an IMO functional equation.

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[@quite_unknown_1]

Note that you are not done there: there could be a horribly convoluted and nasty function which maps x to x for some real x and x to 1/x for others. You still have to prove that you cant have ether combination of both to satisfy the original equation.

[@oinkityoink]

i did that work in my own comment but i figured id share it here too??? suppose pointwise trap exists, let there be distinct positive reals a,b not equal to 1, st f(a) = a, f(b) = 1/b. substituting P(a,b,ab,1), (f(a) + f(b))/(f(ab)+1) = (a+b)/(ab+1) (a + 1/b)(ab+1)/(a+b) = f(ab) + 1 note that ab is distinct from a and b if f(ab) = ab, (a+1/b)/(a+b)=1 b = 1, contradiction if f(ab) = 1/ab (a+1/b)/(a+b)=1/ab a(ab+1)=a+b a(ab) = b a = 1, contradiction therefore pointwise trap does not exist

[@nirajmehta6424]

@@oinkityoink nice proof, well done

[@ShaunakDesaiPiano]

I suppose they should have added a condition for f to be smooth, so that we have the nice and clean f(x) = x and f(x) = 1/x, no combination thereof.

[@ethanbartiromo2888]

@@oinkityoinkI believe you might have made a mistake going from ((a + 1)/b)/(a + b) = 1/(ab) to a(ab + 1) = a + b, it should be a(a + 1) = a + b which gives a^2 = b, if I’m not mistaken

[@MyOneFiftiethOfADollar]

“horribly convoluted and nasty” oh my. Please tell us more about your childhood.

[@oinkityoink]

dont forget to check for pointwise trap

[@oinkityoink]

suppose pointwise trap exists, let there be distinct positive reals a,b not equal to 1, st f(a) = a, f(b) = 1/b. substituting P(a,b,ab,1), (f(a) + f(b))/(f(ab)+1) = (a+b)/(ab+1) (a + 1/b)(ab+1)/(a+b) = f(ab) + 1 note that ab is distinct from a and b if f(ab) = ab, (a+1/b)/(a+b)=1 b = 1, contradiction if f(ab) = 1/ab (a+1/b)/(a+b)=1/ab a(ab+1)=a+b a(ab) = b a = 1, contradiction therefore pointwise trap does not exist

[@VaradMahashabde]

​@@oinkityoinknoice

[@mstarsup]

i solved it on my own (this one is not really difficult), so i just went through the rest of the video rapidly, but in your solution you still need to show that it's either f(x)=x for all x, or f(x)=1/x for all x, and you can't have any "hybrid" function like f(x)=x for some x and f(x)=1/x for some other x.

[@lorenzord7110]

It is not over, you have proved that for every x, then f(x) = x or f(x) = 1/x, not that these are the solutions. Maybe there could be a solution that such that f(x) = x for some x and f(x) = 1/x for the rest of x and satisfies the hypothesis. There is not btw hahah

[@jamiewinter2134]

That's true. However, if you suppose that f(a)=a and f(b)=1/b for some a,b>0, then depending on whether or not f(ab)=ab or f(ab)=1/ab you can show using the identity at 7:50 that either a=1 or b=1, hence when a and b are not equal to 1, f(a)=a iff f(b)=b, i.e. once you've settled on a choice between the two types of functions x and 1/x in a point that is not 1, f(x) is forced to stay that function on the entire domain.

[@divyakumar2207]

amazing question explained by one the most amazing prof, what more could be asked thanks for the video

[@Maths_3.1415]

More advanced problems can be found in CHINA TST ( Chinese Team Selection Test) for IMO China TST includes extremely difficult problems

[@HagenvonEitzen]

13:54 Actually, so far we only have that f(x) = x or = 1/x *pointwise*, i.e., we did not yet exclude a (discontinuous) solution where f(2)=2 and f(3)=1/3. To get rid of this problem, suppose thre are a,b (both not equal to 1) such that f(a)=a and f(b)=b (and f(ab) is either ab or 1/(ab)) and then use (f(a)+f(b))/(f(ab)+1) = (a+b)/(ab+1)

[@Monolith-yb6yl]

So technically we have to check that x and 1/x are solutions by substituting them into equation. And only after that will be a good place to stop

[@FF-ms6wq]

No it’s still not sufficient then. He hasn’t shown that there’s no ‘pointwise trap’.

[@HagenvonEitzen]

11:08 Actually, f(x)=x and f(x)=1/x are obvious solutions (and hence are *the* solutions) - e.g., because the equation can be rewrittem as f(x) + 1/f(x) = (f(x)^2+1)/f(x) = (x^2+1)/x = x + 1/x, which is symmetric in x 1/x

[@singious]

Great! I saw the piezo and LN2.

[@goodplacetostop2973]

13:55

[@paslot]

Maybe this is a slightly stupid question, but i genuinely don't understand why you are just able to substitute t in for w, x, y and z as they are all independent of each other if I am not mistaken. The same goes for why you substitute c with ab. Aren't you now looking at a particular case instead of all cases like the question states, because you are making the variables dependent of each other? I know that wx=yz/ab=cd, but that doesn't immediately mean that w=x=y=z or c = ab. Could someone please explain this to me? Again, maybe this is a slightly stupid question as I see no one else asking this in the comment section. Maybe the problem is that I don't understand, because haven't started my mathematics study at uni yet and thus don't have all the knowledge needed to understand, but maybe also not. Anyway, I hope someone can explain.

[@paunb8550]

I will try my best, but explaining things is not my best. Because it is a functional equation, the solutions we seek must make true the initial statements for all the numbers in the domain and in the range (In this video is r+). Because of that, you can pick any number to plug it in and see if you find something. For exampme, when he lets w=x=y=z=t, this is because if t is a positive real number, then I can plug in t for w. The same can be said for x, y, z. Hope that helped. If not, tell me and I will try again.

[@paslot]

I think I get it now yeah. Thank you very much.

[@jilanwicaksono]

2:06 that's so cursed. Didn't know someone can write xwyz in this order instead of wxyz or xyzw

[@synaestheziac]

Lol, I’m guessing his thought process was something like: I’ll write the “most common” variables t and x first (in alphabetical order), and then write the “less common” variables w, y, and z next (also in alphabetical order). But who knows

[@BridgeBum]

@@synaestheziac Probably even simpler still, x and y are typically paired and it isn't that unusual to see w and z paired around them. He probably started to write x and y but realized x and w are paired in this problem mid-sentence.

[@robotech2566]

well a computer list many equation, you just found two to the question

[@jaxoncr]

for these questions do you must have correct work or can you just write the solutions down by inspection?

[@rockinroggenrola7277]

Even if you write down solutions by inspection, you have to prove that those are the only solutions.

[@jaxoncr]

@@rockinroggenrola7277 thanks!

[@MyOneFiftiethOfADollar]

Want to see a proof of this where the variable choices are not so “alphabetically pleasant” For example, change wx=yz to rb=kf ….

[@GreenMeansGOF]

1/x is the reason that we need wx=yz.

[@FineFlu]

Is there anything special to be said about the fact that multiplying these two functions equals 1? They are sort of like multiplicative inverses, yes? If so, is that a special property in some form of algebra?

[@jantarantowicz1306]

I also think that's a functional equation!

[@jrb0580]

I don’t understand how we can conclude that (f(t))^2 = f(t^2)) by setting t=x=y=z=w. The algebraic steps make sense, but the underlying logic escapes me.

[@ConManAU]

Because the functional equation has to be true for any possible valid choices of the variables, it has to be true for specific choices of their values. So we choose that the values are all equal, and that tells us that any function that satisfies the original equation must also satisfy this new condition.

[@TheSeventhCellist]

@@ConManAU I might be dumb but I'm still no sure to be convinced. Since nothing is said about de type of functions we are working with except their domain, why can't it be functions defined by parts that happen to satisfy f(t)^2=f(t^2) only when w=x=y=z and do something else for other values of w, x, y and z that satisfy wx=yz ? I have the feeling that setting w=x=y=z=t is a wishful bet.

[@ConManAU]

@@TheSeventhCellist If you think about the sets “Functions that satisfy the original functional equation” and “sets that satisfy f(t)^2=f(t^2)”, then the former must be a subset of the latter - because choosing specific values of the variables led us to the new equation. Compare the sets “functions satisfying f(x)=x^2” and “functions satisfying f(2)=4”.

[@TheSeventhCellist]

@@ConManAU So I was being dumb but thank anyway for having taken the time to explain this obvious fact that was escaping my mind. 😅

[@pyaniy_abba577]

Oh, I solved it without the condition of wx=yz, so I got the |f(x)|=|x| answer.

[@FF-ms6wq]

You haven’t proven it completely. It’s not complete. You’ve fallen into the ‘pointwise trap’.

[@JoopWilkens]

This is again not a complete solution, Michael. As a math professor myself, I'd gave a failing grade for this 'solution'. Why? There is not even a mention of the "pointwise trap": Namely, as others have pointed out in the comments before me, there can be terribly convoluted functions that are such that some x's are mapped to x while other x's are mapped to 1/x. You did not even mention that one should check that such "mixed form solutions" cannot happen here---suggesting that, perhaps, you do not even realize that yourself. In short, the reason I'd fail your solution is the lack of any mention of the so-called "pointwise-trap". This is a very serious mathematical (or logical) error, and cannot deserve but a failing grade. This would be the case if I were to 'grade' your solution, you being the metaphorical 'student'. However, you are here not as a student, but rather the opposite: You are the educator here. My judgment must be much more severe in this case, for you do not only not provide a mathematically correct solution, but people watching this who don't know better (it's not a shame, of course, not to "know about" the 'pointwise-trap'), might well be mislead by this "solution". Don't get me wrong, I like how you solve problems and generally also your explanations. I just think that in certain cases where absolute mathematical rigor is required, that rigor is lacking.

[@alielhajj7769]

It can’t be 1/x since it is only defined for natural numbers 😅

[@chaosredefined3834]

He only ever said positive real numbers. He never said natural numbers. pi is a positive real number, but definitely not a natural number.

[@BoxyHD65]

I really read "how do we we the wx = yz condition?" 🥲

[@mcbeaulieu]

That's what was there indeed 😅

[@Macieks300]

how do use we the