IMO shortlist 2018 A1
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- Опубліковано 28 вер 2024
- We present a solution to a problem that was shortlisted for the 2018 International Mathematics Olympiad. This problem involves a functional equation for a function defined over positive rational numbers.
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My issue with functional equations is that usually the answer is so... disappointing :p
Exactly... you try and you try and then nothing.
functional equations arise naturally in dynamical systems and they're not at all boring. invented ones like these probably are, otherwise they become impossible to do anything with.
but naturally occurring ones are rich, i can assure you.
I agree. I like solving them and am good at it but, in the end, they just leave me disappointed. All this hardwork for what exactly? I’d rather be doing integrals or geometry to be honest
I find it astonishing that people can come up with these things in the exam with no warning of what was coming. These people are human??
it is actually pretty easy after a bit of practice
Relativity Rocks ! Have u been to the imo?
These people are humans and they practice a lot. It’s not only IQ that would help here, practice matters a lot in Functional Equations and Number Theory IMHO
btw I think I heard that the shortlist problems are sorted roughly in order of difficulty, with 1 being easiest, so this is the easiest algebra question on the shortlist for 2018 lol
It basically comes down to cleverness and being used to it. I've seen and done many multivariable functional equations and, most of the times, it comes down to making clever substitutions to simplify it to one single variable.
I started by considering the bigger case of x being any real # >0 and made x=1:
f(x²f²(y))=f²(x)y(y)
x=1
f(f²(y))=f²(1)f(y)=f²(1)sqrt(f²(y))
Since f(1) is just a constant and we know that the domain of the function is equal to the image, we can make the following substitutions:
f(1)=c => f²(1)=c²
f²(y)=t
=> f(t)=c²*sqrt(t)
then, if we set t=1
f(1)=c²*sqrt(1)=c²
c=c²
c= 1 or 0
if c=0, we get the constant function f(t)=0.
if c=1, we get the function f(t)=sqrt(t).
Now, there's a third case. Remember when we set x=1 in the beginning? Let's try x=0 this time:
f(0²f²(y))=f²(0)f(y)
f(0)=f²(0)f(y)
set y=1
f(0)=f²(0)f(0)=f³(0)
which, the only solutions are 1 and 0
f(0)=0 leads us to the same solution f(t)=0
f(0)=1 leads us to f(t)=1, the one in the video.
Thanks for this nice problem.
I did it a bit different though. First demonstrated that f(1)=1:
- a=f(1)
- y=1 => f((xa)^2)=a*f(x)^2 then replacing x by 1/a => f(1)=a=a*f(1/a)^2 => f(1/a)=1
- x=1/a => f(x^2)=f(x)^2 which leads to f(1)=1 with x=1
- x=1 => f(f(y)^2)=f(y) => f(f(y))^2=f(y) using above identity
this means f(y) is a square of some rational = p^2/q^2
- x=q/p => after few manipulation I get
f(q/p)^2=q/p
- so q and p are also squares with
x = q'^2/p'^2 = f(q'^2/p'^2)^2 = f(q'/p')^4 which means q'/p' is also a square of some rational.
- by infinite ascent, we conclude q'=p'=1 => x=1 => f(y) =1 for all y
I misread the problem, and looked at the functional equation f(x^2 f(y)^2 ) = f(x^2) f(y) instead. Slightly surprisingly, that also turns out to be a tractable and interesting problem.
A quicker way but with similar thinking: Show first that f(1)=1 by computing it in two different ways. Then, by the given equation, this implies that f(x^2) = f(x)^2 and f(f(y)^2) = f(y), for every x,y in Q_+ . Combining the last two observations we obtain that: (1) every y in the image of f is a perfect square of another rational and (2) for every y in the image of f we have f(y^2) = y. By the first conclusion if y is in the image of f and is not 1, then y=q^2 for some rational q. Also f(q)^4 = f(q^4) = f(y^2) = y = q^2 that is f(y) = f(q)^2 = q. But then q belongs in the image of f and therefore it is a perfect square of another rational, say q_1. A finite repetition of this process, writing q= q_1^2 and arguing that q_1 is in the image of f and so forth, would result in a rational number q_n that is not a square of another rational but is in the image of f. This is a contradiction and therefore the image of f is the singleton 1.
Sorry for late question but how did you prove that f(1)=1? I tried a bunch of different ways but couldn't get it. Feels like I am missing something obvious.
f(x) = sqrt(x) works, though I don't know if this is an exhaustive list.
It doesn't give a positive rational for x=2. Does work for the question at the end though.
This is an overcomplicated way of showing that irrational numbers exist.
Very true!
I found f(x) = +- sqrt(x) using approximation to finite polynomial of order M. And M could be 0 (f(x) = -1, 0, 1) or M could be 1/2 for f(x) = +-sqrt(x)
so +/- sqrt is the additional solution over the real numbers.
I just learnt about functional equations. love these video :)
take x=1 in the equation, you will get f(0)=f(0)^2f(y), or f(y)=C; take y=0, you get C=C^3, or C=1 for +R.
I didn't understand the 3rd line in particular the rhs how is f(f(a)^2f(b)=f(f(b)^2)f(a)?
F(x) = sqrt(x) was my solution. It is for real positive numbers. Is their another solution Michael Penn??
how did you come to this conclusion?
but that is true iff x = y. as a line. well let f(x) = sqrt x then f(y) =sqrt y is a different thing than letting x = y then f(x) = sqrt x
Why you can always find a such that f(a) =x?
That's not what he's doing, he's setting x to f(a) for any a. x is just whatever a happens to be.
Hr dosent he just works only with the f(a)s
I had the same thought ("aren't we assuming `f` is surjective here?") but I realized it's not actually _finding_ `a` such that `f(a) = x`, we are taking an arbitrary `a` and _plugging in_ `f(a)` for `x` in the original equation.
I found the official IMO solutions and they use the phrasing "Take any `a \in Q_>0`. By substituting `x = f(a)`, ..."
@Timur Pryadilin
You're right, being a function from pos rational to pos rational only means that for each a, there is an x=f(a). If f is onto, then we can find an a for any x such that f(a) = x, but we would have to assume that, so perhaps the other functions that are not derived in this video but are solutions are of this nature.
Only if f is surjective over Q+, which is doesn't show. His proof is then incomplete.
Great video
I have on slight suggestion
I would prefer to see the problem first without visible hints.
Keep a up the good work
I have a slight suggestion. Pause the video if you do not want the hints. :)
f(x)=sqrtx is a solution (guess) so clearly there has been some mistake that I couldn't really see
In reals+: f(x)=sqrt(x)
I don't understand why we couldn't just plug in x=0 from the start? We would get that f(0)=f(y)f(0)^2 so f(y) = 1/f(0)=c. 1/f(0) will still be rational so everything is ok. Plugging f(x)=c back again we have that it must be that c=c^3 so c=+/-1 because f(0) can't be 0 and so also f(x) = +/-1.
That's what I thought, although it does say x,y > 0...
Ok I'm really struggling to understand this concept. If you could choose from an infinite number of integers for the numerator of a fraction and an infinite number possible numbers for the denominator of a fraction how can that not represent any real (rational/irrational) numbers? I know you can't write down infinite numbers but you also can't write down irrational numbers. Could someone actually explain this?
See the proof of why sqrt(2) is not rational for a start.
Between two rationals, there's always a real number (and vice-versa). The rationals are dense in the set of real numbers (and vice-versa). That's probably why it's difficult to get past your understanding.
Also search for the proof of the uncountability of the real numbers set (Cantor's diagonal), it will show you that real numbers are very different than integers and rational (btw, did you know that the integers are as numerous as the rationals ?)
@@speedsterh thanks for your reply. Yes I did know that 😁. We did Cantor's diagonals proof at University in the discrete mathematics course. One way I thought about it was trying to print all of the real numbers in a computer terminal (I mean in order obviously it's impossible) with the rationals you can find an algorithm but with the reals you could always divide by 10 and make that the starting point so there's no obvious starting point.
f(x) = 0
Good
aren't you making a hypothesis when setting x = f(a)? the full way of writing that would be "for every x such that there exists a s.t. x = f(a) etc"
i don't thing that f being onto is part of the hypotheses as you've written them down. moreover, it contradicts your conclusion that f is a constant function
No. He's not choosing an x that is required to be f(a) for some a. He's simply saying that for all a in the domain, there (trivially) exists an f(a).
Can we not put f(y) as z?
Please stop including hints of the first frame of the problem. I want to see you introduce the problem, and have the opportunity to solve it without hints. It kills the fun of most of your videos.
Missing the: That's a good place to stop
that explains why I didn't stop when the video finished... 😁
Because it's not
Funny story
This problem was in the last years german imo qualification competition but with a typo
They got one square wrong, so the question they ask was
f(x^2 f(y)^2) = f(x^2) f(y)
Which was a lot harder than the original question (you need group theory, in particular subgroups of Q+ and cosets, to even write down all possible solutions, which isn't obvious at all when given a functional equation)
Can you please outline the solution using group theory?
Hast du mitgemacht damals ? Ich schon
@@kadirdumm8572 ja, aber ich war nicht wirklich gut 😂😅
@@________6295 the prove is really complicated. (Maybe you can find it online somewhere) but the main idea was, to show that the range of f is a subgroup of Q_>0 (the multiplicative group)
Tou can probably guess two solutions: f(x)=1 and for real numbers we would have f(x)=c*sqrt(x) for a fixed c, but this doesn't work in Q. We can however take f(x) to be c*sqrt(x) if sqrt(x) is in Q and be arbitrary for all x where sqrt(x) isn't in Q. This is also a solution. But these aren't all solutions.
The idea to get all solutions is, that the strange solution with sqrt works, because the squares in Q are a subgroup of Q and therefore we can "seperate" them from the other numbers in the functional equation.
If you take an arbitrary and fixed subgroup U of Q, then you can construct other similar solutions using representatives of the cosets of U.
I thinkt at 16:20 it should be C^3 - C = 0
He immediately fixed it with the next equation lol
Yes, but in the next expression he writes C(C - 1)(C + 1) = 0, which expands to C³ - C = 0. A simple typo.
I love functional equations. So deceptively simple, so much to learn!
And it's always the plus minus identity, or the constant function. :)
Please teach me I wanna learn these functional equations I suck at them do you have discord account? Maybe you can teach me there I want to learn how to solve these equations please teach me
Me when 2012 IMO A1:
the solution, with weird functions that i didnt think existed:
ok but seriously, have a look at Evan Chen's Monsters handout.
"Ok, now we're done" ≠ "and that's a good place to stop"
Amazing man. You are explanations are so easy and clear :)
lol
@@assini1141 POV: Te cansaste de ser espectador.
Neat reasoning, but I'm puzzled about something. Putting x = f(a) for any x in Q+ seems to require the function to be bijective. And since the answer certainly isn't, I'm not sure how that works.
It works because you can pick any arbitrary a in Q+, and set x = f(a). Since the function is defined for all Q+, this x is defined, as well.
Could you Pls do IMO 2006 problem 5 Thx for your awesome work
my first guess before seeing the video is f(x)=1
Edit: I guessed right
For the function over the real numbers it is f(x)=+-sqrt(constant * x)
as long as the constant is 0 or 1
@@wolffang21burgers Well it could be -1 if x
@@markerena2274 Ahh I didn't see this was over the reals: The challenge at the end of the video was to find for x in the positive reals.
In this case
Over the reals, + - sqrt(abs(x)) works,
The codomain is just the set of the value the function _might_ take. The set of all the values effectively taken is called the _image_ of the domain upon the function.
So far, the solutions f: R --> R are:
f(x) = + - sqrt(abs(x))
f(x) = + - 1
f(x) = 0
where only the constant ones are such that f(Q>0) in Q>0
The solutions for real numbers are:
f: x -> 0
f: x -> 1
f: x -> -1
f: x -> sqrt(abs(x))
For the first three, just assume that f is constant and solve.
For the last one, assume f is not constant, show that it's symmetric, then replace x with 1/f(y) (for a given y where f(y) 0), and continue the reasoning from that point.
The negative of your 4th solution works too, doesn't it ?
There is another strange solution :
f(0)=0
for all x=/=0, f(x)=1
Or
f(0)=0
for all x=/=0, f(x)= -1
@@volodyadykun6490 we're talking about the solutions that are R ->R
And there is a lot of them.
@@20-sideddice13 oh yes, you're right, didn't see it
I got to f(1) = 1 but that's about as close as I could get lol. Naive guess is that would hold for all x in f(x) but quite a different matter to prove it!
Thanks for sharing
To whoever did this on their own ... you are a KING!. 🙌🙌
I did it! Feeling really proud. I haven't watched the video yet but I got that the answer is 1 (because the output of the function has to be square-rootable an arbitrary amount of times and in the rationals greater than zero that can only happen if the output of the function is 1)
@@marcusrees5364 that doesn't prove that f(x) = 1 is the only solution
Hey micheal. I wish you to help me in this question
If m greater than 4 find m+n if 5!m!=n!
English is not my first language, so parts of my answer might not be as comprehensive as i want them to, but i hope you'd understand my proof.
If we divide both sides by m! we get that n!/m! = (m+1)(m+2)...(n-1)(n) = 5! = 120.
Now, we need to consider how can we write 120 as a multiple of consecutive numbers: m+1, m+2,..., n.
If there are three consecutive elements, then it would be (m+1)(m+2)(m+3) = 4*5*6 but that suggests that m+1 = 4, but it contradicts the constraints that m>4, that also suggests that we can't have more consecutive elements than 3. if we have 2 elements, then there is no integer such that (m+1)(m+2) = 120, so that leaves us with one option, if there is only one consecutive element, or in other words, n = m + 1.
Plugging it in we get 5! = (m+1)!/m! = m + 1 = 120. therefore, m = 119, n = 120, their sum is 239.
Answer: m = 119, n = 120, m+n = 239
Itay s thank you for this solution
@@aliakkari5742 glad to help
May you suggest me a book for functional equation. I am begginer in this topic! Thanks a lot, for your lectures!
Try this problem about functional equations, from the 2019 Brazilian Mathematical Olympiad (OBM):
Find all functions f: (0, ∞) ----> (0, ∞) such that
f(xy + f(x)) = f(f(x)f(y)) + x , for all x , y in the interval (0, ∞)
where (0, ∞) is the same as a set A = {x | x > 0 , x is a real number}
Can anyone tell me the solution of this functional equation or its source please :)
The problem:
Let α,β ∈ Q* . Find all functions f: R ->R
Such that:
f([x+y]/α)=[f(x)+f(y)]/β
Problem:
Let α,β ∈ Q* and find all functions f from
R to R such that f([x+y]/α)=[f(x)+f(y)]/β.
Solution:
Note that for x=y=0: f(0)=f([0+0]/α)=[f(0)+f(0)]/β=2f(0)/β,
so f(0) must be 0, but also for x=t and y=-t:
f(0)=f([t+(-t)]/α)=[f(t)+f(-t)]/β=0, so f(-t)=-f(t) for all
t ∈ R, so f is an odd function. Furthermore, for x=αt and y=0:
f(t)=f(α*t/α)=f([α*t+0]/α)=[f(α*t)+f(0)]/β=f(α*t)/β,
so ^^^f(α*t)=β*f(t)^^^ for all t ∈ R, and also for x=y=α*t:
2*f(α*t)/β=[f(α*t)+f(α*t)]/β=f([α*t+α*t]/α)=f(2*α*t/α)=f(2*t),
so 2*f(α*t)=β*f(2*t) for all t ∈ R, but with ^^^, we have:
2*β*f(t)=2*f(α*t)=β*f(2*t), so 2*f(t)=f(2*t) for all t ∈ R,
but t is just a dummy variable, so we can replace t with x:
2*f(x)=f(2*x) for all x ∈ R, but this implies that if we take
f(x)=its infinite series=sum(n>=1){c_(2*n+1)*x^(2*n+1)}, [since f(x) is odd (see above), all the even terms are 0], we have: 2*sum(n>=0){c_(2*n+1)*x^(2*n+1)} = 2*c_1*x+2*c_3*x^3+... = c_1*(2*x)+c_3*(2*x)^3+...=sum(n>=1){c_(2*n+1)*(2*x)^(2*n+1)}, so 2*c_1*x+2*c_3*x^3+... = 2*c_1*x+8*c_3*x^3+..., but since 2=/=8, and in general 2=/=2^(2*n+1) for all n>=1, then 2*c_3*x^3=/=8*c_3*x^3 unless c_3=0, and in general 2*c_(2*n+1)*x^(2*n+1)=/=2^(2*n+1)*c_(2*n+1)*x^(2*n+1) for all n>=1 unless c_(2*n+1)=0, but the 1st term works out, since 2*c_1*x = 2*c_1*x for all c_1 ∈ R. Since c_1 is just a dummy variable, then we can replace it by c, so we have:
f(x)=c*x, where c ∈ R, but we must find all c that work with c * [x+y] / α = f([x+y]/α) = [f(x)+f(y)]/β = [c*x + c*y] / β, so c * (x + y) / α = c * (x + y) / β, so either α = β, or c=0. Therefore, all functions f from R to R that satisfy
f([x+y]/α) = [f(x)+f(y)]/β, are the following: f(x) = c*x, with c ∈ R and α=β, OR with c=0 and α=/=β.
Is there a function f:(0,oo) -> (0,oo) satisfying the equation which doesn't satisfy f(xy)=f(x)f(y) ?
When you set "x=f(a)", doesn't that assume that f is surjective? And, like, given the solution we know it isn't, right?
The formula is true for every x, hence it must be true for x = f(a), with a a positive rational. You just "plot" the formula at this particular value.
Wait a minute he makes a miatake at 2:56..you can't just interchange xand y like that ..they are not the same..so that equality doesnt hold..
@@kossarocroft4024 function not being surjective implies that not for every positive rational x there is positive rational a, such that x=f(a)
@@leif1075 It is now. If you observe the formula he derived above, you'll notice that if you write the equation for the pair (b, a) instead of the pair (a, b) (the order matters), you will get the exact same left hand side, because multiplication is commutative. The b term will look like the a term in the other equation and viceversa, so effectively you have the same lhs. And that's why he can claim the other equality.
In other words, the implied step is that he wrote the equation for (a, b) and then for (b, a) and noticed the lhs are the exact same expression, so you have the equality.
Not necessarily, if you just fix "a" you can consider "x" defined as "f(a)" and since the hipothesis property is hold for all possible values, in particular it holds for "f(a)" where the variable we will work with is "a"
Can you please make a video on problem C1 from IMO Shortlist 2010? Really thankful for your video...
There is a solution satisfying the above conditions for f(x)= root(x), I have the proof but I'm cba typing it out on here.
Ah, I see you are a modern-era reincarnation of Fermat 😛
The proof is trivial: just substitute f(x)=sqrt(x) into the original functional equation and you can see that it is satisfied. It's the derivation of how you might *find* that solution that's more interesting. (Note that there was no claim above that this and the rational solution are the only real solutions, which would indeed take some proving)
4:00 why you set b=1
He's not setting b to 1, he's saying that the equality holds for all positive rational b, so it holds for b = 1.
Separate observation, but if we _don't_ disallow 0 in the domain of f, it is straightforward to show f(x)=1. (or f(x)=-1 if we allow negatives)
Simply put x=0 and leave y free, to get f(0)=f(0)^2*f(y), so f(y)=1/f(0), so f(0)=1/f(0), so f(0)=+-1.
... or f(0) = 0 when you can't do your division. Otherwise, nice shortcut over all rationals.
@@richardfarrer5616 The range of f was defined as Q>0 so f(0) is not 0.
@@kwea123 I know, but I was addressing a comment where 0 was allowed in the domain and, implicitly given the suggested possible answers, 0 was also allowed in the range. All I was doing was adding the third solution.
Since x is rational, f(x)^2 is rational and f(x^2*f(y)^2) must also be rational let A = xf(x^2 * f(y)^2) / (f(x))^2. A must be a positive rational number and it turns out f(y) = f(A^2) now let x = A and we get A = A * f(A^2 * f(A^2)^2) / (f(A)^2).
Divide by A and multiply by the denominator to get f(A)^2 = f(A^2 * f(A^2))^2). Now replace A with x and f(A^2) with f(y). f(x)^2 = f(x^2 * f(y)^2) using the original equation f(y) must = 1 and therefore f(x)^2 = f(x^2) = 1 so f is just 1
Extend task to include full R. Then assume x=0: f(0)=f(0)^2*f(y) => f(0)=0 or f(y)=1/f(0), which gives us f(y)=±1.
Then, x=0 doesn't fit Q+, but f(y)=1 does. Thats all.
For reals. f(x)=x^(1/2) would work.
c^3-c=0, just a small mistake 🙂 Absolutely amazing problem thought,l and the solution!
It is true that sqrt(x) is a solution to the problem on positive real numbers, but you can describe some solutions in terms of a hamel base of R/Q. not sure about all the solutions though. you need to replace f(x) with g(ln(x))
Your "homogenizing step" is a lot like separation of variable methods for wave and heat PDEs on rectangular domains
I didn't understand the 2nd line in particular the rhs how is f(f(a)^2*f(b)=f(f(b)^2)*f(a)?
Since f: Q>0 -> Q>0, so f(x) > 0
x=0, y=0 → f(0)=0,1,-1 → f(0)=1
x=f(0), y=0 → f(1)=0,1 → f(1)=1
x=0, y=y → f(y)=1
So f(x)=1, easy.
I don't think assuming x=f(a) is correct. There may be some rational x that is not a value of f(a) for any a
Over the positive real numbers, f(x) = sqrt(x) is a solution.
I like how the solution used the fact that the multiplicative positive rationals is free abelian on the primes.
Did you make a mistake at 16:20? It seems like you can just deduce immediately that C=1.
Why did you write c=c^3 so c^3 - c=1 ????
Excuse me, i am a chemist
Sqrt x also works
Sorry I don't understand the contradiction.
Cool problem!
f(x)=sqrt(x) for all x>0 is an answer, too.
This one was a bit painful.
Why "a" such that x=f(a) does exist?
we do not know the function is onto , so how can we set x=f(a)
You don't need the function to be onto. All that Michael is doing is saying that IN PARTICULAR, the functional equation is true for positive rational numbers of the form f(a). This is fine to do because the range of f is a subset of its domain.
Just For Fun. Both are given as Q+.
@@tracyh5751 ok. got it
@@tracyh5751 Why wouldn't this then restrict this proof to finding f on the codomain?
@@tracyh5751 No, it's not fine. f(a) = x iff f is surjective (over Q+ here), that's, for every x € Q+, it exists a€ Q+ such that f(a) = x. He did not show that if f is a solution, then f is surjective (we don't know if f maps all Q+). His proof is incomplete
Just in the original equation f(x²f²(y)) = f²(x) f(y) assume x=1, f(y) = √(b) (here is where we use Q+ -> Q+).
We get f(b) = f²(1)√b, or f(b) = c√b where c=f²(1)>0. Replacing b with x, gives: f(x)=c√(x) which indeed maps Q+ to Q+ when c>0.
Evaluate both sides of the given equation for f(x)=c√(x):
LHS: f(x²f²(y)) = f(x²c²y) = c√(x²c²y) = c²x√(y)
RHS: f²(x) f(y) = c²xc√y = c³x√(y)
The only difference is c² at left and c³ at right. Assuming x=y=1 gives c³=c², which divided by c² (c > 0) yields c=1,
where both sides become identical. Thus f(x)=√x satisfies the condition of the problem.
Look too simple for IMO, but... this approach has a glitch: replacing b by arbitrary x assumes that every x>0 has a y, so that f(y) =√(x). This is true only if f(x) maps Q+ to the WHOLE Q+ (i.e is surjective) Therefore such an assumption results in dropping non-surjective solutions, like f(x)≡1. So, there can exist a non-constant solution apart from f(x)=√x, which is certainly non-surjective.
Square root of x
The first step was assuming that x could be written as f(a) for some a. But that is not true for all x. Is that an issue?
I guess it is not since, at the end, we work with all the a in the rational numbers....
Another question, can this be solved in the real numbers?
That is not the first step at all. The first step is that for _any_ value a in the domain, f(a) is a valid value of x. x is only required to be some positive rational number, which f(a) plainly is for any value of a.
f(x) = x^(1/2)
It is quite easy to see that f(x^2) = f(x) : set x = 1/f(1), y =1, to get f(1) = f(1/f(1))^2 f(1) to get f(1/f(1))^2 =1 hence f(1/f(1)) = 1. Then setting x = x, y = 1/f(1) you get f(x^2f(1/a)) = f(x^2) = f(x)^2f(1/f(1)) = f(x)^2. It is then immediately clear that f(1)^2 = f(1) and so f(1) =1. This kills the constant C right from the start.
for real positive numbers the only additional solution is f(x)=sqrt(x)
solution: Lets take arbitrary y and x=1/f(y)^2. Plugging it in we get f(y) = 1 / f(1/f(y)^2). If we set f(y) = K, we get K = 1/ f(1 / K^2). That hold in general, so f(1/y^2) = 1 / f(y). Plugging in y = 1, we get f(1) = 1. The when we plug y=1 into original equation, we get f(x^2) = f(x)^2, which has the solution f(x) = sqrt(x).
Ok, so I think he could have been done much sooner. When he noticed that f²(f(a))=C*f(a), he was practically done. Let f(a)=p₁/q₁, where p₁,q₁∈ℤ⁺:p₁≠q₁ and p₁ and q₁ are relatively prime and substitute it in we get f(p₁/q₁)=√(Cp₁/q₁). This forces C=p₁/q₁ (or C=q₁/p₁, but will stick with first C, WLOG). Now let f(b)=p₂/q₂, p₂≠q₂, relatively prime, and substitute it in, using our new C value, and we get f(p₂/q₂)=√(p₁p₂/(q₁q₂)). But f(p₂/q₂)=√(p₁p₂/(q₁q₂))∉ℚ⁺, which is a contradiction with our initial conditions. But we excluded one rational number from our check, which is 1 (because 1/1 means p₁=q₁), and if we substitute that in, we are saying that f(a)=1, which makes C=1, and holds for all values in the original equation, so f(a)=1 ∀a∈ℚ⁺.
I found some amusing identities that didn't amount to much by letting y=a and x=1/f(a), since Q+ is closed under reciprocal. The defining equation gives you f(f(a)²/f(a)²) on the LHS, which is f(1), and a constant regardless of a. Putting in the reciprocals of those, I got that f(f(b))=√(f(1)/f(1/b)). f(x) gives perfect squares, and f(f(x)) takes square roots of them? Time to guess f(x)=1 and try to prove it.
for positive reals...
square root?
That’s right!
Hey, can i recommend a problem? is one of the hardest ones in brazialin math olympiad and a functional equation: problem 3 from 2019 level 3, here is a link to the test www.obm.org.br/content/uploads/2019/11/Prova_Nivel_3_OBM_2019.pdf , it is in portuguese but the question is short so if you dont get the idea you can just use a translator, it is a pretty tricky question, give it a try please!
Hi!! I actually didn't come up with a solution for real positif numbers with a proof, but a saw that the function sqrt(x) is a solution
I used his usual suggestions (not given here ) to get f(0) = f(0)**3 so I figured that it was going to be something simple.
It's always fun seeing how different my solutions are to yours. What a dissapointing answer!
If Q>0 wasnt a requirement, would f(x)=sqrt(x) satisfy the equation?
Actually I solved without using the fact that was in rational numbers. But I needed to be in positive real numbers
f(x) = √x would be an example of a function from the positive reals to the the positive reals that meets satisfies the functional equation.
The few, the proud, the Marines.
The infinitesimal few, the brilliant, the mathematicians!!!
Square root actually works for the real numbers scenario
Sqrt(x) is a reel solution
But not sure if it is unique
sorry mate, you get a 6, tiny mistake near the end on the cubic
How much time does it take you to solve it?
For real numbers, the equation can be reduced to f(f(a))² = C f(a) as in the video. Substitute f(a) = x, and you get f(x)² = C x. Solve for f(x), and you get f(x) = ±√(C x). Plugging both the positive and negative branches into the original equation, you get either C x √y = C^(3/2) x √y or equivalently -C x √y = -C^(3/2) x √y. Only C = 0 and C = 1 works for all x and y. Thus f(x) = √x and f(x) = 0 are solutions.
But where you lost the solution f(x)=1?
good job
how can we be sure that for every rational "x" there is a "a" sush as f(a)=x ? this is essential to set x = f(a). but we dont know if f is surjective . please explain if i get it wrong thanks
No. For all a there exists an x=f(a), staying within the positive rationals. You seem to think he's saying the opposite (that for all x there exists an a satisfying x=f(a)). But that is not required at all.
Go on problem solving pls.
Simplest functions I've found over the reals to the reals are: f(x) = 0, f(x) = 1, and f(x) = sqrt(|x|). I haven't ruled out there being more though.
@@angelmendez-rivera351 How do you do it?
Is it just me, or does this problem seem a bit hard for IMO question 1?
But it's fitting that the answer to question 1 is itself 1
Functional equations are, in general, fairly easy for IMO contestants, especially compared to the horrors combinatorics can give birth. So no, it's not at all hard for IMO A1
is f(x) = sqrt(x) not a solution? It works, am I missing something?
Not on the rational numbers, where sqrt(x) is not defined over the entire specified domain. It works for the question he posed about the positive real domain, though.
why in 6:20 he can replace a for f(a) i mean he's saying that f(a)=a , so he could say it without problem? or is like replace a for f(n)?
I think he's saying to replace a with f(n). Letting a=f(a) would become a special case of the function where input equals output.
@@ansmw yeah but i think that for example the fuction is f(x)=1+x the input and the output always will be different, that's why the doubt arose :p
Let x = y = 0. Then we get f(0) = f(0)^3. As f > 0, this implies f(0) = 1. Then let x = 0 and let y be free. Then f(0) = f(y). Thus f(x) = 1.
Supposedly this can fail over the reals, although I can't see how. Any ideas? Letting x = 0 and letting y be free should be enough to conclude the function must always be constant.
We're sadly not able to set x and y equal to 0 as the domain is only for positive rationals.
@@duskhound2883 ohhh no, you're right! Thanks
Hey, I am sitting on a solution to a problem I wrote and was wondering if you could try it I’m pretty proud of it but it’s pretty difficult but has a beautiful result: find an funtion f(x) over the positive quotients such that for S(n) = (f(S(n))+f(S(n)-1))/f(S(n)), where n is a natural number and s(1) = 1, that for all values of n S(n) = (Pn+1)/Pn where Pn is the nth pascal number.
when you set x=f(a), how do you know that f is surjective?
For all a there exists an x=f(a), staying within the positive rationals. You seem to think he's saying the opposite (that for all x there exists an a satisfying x=f(a)).
f of f of f of f of a
I like this song lmao
I think there is another family of functions apart from sqrt(x), since f being continuous is not a requirement. Have f(0)=0. f(a) is either 0 or 1; such that f(a^2)=f(a). For example, f(a)=1 if a=3^(2^n), where n is a integer; f(a)=0 otherwise.
hold on: f(x) = 0 is a solution
The output of the function is positive rational, so f(x)=0 is not a solution.
0 is not a positive rational number though. .. . .
needs to be positive