)+f(3/2)=1/3 And then x=3/2 --> f(3/2)+f(-2)=3/2 I then calculated solutions for f: f(-2)=-5/12, f(1/3)=-19/12 and f(3/2)=23/12 because getting some values to look at was what I was really trying to do, but then the big 'Aha!' moment was realizing that the process cycled back to the original input again and in the same number of steps. Would it do this every time? I then used the same process but instead of using real numbers, I just left x as it was, and followed the algebra of cycling through inputs x, 1/(1-x), and then of course you are led to 1/(1-(1/(1-x))=(1-x)/(-x)=(x-1)/x which in turn cycles to 1/(1-((x-1)/x))=x. The system of 3 equations gives solutions by adding any two and subtracting the other. For f(x) you choose to add the two equations with f(x) and the subtraction removes the f(1/(1-x)) and f((x-1)/x) terms. Many of these problems are solved by first messing around with them in very messy ways before you recognize what approach will yield a neat solution.

@takyc7883 3 года назад

How do you even think of compositing functions to get the identity function?!

@bourhinorc1421 3 года назад

that's a common thing in that type of question (either assume x=0, x=1, repeat the fonction, try to get out the thing that is not easy (here g(x)) and see properties and deduce a general formula, (that shows the unitcity too) then show that the general formula satisfies what is asked (show that the fonction exist) )

@michalbreznicky7460 3 года назад

You don't necessarily. You can start with assuming let's say, that f(2) = a for some unknown number a. You can use that fact to get f(-1), since f(2) + f(1/(1-2)) = 2. You get f(-1) = 2 - a, so you now know 2 values of the function (discounting the unknown "a" for a moment). You can then apply the trick again to get a third value of f, f(1/2), since f(-1) + f(1/(1-(-1))) = -1. The idea is to continue doing this getting more and more values of the function, thus learning more about what it could possibly look like. Luckily, it turns out that if you apply the trick the third time, you get back where you started, and figure out the values f(2),f(-1),f(1/2) directly using linear algebra. Replacing f(2) by f(x) then gives you the general solution.

@takyc7883 3 года назад

@@bourhinorc1421 I can never find the solution for these function questions

@ryanchin5232 2 года назад

can you show us the thought that how to figure out using that tools: g(x)=1/1-x, and g(g(x)), and g(g(g(x))). Math is more of thought/logic than skills.

@varadarajm.s.648 3 года назад

I think this is the only channel where i learn equally good mathematics by reading the comments section.

@azharlatif6228 2 года назад

Perhaps you have difficulty following Prof.Michael Penn's hasty arguments. His enthusiam to use all the surface of Chalk Board make me admire his hard work, which was beyond me.Following the comments, it is possible to grade his work, though.

@alexandermorozov2248 8 месяцев назад

Вам очень повезло, что на 2-м шаге пришли обратно к f(x) и можно составить простое уравнение. В более общем случае непонятно, как решать такие функциональные уравнения. ~~~ You are very lucky that at the 2nd step you came back to f(x) and you can make a simple equation. More generally, it is not clear how to solve such functional equations.

@ExcelEveryDayv57 3 года назад

Is the method still rigorous when approaching x=1? Do we need to be careful in that situation?

@zealiskander 3 года назад

That's not the only problem, x=0 is also an issue. That said, it's not really something you could have prevented. By necessity f(1) has to be undefined, because f(1) = 1 - f(1/0), and f(0) has to be undefined, because f(0) = -f(1) which is itself undefined. Your resulting function is indeed undefined on 0 and 1, which is completely logical. You could wonder about problems at the limits, but since your function is defined by f(x)+f(1/(1-x) = x, you'll find that, for example, lim x->0 f(x) +f(1/(1-x)) will also trivially equate 0, because f(x) - f(1/(1-x)) is equal to x for every x except 0 and 1, so going back to the definition of the limit, you're always capable of approaching 0 or 1 by less than epsilon.

@zealiskander 3 года назад

@@angelmendez-rivera351fair!

@muckchorris9745 3 года назад

FUNctional equation

@backyard282 3 года назад

Showing the uniqueness was unnecessary because technically you have shown that if a function were to satisfy the original equation it must necessarily be of the form you arrived in the middle of the video. There was no ambiguity, it was straightforward sequence of "if a function satisfies this, then it MUST satisfy this and then it must (...) be of the following form.)

@mryip06 3 года назад

a+b = p b+c = q a+c = r a = ½(p+q+r)-q would it be easier in this way?

@mayurathavale2791 3 года назад

Sir! 😇

@MarcoMate87 3 года назад

The video is very interesting but full of mistakes. 1) At 10:16 and later, you meant f_h((x-1)/x) and not f_h(x/(x-1)), but this is not really a big mistake, only distraction, and it doesn't affect the final result. 2) The process we used to find f guarantees itself that f is unique, you used all forcing implications: IF f respects that functional equation THEN f has to be f(x) = (x^3-x+1)/[2x(x-1)] for every x different from 0 and 1. Instead, what we really have to do after 7:52 is the exact opposite implication: we have to check that IF f(x) = (x^3-x+1)/[2x(x-1)] for every x different from 0 and 1, THEN f respects the initial functional equation. I checked by myself, and also this implication is true. 3) Indeed, f is NOT unique. that functional equation defines f and makes sense only when x is different from 1, it loses his meaning if we substitute x=1. Thus it tells us nothing about f(1), and it tells us that f(0) + f(1) = 0, so f(0) = -f(1) if this f(1) value exists. Thus we can construct an infinite family of solutions of that functional equation, defined in this way: f_a(x) = (x^3-x+1)/[2x(x-1)] for every x different from 0 and 1; f_a(0) =a; f_a(1) = -a. We can also insert "special solutions" assigning the "value" infinity to both f(0) and f(1), maybe doing the right and left limits of f, for x approaching to 0 or to 1 from the left and the right.

@MarcoMate87 3 года назад

@@angelmendez-rivera351 No, the functional equation simply doesn't make sense if x = 1, but this absolutely doesn't imply that x = 1 isn't in the domain of f. The request is to find any function f, with a domain not explicitly expressed, which respects the functional equation for every x different from 1. In other words, the initial functional equation gives infinitely many conditions for f: one condition for every x different from 1. Your funny equation f(1) = 1 - f(1/0) is NOT derived from the functional equation, because the functional equation doesn't exist for x=1 and so it doesn't give your funny equation. Does my family of functions f_a respect the initial functional equation for every x different from 1? Yes, for all a. Do they also respect f_a(0) = -f_a(1)? Yes, for all a. Is f_a (1) = -a in contradiction with the functional equation? No, because you can't insert x=1 in the functional equation. And yes, his proof forces f to be of that form, for every x different from 0 and 1. So you are wrong, not me.

@hybmnzz2658 3 года назад

@@MarcoMate87 agreed. But it is the fault of functional analysts and their conventions.

@MarcoMate87 3 года назад

@@angelmendez-rivera351 I'm happy that you will not reply anymore, because it means that you understood how you were wrong. Bye bye, poor man.

@empathy800 3 года назад

How did you know to use the composition of g three times would yield x?

@ga35am 3 года назад

It is not that much experience. It is the fact that when one mathematician has to deal with a problem, he or she tries to take advantage of little informations. In this case, he tried to see what happens with the function that it is involved in the question, which properties he can take advantage of. If it was some other function, the solution could lead to a very different reasoning. It could even make the problem impossible to be solved by algebraic methods. Verifying that a function have "finite order" relative to the operation of composition is very natural to mathematicians. As it is other properties.

@ga35am 3 года назад

@@angelmendez-rivera351 I said "It is not that much experience". Experience is sufficient, but not necessary. I expect this verification from students with neither experience nor hints.

@ga35am 3 года назад

@@angelmendez-rivera351 also, as I said, this reasoning would not work if the function doesn't have finite order. And don't be unnecessarily dramatic.

@ga35am 3 года назад

@@angelmendez-rivera351 well, I proved that what you said is vague and unimaginative. I didn't intend to disprove anything you said. Relax.