A nice functional equation from Romania

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[theantonlulz]

So I spent around half an hour on this before giving up only to find out there is an exponent missing in the thumbnail image of the equation...

[hakerfamily]

In the equation starting at 11:30, Michael meant to write t^2 on the right. Nice video!

[kevinmartin]

Around 17:00 things would be simpler if he cancelled the x^2 properly in both cases. In the +/+ case you ultimately get y=0 which contradicts the assumption that b>0. In the +/- case you get y(x-y)=0 which means either y=0 or y=x so b=0 or b=a both contradicting the assumptions.

[lexinwonderland]

interesting and very well explained! though i wish we got more functional equations that weren't just the identity or simple involutions (1/x, -x, etc), it would be great to see some other functions pop up!

[crazyhitman]

Thank you! These are my favorite type of problems, when you can’t see how to solve it straight away, but after some analysis and trial and error, you eventually hit the nail on the head.

[Хорошийшахматист]

Please make a functional equation playlist. Ur questions of func eqn are amazing. I need to watch them all.

[hyperboloidofonesheet]

16:09 The x² cancels on both sides as well, leaving 2y²=0. This simplifies to y=0, which is a contradiction.

[andreben]

For the case work you could have deduced that from f(y²)= -y² then f(y)= -y by simply using the fact that
f(a²)=af(a) for all a real.
Indeed, we then have by equating a=y
-y²=f(y²)=yf(y) thus, since y>0, so non-zero, f(y)= -y
You thus have only 2 cases to work on.

[abrahammekonnen]

16:24 I think it's supposed to be 2y^2 =0 not 2(x^2 -y^2)= 0.

[abrahammekonnen]

Same thing 17:25 the x^2's cancel out.

[riccardofroz]

There is a Typo in the title image; f(b) should be f(b^2). I was getting stuck at trying to solve it, so I watched the video and spotted the issue.

[johndougherty]

For the last case, how is it that the x^2 term on both sides of the equation doesn’t cancel out?

I’m seeing it become
2xy - 2y^2 = 0

Then either y = 0 or x = y, both of which are contradictions.

[xxsrez]

Simpliest way to finish it is to take a=-f(b)=-f(b^2)/b, so we can calculate b as b=-f(a)=-f(a^2)/a. With all simplifications that gives us for initial formula f(a^2)=-a^2+f(a)^2+af(a). But f(a^2)=af(a). So f(a)^2=a^2 for any a.

[fartoxedm]

I simply took a = -b which lead to f(f(b^2)) = -b * f(b) + b^2 + f(b^2). Knowing that f(f(b^2)) = b^2 + f(0) we deduce that f(b^2) = b f(b) + f(0) = bf(b) and also we know that bf(b)=-bf(b) so the function is odd. Hence f(b) = sqrt(b)f(sqrt(b)) for positive ones. via limit we get f(b) = b for positive and due to oddness we get the same with negatives.

[domiswan]

f(a²+ab+f(b²))=af(b)+b²+f(a²)
Let us use symetries of the parabola. For real c, setting a=(-b+c)/2 and a'=(-b-c)/2 we have a²+ab=a'²+a'b, so f(a²+ab+f(b²))=f(a'²+a'b+f(b²)), so af(b)+b²+f(a²)=a'f(b)+b²+f(a'²), so af(b)+f(a²)=a'f(b)+f(a'²).
Replacing a and a' by their values we get (-b+c)/2 f(b)+ f(((-b+c)/2)²)=(-b-c)/2 f(b)+f(((-b-c)/2)²), which becomes after simplification
Since the right term remains unchanged by swapping b and c, so does the left term, so cf(b)=bf(c).
Taking c=1, we get that f(x)=kx for all x, where k=f(1). Substituing this expression of f in the initial equation, we get by identification k²=1, so k=1 or -1, and so f=Id or -Id.

[AntonioLasoGonzalez]

In the last step, I proceeded a bit differently. I distinguished 2 cases. Case 1 if there exists a b such that f(b)=b, in which case, for all a you get a contradiction if you assume f(a)=-a, and so f(a)=a for all a. Case 2 would be that there doesn't exist a b such that f(b)=b, that is, f(b)=-b for all b. Finally, you can easily check that both only possible solutions are in fact solutions to the equation.

[laprankster]

I found f(a)=a and f(a)=-a to be solutions very quickly. Couldn’t find any other solutions though.

[MrRyanroberson]

since it is R -> R this time, meaning it includes 0...
1. equation(a, 0) : f(a^2+f(0)) = a f(0) + f(a^2), very odd.
2. equation(0, b) : f(f(b^2)) = b^2 + f(0), which is ALMOST self-inverting over the positive reals.
3. equation(0, 0) : f(f(0)) = f(0), which means f(0) is a fixed point.
4. equation(1, -1) : f(f(1)) = f(-1) + 1 + f(1).
5. 4 minus 2 with b=1 : f(0) = f(-1) + f(1). I suppose it's obvious here that the identity function is a solution to this.
5b. 2 with b = 1 gives f(f(1)) = f(0) + 1
6. equation(a, -a) : f(f(a^2)) = a f(-a) + a^2 + f(a^2).
7. 6 minus 2 with b=a : f(0) = a f(-a) + f(a^2).
8. f(f(a))-a = f(0) for all a >= 0
9. 1 with a=1: f(1+f(0)) = f(0) + f(1) = f(f(f(1)))
10. equation(-1, 1) : f(f(1)) = 1.
10b. from 4 : f(-1) + f(1) = 0
10c. from 5 : f(0) = 0
10d. equation(a, 0) is redundant
10e. equation(0, b) : f(f(b^2)) = b^2, which means for all b >= 0, f is self-inverting
10f. from 7 : f(a^2) = -a f(-a). apply f: a^2 = f(-a f(-a))
11. equation(-1, -1) : f(2+f(1)) = -f(-1) + 1 + f(1) = 1 + 2f(1)
just from looking at this, f(x) = -x also works : -a^2 - ab + b^2 = -ab + b^2 - a^2.
i can't seem to rule out much, or find any explicit result either.

[MrRyanroberson]

you don't really need to use injectivity or surjectivity as arguments (8:00 to 11:30), you could just set t = f(x), and at 11:27 just apply f to both sides. knowing f is bijective is sufficient to declare that you can apply f to both sides like that, and the equation f(f(x)) = x defines t as satisfying f(t)=x. bijectivity does contain injectivity and surjectivity, but it's simpler to just stick with the one label

[cmilkau]

11:25 actually, f²(t) = t² follows from injectivity.