A nice functional equation from Romania

[@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...

[@JM-us3fr]

Yeah definitely don’t start the problem until you’ve seen how he writes it in the video. He has a tendency of writing it down wrong in the thumbnail

[@Aditya_196]

🥲 I understand your pain bro

[@hakerfamily]

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

[@MatthewBouyack]

To clarify for others who were as confused by that step as I was, it should have been (f(t))^2 = t^2 => f(t) = +/- t

[@mathieulemoine1294]

​@@MatthewBouyackThanks, I was wondering as well

[@kevinmartin7760]

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.

[@Хорошийшахматист]

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

[@hyperboloidofonesheet1036]

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

[@lexinwonderland5741]

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!

[@abrahammekonnen]

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

[@crazy4hitman755]

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.

[@abrahammekonnen]

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

[@andreben6224]

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.

[@MrRyanroberson1]

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

[@MrRyanroberson1]

i suppose for educational purposes, to reinforce the terms to students, you would find that repeating these terms improves retention though

[@xxsuper99xx]

Well he literally says bijectivity implies infectivity

[@cmilkau]

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

[@johndougherty7216]

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.

[@inigoverafajardo7245]

Also, for the ++ case, x^2 cancels out, and what's left is 2y^2=0, so y=0.

[@ojasdeshpande7296]

It's a typo but it will be one of the +- or -+cases only so it's fine :)

[@Alex_Deam]

Yes, also the reason given in the (incorrect) case 2 made no sense anyway. x^2 +xy -y^2 =0 clearly has an infinite family of positive real solutions just by using the quadratic formula, not none!

[@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.

[@tonyha8888]

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

[@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.

[@fartoxedm5638]

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.

[@MrRyanroberson1]

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.

[@goodplacetostop2973]

18:33

[@ericthegreat7805]

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

[@atikshagarwal5147]

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

[@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.

[@thomaslangbein297]

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

[@konraddapper7764]

There are more involution, eg. f0(x)= c-x or more generally, for a given invertible function, g f(x)= g^-1( f0(g(x))) is also a solution But I am not sure if there are even more different solution

[@thomaslangbein297]

Yes, you are right. I was too fast. Every function symmetrical to the angle bisector of the first quadrant fulfills this property. E.g. if you restrict the domain to (0;1) even the quarter unity circle works.

[@Bazzzzz93]

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

[@den41k2204]

also curious about it. I see only that we can replace f(t^2) with t * f(t) to get: (f(t))^2 = t * f(t) => f(t) * (f(t) - t) = 0 => f(t) = t or f(t) = 0

[@siyuanhuo7301]

It's f((f(t))^2)=f(t^2) which would imply |f(t)|=|t|

[@simonreiff3889]

I think you're missing an f. It's f[f(t)]^2)=f(t^2). By injectivity, f(x)=f(y) implies x=y. Hence, [f(t)]^2=t^2. Taking the square root of both sides, we have that f(t)=+-t.

[@reeeeeplease1178]

@@simonreiff3889 ye, michael penn made a typo there

[@Epyxoid]

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

[@jongyon7192p]

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...

[@laprankster3264]

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

[@m2a2x2000]

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.

[@축복-l1l]

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

[@Domiswag]

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.

[@nerdgonewild]

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?

[@FrozenArtStudio]

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

[@ojasdeshpande7296]

-Typos-

[@zrksyd]

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

[@BerfOfficial]

But isn’t that f(f(0))=f(0)?

[@dutchie265]

Indeed, only means f(0) is a fixed point. Doesn't say anything about its value.

[@zrksyd]

@@BerfOfficial oh shoot I guess you're right since one to one wasn't proven yet. My bad.

[@dopo8333]

f(x)=x

[@williamhogrider4136]

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

[@theartisticactuary]

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?

[@abrahammekonnen]

Also cool problem.

[@charleyhoward4594]

i'm lost...

[@alainleclerc233]

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

[@Notthatkindofdr]

The function defined by f(x)=x for rational x and f(x)=-x for irrational x is also odd, so the final analysis (though it could have been much simpler) is needed to eliminate possibilities like this. The point is that you have to go back to the original equation to eliminate these discontinuous examples.

[@alainleclerc233]

@@Notthatkindofdr Hi Wayne! Right you are! I incorrectly assumed f was continuous. Having said that, the final Analysis is much simpler as the asumptions f(a)=a et f(b)=-b imply f(a2)=a2 and f(b2)=-b2. Only two cases must be looked at, not 4. Nice video by Michael!

[@taongandolo2332]

💯

[@ayoubelouafy6174]

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.

[@juliang8676]

Yeet

[@mathcanbeeasy]

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 KZbin without such miserable tricks. Unsubscribe.

[@patricius6378]

You know, nothing prevents you from checking the start of the video and doing *that* version of the problem :D

[@tonyha8888]

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!