Canadian Mathematical Olympiad | 2018 Q2

[@drpkmath12345]

I believe that f being a bijective is a key and the most important part to proceed for a solution on this question.

[@kevinmartincossiolozano8245]

@Adam Romanov Dude, if you are pointing errors in a video, it's logical to erase it or people would get confuse or could make same mistakes. Is a personal issue, he doesn't owe you any apologies nor to keep a video that is not worth having. (Even though I'm not completely aware of the situation). Please, don't make drama out of something that isn't necessary.

[@anushrao882]

Indeed it is.

[@drpkmath12345]

Anush Rao yes right! Very interesting to solve haha

[@jamesgoodman5102]

Some comments have asked about why the problem is posed over f: Q-->Q, instead of f: R-->R. The motivation is how difficult it is to solve f(x+y) = f(x) + f(y) (Cauchy's functional equation) over Q and R respectively. Simply put, Cauchy's functional equation has two classes of solutions: the nice, simple, linear ones and the difficult, complicated, nonlinear ones. If we restrict our scope to just Q, we're guaranteed to have only the nice, linear solutions. Loosely, since we can tidily move about through Q with just multiplication/division by integers, that extends to the solutions too, which are also tidy and linear over Q. More concretely, we can prove that f(qx) = q*f(x) for all x, for rational q, which, together with c = f(1), gives us f(q) = c*q, and therefore f is a linear function over Q. On the other hand, R has a much finer structure than Q and is microscopically vaster and more "filled out" where Q had "gaps" (i.e., R is a complete metric space). And, unlike in Q, multiplication/division by integers in R does *not* allow us to easily move about since we would miss out all the irrational numbers (which make up "most" of R, in the sense of measure). So because of that "finer detail" of R, nonlinear solutions *might* also exist. In fact, whether or not they *do* exist depends on whether our chosen logical toolbox would allow us to construct them, in particular, whether we can use the axiom of choice (AC). So the issue goes all the way down to the logical framework we're using to support our math. And what would these nonlinear functions look like? They would roughly just look like TV static or a cloud over the plane. Everywhere in the plane would have points making up the function, stippled arbitrarily close to one another or located arbitrarily far away in the plane. More technically, it would be dense and discontinuous everywhere in the plane. Likewise, the functions would also be indescribable since we would have to use AC to "pick" at least an uncountably infinite set of points in the function to form a basis. And good luck specifying the members of an uncountable set! By limiting the scope of the question to functions over Q, the question avoids those difficult, problematic solutions. Otherwise, the participants would need to be comfortable with manipulating some fairly advanced tools like AC that would be beyond what would typically be expected for an IMO (or Canadian MO). For more details, please refer to Cauchy's functional equation and see math.stackexchange question 423492 "Overview of basic facts about Cauchy functional equation".

[@abebuckingham8198]

I would also note that first establishing a finite number of values and then calculating other values based on them can at most give you a countable collection of solutions. This is because the set of finite length strings over some non-empty alphabet is countable. An immediate consequence is that such methods are insufficient to characterize real valued functional equations.

[@meiz1795]

It's one of the questions where as soon as you see it you immiedietaly know its going to be linear

[@hach1koko]

7:40 to get that result one can also plug in y=0 in 2f(x)+f(y)=f(2x+y)

[@warmpianist]

This is a very direct Cauchy's functional equation after f(2x)+f(y) = f(2x+y), and I believe you can directly state it for f: Q->Q.

[@factsverse9957]

Yep cauchy for f:Q -> Q works only for y = kx.

[@elfek009]

at this point you re proving the cauchy equation over the rationals which I think you don't need to as your competing in the math olympiad

[@AnlamK]

Congrats on the new camera angle.

[@erikr007]

It would be interesting to explore why they choose the domain of this problem to be just Q and not R.

[@sugarfrosted2005]

Red herring I'd guess.

[@sugarfrosted2005]

Apparently there is an overpowered theorem you can use if it's rational from the other comments.

[@andrewkarsten5268]

Because there are essentially an infinite number of solutions to this equation that are not “nice.” This proof he used only works because we’re dealing over the ration numbers, so adding, subtracting, multiplying, and dividing by integers is enough to move around through Q. That is not enough for all real numbers, and you get a lot of problematic things when it comes to that.

[@ericzhu6620]

that's because Cauchy's functional equation is evolved in the solution and it only works well under Q

[@georgemissailidis3160]

I just looked at the given equation and the answers immediately jumped out to me. It's pretty obvious actually. But the problem I had was proving rigorously that these were the only solutions. Thanks Michael for the outstanding proof.

[@pooydragon5398]

You can clear see that f(x) = x is a solution even before you start solving the question. Now all you have to do is check if it's the only solution or you can generalise it

[@kerimbendov5017]

15:40 we get f(x)=ax and a^2=1, so Square the both side of f(x)=ax to get f(x)^2=a^2*x^2 and from a^2=1 we get f(x)^2=x^2 and take the square root of both side and we get f(x)=±x so, for some x we have f(x)=x and for some x we have f(x)=-x you can't just say f(x)=x and f(x)=-x are the only solutions of the equation. What if for some x , f(x)=x and for some x , f(x)=-x ? You must give a contradiction that for some x , f(x)=x and for some x , f(x)=-x is not a solutuion to the equation. We can give a contradiction for this. We have assume that for some x , f(x)=x and for some y , f(y)=-y. Now we have f(x+y)=f(x)+f(y), from our assumtion, f(x+y)=x-y , and f(x+y) is equal to x+y or -x-y , so there are 2 cases, x-y=x+y or x-y=-x-y but both of them leads to the contradiction. And from here we can say f(x)=x and f(x)=-x are the only solutions to the equation.

[@wojciechwisniewski6180]

Love that guy's way of ending videos :D

[@zygoloid]

Interesting. The result is the same (but the proof is a little different) if we take f: Z -> Z, but if we take f: V -> V where V is any vector space over Q (such as R), then f satisfies this equation if and only if it is a linear involution. (That is: you can pick any basis for V over Q and then allow f to swap over any pairs of basis vectors, and you can choose to negate any of the remaining basis vectors. For example, the unique linear function for which f(1) = sqrt(2), f(sqrt(2)) = 1, and all other basis vectors map to themselves, satisfies the equation.)

[@roboto12345]

At the end is a Cauchy equation and I think that the bijection is sufficient to prove the claim

[@drpkmath12345]

Exactly!

[@klevap]

Since f and f-1 are symmetrical against y=x, equality f=f-1 immediately means that f=x or f=-x

[@mairisberzins8677]

Yet, it's not the only answer. f=x AND f=-x

[@klevap]

@@mairisberzins8677 yeah, f=-x is also symmetrical against y=x, i just missed it, so my logic actually gives both f=x and f=-x, and there is no other function which satisfies both conditions

[@TechToppers]

8:22 Well prepared students be like: *It's over!* This is Cauchy's Functional equation over Rationals.

[@digxx]

Once you have the involution and f(0)=0, then f(2x+y)=2f(x)+f(y) and you first set y=-x to get f(-x)=-f(x) and then set y=0 to get f(2x)=2f(x) and y=x to get f(3x)=3f(x). Inductively, y=(n-1)x gives f((n+1)x)=2f(x)+(n-1)f(x)=(n+1)f(x) i.e. f(nx)=nf(x) for all integers n, or f(x)=f(nx)/n. Letting x=m/n, this becomes f(m/n)=f(m)/n=f(1)*m/n.

[@Robert-jy9jm]

15:45 Why do we still have to show that the solutions satisfy the functional equation? Hasn't it already been proven that they satisfy all the necessary conditions? If that's not the case could you give a counterexample? Is there a functionalequation for which you could deduce a "false" solution? Great video! The explanation was so clear and easy to follow!

[@ericzhu6620]

it's always a necessary step in diophantine equations and functional equations since you won't know if anything wrong could happen, like there maybe some conditions that the equation could not hold but we didn't considered that when solving

[@Robert-jy9jm]

@@ericzhu6620 Could you give an example of such an equation? That would clarify it a lot for me.

[@adamnevraumont4027]

The logic was showing "any function f implies these things true". Take the same problem, add an extra constraint f(1)=7, and every step done remains logically valid ... until you check to ensure your f(x)=x and f(x)=-x, when it no longer obeys contraints. Similarly the constraint f(1)>0 would not prevent any of the logic shown except the final check. Up to the check, Mike showed implications; which restricted possible solutions. Showing the constrained results are actually solutions is not proven by what he did earlier.

[@andrewkarsten5268]

An example of where you may get false solutions is something like this; Find all solutions to x²−1=0 on [0,∞) Since x²−1=0⇒x²=1⇒x=-1 or x=1. Since -1 is not in the domain, it is a false solution. I have a really simple example, but the point is each implication only restricts the possible solutions, but the backwards is not necessarily true that every solution at the end solves the original question. Another example is finding local max and min in calculus questions. You find candidates by checking where the derivative is zero, but not every one of those candidates are actually maximal and minimal because they could be inflection points. In general, at the end of these deductions you only get candidate solutions, but they might not all be solutions. That’s why you always check they satisfy the original question.

[@nitayderei]

You could see that the function is linear transformation (above Q, for course) and really make your solution a lot shorter. Great video!

[@Miguel-xd7xp]

Can you explain it?

[@abebuckingham8198]

@@Miguel-xd7xp A linear function is one such that for vectors x,y f(ax+y)=af(x)+y for some scalar a. You can probably see that this function is linear but it's worth checking. The only linear functions over Q are of the form f(x)=cx for some constant c. However you still need to figure out that f(0)=0 to establish that linear functions satisfy the functional equation.

[@MyMathsAdventure]

One day I’ll be able to keep up and understand what’s going on 😂

[@tomatrix7525]

Keep going, you’ll get there no bother. I was in your position not so long ago, crazy how fast I improved.

[@tanmayroy6372]

Buy functional Equations ..bj venketchala

[@hussainfardan1577]

Easy explaination, organized that made a dificult problem sounds easy. Very very useful. Great job. Thank you very much.

[@pendronator]

Once you prove that f(f(x))=x, you can immediately show that f(x)=x is a solution by using the geometric interpretation of an inverse function (reflecting across y=x).

[@運慶-w3s]

How about f(x), f(x)=-x (|x|≦1) f(x)=x (|x|＞1) or f(x)=1/x (x≠0) f(x)=0 (x=0)

[@meiz1795]

Would you mind explaining how does inverse function help here?

[@mairisberzins8677]

@@meiz1795 Graph of an inverse function is the same function flipped about an axis that is at 45 degrees. or flipped around the graph of y=x. But since you flip something around the axis and claim that f(x) = f-1(x) which means that their values don't change, it indicates that the function itself lies on the axis of symmetry or that all of the points on said function lie on the same line after fliping. In other words: The only way to have a function not change after flipping it about some axis of symmetry is for it to BE the very axis or for it to be perpendicular to it. Since our axis of symmetry is f=x, it comes down to 2 answers: f=x and f=-x (where f=-x is perpendicular to f=x) It is a good way to "guess" the answer, but i don't think this counts as a proof.

[@icfj77]

all of this is valid also for differentiable f(x) from R to R

[@jölwritesmusic]

Wait couldn't we stop the solution at 5:12? Because if f is its own inverse than f can only be x or -x?

[@behnamashjari3003]

Prof Penn, I worked out y=x^x (x to the power of x) and found for x=0, y approaches 1. Also for x=1, y will be 1. For x=1/e=~0.3678, y will have its minimum which will be 0.6922. beyond x=1, y increases continuously until +infinity. Here's my problem and I need your help, perhaps a video: for x= -infinity, y=0. For all even negative integers, y>0. For all odd negative integers, y

[@franciscomorilla9559]

I know this comment is old but maybe this is still of help. Your problem is that in the complex numbers x^ab can be different from (x^a)^b or (x^b)^a, you have to take into account how you define the argument and the branch cut it generates (also you may introduce false solutions). For example, (-1)^(1/2) is + or - i, but if you do ((-1)^2)^(1/4) you get the 4th root of 1 which is usually taken to be 1 (though + - i are still 4th roots of unity). When taking noninteger exponents of negative numbers things can get ugly and arithmetic rules that we apply in the real numbers no longer hold, that's why your manipulation there is wrong: to answer your question about how does x^x behave if x

[@AlecBenzer]

Out of curiosity, are there functions over the reals that satisfy f(x+y) = f(x)+f(y) but aren't linear? -I guess you can easily define like f(x) = ax if x is rational and f(x) = 0 otherwise.- (nope nvm that doesn't hold lol). If f(x+y) = f(x) + f(y) and f is continuous, would f then need to be linear? I guess so, because every real is arbitrarily close to some rational?

[@duskhound2883]

en.wikipedia.org/wiki/Cauchy%27s_functional_equation There are non-linear solutions over the reals. However having it continuous does force it to be linear as a limiting sequence of rationals can be taken to any real number.

[@mushroomsteve]

Similarly, if f is continuous and f(x + y) = f(x)f(y), must f be an exponential function? I think there is a counterexample, but it's not easy to construct.

[@cenaalan5825]

how do you make that crossing out effect (with line over text) on your comment?

[@AlecBenzer]

@@cenaalan5825 Put dashes around the text you want to strikethrough.

[@duskhound2883]

@@mushroomsteve Having f continuous does force it to be all functions of the form a^x. If it isn't then there are non-linear solutions. This problem is equivalent to the Cauchy equation by taking logs both side. (First notice f is non negative for all inputs because f(x)=f(1/2x)^2>=0 and if f is 0 anywhere (Let's say at z then f(x)=f(x-z)f(z)=0 for all x) So f(x) = 0 for all x is or x is strictly positive everywhere. Hence log(f(x+y)) = log(f(x))+log(f(y)) Let g(x)=log(f(x)) Hence g(x+y)=g(x)+g(y) g is continuous if f is. Hence g(x) = mx (Cauchy functional) for some real m log(f(x)) = mx f(x)=(e^m)^x f(x)= a^x Another interesting one is f(xy)=f(x)f(y) with f continuous and R+ -> R+ implies f(x)=x^a (math.stackexchange.com/questions/43964/if-fxy-fxfy-then-show-that-fx-xt-for-some-t)

[@shafikbarah9273]

We can solve it in an easier way.. P(0,0) P(3f(0),3f(0)) yields f(0)=0 P(0,x) yields ff(x)=x P(f(x),0) yields f(2x)=2f(x) Finally P(f(x),f(y)) yields f(2x+y)=f(2x)+f(y) and its cauchy So f(x)=cx and we substitude to find that c=1,-1

[@師太滅絕]

is there functional equation involving trigonometry? D=For example, xf(x) + [sin x] f[f(x)] = 2

[@kriswillems5661]

I don't get it. He's claiming a linear function is a solution in Q (and N) and he proves this, and even calculates which linear function (he calculates a). But he never proves there are no other solutions. What am I missing?

[@loltroll78]

Other way round. He claims that solution is a linear function, which means that any possible solution is a linear function, but not the opposite way.

[@edindrevataj9284]

f: Q to Q, f(x+y)=f(x)+f(y) is a known equation, with all its solutions being of the form y=kx. Now k is to be determined.

[@kriswillems5661]

@@edindrevataj9284 Thank you. kzbin.info/www/bejne/jZ_KnXmkm8-dfNU

[@AlecBenzer]

He doesn't just assume that f(x+y)=f(x)+f(y) implies f(x)=kx though, he does actually prove this.

[@geometrydashmega238]

I also thought that before, but we misunderstood his proof. He is proving the opposite, that given f(x+y)=f(x)+f(y), with only this assumption and making several manipulations from rational numbers, he ends up with this f(p/q)=(p/q)*f(1), which must be the case for the function f(x)=ax evaluated at x=p/q (all our domain) and a=f(1).

[@sujitsivadanam]

Fun fact: if a function "f" satisfies the equation "f(f(x))=x" for all x in its domain, then "f" is what's called an involution.

[@СергейПупчин-ж7т]

What book would you recommend to start studying functional equations?

[@Wurfenkopf]

I see you didn't get any answers, so I'll try to hint. There are two completely different kinds of functional equations. The ones you study for scientific purposes are based on derivatives and imply that the solutions should have some degree of regularity. This other kind of functional equation that we see in the video is unique, and as far as I know is only used in competitions like the IMO. I can't recommend lectures for either one because I have never referred to books, but if you are interested in the second category, I suggest you check math olympiads websites.

[@eusterich3035]

Ur channel growth speaks how good ur videos are

[@Alan-zf2tt]

Unless I am very much mistaken then 9:37 holds for almost anything satisfying commutative multiplication. A tautology All very amateurishly speculative on my part and putting it naively as professional as my abilities allow: Principle of nested sub-spaces supra-spaces. - all have a common zero of various dimensions so what has solution in a subspace remains a solution in all of its sub-spaces and all of its supra-spaces. Note: I think there may be something topologically interesting going on there. As stated in problem collapsing a 2-space problem to a 1-space solution actually generates a topological thing in those spaces. That is tautological property holding in all supra-spaces holding those events. Interim conclusion: set X and Y= as some polynomials of any rank then the solution will also hold including for quotients reals but not sure about complex numbers or non-commutative algebras So it should work for spaces ℚ ℝ ℕ and ℤ maybe for ℂ but not sure about ℍ

[@DragonKidPlaysMC]

I recently encountered your discrete calculus video. It made me really interested in this subject. But I’m stuck on finding the anti-difference of ttrig functions like sinx for example. Can you make a video about that? It’s really interesting! Nice video as always!

[@othman31415]

It's "Canadian Mathematical Olympiad 2008" NOT 2018.

[@long8398]

Othman Slassi he even wrote that on his blackboard

[@tighemcasey7589]

Such a clear solution to a cool problem! Thanks Michael Penn!

[@sakanagakyoko]

Mine wasnt very rigorous, i derivated two times and got that f"(2f(x)+f(y))=0, f(x)≠0, f(y)≠0. That let me think that f is linear. Since i assumed 2f(x)+f(y) could give you any number. Then plugged the general linear f(x)=ax+b in the original ecuation and got that a=1 and b=0. Then f must be the identity function f(x)=x.

[@thaddeusolczyk5909]

That only works if the function is continuous. But this is only defined on Q so you have to show that there exists a continuous and differential function on R with is an extension of f. Well I can create an infinite number of functions on R that can be f on Q, just chose any number times the characteristic function of the irrationals, but are any iof them continuous?

[@vladimirkobarov7154]

How about f(x) = k (k is the constant value)

[@JamesD2957]

At 2:40 why does x=3f(0)?

[@050138]

It was very intuitive that the function has to be linear, as the first order function is equal to a second order function....

[@4fecta353]

Hi, could you please make a video on the fifth problem of the 2020 Canadian Mathematical Olympiad? As a competitor who couldn't solve it at all (0/7), I would love to see how you tackle it as you do in all your other videos! Here is the problem statement for your convenience: There are 19,998 people on a social media platform, where any pair of them may or may not be friends. For any group of 9,999 people, there are at least 9,999 pairs of them that are friends. What is the least number of friendships, that is, the least number of pairs of people that are friends, that must be among the 19, 998 people?

[@victorjatoba6050]

Hum, I did not solve this problem, but one common strat when dealing with symmetric relations on social media is to use matrices. You can think of this situation as being a 19,998 x 19,998 matrix with entries (i,j) = 1 if i is friend with j and 0 otherwise. This create a symmetric matrix, since (i,j) = (j,i). Also, the diagonal is 0, since you dont consider one being friend with themselves. The hypothesis then is that any 9,999 x 9,999 submatrix that contain part of the diagonal, must be symmetric and has at least 9,999 entries as 1 out of the (9,999^2 - 9,999)/2 possible non-zero entries; the (-9,999) is the zero diagonal entries and the divided by 2 is from the symmetry. The question: what is the MAXIMUM number n of entries equal to 1 on the upper diagonal such that one square submatrix 9,999 x 9,999 containing part of the diagonal doesn't contain at least 9,999 entries equal to 1 on its upper diagonal? The final answer will be n+1

[@myxail0]

thats a graph theory problem, read some book on those. My teacher recommended Introduction to grapht theory by Robin J Wilson

[@mrhatman675]

@@victorjatoba6050 I doubt that they wanted them to solve it with matrices

[@elyades2480]

This is the first of your videos where I managed to find the solution on my own. I got f(x) = x, but forgot f(x) = - x. I loved the video, thanks you professor.

[@podir47]

Keep the good work etc, but if you missed f(x)=-x , then you did NOT find THE solution on your own, as the question's phrasing states 'determine ALL functions' . What you found is a partial solution.

[@elyades2480]

@@podir47 right ! Well, at least I was on the right track !

[@newkid9807]

Dotan Perlstein you’re a buzzkill

[@podir47]

@@newkid9807 I'm sorry, but I tried to state my comment in a delicate and hopefully constructive manner. This is mathematics here, not an 'everyone is a winner' playground. If you share your mathematics achievements online, it is not too much to ask to straighten the facts out, for pedagogic purposes if not anything else

[@elyades2480]

Alright everyone, I wasn't offended by his comment. Constructive criticism is more important than convincing yourself that you got the solution !

[@yusufa.ratleh467]

Can we think about it in term of isomorphisims and homorphısims

[@zonico5826]

Oblivious student, with a very general question here: Why can we set a variable to one number, arrive at an identity using that resolution, but then conclude that said identity is true for all values of the aforementioned variable?

[@drdca8263]

If we have proven something is true, then it doesn’t matter how we did so. it is true, and we can use this however. If it is true that for all rational x any y, that f(f(x)+f(2y))=x+2y, then in particular it is true for if y=0, and so we get a statement from that which is about f and doesn’t mention y. We now have both that statement and the original statement. We can use them together and there is no problem. I suppose that there is probably some variation of “linear logic” in which the idea of being able to pick any value of x or y, but only once, could work?

[@Mutlauch]

At 7:35 you state that f is it's own inverse because it is the case on the left side, but on the right side you use different cobditions. Why must f in the second case also be it's own inverse? And for 2f(x) = 2x to be the case you have to assume linearity and therefore can not conclude it or am I mistaken?

[@kriswillems5661]

2x+y represent the domain of all rational numbers. By taking x=0 and y free or x=free and y=0 you can also go over the domain of all rational numbers (by changing x or y). So, the calculations on both sides regarding to f(x) or f(y) are valid for all rational numbers x or y. So, you can use conclusions from one side on the other side. Second question: He says f(2f(x))=2x at that point. He started out with the equation in the question.

[@stvp68]

Is that a door handle to the left of the board?

[@hocky-ham324-zg8zc]

Wait for around 4:09, if x=y=0 and x=y=3f(0) then couldn’t you have skipped all of the stuff you did because 3f(0) = 0, so f(0) must = 0?

[@duskhound2883]

We have f(3f(0)) is 0. So a function like f(x) = -1/3x+1 satisfies that. Theres another f wrapped around it so sadly we can't assume that.

[@geometrydashmega238]

x and y are just variables which can take values on the rationals. They are not unknowns from an equation, so it doesn't make sense to say 3f(0)=0 , you are just putting your variables equal to different values

[@imobusters4049]

But, Micheal sir, you did not check if f(a)=a for some a in the rationals and f(b)=-b for some b in the rationals happening at the same time.

[@danallan8526]

Could not the following argument take care of this case? if f(a) = a and f(b) = - b, f(a + b) = f(a) + f(b) but f(a) + f(b) = a - b however, f(a+b) can only be equal to either a + b or -(a + b). In either case, either a or b are 0 and if a or b is 0.

[@imobusters4049]

@@danallan8526 it will, but he must explain properly, as it is very important to do so

[@16sumo41]

Hey! Nice problem. When I tried solving it I used the fact that x=y implies f(3f(x)) = 3x and then proceeded with an ansatz that the equation has the form f(x) = Ax+B where A and B are rational coefficients. This implies that f(3f(x)) = ... = 3A^2+3AB+B = 3x and since the equation should be true for all x the coefficients and constants on the LHC should be equal to the coefficients and constants on the RHS, we get a system of equation which yields the same solution as you have, namely f(x) = x or f(x)=-x, which is pretty easy to check. Now, is there a problem with this solution? Is it okay to solve it using an anzats? Or is it an illegitimate solution?

[@cenaalan5825]

Why f(r+s) = f(r) + f(s) ??? consider function x^2 + 3 and f (1+4) is not equal f(1)+f(4) because 5^2+3 is not 1^2+3+4^2+3

[@factsverse9957]

For the particular function he's explaining, he has proven that the function indeed satisfies f(2x + y) = f(2x) + f(y) (this requires proving, I believe the video has shown the reasoning). Let 2x = m, then it fulfills f(m + y) = f(m) + f(y). This functional equation is Cauchy's Functional Equation (you can see Wikipedia). Since f:Q --> Q, the only solution is f(x) = kx. So no other functions fulfill. On your example, I believe it is assuming all functions fulfill. In fact, only linear functions (i.e. f(x) = kx) fulfill the equality over rationals, not all functions.

[@AlecBenzer]

Did any part of the argument at 11:30 depend on the fact that the rational was positive?

[@hoodedR]

Splitting the f(p/q) "p times" and "q times" don't really make sense when one of them is negative. Of course you can do it with a little bit of extra steps though

[@whythosenames]

if f:R->R would be continuous then it would work for R

[@skylardeslypere9909]

At 6:26, can't you now say that f is linear? And f(0) = 0 so you can just say f(x) = a.x with some rational a

[@The_Aleph_Null]

Because there aren't non linear functions that pass through the origin P(0,0)?

[@33220916614312811811]

I wonder why it matters that f goes from Q to Q. If f would go from R to R, what are the solutions for f then?

[@s.m.m99203]

It matters when you wanna prove f(ax)=af(x)

[@_Ytreza_]

@@s.m.m99203 It matters when you prove f(x) = ax (you need that f is continuous)

[@benjaminguarda2505]

at 4:56, why f (0) = 0, if x is now assumed not to be equal to y? how do you infer that value? Good video btw

[@raptor9514]

X and y here are just variables - if once we got that f(0)=0 it'll be true always

[@benjaminguarda2505]

@@raptor9514 you're right, thanks bro

[@rigardlopez6290]

Ahh...now i understand this point ...thank

[@mathssolverpoint6059]

It can be more easy by comparing of polynomial orders

[@saikat93ify]

Amazing video ! I have a doubt. How to prove that the function is always linear ?

[@Ennar]

The whole video is a proof that the function is linear.

[@sujals7108]

He proved it at around the end of the video

[@comingshoon2717]

Estoy problemas son muy típicos de olimpiadas 💪💪💪 buen trabajo, Dr

[@JM-us3fr]

Does this strategy generalize to f(ax+by)=cx+dy?

[@alejandrogil198]

What is the issue if you consider the real numbers as range and domain? Is the result incorrect or just harder to prove?

[@LIVIU2003S]

If the function is continuous the result is correct. You will find the proof for the real numbers if you google cauchy functional equation, it's a classic problem

[@David__U]

At 7:35 you say f is bijective, but at that point you had only shown that result in the special case of x=0, which is not the case here.

[@_Ytreza_]

He showed earlier that f(f(x)) = x for any rational x, that's what he is using here

[@s.m.m99203]

What if the domain and range are both the set of real numbers?

[@Taurwathwylth]

Then the Cauchy problem f(x+y) = f(x) + f(y) has pathological solutions that are not of the form f(x) = a*x

[@Ιωάννης-Ν.Ξανθόπουλος]

hello! I have a question. I solved it, but I used something else. I said that any polyonimic function of degree n when composed to itself will yield a polynomial of degree n^2. So I let y be equal to x and said that since the right hand side is a linear function the composition must yield a linear polynomial and the only way for this to happen is when f is of the form f(x)= ax +b . So I substituted that in and solved a system of a and b and found the same result. Is this way correct?

[@joshuastueck1899]

If you assume f is polynomial then a straightforward argument like this would work, but all we know for sure at the start is f is a function from Q to Q. That includes non-polynomials (there are uncountably many such functions, for example, which is more than polynomials). In order to prove f is polynomial would likely require similar steps to proving it is linear.

[@caesar_cipher]

Professor, I understand you prove f(x)=0, f is bijective and f(x+y)=f(x)+f(y) My question is how do u prove that f(x)=ax is the only possible family of function that can satisfy this ? Thanks

[@shubham-sc3jn]

13:12 Didn't he derive f(p/q)=p/q*f(1), only using the property f(x+y)=f(x)+f(y)? That proves that f(x)=ax

[@typeswitch]

He proved it by going through every possible case of the rational numbers. First, the natural numbers (just to get a feel for it). Then the positive rational numbers. Then the negative rational numbers. He proved it using the f(x+y) = f(x) + f(y) equation.

[@typeswitch]

Crucially, he didn't assume that it was f(x) = ax, instead, he showed that it must be f(x) = ax, because of the equation f(x+y) = f(x) + f(y).

[@raptor9514]

Aren't f(x)=× and f(x)=-x the only functions that are equal to their inverses? Then the proof would be much shorter

[@LIVIU2003S]

No. f(x)=k-x, g(x)=1/x, ...

[@Zondeman0]

Functions with asymptotes can also be symmetrical around y=x

[@raptor9514]

Okay, nevermind. When it's late one's mind doesn't work properly

[@Zondeman0]

Angel Mendez-Rivera You are right, thank you. I’ve read up on the involutive functions x This is a nice article about involutions eqworld.ipmnet.ru/en/education/wiener.pdf

[@PFYannik]

damn... After like 5 min I was like... well it must be unity! But how to proof it is the only solution? I too me unti 11:55 to see the other solution

[@gatoradeee]

This is really good. You might want to consider record on a tablet a la khan academy.

[@serkanbasatlk3322]

You could just say f is an odd function

[@김진우-l7l]

How do you know that solution is unique?

[@georgemissailidis3160]

It obeys the equation f(x)+f(y) = f(x+y) which, by Cauchy, was proven to only be obeyed by the function f=cx for some c (in the video's case, c must be rational since the morphism of f maps to and from the set of rationals).

[@AllanKobelansky]

Your videos are excellent. Thank you.

[@avrajitsarkar8563]

x=y would give f(f(3x))=3x so now u can conclude f^-1t=ft. and the only solution is x=+_y

[@pan_nekdo]

Nope. There are other "weird" functions that are their own inverses. For example: f(x)=-x/2 for x>=0 f(x)=-2x for x

[@itsomar6289]

In f(2x)+f(y)=f(2x+y) You can just put x=x/2 And boom you will get cauchy So f(x)=cx

[@arnaudvanuden9495]

The board says that the question is from 2008 and the title says 2018 which one is it?

[@garrethutchington1663]

Get a function

[@tomatrix7525]

Very good

[@PANKAJKUMAR-nq1qr]

Very Interesting functional equation!!

[@moslemasultana9388]

putting f(0)=c may make things look easier...

[@NavyBlueMan]

Don't you need to prove there are no other functions that satisfy the equation as well?

[@ilonachan]

He did. What he proved was that if a function satisfies the conditions, it can be written as f(x)=ax. And that even then, for it to be a solution, a needs to be ±1.

[@ilonachan]

In other words: If a function is not one of the solutions he found, it can not satisfy all the conditions we know f has.

[@jurjalin3690]

Can someone explain me why it s must that function to be linear? 10:30

[@IkeMaster11]

Couldn't you just make a system of equations assuming the solution is of the form f(q)=aq+b of the following form: 2a^2=1 a^2=1 b(3a+1)=0

[@KostasOreopoulos]

No.because that way you assume the form of the equation. Here he proves a general point, that every linear function is of type ax. I think in the competition you can assume that, but it is always nice to see the proof of something so “obvious”

[@KostasOreopoulos]

Also ax+b is not linear since a(x+y) + b !== ax+b + ay+b

[@oliverherskovits7927]

you can't assume f is in that form without proving it first

[@geometrydashmega238]

@@KostasOreopoulos Speaking of obvious things, I don't see how he proved that every linear function is of that type. I see he proved that if f(x)=ax then f is linear. What about the other way? How did he prove that if f is linear then f is of the form f(x)=ax?

[@mushroomsteve]

@@geometrydashmega238 f linear implies f(ax + y) = af(x) + f(y). If we calculate [f(x+h)-f(x)] / h, we have [f(x+h)-f(x)]/h = [f(x)+f(h)-f(x)]/h = (1/h)f(h) = f(1). Therefore, [f(x+h)-f(x)] / h is constant, which implies f has the form f(x) = ax + b. However, by the comment from @Kostas Oreopoulous, b must equal zero. Therefore, f(x) = ax for some real number a.

[@ImaginaryMdA]

I found out f was linear before I figured out that f(0) = 0... lol. It was a pain, especially compared to this solution!

[@Blaze7439]

8:43 can someone explain this step to me? I dont understand. what if f(x) = x^2, then f(r+s) =! f(r) + f(s). where does he get the proof from?

[@ogasdiaz]

2f(x) = f(2x) applied to f(2f(x) + f(y)) = 2x + y gives us f(f(2x) + f(y) = 2x + y let r = 2x and s = y f(f(r) + f(s)) = r + s then f(f(f(r) + f(s))) = f(r + s) but f(f(x)) = x so: f(r) + f(s) = f(r + s) f is associative over the rationals.

[@Supernova799]

A very great video. !! Hey can u make a video on Ramanujan Master theorem ? U can get the proof on academia.edu .

[@vnknovn]

Pretty nice video

[@giabao576]

very helpful! thanks!

[@ghanshamchandel1854]

Can't we partial differentiate top right equation in 9:30 to prove for all real numbers?

[@ElchiKing]

Then you'd need to assume f to be continuous (even differentiable). That's not a given.

[@sergiomanuel2206]

F(X) = X

[@devarunbhattacharya8438]

somebody tell me why am i wrong? domain and range of f are the same. So we can conclude 2f(x)+f(y)=2x+y ; f(x)=x ; f (y)=y so f =x=y.

[@rigardlopez6290]

Great!!!

[@SolutionsForMath]

Isomorphism!

[@uberjhonny9758]

❤❤❤❤❤❤❤

[@josephhajj1570]

But plz don't use side camera

[@sripuranam]

Don’t you have to prove that this is both necessary and sufficient? I think you only proved that your solution is a sufficient condition and not necessary.

[@KostasOreopoulos]

No. He proves that being linear is both sufficient and necessary since all steps are . What follows is just proving that every function that has those properties is of type ax with a==1

[@sripuranam]

Thank you for the replies, you are right, all steps are . I had to work it out on my own to figure out what I was missing.

[@emanuellandeholm5657]

Me before watching the video: identity, duh! Me after watching the video: f(x) = -x, yeah I kind of missed that. Also I never proved anything. :(

[@eusterich3035]

Ur channel growth speaks how good ur videos are