an absolutely surprising final solution to this functional equation.

[@farfa2937]

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.

[@hasch5756]

f(x) = 0

[@Nico-le5bb]

f(x) = 1

[@TheEternalVortex42]

Those aren’t functional equations, they are just functions

[@tompazourek5150]

I like this one. f: R -> R such that for all x in R except 0, 1, -1 we have f(x)² f( (1-x)/(1+x) ) = 64x solution: ∛[ (64x²(x+1))/(1-x) ]

[@coc235]

​@@TheEternalVortex42These are precisely functional equations

[@goodplacetostop2973]

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

[@GnarGnaw]

It's always a constant or identity function

[@bot24032]

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!!!

[@orionspur]

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?

[@wynautvideos4263]

Thats because the ones with nontrivial solutions are pretty much impossible to prove that they are the only ones

[@ryanstaal3233]

I participates in the Dutch team selection test for the IMO and the last problem was a hard functional equation. The problem was : Find all functions of Q+ to Q such that f(x)+f(y)=(f(x+y)+1/(x+y))(1-xy+f(xy)) for all positive rational numbers x and y.

[@nirajmehta4264]

@@ryanstaal3233 do you recall the answer?

[@ryanstaal3233]

@@nirajmehta4264 I found (all) the function(s) that satisfy the given equation diring the test. If I remember correctly it was f(x)=x-1/x

[@mathematicskid]

at 7:10 you're basically done, just let -zf(-1)+f(-z)=k so that f(k)=0, then plug in x=k, y=1, f(kf(1)-1f(k))=f(1*k)-1*k which simplifies to f(k)=f(k)-k so k=0, now -zf(1)+f(-z)=0 so f(-z)=-zf(1), boom

[@arvindsrinivasan424]

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.

[@maelhostettler1004]

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

[@minamagdy4126]

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)?

[@petersievert6830]

I thought this as well. Apparently, the issue is, as I read below in a comment, that setting e.g. x=-1 only gives you f((-1)²)=(-1)² , thus f(1) = 1. So it tells us absolutely nothing about the negative numbers.

[@mokouf3]

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!

[@DeJay7]

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

[@kristianwichmann9996]

It's called "analytical thinking". Surprisingly useful out in the real world.

[@DeJay7]

@@kristianwichmann9996 I bet, and I was being serious, since I'm probably gonna pursue something math related.

[@radadadadee]

why? that's such an utilitarian view of life. You can enjoy doing things without a need for them to become handy.

[@DeJay7]

@@radadadadee Well, I do certainly enjoy things without practically using them. What I meant in THIS particular case is that I WANT to use these ideas in my life, as in I hope I find myself in a place where these are useful. :D

[@technolan6015]

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

[@manucitomx]

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

[@awesomechannel7713]

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

[@gigantopithecus8254]

i did a simular thing but i set y to -y and x to -x and then i could just di sum stuff and take rhe inverse and then plug in

[@digxx]

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.

[@colbyforfun8028]

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

[@feltabbar]

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))

[@Notthatkindofdr]

I got that far too. The difficulty was extending this to f(x). I think the substitution at 7:30 was the key step.

[@MrRyanroberson1]

(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.

[@willemesterhuyse2547]

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

[@joshuajacobs8061]

f is a function that takes in a real number and spits out a real number. If you plug in 0, you better hope you get a number out, or something went wrong!

[@user-en5vj6vr2u]

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?

[@ConManAU]

It’s fairly easy to set a functional equation whose solution isn’t linear - for example, any equation of the form f(g(x))=h(x) where g is invertible has the solution f(x) = h(g^-1(x)), which you can make arbitrarily fancy based on your choice of g and h. The hard part is creating an equation where finding the solution and proving its uniqueness is at a suitable difficulty for a maths competition. I’m sure it’s possible, but also quite difficult.

[@gloid4051]

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

[@sarthakpoddar6068]

This function seems to be of the form f(x) = k tan^-1(x), where k€R

[@megauser8512]

@@sarthakpoddar6068 Why k in Z and not k in R, because this functional equation is given for all real numbers >= 0?

[@sarthakpoddar6068]

@@megauser8512 ahh yes sry it should be k in R

[@aadfg0]

Define g : (-pi/2, pi/2) -> R by g(x) = f(tan x), then replacing x with tan(x) and y with tan(y) in the FE we get g(x)+g(y) = g(x+y) for |x+y|

[@hba12]

good thanks

[@nicholasparris7989]

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.

[@alvinuli5174]

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

[@tenebrae711]

It is written two lines above that c≠0 😂

[@CglravgHRjsksgS]

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

[@icew0lf98]

i somehow missed the absolute value solution

[@ivan5595]

Based term

[@jane-harrietglessner808]

an absolutely surprising what

[@GreenMeansGOF]

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

[@candamir26]

Yeah, so f(x)=|x| is not a valid solution

[@jurem2978]

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.

[@random19911004]

absolute value

[@idjles]

-e^i*pi!!

[@thomashoffmann8857]

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 🤯

[@btd6vids]

You can't assume that an inverse exists like that. For example plugging in x=-2 tells you f(4) = 4 which doesnt tell you anything about f(-4)

[@thomashoffmann8857]

@@btd6vids you are right. With complex number ix we would at least get one solution 🙂

[@rozpiotr]

5 ads in 18 min........

[@laprankster3264]

Speaking of final solutions /jk.

[@zygoloid]

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

[@DanChereches-e8q]

I was just looking at the equation and the first thing i thought was to check if f(x) = x ; 1/x or x+(-)1/x are solutions. by verification f(xy-xy)=f(xy)-xy - 0=0 if f(x)=x f(x/y-y/x)=1/xy-xy so xy/(x^2-y^2)=1/xy-xy does not verify so 1/x is not solution f(xy+x/y-xy-y/x)=xy-1/xy-xy x/y-y/x+xy/(x^2-y^2)=-1/xy also does not verify Time to see how close my hunch is to some rigorously found solutions. Let x=0 then f(0f(y)-yf(0))=f(0)-0 f(-yf(0))=f(0) since right side is independent of y then f(0) must be 0 to hold this identity Let x=y then f(f(x*f(x)-x*f(x))=f(x^2)-x^2 f(x^2)=x^2 so for any positive value f(x)=x Let y=-x then f(x*f(-x)+x*f(x))=f(-x^2)+x^2 Let x=-y then f(-y*f(y)-y*f(-y))=f(-y^2)+y^2 we can change variable from y to x and factor out a -1 in the function f((-1)*(x*f(-x)+x*f(x)))=f(-x^2)+x^2 from previous equation => f((-1)(x*f(-x)+x*f(x))=f(x*f(x)+x*(f(x)) so either f(x) is even f(-t)=f(t) or x*f(x)+x*f(-x)=0 for any x which implies f(-x)=-f(x) so function is odd Case I function is even coupled with previous findings then f(x)=abs(x) this would imply the equality |x|y|-y|x||=|xy|-xy | xy*sgn(y)-xy*sgn(x)| which is identical 0 if and and y have the same sign or 2|xy| if xy have opposite signs the right hand side will be 0 if x,y have same sign and 2 |xy| if signs are opposite so absolute value function works case II function is odd so f(x)=x for any x will result in f(xf(y)-yf(x))=xy-yx=f(xy)-xy which also holds equality Nice problem ... wonder if i missed any solutions gonna check video soon Thank you professor