A nice functional equation from Romania 

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...

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.

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

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!

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.

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

can't tell if the functional equation is intentionally different to the one presented in the video (LHS in thumbnail has f(b) term, but LHS in video has f(b^2) instead) so that the guess f(x) := (+ or -)x seems wrong when just looking at the thumbnail alone. probably isn't but thought i'd point it out anyway.

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

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.

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

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.

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

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.

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.

The second part of the video simplifies a lot: after we have f(f(x))=x follows: taking the inverse function on both sides leaves us with f(x)=f^-1(x). There are only two functions that fulfill this equation: f(x)=x and f(x)=-1

You could do an iff thing. It looks like if you substitute f(b) = b = 2x you get f(a^2+2ax+4x^2) = a^2+2ax+4x^2 f((a+2x)^2) = (a+2x)^2 f(y) = y = +-(a+2x) = +-|a+2x| Then since a and x = b/2 are arbitrary label |a+2x| = t to get f(t) = +-t

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.

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.

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 cf(b)=f((b+c)²/4)-f((b-c)²/4). 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.

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

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.

Thanks for a very nice solution. please can you solve "British Mathematical Olympiad Round 2" 2012 problem 2. Thanks in advance!

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

12:50 pausitive! Ah, yea 🧐 Very pausitive!

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

18:33

these are very fun to watch, but hard to solve, and hard to compose . My initial guess was f(x) = x because it has to be something simple and there was f(x) = 1/x in the last video already and in both cases f(f(x)) = x.

If the function is odd, why would we need to check for some values being x and others -x? Is this just to rule out wacky discontinuous combinations of y=+/-x?

Don't tell me the answer but I got to f(f(x))=x^2 and f(x^2)=xf(x) What do I do from here? I haven't watched the video yet Taylor does gets me ax, but that obviously can't be from the starting equation...

asnwer=1 (a+b)-(a+c) isit

💯

Damn... This one's hard, I had no idea of doing it 🍺🍺🍻.

Also cool problem.

-Typos-

14:55 f(b) should be f(y²) and not f(y)..

equation in the thumbnail is wrong, on the LHS there should be f(b^2) not f(b)

f(x)=x

i'm lost...

In my head, I had set both a and b equal to 0 and got that f(0) = 0 fairly quickly.

11:44 What? How does this imply? (f(t))^2 = f(t^2) implies |f(t)| = |t| ???

Everyone of us can make some mistakes in this kind of equations just take a pen and try to solve it then share the answer that would be helpful for everyone.

You're going to be losing so many marks for confusing the examiner by mixing up the notation. Are there really only six letters in the Romanian alphabet?

The final Analysis is not required as f(t)=+-t is the only solution in the real numbers as f is an odd function.

Yeet

Seems like a lot of mistakes in this video.

This is not the first time when you have intentionally "a mistake" in thumbnail. This already shows the lack of seriousness, a much disrespect and you force the followers to see your "solution" to a practically non-existent problem, after they have struggled with another. I'm sorry, but there is anough quality content on RU-vid without such miserable tricks. Unsubscribe.

Thanks for a very nice solution. please can you solve "British Mathematical Olympiad Round 2" problem 2, bmos.ukmt.org.uk/home/bmo.shtml#bmo2, thanks on advance!