Do you have to explicitly check the solutions in your text? Or just writing "It is trivial to check that both solutions satisfy all conditions" enough?
i feel like linear functions are the most complex functions that ever occur in these contest functional equation problems. and the difficulty only comes from proving that they are the only solutions ^^
@@cloudwalker2730 surjective*! No, but you can restrict the codomain of all functions so that they are surjective wrt that space. For example, f:R-->R where f(x)=0 for all x is certainly not surjective as there are a fair few examples of real numbers which are not 'reached'. However, this can also be considered a function f:R--->{0}, in which case it is surjective as all elements of the set {0} (ie just 0) are 'reached'.
Another solution is, once knowing that f(f(x))=x, replacing x with f(t) for t in R. This gives f(f(t)*f(f(t))+f(y))=f(f(t))^2+y which is equivalent to f(f(t)*t+f(y))=t^2+y, which can be rewritten as f(x*f(x)+f(y))=x^2+y. Comparing this to the initial equation gives f(x)^2=x^2. Then f(x) can be either x or -x. Only thing left is to show they can't be "mixed". Supposing a,b in R^2 s.t. f(a)=a and f(b)=-b, replacing x=a and y=b gives f(a^2-b)=a^2+b which is implies a=0 or b=0. Therefore the only solutions can be f(x)=x and f(x)=-x, and they trivially work. It's kinda the same but doesn't need the injectivity proof.
Professor Penn, do you even sleep? :D Thanks for the never-ending content! As for the problem, standard stuff where Id is obviously one of the solutions. I like it!
Shorter way: After you get f(f(y))=y, you can put f(x) for x in the first equation. Because of f(f(x) ) = x nothing changes on the left hand side and on the ride hand side you get x^2 +y. Therefor you directly get f(x)^2=x^2.
@@Kartik-yi5kiIn this part it's not necessary to memtion that function ist surjective. But since f(f(x)) =x implies that the function is surjective, it's right that we can only do this step if the function is surjective.
@@germanyop6063 That is close to what I thought when I saw the involution. I thought to myself, "no way this would be some crazy function", then I tried x, -x, 1/x, and -1/x since those are the simplest involutions.
If you know a little calculus, you can take the derivative of both sides instead of assuming f is surjective and obtain d^2(f)/dx^2 = 1, which implies f = +- x given f(0) = 0. Of course, this method instead assumes f is differentiable which is a whole different problem
Here's a more broad functional equation hint: choose variables such that terms cancel. Alternatively, choose variables such that terms become equal. Often x in +/- {0, 1, y, f(y)} does that, but look at your particular equation. Solution summary: We known f: R -> R such that f(xf(x) + f(y)) = f(x)^2 + y (*) f is surjective: let c given, pick an x, let y = c - f(x)^2 and z = xf(x) + f(y), then f(z) = c [apply (*) and simplify] Let f(x) = 0, then f(f(y)) = y since f(xf(x) + f(y)) = f(f(y)) = f(x)^2 + y = 0^2 + y = y Let x = 0, then f(0) = 0 since f(0f(0) + f(y)) = f(f(y)) = y = f(0)^2 + y implying 0 = f(0)^2 Let y = 0, then f(zf(z)) = f(z)^2 by canceling f(y) = f(0) = 0 = y and letting x = z But then f(f(z)f(f(z))) = f(f(z))^2 so f(f(z)*z) = z^2 since f(f(z)) = z, by letting x = f(z) Combining the two, f(z)^2 = f(zf(z)) = z^2 so f(z) in {-z, z}, for each z individually. Let f(s) = s and f(r) = -r; then s^2 + r in {s^2 - r, r - s^2} by (*) and thus 0 in {r, s}. Ergo f(x) = ax for some a in {-1, 1}. Both work, so {x -> x, x -> -x} is the set of all such functions.
In IMOs, there is quite a lot of emphasis on these "functional" problems of the type, "find all functions that satisfy ". But then, when actually studying university math, they don't seem to ever play a particularly important role. Seems like a "puzzle" type of a problem that is later in life never connected to any deep theory in "actual" professional mathematics. I still like this problem and love the thought process to solve it; it's just an observation.
Most of contest math is just puzzle solving. And personally that’s why I enjoy it so much. Don’t get me wrong university math is fun too but I feel like it is more “professional” and research oriented as opposed to just solving hard problems/puzzles like you do in contests
I always wondered how these problems are created to begin with. Do some jury members just come up with some "random" combinations of x, y, f(x), f(f(x)), and so on; and then try to prove that there is only one function such as f(x)=x, or similar. Or, how exactly are these "functional" IMO problems created?
I tried to solve the problem before watching the video, and I managed to show that |f(x)|=|x| for all x. Obviously, if we assume f to be continuous, then f(x)=x and f(x)=-x are the only two solutions. But what if we do NOT assume f to be continuous? Then, |f(x)|=|x| only implies f=sigma*id for some map sigma : R -> {-1,1}. BUT in this case the identity f(xf(x)+f(y))=f(x)^2+y expands to x^2+y=c*(sigma(x)x^2+sigma(y)y), where c is +1 or -1. This implies sigma(x)=sigma(y) for all x,y. In other words, sigma must be constant. Therefore, the continuous solutions (f(x)=x and f(x)=-x) are the only solutions.
Great，thanks! I think in order to make things easy, Dr. Michael knows but do not use this conclusion：If the composite function f(g(·)) is surjective, then the function f(·) must be surjective. If the composite function f(g(·)) is injective, then the function g(·) must be injective. So if the composite function f(f(·)) is both surjective and injective(bijective) , then the function f(·) must be bijective, such as the function f(f(·))=x+c that appears in this problem.
Was thinking of exactly this too! We only showed that equation => f(x) = +x for all x or f(x) = -x for all x but not vice versa. Granted it's just two lines, but..
Agreed. It's easy to check: If f is the identity function (for all z: f(z) = z) then f(xf(x) + f(y)) = f(x*x + y) = x*x + y = f(x)^2 + y If f is the negation function (for all z: f(z) = -z) then f(x*f(x) + f(y)) = f(x*(-x) + (-y)) = f(-(x*x + y)) = -(-(x*x + y)) = x*x + y = [-f(x)]^2 + y = f(x)^2 + y So those two functions are valid solutions. What the video proves is that no other function is a valid solution, so now we have a complete description of the solution set.
I believe that having proved that f(f(x)) = x it is easier to plug x -> f(x) in the original equation. We notice that the LHS stays the same and the RHS becomes x^2 + y. Thus the 2 RHS are equal, so (f(x))^2 + y = x^2 + y => f(x) = +-x
Little hint: as soon as you look at the condition try f(x)=x because in most problems it's a solution. I tried it as soon as I saw the problem because at this point it's kind of trivial procedure
Alternative proof by contradiction for 8:52. assume f isn't injective, then there is z in R s. t. f(z) = y1 and f(z) = y2. Further down one can argue y1 = f(f(y1)) != f(f(y2)) =y2, but that can't be true since f(f(f(z))) = f(f(y1)) =f(f(y2)). So f is injective.
Easier way, differentiate wrt y, and you get f'(xf(x)+f(y))f'(y)=1. Differentiate wrt x, and you get f'(xf(x)+f(y))[f(x)+xf'(x)]=2f(x)f'(x). Now substitute the first into the second to obtain [f(x)+xf'(x)]/[f'(y)]=2f(x)f'(x). the LHS is dependent on y while the RHS is not. this means the denominator of the LHS has to be a constant=> f is linear. Either this or the numerator both have to be zero, this case doesn't work out. The exact form of the linear term is easy to figure out from there on.
Perhaps an alternative method might be to assume f() is linear. Put f(x) = ax + b. Use this form in the given equation. One rapidly finds that a = 1 and b = 0.
that's actually a logical error. you ASSUME that f(x) is of the form ax+b. this of course leads to a valid solution, but you don't know that this is the only solution. what if there's another solution of the form f(x) = ax^2 + bx + c? you wouldn't know until you checked. and now you have to check every possible form for f(x)!
These questions frequently have solutions in the form of a linear or quadratic function. It is often worth trying these to see if they solve the question. These are artificial equations designed to be solved. Assuming a hypothetical solution exists and then proving that it does is a common techique in mathematics. One can reasonably ask if the solution so found is the only solution. In simple cases there is usually only one solution. But the point is valid.
@@alganpokemon905 These equations frequently have solutions which are linear or quadratic equations. These are questions which have known solutions. For this reason they tend to have fairly simple functional forms. The use of a test functional form to solve an equation is common in mathematics. On example is when we assume the existence of a solution in a partial differential equation and then solve it by separation of variables. Another example is in the method of variation of parameters also to solve differential equations. I agree that it is difficult to prove that there is a unique solution to the functional equation in general. This is normally something we can rey on in differential equations. At least unless we admit stochastic solutions where there is a continuum of solutions. For a prize exam like this where many people cannot solve even one question its a worthwhile strategy to at least try a simple functional form such as a linear or quadratic one initially to see if a solution can be found at all. This frequently works
At 6:52 why can you use your first calculation when you had asumed that x=0 to show that x0=0? Hadn't you left that asumption behind to see what happened with f(0)=0?
The functional equation must be valid for ALL x, in particular for x=0 and for x=x0 (such that f(x0)=0). The equation could have two solutions, but it doesn't, therefore the conclusion is that x0=0.
Once you found f(f(x))=x f(xf(x)+y)=f(x)²+y = f(f(x)f(f(x))+y) = f(f(x))²+y=x²+y implying f(x)²=x² squaring both sides of the original eq f(xf(x)+y)²=(x²+y)² (xf(x)+y)²=(x²+y)² x²f(x)²+y²+2xf(x)f(y)=x⁴+y²+2x²y 2xf(x)f(y)=2x²y for all non-zero x f(x)f(y)=xy so either f(x)=x or f(x)=-x (cannot be both, hence the use of "either")
Wasn't it just visible? I just plugged in f(x)=x as hit and trial and then found out even -x is satisfying. But the proof that only these two exist is really good
With only a bit of exaggeration you can say that almost every time functional equations in mathematical contests come up, the solutions are linear functions. Of course, proving this for the problem at hand is the interesting and hard part.
@@keedt Yup, but the exaggeration is just too much. I've solved so many questions, which do not involve linear functions as solutions but they always turn out to be some general function like e^x or ln(x) or 1/x.
@@danishjuneja The question says “Find all f” but it doesn’t mean it. It means “Find all f and prove that there are no more”. In pure math, the “prove it” is always part of the question.
...That proof at the end could've been BWOC if you'd simply stipulated that x0 and x1 were in R≠0. At which point x1 can only be 0, thus contradicting the assumption that the function could swap between both.
You can actually prove that f(x) is surjective by considering this suppose there is a value a where f(y)≠a for all y belonging to R let f(y)=z where z can take values from the set R-{a} this means that f(z)=(f(0))^2+y but this means that for all values z belonging to R-{a} there s a value y where f(z)=a but since z belongs to a subset of R and we assumed that for all values belonging to R we can t have f(y)=a this means we are led to a contradiction and thus f(x) is surjective
How about this: f(f(y) = A^2 + y, for all y. (A is f(0).) Call this g(y). It is defined for all y, so by its rule (a linear function), g(y) is invertible. Now do a side lemma - if a composition of a function with itself is invertible, that function is invertible. So as g is invertible, so then is f. As f is invertible, f is onto, so x* such that f(x*)=0 exists. Then we do as Michael did, to show A=0. This gives f o f = id. I think this is a weaker requirement of f than the original functional equation, because it still admits the piecewise functions, not just id and -id. Basically, a function that satisfies the FE must be an involution, but not all involutions satisfy the FE. So the last bit of work, ruling out piecewise functions, is still necessary.
At first, he showed that f maps a to -a or +a for all a in R. We know that there are infinitely many bijective functions satisfying that property. However, he claimed that only 2 functions can satisfy the conditions in the problem. To prove this claim, he used contradiction technique. He assumed that there is a function different from x -> x and x -> -x such that this function satisfy the property in the question. Then, he got that that function map a real number a to another number which is not a or -a. Contradiction occured!
Around 9:30, you state that f(f(y)) = y implies that f is involutive. But it actually implies that f is bijective, so why bother with proving that f is injective in the first place? Great video, as always!
1) fix any point x0, take y=-(f(x0))^2 and get that there exists point a such that f(a)=0 2) let x=a, f(f(y))=y 3) let x -> f(x), f(f(x)f(f(x))+f(y))=x^2+y, but f(f(x)f(f(x))+f(y))=f(f(x)x+f(y))=(f(x))^2+y this implies (f(x))^2=x^2 => f(x)=+-x
The proof goes by contradiction. The inverse of the claim would be that there exist some x0, x1 such f(x0) = x0 != -x0 (not equal to) and f(x1) = -x1 != x1 since if x1 = 0 for example, we could equivalently just have f(x0) = x0, f(x1) = x1 and this would not contradict the claim in the first place. The proof then shows that you in fact do need x1 (or x0) = 0.
Wait, if f is its own inverse and inversing a function R to R is mirroring it over a 45 degree line, then f can only be that line, f(x) = x, or a line orthogonal to it, which is f(x) = -x. And it has to be either one or the other and not both at the same time because they are orthogonal. Isn't this a valid argument?
I don't understand why the last part is necessary. Since we know f(x) is bijective, and f(x) = x covers all real numbers, adding f(x) = -x cannot be a simultaneous solution. Therefore, they must be separate solutions. Or in simpler terms, since f(x) is bijective, then it is also just what we think of when we normally think of a classic function: for each input, there MUST be only one output; thus, f(x) cannot give both x and -x, and must be one or the other... obviously leading us to the conclusion that there are two answers, f(x) = x and f(x) = -x. Going through that entire last board was sort of interesting, but reeaallly kind of overkill, given the amount of time/room the rest of the problem took up. Is that not rigorous enough?
I don't think you fully understand what "function" and "bijective" means. Of course he is not suggesting f(3), for example, could be both 3 and -3, at the same time; he is checking that it is not possible that the function f associates x to SOME of the values of x and -x to SOME OTHERS values of x. This is needed, for example, to exclude that |x| is a solution or that -|x| is a solution, or many other discontinuous functions that have a graph contained in the union of the quadrant bisectors are solutions.
The following two statements are not equivalent. In particular, the second statement in itself is not yet enough to prove the first statement. Either f(x) = +x for all x in R, or f(x) = -x for all x in R For all x in R, either f(x) = x or f(x) = -x For example, the function f(x) = |x| satisfies the second condition, but not the first.
Honestly, it could show up in a number of courses. Could be set theory and proofs (sometimes combined into one course), could be a number theory class, could be abstract algebra... it's pretty fundamental to a lot of proofs but isn't really needed for most of the calculus stuff that people learn earlier on
These topics are thought in some basic courses in mathematics. I mean all mathematic students should be familiar with them. (we learned these topics in high school in Turkey.)
From the result at 6:52 we have f(f(y)) = y, which implies that f is an involution. Hence, it is bijective. It was not necessary to show injective but also no loss in doing that :)
Can I have some help? I did this: X=0 and y=0 So F(0f(0)+f(0))=f(0)²+0 So F(f(0))=f(0)² Now I put F(0)=T So F(t) =t² So, also f(x) =x² What I am getting wrong? BTW, sorry for my bad English
First of all, it feels weird that you can find a whole function just by evaluating an expression at a single point, the point (0, 0). Plus, if you set F(0)=t, that means that t is a concrete number (not a variable), so you cannot say: “also, f(x)=x^2”. Why? You “showed” (I am not sure if that is a proof, I’m not even an expert 😅) that there is *one* number (t) such that F(t)=t^2, but that does not mean that it will happen to every number. As I said, I’m not an expert, but I hope I could help a bit :) Edit: Nice try, though!
Interesting problem. I got f(x)=+x without issue while just looking at it. I missed the f(x)=-x as an alternate possibility. I would have cocked it up because it asked for all f(x). I would have said, by inspection, I guess that f(x)=x. And then I would show that satisfies the property and move on having missed the -x. It looks like both of these functions hold for complex input and output. But if they said that, it would probably be harder as there may be other functions that also satisfy.
Even surjectivity was not required. All that was required was a real number for which the function gives zero. Which can be done by selecting x = 0 and y = - (f(0))^2.
You want to show that f can hit any b in the real numbers - so set f(f(y) = f(0)^2 - y = b hence y = b - f(0)^2. You now know that f(f(y)) = f(f(b-f(0)^2) = b, so it appears that exactly the choice a = f(b-f(0)^2) gives you that f(a) = b, and therefore f is surjective since b was arbitrary.
I really like how you teach but as someone with very limited math background, many of what you're saying is quite inaccessible. Trying to solve one of these for something and I'm still stuck lol
I feel like I'm about to ask a really dumb question. What does "quantity" mean? I've never heard that term before and I've done a lot of mathematics :grin: I assumed it was the closest integer when I tried this puzzle, but it's clearly not given the mess I got into and given what the answer is. On the plus side, it made the puzzle more challenging. I believe there are no solutions if you interpret that to mean closest int :)
"quantity" just refers to a grouping. Like the difference between (1+x^2) and (1+x)^2, for the former you would say "one plus x squared", but for the latter you'd say "1 plus x, quantity squared" to indicate that you are squaring the whole term and not just the x.
It simply means that both x and f(x) are real numbers. R is the set of real numbers, the notation f:E --> F means that f takes an element of the set E and outputs an element of the set F, for instance if i write f:N --> R it means that my function takes an integer and outputs a real number
If you put f(x)=2+x, on the left side you get f(x(2+x)+2+y)=f(x^2+2x+y+2)=x^2+2x+y+4 . On the right side, you get (x+2)^2+y=x^2+4x+y+4 . It is close but not quite. As he has proven, f(0)=0 in every case.
@@mattrough6305 Lol calculate again right side, please. (x+2)^2= x^2+2x+ 4, not x^2+4x+4. Itachi right, and there are two solutions f(x)=x, and f(x)=x+2.
Maybe my mind is impaired but -x seems not to be a solution. One gets a contradiction when plugging it into the functional equation. So x seems to be the only solution.
![KSI - Dirty [Official Visualiser]](http://i.ytimg.com/vi/VoDh1h1dPUE/mqdefault.jpg)
Don't forget to check that both of your solutions are solutions to the problem !
When this very simple step is forgotten, you get a 5/7, or 6/7 at IMO
@ABHIKALP SHEKHAR I think he is making a joke about schools
Testing doesn't count in creative problem solving but we don't know until we see
@@justaguy9847 I don't think that is joke. My teacher has silver medal and he always say that we must check our solutions.
@@justaguy9847 He is serious as we sometimes get extraneous solutions.
Do you have to explicitly check the solutions in your text? Or just writing "It is trivial to check that both solutions satisfy all conditions" enough?
Your shirt has a undefined region.
I LAUGH SO HARD
i feel like linear functions are the most complex functions that ever occur in these contest functional equation problems. and the difficulty only comes from proving that they are the only solutions ^^
Sometimes (especially in harder problems) some funktions have a form like f(x) =1 for all x>=0 and f(x) =-1 for all x
Helpful note:
f(f(x)) = x implies f is bijective (that is, both injective and surjective)
Aren’t all functions surjective
@@cloudwalker2730 surjective*! No, but you can restrict the codomain of all functions so that they are surjective wrt that space. For example, f:R-->R where f(x)=0 for all x is certainly not surjective as there are a fair few examples of real numbers which are not 'reached'. However, this can also be considered a function f:R--->{0}, in which case it is surjective as all elements of the set {0} (ie just 0) are 'reached'.
@@cloudwalker2730 If all functions were surjective, then N and R would have the same cardinality.
@@cloudwalker2730 Look what you just did. You got us all riled up.
Its easy to see that is inyective but how do you prove that is surjective?
Another solution is, once knowing that f(f(x))=x, replacing x with f(t) for t in R.
This gives f(f(t)*f(f(t))+f(y))=f(f(t))^2+y which is equivalent to f(f(t)*t+f(y))=t^2+y, which can be rewritten as f(x*f(x)+f(y))=x^2+y.
Comparing this to the initial equation gives f(x)^2=x^2. Then f(x) can be either x or -x. Only thing left is to show they can't be "mixed".
Supposing a,b in R^2 s.t. f(a)=a and f(b)=-b, replacing x=a and y=b gives f(a^2-b)=a^2+b which is implies a=0 or b=0.
Therefore the only solutions can be f(x)=x and f(x)=-x, and they trivially work.
It's kinda the same but doesn't need the injectivity proof.
Professor Penn, do you even sleep? :D Thanks for the never-ending content!
As for the problem, standard stuff where Id is obviously one of the solutions. I like it!
10:05
Syntax error on line 4. ")" expected.
lol
Isn't it the opposite? He has one too many closing symbols.
Shorter way:
After you get f(f(y))=y, you can put f(x) for x in the first equation. Because of f(f(x) ) = x nothing changes on the left hand side and on the ride hand side you get x^2 +y. Therefor you directly get f(x)^2=x^2.
That is assuming that the function is surjective, because otherwise there is a possibility that this only holds for domain= range of the function
@@Kartik-yi5kiIn this part it's not necessary to memtion that function ist surjective. But since f(f(x)) =x implies that the function is surjective, it's right that we can only do this step if the function is surjective.
@@germanyop6063 That is close to what I thought when I saw the involution. I thought to myself, "no way this would be some crazy function", then I tried x, -x, 1/x, and -1/x since those are the simplest involutions.
If you know a little calculus, you can take the derivative of both sides instead of assuming f is surjective and obtain d^2(f)/dx^2 = 1, which implies f = +- x given f(0) = 0. Of course, this method instead assumes f is differentiable which is a whole different problem
Here's a more broad functional equation hint: choose variables such that terms cancel.
Alternatively, choose variables such that terms become equal.
Often x in +/- {0, 1, y, f(y)} does that, but look at your particular equation.
Solution summary:
We known f: R -> R such that f(xf(x) + f(y)) = f(x)^2 + y (*)
f is surjective: let c given, pick an x, let y = c - f(x)^2 and z = xf(x) + f(y), then f(z) = c [apply (*) and simplify]
Let f(x) = 0, then f(f(y)) = y since f(xf(x) + f(y)) = f(f(y)) = f(x)^2 + y = 0^2 + y = y
Let x = 0, then f(0) = 0 since f(0f(0) + f(y)) = f(f(y)) = y = f(0)^2 + y implying 0 = f(0)^2
Let y = 0, then f(zf(z)) = f(z)^2 by canceling f(y) = f(0) = 0 = y and letting x = z
But then f(f(z)f(f(z))) = f(f(z))^2 so f(f(z)*z) = z^2 since f(f(z)) = z, by letting x = f(z)
Combining the two, f(z)^2 = f(zf(z)) = z^2 so f(z) in {-z, z}, for each z individually.
Let f(s) = s and f(r) = -r; then s^2 + r in {s^2 - r, r - s^2} by (*) and thus 0 in {r, s}.
Ergo f(x) = ax for some a in {-1, 1}. Both work, so {x -> x, x -> -x} is the set of all such functions.
Actually from f(f(y))=y+c you get f is bijective.
😀
17:34
In IMOs, there is quite a lot of emphasis on these "functional" problems of the type, "find all functions that satisfy ". But then, when actually studying university math, they don't seem to ever play a particularly important role. Seems like a "puzzle" type of a problem that is later in life never connected to any deep theory in "actual" professional mathematics. I still like this problem and love the thought process to solve it; it's just an observation.
Most of contest math is just puzzle solving. And personally that’s why I enjoy it so much. Don’t get me wrong university math is fun too but I feel like it is more “professional” and research oriented as opposed to just solving hard problems/puzzles like you do in contests
I always wondered how these problems are created to begin with. Do some jury members just come up with some "random" combinations of x, y, f(x), f(f(x)), and so on; and then try to prove that there is only one function such as f(x)=x, or similar. Or, how exactly are these "functional" IMO problems created?
I hope this channel never ends
Love your content Prof. Penn! Keep it up
It's all look so satisfying when you do everything thing technically and mathematically.😄
I tried to solve the problem before watching the video, and I managed to show that |f(x)|=|x| for all x. Obviously, if we assume f to be continuous, then f(x)=x and f(x)=-x are the only two solutions. But what if we do NOT assume f to be continuous? Then, |f(x)|=|x| only implies f=sigma*id for some map sigma : R -> {-1,1}.
BUT in this case the identity f(xf(x)+f(y))=f(x)^2+y expands to x^2+y=c*(sigma(x)x^2+sigma(y)y), where c is +1 or -1. This implies sigma(x)=sigma(y) for all x,y. In other words, sigma must be constant. Therefore, the continuous solutions (f(x)=x and f(x)=-x) are the only solutions.
Technicaly, continuity also allows f(x) = |x|. So even if you assume continuity you get a fake solution.
Set x=0. Differenentiate both sides by y twice. We find f''(y)=0. So f(y)=a+by. Then substitute this back in to the original equation and solve.
Nice one but you are assuming f to be continuous. which it's a non trivial assumption
There should be more channels like this
Plz follow rz mathematics for that
ruclips.net/channel/UCvZCNCHoOINkG24aRruHtcQ
Totally agree
Great，thanks! I think in order to make things easy, Dr. Michael knows but do not use this conclusion：If the composite function f(g(·)) is surjective, then the function f(·) must be surjective. If the composite function f(g(·)) is injective, then the function g(·) must be injective. So if the composite function f(f(·)) is both surjective and injective(bijective) , then the function f(·) must be bijective, such as the function f(f(·))=x+c that appears in this problem.
Would you also need to plug in the two solutions to show that these are indeed true solutions? Wouldn't hurt to do so right?
Was thinking of exactly this too! We only showed that equation => f(x) = +x for all x or f(x) = -x for all x but not vice versa. Granted it's just two lines, but..
Agreed. It's easy to check:
If f is the identity function (for all z: f(z) = z) then
f(xf(x) + f(y)) = f(x*x + y) = x*x + y = f(x)^2 + y
If f is the negation function (for all z: f(z) = -z) then
f(x*f(x) + f(y)) = f(x*(-x) + (-y)) = f(-(x*x + y)) = -(-(x*x + y)) = x*x + y = [-f(x)]^2 + y = f(x)^2 + y
So those two functions are valid solutions. What the video proves is that no other function is a valid solution, so now we have a complete description of the solution set.
What a Nice video and elegant solution!
One small suggestion. When you say to give this problem a try, could you move out of the way of the problem, please?
@@angelmendez-rivera351 I like convenience and ease.
I believe that having proved that f(f(x)) = x it is easier to plug x -> f(x) in the original equation. We notice that the LHS stays the same and the RHS becomes x^2 + y. Thus the 2 RHS are equal, so (f(x))^2 + y = x^2 + y => f(x) = +-x
Little hint: as soon as you look at the condition try f(x)=x because in most problems it's a solution. I tried it as soon as I saw the problem because at this point it's kind of trivial procedure
Alternative proof by contradiction for 8:52. assume f isn't injective, then there is z in R s. t. f(z) = y1 and f(z) = y2. Further down one can argue y1 = f(f(y1)) != f(f(y2)) =y2, but that can't be true since f(f(f(z))) = f(f(y1)) =f(f(y2)). So f is injective.
আপনি ভালো ভিডিও বানান.... ভারতবর্ষ থেকে ভালোবাসা জানাই...(You make good videos....love from India)
khati kotha bolli
Easier way, differentiate wrt y, and you get f'(xf(x)+f(y))f'(y)=1. Differentiate wrt x, and you get f'(xf(x)+f(y))[f(x)+xf'(x)]=2f(x)f'(x). Now substitute the first into the second to obtain [f(x)+xf'(x)]/[f'(y)]=2f(x)f'(x). the LHS is dependent on y while the RHS is not. this means the denominator of the LHS has to be a constant=> f is linear. Either this or the numerator both have to be zero, this case doesn't work out. The exact form of the linear term is easy to figure out from there on.
12:28
Why we don’t use quantity sign?
2:40 spoiler! 😱
Awesome question. I was happy with my +/- x and completely forgot about the final part :)
Perhaps an alternative method might be to assume f() is linear. Put f(x) = ax + b. Use this form in the given equation.
One rapidly finds that a = 1 and b = 0.
that's actually a logical error. you ASSUME that f(x) is of the form ax+b. this of course leads to a valid solution, but you don't know that this is the only solution.
what if there's another solution of the form f(x) = ax^2 + bx + c? you wouldn't know until you checked. and now you have to check every possible form for f(x)!
These questions frequently have solutions in the form of a linear or quadratic function. It is often worth trying these to see if they solve the question. These are artificial equations designed to be solved.
Assuming a hypothetical solution exists and then proving that it does is a common techique in mathematics.
One can reasonably ask if the solution so found is the only solution. In simple cases there is usually only one solution. But the point is valid.
@@alganpokemon905 These equations frequently have solutions which are linear or quadratic equations. These are questions which have known solutions. For this reason they tend to have fairly simple functional forms.
The use of a test functional form to solve an equation is common in mathematics. On example is when we assume the existence of a solution in a partial differential equation and then solve it by separation of variables. Another example is in the method of variation of parameters also to solve differential equations.
I agree that it is difficult to prove that there is a unique solution to the functional equation in general. This is normally something we can rey on in differential equations. At least unless we admit stochastic solutions where there is a continuum of solutions.
For a prize exam like this where many people cannot solve even one question its a worthwhile strategy to at least try a simple functional form such as a linear or quadratic one initially to see if a solution can be found at all. This frequently works
2:00 What if f(x)=1/x? Then 0*f(0) would not be 0, because that would be undefined...
At 6:52 why can you use your first calculation when you had asumed that x=0 to show that x0=0? Hadn't you left that asumption behind to see what happened with f(0)=0?
The functional equation must be valid for ALL x, in particular for x=0 and for x=x0 (such that f(x0)=0).
The equation could have two solutions, but it doesn't, therefore the conclusion is that x0=0.
Once you found f(f(x))=x
f(xf(x)+y)=f(x)²+y
= f(f(x)f(f(x))+y)
= f(f(x))²+y=x²+y
implying f(x)²=x²
squaring both sides of the original eq
f(xf(x)+y)²=(x²+y)²
(xf(x)+y)²=(x²+y)²
x²f(x)²+y²+2xf(x)f(y)=x⁴+y²+2x²y
2xf(x)f(y)=2x²y
for all non-zero x
f(x)f(y)=xy
so either f(x)=x or f(x)=-x (cannot be both, hence the use of "either")
Wasn't it just visible? I just plugged in f(x)=x as hit and trial and then found out even -x is satisfying.
But the proof that only these two exist is really good
With only a bit of exaggeration you can say that almost every time functional equations in mathematical contests come up, the solutions are linear functions. Of course, proving this for the problem at hand is the interesting and hard part.
@@keedt Yup, but the exaggeration is just too much. I've solved so many questions, which do not involve linear functions as solutions but they always turn out to be some general function like e^x or ln(x) or 1/x.
@@danishjuneja The question says “Find all f” but it doesn’t mean it. It means “Find all f and prove that there are no more”. In pure math, the “prove it” is always part of the question.
Danish Juneja You're right, but you'd be lucky if you even got 1 point for finding the functions on a math contest.
...That proof at the end could've been BWOC if you'd simply stipulated that x0 and x1 were in R≠0. At which point x1 can only be 0, thus contradicting the assumption that the function could swap between both.
You can actually prove that f(x) is surjective by considering this suppose there is a value a where f(y)≠a for all y belonging to R let f(y)=z where z can take values from the set R-{a} this means that f(z)=(f(0))^2+y but this means that for all values z belonging to R-{a} there s a value y where f(z)=a but since z belongs to a subset of R and we assumed that for all values belonging to R we can t have f(y)=a this means we are led to a contradiction and thus f(x) is surjective
How about this: f(f(y) = A^2 + y, for all y. (A is f(0).) Call this g(y). It is defined for all y, so by its rule (a linear function), g(y) is invertible. Now do a side lemma - if a composition of a function with itself is invertible, that function is invertible. So as g is invertible, so then is f. As f is invertible, f is onto, so x* such that f(x*)=0 exists. Then we do as Michael did, to show A=0. This gives f o f = id.
I think this is a weaker requirement of f than the original functional equation, because it still admits the piecewise functions, not just id and -id. Basically, a function that satisfies the FE must be an involution, but not all involutions satisfy the FE. So the last bit of work, ruling out piecewise functions, is still necessary.
I don't see why the proof of the last claim around 14:00 is actually a proof of the claim.
Rewrite the claim: Let x0 and x1 be two real numbers different than 0. [then copy paste the proof]. But that implies x0=0 or x1=0, which is absurb
He showed that only the number in R where f(x)=-x is 0
@@ThePiotrekpecet exactly
At first, he showed that f maps a to -a or +a for all a in R. We know that there are infinitely many bijective functions satisfying that property. However, he claimed that only 2 functions can satisfy the conditions in the problem. To prove this claim, he used contradiction technique. He assumed that there is a function different from x -> x and x -> -x such that this function satisfy the property in the question. Then, he got that that function map a real number a to another number which is not a or -a. Contradiction occured!
Someone please explain why f is surjective. I didn't understand that part
I don't understand what part of your proof required the injectivity
I love your videos!
Around 9:30, you state that f(f(y)) = y implies that f is involutive. But it actually implies that f is bijective, so why bother with proving that f is injective in the first place?
Great video, as always!
Two possible answers:
1 he thought that showing us that way could be helpful
2 he's not so good in solving FE and he didn't notice that
1) fix any point x0, take y=-(f(x0))^2 and get that there exists point a such that f(a)=0
2) let x=a, f(f(y))=y
3) let x -> f(x), f(f(x)f(f(x))+f(y))=x^2+y, but f(f(x)f(f(x))+f(y))=f(f(x)x+f(y))=(f(x))^2+y this implies (f(x))^2=x^2 => f(x)=+-x
But any typical function "x->y" is surjective, isn't it?
おお、日本の数オリの問題キタ。
Such a good video, such a good channel.
I don't see why the proof of the last claim (around 14:00) is actually a proof of it.
The proof goes by contradiction. The inverse of the claim would be that there exist some x0, x1 such f(x0) = x0 != -x0 (not equal to) and f(x1) = -x1 != x1 since if x1 = 0 for example, we could equivalently just have f(x0) = x0, f(x1) = x1 and this would not contradict the claim in the first place. The proof then shows that you in fact do need x1 (or x0) = 0.
Please do some geometry problems
Wait, if f is its own inverse and inversing a function R to R is mirroring it over a 45 degree line, then f can only be that line, f(x) = x, or a line orthogonal to it, which is f(x) = -x. And it has to be either one or the other and not both at the same time because they are orthogonal. Isn't this a valid argument?
Not really, since f does not have to be a straight connected line, the argument that f is its own inverse alone is not sufficient.
I don't understand why the last part is necessary. Since we know f(x) is bijective, and f(x) = x covers all real numbers, adding f(x) = -x cannot be a simultaneous solution. Therefore, they must be separate solutions. Or in simpler terms, since f(x) is bijective, then it is also just what we think of when we normally think of a classic function: for each input, there MUST be only one output; thus, f(x) cannot give both x and -x, and must be one or the other... obviously leading us to the conclusion that there are two answers, f(x) = x and f(x) = -x. Going through that entire last board was sort of interesting, but reeaallly kind of overkill, given the amount of time/room the rest of the problem took up. Is that not rigorous enough?
I don't think you fully understand what "function" and "bijective" means. Of course he is not suggesting f(3), for example, could be both 3 and -3, at the same time; he is checking that it is not possible that the function f associates x to SOME of the values of x and -x to SOME OTHERS values of x. This is needed, for example, to exclude that |x| is a solution or that -|x| is a solution, or many other discontinuous functions that have a graph contained in the union of the quadrant bisectors are solutions.
The following two statements are not equivalent. In particular, the second statement in itself is not yet enough to prove the first statement.
Either f(x) = +x for all x in R, or f(x) = -x for all x in R
For all x in R, either f(x) = x or f(x) = -x
For example, the function f(x) = |x| satisfies the second condition, but not the first.
@@angelmendez-rivera351 Thanks. This was the boost I needed to see what I was missing. Cheers.
3:40. Still not clear with it😂. Con someone tell me the reason
for the case f(x)=-x, f(xf(x)+f(y))=f(x*(-x)-y)=f(-x^2-y)=x^2+y=(-x)^2+y=f(x)^2+y is also valid 🤔
What branch of mathematics deals with surjective, bijective, image, pre image and the like? And does he have videos on the subject?
You learn this stuff in set theory when you learn about functions.
I first learned it in Linear Algebra
Honestly, it could show up in a number of courses. Could be set theory and proofs (sometimes combined into one course), could be a number theory class, could be abstract algebra... it's pretty fundamental to a lot of proofs but isn't really needed for most of the calculus stuff that people learn earlier on
These topics are thought in some basic courses in mathematics. I mean all mathematic students should be familiar with them. (we learned these topics in high school in Turkey.)
So thorough!
From the result at 6:52 we have f(f(y)) = y, which implies that f is an involution. Hence, it is bijective. It was not necessary to show injective but also no loss in doing that :)
I appreciate the last step, I wouldnt have thought of that (that it can be x some times and some times -x)
One of my most favourite FEs ever.
0.f(0) is null only if f(0) is not infinity, which, at the time of the reasoning is not proved.
No, the assumption of the exercise is the function is defined everywhere, in particular in 0.
Can I have some help?
I did this:
X=0 and y=0
So
F(0f(0)+f(0))=f(0)²+0
So
F(f(0))=f(0)²
Now I put F(0)=T
So F(t) =t²
So, also f(x) =x²
What I am getting wrong? BTW, sorry for my bad English
First of all, it feels weird that you can find a whole function just by evaluating an expression at a single point, the point (0, 0).
Plus, if you set F(0)=t, that means that t is a concrete number (not a variable), so you cannot say: “also, f(x)=x^2”. Why? You “showed” (I am not sure if that is a proof, I’m not even an expert 😅) that there is *one* number (t) such that F(t)=t^2, but that does not mean that it will happen to every number.
As I said, I’m not an expert, but I hope I could help a bit :)
Edit: Nice try, though!
@@nadiie1nadiie222 l also thought like you, but I think "t is a variable", so who not?. But I Think you are right, so thank you
Vittorio Lufrano because t is not a variable. t is a certain value and that is f(0)
Interesting problem. I got f(x)=+x without issue while just looking at it. I missed the f(x)=-x as an alternate possibility. I would have cocked it up because it asked for all f(x). I would have said, by inspection, I guess that f(x)=x. And then I would show that satisfies the property and move on having missed the -x.
It looks like both of these functions hold for complex input and output. But if they said that, it would probably be harder as there may be other functions that also satisfy.
His proof is fine for complex numbers.
That's a beautiful problem. Nice video! But why did you need the injectivity? I didn't see where you used it.
He didn't need it lol
Even surjectivity was not required.
All that was required was a real number for which the function gives zero.
Which can be done by selecting x = 0 and y = - (f(0))^2.
Because he could! :)
I think to show that f is its own inverse (inverses must be bijective).
@@angelmendez-rivera351 But he already showed f(f(x)) = x at 6:45 without using injectivity.
I can't figure out where the inverse function thing expression for a comes from at 4:00.
You want to show that f can hit any b in the real numbers - so set f(f(y) = f(0)^2 - y = b hence y = b - f(0)^2. You now know that f(f(y)) = f(f(b-f(0)^2) = b, so it appears that exactly the choice a = f(b-f(0)^2) gives you that f(a) = b, and therefore f is surjective since b was arbitrary.
Thank you, it is amazing!
Look at 1992 IMO problem... If you know it, this is basically a win!
I really like how you teach but as someone with very limited math background, many of what you're saying is quite inaccessible. Trying to solve one of these for something and I'm still stuck lol
Surjective! Beautiful trick
Can u give such idea to solve that type functional problem to approach injective and subjective functions!!
Anyone knows where can I learn functional equations from scratch? The material I find online are way too advance
have you found out? Late for 3 years, but if you're still wondering, mine would be the one by Christoper Small
Nice shirt 🙂
So, what are the functions that satisfy the condition?
last 5 seconds of the video
What level of mathematics is this ?
I wanna know that too, should i know this stuff in precalc level? 🤔🤔
olympic level
@@leovl1041 lol its olympiad
@@chetan6984 what do you mean lol? If this is easy for you(im not saying it's hard) then go solve reimann hypothesis lol.
The presentation is more complicated than necessary.
I feel like I'm about to ask a really dumb question. What does "quantity" mean? I've never heard that term before and I've done a lot of mathematics :grin:
I assumed it was the closest integer when I tried this puzzle, but it's clearly not given the mess I got into and given what the answer is. On the plus side, it made the puzzle more challenging. I believe there are no solutions if you interpret that to mean closest int :)
"quantity" just refers to a grouping. Like the difference between (1+x^2) and (1+x)^2, for the former you would say "one plus x squared", but for the latter you'd say "1 plus x, quantity squared" to indicate that you are squaring the whole term and not just the x.
@@phiefer3 thanks, yeah, I figured it out over time. Maybe more of an Americanism? We used to clarify in the UK by saying "all squared"
I like your shirt, I’m in Bowdoin’s Class of 2024!!
How about y equals lxl?
this is not injective
Did you go to Bowdoin College?! That’s my alma mater!
The Odds
What are they?
You lost me at the part when you started using a & b. Could someone explain that to me?
He showed that for every b in R there is a such that f(a)=b
Hope that helps (he chose a in a way that was convenient)
I am almost sure that this problem appeared in Argentina IMO TST 1997.
This is how long it takes once you know how to work it out.
Must look up those -ectives.
Cool video :)
Well assuming f(x)=-x shouldn't this result in trouble? Let's see: f(xf(x))+f(y))=f(-x^2-y)=x^2+y=(-x)^2+y=(f(x))^2+y nope it resolved itself.
Thanks sir
Bài toán phương trình hàm từ một kì thi. Cảm ơn.
Can someone explain what f: R -> R means
It simply means that both x and f(x) are real numbers. R is the set of real numbers, the notation f:E --> F means that f takes an element of the set E and outputs an element of the set F, for instance if i write f:N --> R it means that my function takes an integer and outputs a real number
f(x)=2+x is also a solution, or am I wrong?
If you put f(x)=2+x, on the left side you get f(x(2+x)+2+y)=f(x^2+2x+y+2)=x^2+2x+y+4 . On the right side, you get (x+2)^2+y=x^2+4x+y+4 . It is close but not quite. As he has proven, f(0)=0 in every case.
@@mattrough6305 Lol calculate again right side, please. (x+2)^2= x^2+2x+ 4, not x^2+4x+4. Itachi right, and there are two solutions f(x)=x, and f(x)=x+2.
Astronavigator Pirx Hahahaha learn your (a+b)^2 identities again... you might be trolling 😄
@@prefabofficiel7842 Oops, I've made mistake ) Happens )
Yes is another solution
crystal clear
tl:dr; answer is f(x)=±x
Just take the derivative of both sides and simplify to get f'(x)=1. You need to assume that f is differentiable.
6:40 не розумію бо не знаю англійську напішіть будьласка по українські чи по російській.
Such an easy problem
That's not exactly easy.what about the level of the best students, decreasing as everybody thinks,the same or increasing
nice vid
I was 1 year old during this competition.
I think we may get the same result uf we applied the partial derivative over x and over y
thanks for the nice problem #mathtutorial
Maybe my mind is impaired but -x seems not to be a solution. One gets a contradiction when plugging it into the functional equation. So x seems to be the only solution.
If the function is continum we can easily prove that f(x)=x or f(x)=-x
Continuity alone isn't enough, consider f(x) = ±|x|
@@ichtusvis derivability works lol, but also without continuity if you're able to find f(x)^2=x^2, you shouldn't find hard the end of the proof.
Bueno , muy bueno Easy!!