A fun functional equation!!

We solve a nice functional equation.

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[lucasdepetris]

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

[stewartzayat]

You put the fun in functional, Michael!

[BrunoVisnadi]

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

[f-th]

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.

[shaiavraham]

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.

[DylanNelsonSA]

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.

[DS-xhfd]

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.

[DanielWalvin]

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.

[goodplacetostop]

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 <= 1991.

[justintroyka]

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.

[ΓιάννηςΗλιάκης-κω]

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.

[jeromepatoux]

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. :-)

[CglravgHRjsksgS]

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...

[biinghannchia]

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).

[bobzarnke]

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.

[ofekn]

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

[starsmasher]

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

[RMGiroux]

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

[carcaperu]

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.

[dmitrystarostin]

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.