an absolutely surprising final solution to this functional equation. 

The real challenge is to find a functional equation that isn't solved by the identity function. Like, if you get rid of all the f's in the original question it's immediatelly a solution.

17:38 My first answer for functional equation is always the identity function 😂

I am so happy to have solved that myself!!! Pretty much all of my knowledge of functional equations comes from your videos. Thank you Michael!!!

I've yet to see a nested functional equation problem with a non-trivial solution. (The proof may be brutal, of course.) Perhaps there is one involving crazy composition of f and its inverse? For example.... f( f(x) + f^-1(x) ) = something?

Let A(x,y) be the argument of the LHS in the original equation, and let B(x,y) be the complete RHS. It is pretty clear that B is symmetric about x and y so that B(y,x)=B(x,y). However, A is anti symmetric so that A(y,x)=-A(x,y). Switching x and y in the original equation therefore gives f(-A)=B and f(A)=B. This, coupled with the fact that f(x)=x for x>=0 and this immediately gives us the solution f(x)=abs(x). The other solution would require A(x,y)=0 for all x,y, and therefore B(x,y)=0 as well. It is not hard to see then that the only solution to this is the identity f(x)=x

First we notice either f is even because we can interchange the role of x and y (if x*f(y) - y*f(x) isnt always null) : Then we take y = x, so f(0) = f(x^2) - x^2 and so for x > 0, f(x) = f(0) + x And because f is even for all x in R, f(x) = |x| + f(0), and by using definiton for x = 2, and y = 1 we have f(0) = 0 and f(x) = |x| either x*f(y) - y*f(x) is always 0 in which case f(x) = x * f(1) and by using definition we get f(0) = x*y*(f(1) - 1) so necessarly f(1) = 1 and f(x) = x

It is clear that f(x)=x works. For f(x)=|x|. | x|y|-y|x| | = sqrt{(x|y|-y|x|)^2} = sqrt{2x^2y^2-2xy|xy|} = sqrt{(|xy|-xy)^2} = |xy|-xy

For anyone saying that they never see functional equations with nontrivial solutions, I would refer you to Abel’s equation and Schroeder’s equation and many similar families. These equations provably have a solution (under certain conditions), but you almost never find the closed form and are usually solved computationally. In general functional equations are like differential equations, if you just put one together, the solution may exist, but it won’t be nice. And similar to differential equations, there is a class of functional equations which have closed form solutions. For differential equations, the closed form friendly problems are usually linear (or can be transformed to be) or separable. For functional equations the closed form friendly problems almost always have the identity as a solution.

Shorter way for everything since finding out that, for x > 0, f(-x) = mx gor m = f(-1); couldn't you use f(x^2) = x^2 (for all x) with x = -1 to show thatm could only be ±1, showing that f(x) can only be to x or abs(x)?

If we solve the same problem but change domain of f to f: C->C When you see f(x²)=x², you can substitute t=sqrt(x), u = i*sqrt(x) The whole thing becomes really easy. Now cut out the imaginary part, the same solution still works with f: R->R The problem is, I don't know if that's wrong!

i permutated x and y in the first equation, and concluded that f(x)=f(-x), and then when we find that f(x)=x for x>=0, we already know that f(x)=f(-x), which gives us f(-x)=f(x)=x for all x

(x + |x|) (y + |y|) = xy + |xy| + x|y| + y|x| therefore |xy| + xy = x|y| + y|x|, and somehow taking the absolute value of the right side negates the xy? (x + |x|) (- y + |y|) = -xy + |xy| + x|y| - y|x| therefore |xy| - xy = x|y| - y|x| and since |xy| >= xy we can say that if |xy|-xy = t, then it also equals |t|. Therefore this equation is true.

Love your videos, only one critic I have... Your thumbnails are sometimes really hard to read... And if you put problem on thumbnail it wouldn't hutr if it would be readable. Like I said, amazing videos.

I plugged in y=1. Then plugged in x=1 (renaming y to x). This changes the sign inside f on the LHS but keeps the RHS the same. It is quick then to show if there exists x st f(x) =/= x then f(x-f(x))=abs(x-f(x))

Gnarly, but it simplifies quite nicely. Thank you, professor

an absolutely surprising what

Or we see that f(0)=0 so we sub f(x)=x g(x) to get g(xy)-1=(g(y)-g(x))g(xy(g(y)-g(x))) for xy=/=0, and sub y=x to get g(x^2)=1. By the original functional, f is either an even or odd function, giving us f(x)=x or f(x)=|x| as the only two candidates.

Is it just me, or does it get rather complicated starting at 10:00? If you set x=-t0, then f(-tf(x)-xf(-t))=f(-tx-mxt)=f(-(m+1)xt)=f(-xt)+xt=mxt+xt=(m+1)xt=m(m+1)xt and thus m(m+1)=(m+1) or m^2=1.

We should not change varables keeping the original names. Additionally, when Michael substitutes x by x/c, may be he is dividing by zero.

Based term

Can you do the functional equation f(x) + f(y) = f((x+y)/(1-xy)) for all x,y>=0, xy≠1?

I seriously hope these tricks for problem solving come in handy some day in my life.

I think, more clever way is to use the following substitutions: 1) x->a, y->0 (like Mike did). Deriving that f(0) = 0. 2) x->a, y->a (like Mike did). Deriving that f(x^2)=x^2 3) x->1, y->a 4) x->a, y->1. a

i somehow missed the absolute value solution

Why is it that whenever a functional equation is solved by a linear function, the only solutions to the equation are linear functions? I have seen so many functional equations solved by linear functions but there has never been a counterexample to this rule. In this problem one solution was piecewise but it’s still a linear function. Does anyone know any counterexamples?

f(0) != c is established but is as a consequence for example f(0) = x valid?

If f(x^2) = x^2 you have only two cases left for negative x. f(x) = x or f(x) = - x.... Wrong thinking as commented 🤯

good thanks

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absolute value

-e^i*pi!!

Speaking of final solutions /jk.

I missed this kind of videos(with functional equations). That was amazing 😍

Set x=y, f(0)=f(x²)-x², so f(x²)=x²+f(0). Set y=0, f(xf(0))=f(0), so if f(0)≠0 then f(x) is a constant function, so f(...)=f(...)-xy reduces to xy=0, a contradiction. So f(0)=0. Setting x²=z, for x²=z≥0, f(z)=z. Finally, setting x=y,y=x gives f(z)=f(-z) where z=xf(y)-yf(x). Set k(x)=1-f(x)/x for x0, then z=ky, so f(ky)=f(-ky) for all y>0. If there exists x with k(x)≠0 then f(-z)=f(z) for all z. Otherwise, k(x)=0 for all x