Hey Presh Talwalkar, very nice solution! I find your video extremely clear and well presented, I should better learn some presentation skill from you. BTW, thank you for mentioning my channel, I really appreciate it. Cheers!
Thanks for the feedback! I really thank you for your video--I didn't solve the problem, and your video helped me understand it. People often request videos for harder problems, so I'm sure they will enjoy your videos!
@@אביב-ת7ל MindYourDecisions has kindly share the (early stage) video with me last week and that's how I got the chance to see the video a bit earlier than you guys. Cheers!
the only thing I understood about this math video is that if there are 6 problems, and each one is worth 7 points, that there is a total of 42 possible points.
I actually paused the video and solved it myself. Took me a few hours and sheets of paper but it's so damn satisfying when everything comes together. I would argue mathematicians live for this moment when their proof is complete. It's so satisfying. My proof that f has to be linear was different and a little less elegant though. Still a very fun problem, but definitely not easy for me.
Same with the chemistry equivalent, I think I got some points from the most basic questions, which were pretty much the hard questions that I'd get in my school's competition
@@sarthakpatnaik65 yeah in my country there are school level competition, then city level, then district level, then the whole country level. Iirc you need to be in top 5 at the country competition to be able to join
IDK where on earth you are but that is a very modern American disposition. If you feel like a dummy where you are, come to America where you will fit right in LOL
Yeah, he did brush over that a bit... The key is to recognize that the integers under addition have the property of commutativity. All that means is that the order in which we add integers doesn’t matter. Formally, we may write (a+b) = (b+a) such that a, b are integers. Since 1 and b are integers, we can say that (1+b) = (b+1)
Its like that all the time, with these equations of functions, right? I never liked this section of maths when i studied for math competitions. Arbitrary and unintuitive, with little application.
worth mentioning that though IMO contestants may have found this comparatively easy, they are constantly training and facing problems of this nature and these types of techniques. most of them will see this type of problem and have a lot to fall back on almost immediately. source: am an IMO silver medalist
Same math olym contestant here too!! Actually the step of putting a = 0 is very common in solving function questions. But I dun think I can think of this immediately if I haven't faced this type of questions before🤣🤣so it's all about training
as someone who’s been doing a lot of linear algebra lately, i immediately picked up on the fact that this equation very closely resembled the definition of linearity, (i.e. f(a) + f(b) = f(a+b) and f(ca) = cf(a)). then assuming f was linear, it was very easy to solve from there. i was just going off my intuition though, so your solution was probably much more rigorous and would certainly stand up better under scrutiny!
You can observe by setting a=0, then b=0, that this is a linear function, which maps values f(a) onto 2f(a)). So at least for any z in the image of f, f(z) is simply 2z. You can then do a simple demonstration that f is surjective and you're done. I don't know why you'd do it as complicated as in the video.
To people who feel bad for not finding it while its supposed to be "easy". Consider this, the people who do these olympiads have seen this type of question before. It's very often the same type of questions, and with the same way of solving them. You have to train for these olympiads and this type of question is a pretty common one
I don't think so. There's a way to solve this problem in an easy way that just requires you to understand the mathematics, not memorize similar problems. Hint: the solution I'm talking about requires you to interpret the equation geometrically. If you can imagine the geometry of this equation, it becomes obvious what the solutions are.
@@peezieforestem5078 It's definitely a matter of having solved similar problems before. I used to compete in olympiads myself, this type of question is very common, it even appeared on my entrance exam. Typically the way to solve this is through substitution as in the video, that's why it's considered easy. I'd love to hear your method of solving it, even with your hint I'm not sure what your method is.
@@rayyansohaib8238 Elaborate too much would give my solution away. I'll give you some stepping stones: 1) Consider what a variable looks like geometrically. Let's say x - what does it describe? 2) Realize that what we call the variable doesn't matter. We can call it a, b or x, it's all the same. 3) Once you have completed point 1, consider what a sum of 2 variables looks like geometrically. Let's say (x + y), or (a + b) - the name doesn't matter, as established in point 2. Perhaps recall complex numbers. 4) Consider that a function can be viewed as a mapping of points to a different set of points. Once again, imagine the geometric meaning of this. 5) Consider which type of transformation multiplication by 2 is, from the point of geometry. What does multiplying by 2 actually does to the points? 6) The composition of 2 functions is just applying the mapping 2 times. 7) Finally, realize that equality in geometric terms means we have the same set of points, or the same geometric object. Voila! If you understand all these steps, you should be able to formulate the problem in geometric terms: "Which mapping, when applied 2 times to a geometric meaning of (a + b), results in the same set of points as (this mapping applied to the geometric meaning of a variable scaled 2 times) + (geometric meaning of the second variable with the same mapping, scaled 2 times)?" That might've sounded confusing, but that's because I'm not giving you the answers. If you work through the steps, the problem will simplify, and the solution should become obvious and straightforward. Also, geometric concepts are hard to put into words.
@@pb9405that's true, these are pretty standard questions in entrance exams and Olympiads. They are always taught to be solved in exactly the way it was described in video, so I can bet everyone in the Olympiad knew the "trick". It was easy.
my prof once said: "when encountering a problem, first you wonder about the problem itself. the question itself - because you dont understand it. and you look at it. suddendly, you understand the question and you doubt it: is that true? is that even possible? and you look closer and try things out until you have a solid understanding on whats going on. and then comes the hardest part: as "clear" as the solution is to you now, you need to write it down in a way that anybody else can both understand the logic and also see that you can prove every step."
compared to what maths has been like for me in high school this is real maths, the maths at school feels like it lacks the aspect of researching and exploring solutions when I watch your videos I learn how to approach problems, but I'm also stimulated
@Michael Darrow he is right. Come to Turkey and take an exam for university entrance, then see all math questions. We solve harder ones and the time is not 4.5 hours, just 1 or 2 minutes for each question..
This problem was way too easy,thats why he uploaded it,for an imo problem way too easy,in my country we have this type of problem at the city level olimpiad
I remember checking out the 2019 IMO problems after they released. I was able to solve this one in 5 minutes. I was surprised because usually I'm not able to solve any IMO problems at all. If you are already into doing olympiad problems, you will see that this one really isn't that hard at all. This type of problem is called a "functional equation" and a common strategy for these problems is to substitute numbers in to get identities about the function and then continue from there, which is what this solution does (plugging in 0 and 1)
It's never occurred to me to substitute only one value though. I would fail this because I would plug in 0 or 1 for both A and B. Giving only one a value just doesn't come naturally.
@@SomeRandomDude821 My friend did these math competitions for years in Canada (he was the Canadian high school co-champion ..). He said after a few years, they just had so much experience from thousands of problems .. I mean yes you and me both, maybe spend a whole hour poking around and not think to try leaving one variable open and plugging in just one value. The "trained athletes" I guess would usually have those ideas.
Me doing: F(2a)+2f(b)=f(f(a+b) =》2af+2bf=f(af+bf) =》2af+2bf is not equal to af^2+ bf^2 Thought this is the answer and was thinking why we need so much time for these question..... After video ... Wait...what is that(surprised pikachu face)
You make the fact that an IMO aspirant considers these easy as something very surprising, but the fact is they've seen hundreds of problems exactly like this and know exactly what to do to solve it, however, a problem like this would be hard for any IMO aspirant if they had never seen functional equations before
Exactly right. With a lot of experience, it's .. really not the same "problem" at all. It's a little bit like situations in chess. A highly experienced player may just walk up to the table and (in a very short amount of time) start making some very good suggestions, or even "the" move (if it's that sort of thing). Less experienced guys are just sitting there, boggling at him.
Exactly, this actually is a generic problem in IMO, once you have done one (or have been shown how to) all others follow a similar approach... Hard stuff is the stuff that you have never encountered before...
@@gabrielbarrantes6946 this, this was why something like 2011 p2 was so hard for people who were well trained in other areas also it's like the easiest imo problem in like the last 10 years so that also helps
I'm an engineer, and this is based on my basic math knowledge... i think my way of approaching, which i call "the lazy bastard" way, is at least interesting because it's done with as little math as possible, but requires you knowing math to some extent to pull it out. I usually did stuff this way in college and it helped me greatly, so i thought i'd share it. My thought process was like this: First, because we have integer to integer, we discard a lot of possibilities before we even start (say, it can't be trigs). Then, let's see if f(x)=0 could be a valid answer (which....should be at least assumed at first and tested, because if that is or is not an answer has some implications). And it is! So... we at least have SOME points because we say f(x)=0 is an answer. Then...if i'm being asked "all functions"there might be a constant value involved... And, if we were to assume that there IS a constant value being added on f(x), then... it's pretty straightforward that said f(x) should be linear, as with any non linear function, there are some values that would make it not true..... This is because, if we were to evaluate f(f(x)) with a and b being 0, the left side would be 3 times the constant, always , and the right side would depend on the value of the constant because of the double evaluation. For any non linear function, there can't be a constant added on the function, so IF there is a non linear function that satisfies that answer, it would have no constants. Now...let's see all the linear functions than satisfy our problem. We have that f(x) is some form of mx+n. If n can be any number, then let's say n=0. m(2a)+2m(b)=2m(2m(a+b))----dividing both sides by 2m and cleaning it a bit------>a+b=2m(a+b), so m=0, and this fits with the information we already know, that f(x)=0 is an answer. Now.... since this is (well, should be) valid for all a and b, let's do it with a=0 and b=0: 0+n+2(n+0)=m(0+n)+n let's clean this a bit: 3n=n(m+1) let's divide both sides by n: 3=m+1 so... m=2 and n is any integer. If we test it with m=2, as shown in the video, leave us with equalities on both cases. We have f(x)=0 and f(x)=2x+n, with n being any number in Z as our current answers. Now, is there any non linear function that satisfies the equality? I would say it CAN'T be. This is because, of the fact there is no way f(f(x)) has the same exponent (grade?) than f(x) other than f(x) being linear (english is not my first language, i don't know if exponent or grade are the correct terms in english). Can the function be something like f(x)=Y^x. with Y being any number in Z? It can't be, because if a=0 and b=0, we have Y^0 +2*Y^0=Y^(Y^0) ----> 1+2=Y. Now, if b=0 and Y=3, 3^2a+2 != 3^(3^a), so it would not satisfy all possible values of a and b. I would claim that it HAS to be linear because of the f(f(x)) and the fact there are defined constants (the 2s) and there is Z to Z stated We can conclude that...the only possible answers should be f(x)=0 and f(x)=2m+n with n in Z. Notice how little algebra was used , and when used, it was basic substitution? This is what i mean, having some concepts more or less clear can lead skip some steps and reach the answer without much work.
The same approach was taken by me too. Analysis of a question before working it out helps a great deal. I believe that's why engineers are the world problem solvers. - by an engineering aspirant.
Your way of figuring out that it has to be a linear relation is interesting to intuitively find the shape of the final answer, but I'm not convinced it's a proof. To me it looked like f has do be affine because it looks a lot like the relation f(a+b) = f(a) + f(b), so I started guessing and saw that f(x)=2x is a solution. Next I would have tried f(x)+n. I continued watching the video from there so I'm not sure how my proof that it was the only type of solution would look like. I think the completeness proof is the most difficult part of this exercice.
"i would say" is not sufficient, you need proof to get full points your way of thinking is a good start to wrap your head around the question but it isn't a full solution
I agree that the way I approached its not sufficient proof, but I'll also would like to point out that it's based on what I remember from my calc and algebra classes 7 years or so ago... And I was mostly showcasing an approach that could be build into something more solid with a more fresh knowledge... I mean, the things I didn't state were the things I remember less and as you see, I did as little math as possible. I think the fact that the function doesn't change the set when you do f(f(x)) might lead to some of proof that the transformation has to be linear, but my career doesn't elaborate on that much and I don't remember much. Also, that fact at least makes easy to proof that the final function can only have a constant if it is linear, and the little of sets I remember made me think I can just state that some family of functions don't go from Z to Z, but I don't know if I have to prove theorems or properties in these kind of events. I also think that this conclusion can lead to proof that the equality can only be satisfied with a linear function , with some theorem or something I don't know. Even if that is not true, I would say (again, not sure) if you don't know how to prove my previous thoughts, you would probably be able to check all possible family of f(x) so that it goes from Z to Z, and prove that que equality won't be true with all values of a and b. I was a bit careful on not stating what I was not mostly sure about, and well, even with my rusty math knowledge I did reach the correct answer, the exact proof escapes me but to be fair, anyone in these events would probably have prepared in some form and would at least have fresh knowledge and know a bit more of the rules and expectations of a competition, so think I at least did decent enough. I'll say that I only dare to claim that my approach is decently intuitive and uses as little math knowledge as possible, keeping things simple, which I believe it's at least a decent way of approaching a problem : how can I solve this as best as I can with as little work as I can? Math (and many subjects, in fact) is abstract, and you don't need to do pages of algebra if you understand some concepts and apply them. Lastly, I would like to say that I thought all of that in like 2-3 minutes tops (I did the calcs and algebra in my mind), and I don't consider my rusty and mostly forgotten math knowledge to be that good because I honestly only remember some basics, and I did reach the answer before watching it... So... I only though, based on general comments, that this would be a decent example, but I won't be brazen enough to claim it is correct.
Also, I did showed some minor "proof" as to why I though i discard some non linear possibilities, I would dare claim that my little and final thought process of showing there are no Non linear expression that satisfy all the conditions is at least somewhat valid? I think showing that there are values of a and b that wouldn't satisfy the expression, based on an assumption that the functions has a certain form is called something like proof of contradiction, or is at least some form of proof? Please do enlighten me if I'm wrong, I'm just checking subjects to read about again and have yet to start so any insight would be greatly appreciated. Thanks to all of you for the feedback, and it fill my heart my joy a would be engineer found that helpful :)
just think: kids are solving problems like these, while there are adults in their 40s struggling to create secure passwords for their accounts that isnt some combination of password and 123
Btw, a cool tip you can get from this video is to always try f(x) = 0 before continuing. I couldn't find any student answer, but I'm almost sure most of the 40% missed f(x) = 0 since it seems so obvious (especially when there's no constant numbers)
The question has infinite answers. n can be any number belonging to Z covering all the possibilities. It arises from the fact that you cant possibly write all the answers. You just write an equation for all the answers.
During the analysis it was made clear that on the right hand side was the differences between points while the left hand side was constant. This means the function has to be a linear one. In order to work with a set containing all linear functions one can simply use the general expression of such function: f (x)=mx+n (where m is the rise/run and n is the shift on the y-axis, meaning the function intersects the y-axis at (0,n).) That's how n entered the original equation.
Noticing the arithmetic progression is the real magic. I tried solving for a=b, for a=0, for b=0, and for both being zero. Had a bunch of equations but nothing I could really use to get to a linear progression.
@@honorinemunezero6866 for that u need to know what is meaning of 'arithmetic progression'. Arithmetic progression means a sequence of numbers which have a common difference. So you can get consecutive terms of ap by adding that common difference to each consecutive term. The general term of an ap is given by (T= a+nd) where a is first term of the sequence and d the common difference, which is a linear equation.
Many years ago I scored 98% on the Oxford Uni entrance exam. The year before, strong candidates were scoring 30%. It is very difficult to get the level of questions right for bright and insanely well prepared candidates. If a paper has questions just a bit too similar to precedents from previous years, then many candidates will get very high scores; and similarly if they are too hard, then even gifted mathematicians (which I am not) will struggle.
Well put. I used to know these mathlete types. They spent their entire lunch hours and after hours doing math problems and puzzles as well as preparing for contests. They were well primed.
Even though I'm a college freshman, this is really not that hard question. We've just begun learning Calculus and Linear Algebra, but have learned way harder things and had to solve harder questions. Edit. Yes, you learn something so that you can do it proficiently and effectively, not 'wasting' time with it.
In school math, had lots of problems finding numbers to satisfy a set of specified equations, but not for finding functions that would work for a specified set of numbers, It becomes more of a logic problem, needing ad hoc creative insights, than a math problem requiring standard techniques. That makes for the most interesting kind of problem. Thanks,
@@mateapaparisto1173 An arithmetic progression is a succession of numbers where *any* two consecutive numbers will always have the same difference or 'distance' between each other. Since on the left side we have two functions evaluated at two consecutive points, and that is equal to a constant value, we determine it must be an arithmetic progression. Example: 2, 4, 6, 8... is an arithmetic progression. You can take any two consecutive terms and the the difference will always be two. That means we can rewrite the whole succession as: 2, 2+2, 2+2+2 = 2, 2 + 2, 2 + 4, 2 + 6... => nth term = first term + d(ifference)n-1 times.
This was actually one of the few times I really paused and tried it for myself. It turned out to be not too difficult, but probably I just was lucky: We first note that f(x)=0 is the trivial solution and remember that for later. For other possible solutions, we assume f(x) being nonzero and consider the following: Basically we're looking for a function that satisfies a certain condition over the two-dimensional (a,b) in Z². Therefore, it also must satisfy it in any subset of Z². Like in the video, I first set a = 0 and b being arbitrary. That results in f(0)+2 f(b) = f(f(b)). Substitute x=f(b) => f(0)+2x = f(x) [note that this requires that f(b) is not 0 for all b, otherwise x is also 0 and we gain nothing]. Therefore, f must have the form f(x)=2x+c. To find c, make a arbitrary again, plug in the form of f in the original RHS and LHS, simplify and compare: f(2a)+2f(b)=4a+4b+3c and f(f(a+b))=2(2(a+b)+c)+c = 4a+4b+3c. Therefore, any value of c in Z² will do.Concluding, f has the form of either f(x)=0 or f(x)=2x+c with any integer c.
Your answer has a flaw. You proved that if x is a value of f then f(x) is given by some formula. So you showed that f(x) is given by this formula for x big enough and of some particular values. What you should do next is come back to the original equation and calculate f from there (using the received formula in the right hand side).
When it comes to something that you ve seen before you will eventually find a way to it it's just a question of time but if it's something like this trust me the idea of trying with numbers to find out the linear equation is actually genius level and unless you re used to exercices like these which are particularly rare I teach maths btw
@@goissilva you dont use this in school lol. this is for mathematical geniuses. theres only like 3-4-5 people per country that participate to this competition yearly.
Thats amazing! I havent done maths in so long now but very good explanation that I could follow. I think the key to solve this problem is the substitution first, and the realisation of the arithmetic progression function, which most of us won't notice.
@@patricksalhany8787 I don't think this is a particularly beautiful functional equation. (of course, my opinion - since this is a fairly routine problem). If you want to see more functional equations, there's lots of problems and suitable collections on the Art of Problem Solving forums.
This is how I solved for 2x + C : if a = b then f(2b) + 2f(a) = f(f(2a)) = f(2a) + 2f(b) meaning that f(2a) = 2f(a) in this case (this already cancels out a bunch of functions, namely anything with powers, leaving linear). It can then be written that f(4a) = f(f(2a)) you can take out a function leaving 4a = f(2a) meaning the function has to be f(x)= 2x, maybe if I plugged in some more numbers I would have gotten that f(x) also can be 0.
I was studying for maths olympics when I was 14 but realized it was pointless because studying maths a lot means less for other classes so I stopped and focused on all of my classes. Olympics might make sense for some countries where you can get in a collage by mentioning this or just doing sports such as usa but in mine, we have to learn all classes and take an exam to get in a uni. And olympics help only if you get important medals at maths which is really hard to get.
Cool Dude where are you from ? I didn’t do integrals until year 11 and certainly they were more simple than this... I went on to do medicine and have quite few practical uses for integrals but still this problem tripped me up a bit after doing advanced mathematics before college/university.
Cool Dude This is 12 years old math btw... What’s hard is having the math confidence to find the process in limited time, and that is more complicated to have
When I saw this problem, my first assumption was: "The solution has to be linear". Just because (a+b) is in the argument of the right side, and separated on the left side.
You still need to show that it has to be linear though. A handwavey argument like that would not get you far (other than as a starting point for what to try).
At 2:57, finish it off by rewriting f(0) + 2f(b) = f(f(b)) into C + 2x = f(x) by introducing x=f(b) and it's basically done: f(x)=2x+C. Also, the trivial solution f(x)=0 should always be your first ansatz.
Now I got it. f(0)+2f(x) = f(f(x)), That means it's true for any integer x Let, x = b+1, b is an integer So b+1 also an integer. f(0)+2f(b+1) = f(f(b+1))
Even if I am studying maths at college I am often so oaf, but you made it so clear ! I can’t help but feel proud of myself although it is just thanks to you
I tried to watch this video a couple of years ago, I didn't understand it at all. That goes for your other videos as well. But today I did get it, And I've been watching some others. Oh joy of understanding, greater than that of imagining or feeling!
@@drewfreese4707 Chinese is 🤮🤮 compared to other oriental/asian languages. find a new word? good luck finding its pronunciation since there's no alphabet and radicals are useless 99% of the time, not to mention the fact that there's 2 versions of it and one of them eliminates radicals entirely lol ALSO THERE'S NO SPACING AND SOMETIMES YOU CAN'T TELL WHEN AN UNFAMILILAR WORD IN A SENTENCE STARTS OR ENDS (though this applies to other Asian languages in general, too) 😭 sorry for rambling but i just hate the chinese language with a burning passion
@@nerd2544 don’t feel bad. Chinese is my native language and I went overseas for like 2 decades and now I’ve forgotten how to write half the words, not even counting the ones i never remembered how to write before.
Hey Presh Talwalkar, I came here to watch this video from your post on RedPig's channel, and your proof is totally valid, since there is a lemma stating that if some arithmetically progressing numbers, with their functions increasing too, then the function must be a linear function. Nice solution. I had the same solution as RedPig, too bad I got stuck at Nationals this year. ( Not from USA)
Thanks for the comment. These problems are challenging for me, and I appreciate feedback. Plus, it's great to know people are working on math during the summer months!
@@MindYourDecisions No problem. Must say, indeed this solution was better than most solutions. I haven't seen problems in IMO in a while where the sum or product of functions of two variables is a constant. Nice for pointing it out
@@rodwayworkor9202 That is partially due to the fact that this is the easiest IMO algebra in a long time, in my opinion - usually, there's a lot more work to be substantiated before arriving at a such a powerful condition. (though here the 'constant term' sum is presented very explicitly)
Hey, I'm training for mathematical olympiads of my country, National Math Olympiad is totally easy, so I actually train for the internationals, so I just wanted to ask you if you know more youtube channels that solve this kind of problems. I hope I get to the IMO next year. And yes, sorry if I make some mistakes in my english, I speak spanish (it would be very helpful is you correct me if I said something wrong)
One further way to know f(x) has linear shape, is to differentiate the whole equation once by a and once by b. The right term is the same on both equations, so you can equate the left terms and get f'(2a) = f'(b). This can only be true if f'(x)=const, so f(x) must be linear.
Very nice. This is the type of problem that Osman Nal often shows on his channel. There's always some kind of obscure manipulation that you need to be able to "see".
You have to find the funtions ( f(x) ) for which this equation is true. f(x) is the general representation of an unknown/unspecified function that is used to state this equation. a and b are just random variables, like x, that replace the x in this unspecified function. For example f(x) can be f(x) = x + 1. If you place in x=2 you write f(2) = 2 + 1 = 3. Like that f(2a) would he f(2a) = 2a + 1. However, you have to find the functions for which this left side of the equation f(2a) + 2f(b) is equal to f(f(a+b)), which is only true in certain forms of function f(x)
@@ivovandeboom6767 still lost on the function of a function, never mind, no need to reply, anyway what happend to the idea that X=1 as fitting in the first function found ?
So I mostly managed to solve it, but only by making some assumptions and I couldn’t determine if my two solutions were the only solutions, showing it had to be an arithmetic progression was amazing.
Solving this problem by progression method was very good. I realised the function to be linear as in LHS f(x) is present while in RHS f(f(x)) so composition of only a linear function will result in a function having same degree or a constant function. Anyways, your method was more thought provoking !
4:32 Here's my intuition: (f(2)-f(0))/2 is the slope of the secant line from x=0 to 2, while f(x+1)-f(x) is that from x to x+1, since both are the same for all x in Z, that means the function has a const. slope. So we can set f '(x) = m, and integrate to get our solution: f(x) = mx+c.
I was learning a lot of competition maths this summer (almost everything I know). And I can tell you that I'm able to easy understand this solution. I'm sure that before personal training I couldn't be able able to understand it. So, keep study, work hard - push your limits!
I can't believe I solved it on my own. I'm preparing for JEE Advanced and every now and then I try solving such questions, not only for experience but also for my own love for Maths. I was so happy to solve it under like 3-4 minutes and I believe that Maths is all about rational approach rather than cramming all those bookish data. Self-approaches unlock incredible brain potential which is hidden in all of us. Maths is truly a gem of an art ❤
Just awesome Presh Talwalkar, do bringing such videos among us, it took my 2days to solve this problem, untill I didn't reach to apex , but i have solve 80% , I have cracked RMO, now I am pursuing B. Sc maths honours,, 🤙🤙👍👍❤❤
I am struggling with the assumption at 5:38 that 2m=m^2 and 3n = mn+n. I can't figure out why that has to me true. It seems like it's one possible case but how do we know there aren't other possibilities?
He's equating the co-efficients on both sides, it's a fairly helpful and common trick. If you were to expand you'd notice there is no other way get those mixture of terms on both sides due to the + sign in the middle. Therefore it's completely reasonable to equate them
@@wakingfromslumber9555 Your the one who also cant do it? So what are you doing? You cant do it yourself? And dont start saying „SaYs ThE RoBlOx PrOfIlE“ Edit: You probably will you son of a OOF
at the beginning if you let Fx be mx+n and do the rest which is putting the values in the equation you get 2m=msquared so either m is zero or 2 so you don't have to do all the changing values stuff. if you think its going to take long no it took like 3 minutes. sorry for my bed england its not my first language
i honestly dont understand how this is a ''Insanely Hard Problem'' im not kidding i could solve this 8th grade im not flexing btw this is seriously not that hard
Here is my attempt at an explanation: 1st) What is a function (from the integers to the integers)? -> You can think of functions as "mapping" or "connecting with a directed line" every integer with some other (no necessarily different) integer. -- Examples are f(n) = n for all integers n, or f(2*n) = n and f(2*n+1) = 0 for all integers n. ----- In the first example you would "map" 0 to 0, 4 to 4, 103 to 103, -20 to -20 and so on ----- In the second example you would "map" 0 to 0, 3 to 0, 8 to 4, -100 to -50, -2021 to 0, and so on. Functions can describe any sort of mapping. The most important bit here is that while *every number maps to ONLY ONE number*, it is also true that *two different numbers ARE ALLOWED to map to the same number* (as is the case in the second example) 2nd) How do you even begin to solve a functional equation like the one in the problem? -> Here one way to think about this is that you are "given a property that holds for all integers" and your goal is to "narrow down what the function can be" Q: How do you "narrow down what the function can be"? A: You do this by plugging in specific values in the function, noticing what properties must hold true once you plug stuff in, and trying to come up with conclusions from there. So you can think of it as: 1) you give me a rule the function must follow for all integers 2) I plug in specific values and INFER what also must be true for such function 3) I keep plugging stuff in and continue doing inference until I've forced the function to be described in a certain way 4) I prove that this function which satisfies all the equations I got from my inferences and plugged in numbers, satisfies the functional equation in general I'll try to make a video about this problem in the next month and hopefully explain the logic a bit better there
Its easy because its normal and participants have seen many many similar problems Difficult ones are those that need creativity and real skill to solve.
@@mironemiron2454 IMO 2018 P1 isn't that easy at all, I'm quite good at geometry I think but I still spent 30 minutes on that problem. Maybe you meant IMO 2017 P1?
@@mironemiron2454 P6 IMO 2011 is also just angle chasing but it's quite hard I think. I have not seen a simple solution P1 2018, but I guess it was considered easy because it has a lot of solution.
I solved by Using f(2) + 2f(x) = f(0) +2(x+1) -->(explained in vid) And then put the values of x as 1,2,3 And found that f(0),f(1),f(2)... Were in ap So fx = nx+k And solved to find fx = 2x + k
Same here. I considered a very general mapping of Z in itself as a linear form f(n)=mn+k and found easely the solutions. But unfortunately then I thought that a integer power could also teasform an integer in another integer and that the most general form should be f(n) = n^r+mn+k (r, m and k all integers) and I lost myself as the calculations become difficult. The focal point was to realize that the function has to be an arithmetic progression (linear). I failed the IMO test :)
@@Sletty73 Although this is an extremely difficult starting point, a small correction: 'r' will have to be a POSITIVE integer and the expression will also have to include r-1, r-2, r-3, .... , 2. When r is negative the result may not be an integer.
I worked backwards assuming it has to be linear. Because I thought if it isn't linear, then f(a+b)has to have a term of ab. But on the left side, 2a and b are not terms of ab, therefore it had to be linear. Once that is confirmed, the rest is easy.
Hey , actually from the first step when we substitute a = 0 , we can differentiate it be assuming b as x . And we'll see the function f '(f(x))=2 has a constant slope , so therefore it can only be a linear solution and the rest follows . Just wanted to show it can be done using calculus .
I would be amazed if a question was given for something like this and the answers were not low single digits, zero or algebra. I don't think I have ever seen an given question like this where the answer would be a multiple digit significant figure with decimils.
I solved it almost immediately by guessing that the function was linear, look: if f is linear then 2f(a+b)=f(f(a+b)) the only way this is true is if f(x) is equal to 2x
I got surprisingly (and frustratingly) close. I knew it had to be linear by a factor of 2, 2x, but I forgot the "+n" and f(x)=0. This is why you always plug in the definitions of equations first. Awesome video!
Hey Presh Talwalkar, very nice solution! I find your video extremely clear and well presented, I should better learn some presentation skill from you.
BTW, thank you for mentioning my channel, I really appreciate it. Cheers!
Thanks for the feedback! I really thank you for your video--I didn't solve the problem, and your video helped me understand it. People often request videos for harder problems, so I'm sure they will enjoy your videos!
I am comfused, didnt he upload this 10 mins ago, because this comment is one week old
@@אביב-ת7ל He probably uploaded the video privately one week ago.
@@אביב-ת7ל MindYourDecisions has kindly share the (early stage) video with me last week and that's how I got the chance to see the video a bit earlier than you guys. Cheers!
Are you a friend of 3 blue 1 brown
the only thing I understood about this math video is that if there are 6 problems, and each one is worth 7 points, that there is a total of 42 possible points.
🤣🤣🤣
Wow wow slow down egghead
David Suarez if you got it, flaunt it
I laughed too hard at this. Well played!
Study more.
"You can pause the video to try yourself." Five weeks later:"Let's just continue the video."
On the first step: "Goddamnit I got it wrong"
I am the 666th like 😈
RaVeN85
I know you're just joking but I really hope this is a true story
Good perseverance!!😉
I actually paused the video and solved it myself. Took me a few hours and sheets of paper but it's so damn satisfying when everything comes together. I would argue mathematicians live for this moment when their proof is complete. It's so satisfying. My proof that f has to be linear was different and a little less elegant though. Still a very fun problem, but definitely not easy for me.
i guess i couldnt even compete in the paralympics version of this
😂😂😂
🤣🤣🤣🤣
😂😂
Very underrated comment.
I’m dying 😂😂😂😂
I attended the IMO in high school cuz I thought I was good at math and scored a 0. Good times lol
Same with the chemistry equivalent, I think I got some points from the most basic questions, which were pretty much the hard questions that I'd get in my school's competition
I don't get it. You attended the IMO without any preparation because you think you are good at math? You did not even checked the exam once? Lol
@@hoangnguyenvuhuy5535 yea... I was in high school. Didn't think much at all
But.....if I am not wrong, you have to go through 2-3 qualifier tests of your country to take the imo
@@sarthakpatnaik65 yeah in my country there are school level competition, then city level, then district level, then the whole country level. Iirc you need to be in top 5 at the country competition to be able to join
I don't even understand the question
LMAO 😂
I think you aren't alone in that case !
Same here 😑
I thought after seeing the solution and might be able to understand the question atleast... But no
Even me
If functions are a new concept to you the question is impossible to understand. No worries.
me after graduating in mathematics: well, i can solve any problem now.
me looking at a high school imo problem: well, better do a master's degree
ma ne?
Essas questões são insanas mesmo
@@bhaia6077 mane nothing
@@bhaia6077 mane=means??
@@kaperskyplays8016 mane literally means "meaning?"
Pffft this question wasn't a challenge to me. I didn't even bother to try it and I knew I couldn't do it
IDK where on earth you are but that is a very modern American disposition. If you feel like a dummy where you are, come to America where you will fit right in LOL
6 pens cost $2.40 Calculate the cost of one pen (show equation).
@@quake4313
Well... Ugh... I would've just divided the cost by the quantity... That would've given us $0.40. Smartass...
@@patrickpettyjr.2487 you asked for the equation. Dont be so toxic it's not worth it
@@quake4313
I was kidding a bit, sorry. But your equation looks pretty badass...
It's pleasant to hear some honest words like "I couldn't solve it myself" instead of "This is how you proceed" Zero ego involved. I appreciate that.
Okay...so 1 + b is the same as b + 1. Got that...
...gonna need you to slow down now.
Yeah, he did brush over that a bit... The key is to recognize that the integers under addition have the property of commutativity. All that means is that the order in which we add integers doesn’t matter. Formally, we may write (a+b) = (b+a) such that a, b are integers.
Since 1 and b are integers, we can say that (1+b) = (b+1)
It’s actually worth to mention, cause if we weren’t solving in Z, the operation + could be non commutative (if the group isn’t abelian)
That's what she said!!!
@@IStMl Let's try on quaternions!
Its like 2x+2(x-1)+2(1-x)=2x+2(x-1)-2(x-1) or (a+b)(a^2-ab+b^2)=(a+b)^3 or (a+b)^2=(a-b)(a+b)
When in doubt, plug in numbers. 1 and 0 usually work out nice.
Its like that all the time, with these equations of functions, right?
I never liked this section of maths when i studied for math competitions. Arbitrary and unintuitive, with little application.
William Zhang was about to say that lol
In fact I love combinatorics the most,it is very fun for me
I always plug in Graham's number.
@@Iocun I see someone's ambitious x')
Exactly on point. It is usually like how you describe it that these kind of problems are solved.
worth mentioning that though IMO contestants may have found this comparatively easy, they are constantly training and facing problems of this nature and these types of techniques. most of them will see this type of problem and have a lot to fall back on almost immediately.
source: am an IMO silver medalist
Gotta include the source right ;)
@@sjsjjf8feirbfjtjfjifofofof417 is it not relevant?
Ngl i would flex it too, just saying its unecessary but totally justified
it was actually a relevant info
Same math olym contestant here too!! Actually the step of putting a = 0 is very common in solving function questions. But I dun think I can think of this immediately if I haven't faced this type of questions before🤣🤣so it's all about training
as someone who’s been doing a lot of linear algebra lately, i immediately picked up on the fact that this equation very closely resembled the definition of linearity, (i.e. f(a) + f(b) = f(a+b) and f(ca) = cf(a)). then assuming f was linear, it was very easy to solve from there. i was just going off my intuition though, so your solution was probably much more rigorous and would certainly stand up better under scrutiny!
You can observe by setting a=0, then b=0, that this is a linear function, which maps values f(a) onto 2f(a)). So at least for any z in the image of f, f(z) is simply 2z.
You can then do a simple demonstration that f is surjective and you're done.
I don't know why you'd do it as complicated as in the video.
@@noimnotniceI’m just learning functional equations so I’m not very experienced but how would you prove surjectivity in this case?
Yeah-would lose a point or two for not showing that the functions found in that way are the *only* ones
@@EgWyps no you would get a 0 lol
To people who feel bad for not finding it while its supposed to be "easy". Consider this, the people who do these olympiads have seen this type of question before. It's very often the same type of questions, and with the same way of solving them. You have to train for these olympiads and this type of question is a pretty common one
I don't think so. There's a way to solve this problem in an easy way that just requires you to understand the mathematics, not memorize similar problems.
Hint: the solution I'm talking about requires you to interpret the equation geometrically. If you can imagine the geometry of this equation, it becomes obvious what the solutions are.
@@peezieforestem5078 wow a recent reply. also can you elaborate on the geometrical solution cause i dont understand
@@peezieforestem5078 It's definitely a matter of having solved similar problems before. I used to compete in olympiads myself, this type of question is very common, it even appeared on my entrance exam. Typically the way to solve this is through substitution as in the video, that's why it's considered easy. I'd love to hear your method of solving it, even with your hint I'm not sure what your method is.
@@rayyansohaib8238 Elaborate too much would give my solution away.
I'll give you some stepping stones:
1) Consider what a variable looks like geometrically. Let's say x - what does it describe?
2) Realize that what we call the variable doesn't matter. We can call it a, b or x, it's all the same.
3) Once you have completed point 1, consider what a sum of 2 variables looks like geometrically. Let's say (x + y), or (a + b) - the name doesn't matter, as established in point 2. Perhaps recall complex numbers.
4) Consider that a function can be viewed as a mapping of points to a different set of points. Once again, imagine the geometric meaning of this.
5) Consider which type of transformation multiplication by 2 is, from the point of geometry. What does multiplying by 2 actually does to the points?
6) The composition of 2 functions is just applying the mapping 2 times.
7) Finally, realize that equality in geometric terms means we have the same set of points, or the same geometric object.
Voila! If you understand all these steps, you should be able to formulate the problem in geometric terms:
"Which mapping, when applied 2 times to a geometric meaning of (a + b), results in the same set of points as (this mapping applied to the geometric meaning of a variable scaled 2 times) + (geometric meaning of the second variable with the same mapping, scaled 2 times)?"
That might've sounded confusing, but that's because I'm not giving you the answers. If you work through the steps, the problem will simplify, and the solution should become obvious and straightforward. Also, geometric concepts are hard to put into words.
@@pb9405that's true, these are pretty standard questions in entrance exams and Olympiads. They are always taught to be solved in exactly the way it was described in video, so I can bet everyone in the Olympiad knew the "trick". It was easy.
the interesting part was showing that the set of solutions had to be linear, actually from that point it was really straight forward
That certainly solves everything.....
It in fact was... a straight forward problem.
Musical Inquisitor [groan].
@@MusicalInquisit beautiful. Or should I say melodic.
Welp I dont even know what f is
Since the high school, the problem of problem solving for me was always understanding the problem.
Half of the problem solution is to understand the problem.
my prof once said: "when encountering a problem, first you wonder about the problem itself. the question itself - because you dont understand it. and you look at it. suddendly, you understand the question and you doubt it: is that true? is that even possible? and you look closer and try things out until you have a solid understanding on whats going on. and then comes the hardest part: as "clear" as the solution is to you now, you need to write it down in a way that anybody else can both understand the logic and also see that you can prove every step."
How can u be an asian but still bad at maths
ruclips.net/video/FffvCM0C3x8/видео.html
Lmao
compared to what maths has been like for me in high school this is real maths, the maths at school feels like it lacks the aspect of researching and exploring solutions
when I watch your videos I learn how to approach problems, but I'm also stimulated
You could have tried for jee (advanced )it would have been perfect for your hunger of exploring all possibilities in a solution
Very true!
Stimulated...
ruclips.net/video/FffvCM0C3x8/видео.html😊
The candidates had like couple hours to solve, in school you only got 45 minutes. You can now question yourself
Ah this is easy, I've scored a positive score on a meth test plenty of times.
I see what you did there
@Michael Darrow go watch breaking bad then👀😂
😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
Inshallah haram
@Michael Darrow he is right. Come to Turkey and take an exam for university entrance, then see all math questions. We solve harder ones and the time is not 4.5 hours, just 1 or 2 minutes for each question..
Viewers: “we want harder problems“
Presh:
Viewers:“surprised pikachu face“
This problem was way too easy,thats why he uploaded it,for an imo problem way too easy,in my country we have this type of problem at the city level olimpiad
May I ask where you’re from?
@@nilsdula7693 romania
60% solved the problem and 60% get a bronze medal, so I guess the problem is still as difficult as getting a bronze medal in the imo
@@luis_musik that is barely a good argument
I remember checking out the 2019 IMO problems after they released. I was able to solve this one in 5 minutes. I was surprised because usually I'm not able to solve any IMO problems at all. If you are already into doing olympiad problems, you will see that this one really isn't that hard at all. This type of problem is called a "functional equation" and a common strategy for these problems is to substitute numbers in to get identities about the function and then continue from there, which is what this solution does (plugging in 0 and 1)
It's never occurred to me to substitute only one value though. I would fail this because I would plug in 0 or 1 for both A and B. Giving only one a value just doesn't come naturally.
@@SomeRandomDude821 why would you instinctively try a and b with the same value? I would've thought of it as intuitive to run them as separate values
@@SomeRandomDude821 My friend did these math competitions for years in Canada (he was the Canadian high school co-champion ..). He said after a few years, they just had so much experience from thousands of problems .. I mean yes you and me both, maybe spend a whole hour poking around and not think to try leaving one variable open and plugging in just one value. The "trained athletes" I guess would usually have those ideas.
ruclips.net/video/FffvCM0C3x8/видео.html
Yeah, i was rly surprised by this
4 years ago, I didn’t understand any of this. Now, studying maths at Uni, it’s so easy to follow
Competitors found this problem easy
Me: try to understand question
Simply use Mathematical Induction
Pyro Tricks Hum... no ? Explain me your bc, hyp and step ?
Me doing:
F(2a)+2f(b)=f(f(a+b)
=》2af+2bf=f(af+bf)
=》2af+2bf is not equal to af^2+ bf^2
Thought this is the answer and was thinking why we need so much time for these question.....
After video
...
Wait...what is that(surprised pikachu face)
@Thunder_Arch i know....I did it umknowingly
Yeah this is beyond my scope...I can get some of it....maybe
You make the fact that an IMO aspirant considers these easy as something very surprising, but the fact is they've seen hundreds of problems exactly like this and know exactly what to do to solve it, however, a problem like this would be hard for any IMO aspirant if they had never seen functional equations before
Exactly right. With a lot of experience, it's .. really not the same "problem" at all. It's a little bit like situations in chess. A highly experienced player may just walk up to the table and (in a very short amount of time) start making some very good suggestions, or even "the" move (if it's that sort of thing). Less experienced guys are just sitting there, boggling at him.
Frrr
Exactly, this actually is a generic problem in IMO, once you have done one (or have been shown how to) all others follow a similar approach...
Hard stuff is the stuff that you have never encountered before...
@@gabrielbarrantes6946 this, this was why something like 2011 p2 was so hard for people who were well trained in other areas
also it's like the easiest imo problem in like the last 10 years so that also helps
I would agree only partly. Knowing the 'typical' problems helps, but this is still extremely challenging to solve such problems in 4 hours
I'm an engineer, and this is based on my basic math knowledge... i think my way of approaching, which i call "the lazy bastard" way, is at least interesting because it's done with as little math as possible, but requires you knowing math to some extent to pull it out. I usually did stuff this way in college and it helped me greatly, so i thought i'd share it.
My thought process was like this: First, because we have integer to integer, we discard a lot of possibilities before we even start (say, it can't be trigs). Then, let's see if f(x)=0 could be a valid answer (which....should be at least assumed at first and tested, because if that is or is not an answer has some implications). And it is! So... we at least have SOME points because we say f(x)=0 is an answer.
Then...if i'm being asked "all functions"there might be a constant value involved... And, if we were to assume that there IS a constant value being added on f(x), then... it's pretty straightforward that said f(x) should be linear, as with any non linear function, there are some values that would make it not true..... This is because, if we were to evaluate f(f(x)) with a and b being 0, the left side would be 3 times the constant, always , and the right side would depend on the value of the constant because of the double evaluation. For any non linear function, there can't be a constant added on the function, so IF there is a non linear function that satisfies that answer, it would have no constants.
Now...let's see all the linear functions than satisfy our problem. We have that f(x) is some form of mx+n. If n can be any number, then let's say n=0.
m(2a)+2m(b)=2m(2m(a+b))----dividing both sides by 2m and cleaning it a bit------>a+b=2m(a+b), so m=0, and this fits with the information we already know, that f(x)=0 is an answer. Now.... since this is (well, should be) valid for all a and b, let's do it with a=0 and b=0:
0+n+2(n+0)=m(0+n)+n let's clean this a bit:
3n=n(m+1) let's divide both sides by n:
3=m+1 so... m=2 and n is any integer.
If we test it with m=2, as shown in the video, leave us with equalities on both cases.
We have f(x)=0 and f(x)=2x+n, with n being any number in Z as our current answers.
Now, is there any non linear function that satisfies the equality? I would say it CAN'T be. This is because, of the fact there is no way f(f(x)) has the same exponent (grade?) than f(x) other than f(x) being linear (english is not my first language, i don't know if exponent or grade are the correct terms in english). Can the function be something like f(x)=Y^x. with Y being any number in Z? It can't be, because if a=0 and b=0, we have Y^0 +2*Y^0=Y^(Y^0) ----> 1+2=Y. Now, if b=0 and Y=3, 3^2a+2 != 3^(3^a), so it would not satisfy all possible values of a and b. I would claim that it HAS to be linear because of the f(f(x)) and the fact there are defined constants (the 2s) and there is Z to Z stated
We can conclude that...the only possible answers should be f(x)=0 and f(x)=2m+n with n in Z.
Notice how little algebra was used , and when used, it was basic substitution? This is what i mean, having some concepts more or less clear can lead skip some steps and reach the answer without much work.
The same approach was taken by me too. Analysis of a question before working it out helps a great deal. I believe that's why engineers are the world problem solvers.
- by an engineering aspirant.
Your way of figuring out that it has to be a linear relation is interesting to intuitively find the shape of the final answer, but I'm not convinced it's a proof. To me it looked like f has do be affine because it looks a lot like the relation f(a+b) = f(a) + f(b), so I started guessing and saw that f(x)=2x is a solution. Next I would have tried f(x)+n. I continued watching the video from there so I'm not sure how my proof that it was the only type of solution would look like. I think the completeness proof is the most difficult part of this exercice.
"i would say" is not sufficient, you need proof to get full points
your way of thinking is a good start to wrap your head around the question but it isn't a full solution
I agree that the way I approached its not sufficient proof, but I'll also would like to point out that it's based on what I remember from my calc and algebra classes 7 years or so ago... And I was mostly showcasing an approach that could be build into something more solid with a more fresh knowledge... I mean, the things I didn't state were the things I remember less and as you see, I did as little math as possible. I think the fact that the function doesn't change the set when you do f(f(x)) might lead to some of proof that the transformation has to be linear, but my career doesn't elaborate on that much and I don't remember much. Also, that fact at least makes easy to proof that the final function can only have a constant if it is linear, and the little of sets I remember made me think I can just state that some family of functions don't go from Z to Z, but I don't know if I have to prove theorems or properties in these kind of events. I also think that this conclusion can lead to proof that the equality can only be satisfied with a linear function , with some theorem or something I don't know. Even if that is not true, I would say (again, not sure) if you don't know how to prove my previous thoughts, you would probably be able to check all possible family of f(x) so that it goes from Z to Z, and prove that que equality won't be true with all values of a and b. I was a bit careful on not stating what I was not mostly sure about, and well, even with my rusty math knowledge I did reach the correct answer, the exact proof escapes me but to be fair, anyone in these events would probably have prepared in some form and would at least have fresh knowledge and know a bit more of the rules and expectations of a competition, so think I at least did decent enough. I'll say that I only dare to claim that my approach is decently intuitive and uses as little math knowledge as possible, keeping things simple, which I believe it's at least a decent way of approaching a problem : how can I solve this as best as I can with as little work as I can? Math (and many subjects, in fact) is abstract, and you don't need to do pages of algebra if you understand some concepts and apply them. Lastly, I would like to say that I thought all of that in like 2-3 minutes tops (I did the calcs and algebra in my mind), and I don't consider my rusty and mostly forgotten math knowledge to be that good because I honestly only remember some basics, and I did reach the answer before watching it... So... I only though, based on general comments, that this would be a decent example, but I won't be brazen enough to claim it is correct.
Also, I did showed some minor "proof" as to why I though i discard some non linear possibilities, I would dare claim that my little and final thought process of showing there are no Non linear expression that satisfy all the conditions is at least somewhat valid? I think showing that there are values of a and b that wouldn't satisfy the expression, based on an assumption that the functions has a certain form is called something like proof of contradiction, or is at least some form of proof? Please do enlighten me if I'm wrong, I'm just checking subjects to read about again and have yet to start so any insight would be greatly appreciated. Thanks to all of you for the feedback, and it fill my heart my joy a would be engineer found that helpful :)
What surprises me is that I actually understood the demonstration. You do a great job, I guess ^^
Back in elementary school I always had 90s in math class. I thought I was good at math.
just think: kids are solving problems like these, while there are adults in their 40s struggling to create secure passwords for their accounts that isnt some combination of password and 123
@@Blox117 kids with above average iq
@@itachi6336 Wayyyy above.
@@georges1055 iq is nonesense
@@lordx4641 It's not nonsense, but it's not reliable either
You lost me at "hey this is presh talwalkar"
You lost me when I read the title.
🤣🤣🤣
lol🤣
You lost me at the thumbnail
Lost me after "Mind your decisions"
Btw, a cool tip you can get from this video is to always try f(x) = 0 before continuing. I couldn't find any student answer, but I'm almost sure most of the 40% missed f(x) = 0 since it seems so obvious (especially when there's no constant numbers)
Give me a million years i still couldnt have come up with a solution
whatever man you would probably invent your own maths bu that time
I think you have the solution in a million years
Make your own solution and copy their answer:)
Even in 1 month you could easily find the answer. You just have to brush up on Calculus and Linear Algebra
Me: Still figuring out how the letter n pops up in the final answer while the question has only a and b.
The question has infinite answers. n can be any number belonging to Z covering all the possibilities. It arises from the fact that you cant possibly write all the answers. You just write an equation for all the answers.
I just realised you could have been joking, i didnt mean to sound like a smartass after failing to solve the question horrendously. My apologies.
also “mx+n” is the equation used to find the slope(rise/run) of a function
During the analysis it was made clear that on the right hand side was the differences between points while the left hand side was constant. This means the function has to be a linear one. In order to work with a set containing all linear functions one can simply use the general expression of such function: f (x)=mx+n (where m is the rise/run and n is the shift on the y-axis, meaning the function intersects the y-axis at (0,n).)
That's how n entered the original equation.
😅😅
He lost me at "let z be the set of integers"
hahahahahahahha
Hahahahahahahqqhhqhqqh
Same
He actually has it written in his intermission the origin of that Z being used. The German word for integer is Zahlen.
@@mike1024. as a German I'm sorry to say that "Zahlen" just means "numbers". Integers are called "Ganzzahlen" or "Ganze Zahlen" ;-)
Instructions unclear, foot trapped in toilet bowl
Noticing the arithmetic progression is the real magic. I tried solving for a=b, for a=0, for b=0, and for both being zero. Had a bunch of equations but nothing I could really use to get to a linear progression.
But why does “there is an arithmetic progression” imply that the function can be written in a linear form?
@@honorinemunezero6866 for that u need to know what is meaning of 'arithmetic progression'.
Arithmetic progression means a sequence of numbers which have a common difference. So you can get consecutive terms of ap by adding that common difference to each consecutive term. The general term of an ap is given by (T= a+nd) where a is first term of the sequence and d the common difference, which is a linear equation.
I did the same thing you can get an arithmetic equation from that
Many years ago I scored 98% on the Oxford Uni entrance exam. The year before, strong candidates were scoring 30%. It is very difficult to get the level of questions right for bright and insanely well prepared candidates. If a paper has questions just a bit too similar to precedents from previous years, then many candidates will get very high scores; and similarly if they are too hard, then even gifted mathematicians (which I am not) will struggle.
Just wanna point out, anyone can "solve" it in 5 minutes when they've had all the time in the world to prepare a solution lol
Well put. I used to know these mathlete types. They spent their entire lunch hours and after hours doing math problems and puzzles as well as preparing for contests. They were well primed.
Even though I'm a college freshman, this is really not that hard question. We've just begun learning Calculus and Linear Algebra, but have learned way harder things and had to solve harder questions.
Edit. Yes, you learn something so that you can do it proficiently and effectively, not 'wasting' time with it.
@Susuya Juuzou should probably learn how to code, you can use a pre-made engine you know, you don't have to solo engineer it
True, but many wouldnt put in the work to understand a problem like this.
I was able to solve the problem in 5 minutes after seeing it for the first time. It truly is an easy problem...
In school math, had lots of problems finding numbers to satisfy a set of specified equations, but not for finding functions that would work for a specified set of numbers, It becomes more of a logic problem, needing ad hoc creative insights, than a math problem requiring standard techniques. That makes for the most interesting kind of problem. Thanks,
"I think my brain just commited suicide"
LMAO 😂
Your pfp fits perfectly in your comment
You put quote marks on your sentence, why?
Lucky I've never come across this problem in my life.. Usually the 0 on my bank balance is the only math problem I have to deal with...
Yea ?
"this means we have an rithmetic progression". there it is. the smart moment that solves everything.
Yea lol. 90% of this problem wasnt too difficult. But its the 1 to 2 steps of logic that makes it SO challenging and difficult to solve.
but how did he determine that it was an arithmetic progression thats the only part where i got stuck
@@mateapaparisto1173 i guess you can solve it without defining an arithmetic progression
@@mateapaparisto1173 An arithmetic progression is a succession of numbers where *any* two consecutive numbers will always have the same difference or 'distance' between each other. Since on the left side we have two functions evaluated at two consecutive points, and that is equal to a constant value, we determine it must be an arithmetic progression. Example: 2, 4, 6, 8... is an arithmetic progression. You can take any two consecutive terms and the the difference will always be two. That means we can rewrite the whole succession as: 2, 2+2, 2+2+2 = 2, 2 + 2, 2 + 4, 2 + 6... => nth term = first term + d(ifference)n-1 times.
Me: "Oh, this will be interesting to watch, I wonder if I can follow along"
10 seconds into the explanation: "Yeah, no"
Awesome solution, you should post more questions from these math olympiads or similar ones.
This was actually one of the few times I really paused and tried it for myself. It turned out to be not too difficult, but probably I just was lucky:
We first note that f(x)=0 is the trivial solution and remember that for later. For other possible solutions, we assume f(x) being nonzero and consider the following: Basically we're looking for a function that satisfies a certain condition over the two-dimensional (a,b) in Z². Therefore, it also must satisfy it in any subset of Z². Like in the video, I first set a = 0 and b being arbitrary. That results in f(0)+2 f(b) = f(f(b)). Substitute x=f(b) => f(0)+2x = f(x) [note that this requires that f(b) is not 0 for all b, otherwise x is also 0 and we gain nothing]. Therefore, f must have the form f(x)=2x+c. To find c, make a arbitrary again, plug in the form of f in the original RHS and LHS, simplify and compare: f(2a)+2f(b)=4a+4b+3c and f(f(a+b))=2(2(a+b)+c)+c = 4a+4b+3c. Therefore, any value of c in Z² will do.Concluding, f has the form of either f(x)=0 or f(x)=2x+c with any integer c.
Your answer has a flaw. You proved that if x is a value of f then f(x) is given by some formula. So you showed that f(x) is given by this formula for x big enough and of some particular values. What you should do next is come back to the original equation and calculate f from there (using the received formula in the right hand side).
You should do more IMO problems, the solutions are always amazing to see
I can’t believe at one point in my life, I could solve this without any issues. Now I look at it with a potato brain.
Same
that's because it's useless for 99.9999% of world population and we never used this outside school
@@goissilva not directly, but indirectly yes, coding, software etc. which affect almost all of us daily life.
When it comes to something that you ve seen before you will eventually find a way to it it's just a question of time but if it's something like this trust me the idea of trying with numbers to find out the linear equation is actually genius level and unless you re used to exercices like these which are particularly rare I teach maths btw
@@goissilva you dont use this in school lol. this is for mathematical geniuses. theres only like 3-4-5 people per country that participate to this competition yearly.
Thats amazing! I havent done maths in so long now but very good explanation that I could follow. I think the key to solve this problem is the substitution first, and the realisation of the arithmetic progression function, which most of us won't notice.
It is always useful to promote maths and you do it nicely. Keep up the good work. Thank you very much for mentioning my blog :)
I'm sure many people will enjoy your blog! Thanks for your explanation which greatly helped me understand how to solve this problem.
One of the most beautiful maths problems I saw in 2019 so far.
vlatko no I haven't yet, I'll check it out. Thanks.
Have you not seen functional equations before?
Clyde S yes, but I didn't solve a lot of problems about them, not at all.
vlatko, I saw the video on 3b1b channel, the problem is mindblowing.
@@patricksalhany8787 I don't think this is a particularly beautiful functional equation. (of course, my opinion - since this is a fairly routine problem). If you want to see more functional equations, there's lots of problems and suitable collections on the Art of Problem Solving forums.
This is how I solved for 2x + C : if a = b then
f(2b) + 2f(a) = f(f(2a)) = f(2a) + 2f(b) meaning that
f(2a) = 2f(a) in this case (this already cancels out a bunch of functions, namely anything with powers, leaving linear). It can then be written that
f(4a) = f(f(2a)) you can take out a function leaving
4a = f(2a) meaning the function has to be f(x)= 2x, maybe if I plugged in some more numbers I would have gotten that f(x) also can be 0.
Genius
@@catherinechan4822 squish squish
I was studying for maths olympics when I was 14 but realized it was pointless because studying maths a lot means less for other classes so I stopped and focused on all of my classes. Olympics might make sense for some countries where you can get in a collage by mentioning this or just doing sports such as usa but in mine, we have to learn all classes and take an exam to get in a uni. And olympics help only if you get important medals at maths which is really hard to get.
What country? You’re talking about the bac.
@@paulblart4551 I was talking about türkiye
@@momofromatla2318 do you call it the baccalaureate? In Romania that’s what we call it.
@@paulblart4551 no that s a different thing
@@momofromatla2318 bro which university are you at?
It is amazing how the very first right steps can turn an insanely difficult problem into a very simple one.
Good choice you get a problem from a very legit and high level math competition.
Your ideas are always excellent, You deserve praise.
Im just here, purely accident.. I don't even understand the problem.
There is no problem.
you know what they say, the problem becomes a problem just after you consider it to be a problem
Its all greek to me
Cool Dude where are you from ? I didn’t do integrals until year 11 and certainly they were more simple than this... I went on to do medicine and have quite few practical uses for integrals but still this problem tripped me up a bit after doing advanced mathematics before college/university.
Cool Dude This is 12 years old math btw... What’s hard is having the math confidence to find the process in limited time, and that is more complicated to have
When I saw this problem, my first assumption was: "The solution has to be linear". Just because (a+b) is in the argument of the right side, and separated on the left side.
Same here. Would have cracked it.
You still need to show that it has to be linear though. A handwavey argument like that would not get you far (other than as a starting point for what to try).
@@asdfghyter and that's all he said. He didn't claim to have it proven, thx
El Zed No, but César did.
Yeah, I figured pretty fast that ought to be 0 or 2x+b, but it probably would've taken me much longer to proof it.
I've never actually done functional equations like these in school. Interesting to see how they might be done.
At 2:57, finish it off by rewriting f(0) + 2f(b) = f(f(b)) into C + 2x = f(x) by introducing x=f(b) and it's basically done: f(x)=2x+C. Also, the trivial solution f(x)=0 should always be your first ansatz.
i got lost at 3:41 i don't understand why f(f(b+1)) = f(0)+2f(b+1)
Me too
Now I got it.
f(0)+2f(x) = f(f(x)), That means it's true for any integer x
Let, x = b+1, b is an integer So b+1 also an integer.
f(0)+2f(b+1) = f(f(b+1))
i dont get it.First we define x to be b and then b+1.How come does this work?@@copyrightmaster238
@@copyrightmaster238ah thanks couldnt see that lol
Even if I am studying maths at college I am often so oaf, but you made it so clear ! I can’t help but feel proud of myself although it is just thanks to you
I'm actually learning how tot solve problems like this in math class! Thank you for explaining this; it was very clear!
I tried to watch this video a couple of years ago, I didn't understand it at all. That goes for your other videos as well. But today I did get it, And I've been watching some others.
Oh joy of understanding, greater than that of imagining or feeling!
Its like watching a guy speaking chinese for 7 min
Lol im taking chinese and its easier to understand then what this guy said
@@drewfreese4707 Chinese is 🤮🤮 compared to other oriental/asian languages. find a new word? good luck finding its pronunciation since there's no alphabet and radicals are useless 99% of the time, not to mention the fact that there's 2 versions of it and one of them eliminates radicals entirely lol
ALSO THERE'S NO SPACING AND SOMETIMES YOU CAN'T TELL WHEN AN UNFAMILILAR WORD IN A SENTENCE STARTS OR ENDS (though this applies to other Asian languages in general, too) 😭
sorry for rambling but i just hate the chinese language with a burning passion
@@nerd2544 don’t feel bad. Chinese is my native language and I went overseas for like 2 decades and now I’ve forgotten how to write half the words, not even counting the ones i never remembered how to write before.
@@drewfreese4707 ironic
@@nerd2544 as a chinese, i totally agree hahah i never passed my chinese test back in junior school
Hey Presh Talwalkar, I came here to watch this video from your post on RedPig's channel, and your proof is totally valid, since there is a lemma stating that if some arithmetically progressing numbers, with their functions increasing too, then the function must be a linear function. Nice solution. I had the same solution as RedPig, too bad I got stuck at Nationals this year. ( Not from USA)
Thanks for the comment. These problems are challenging for me, and I appreciate feedback. Plus, it's great to know people are working on math during the summer months!
@@MindYourDecisions No problem. Must say, indeed this solution was better than most solutions. I haven't seen problems in IMO in a while where the sum or product of functions of two variables is a constant. Nice for pointing it out
@@rodwayworkor9202 That is partially due to the fact that this is the easiest IMO algebra in a long time, in my opinion - usually, there's a lot more work to be substantiated before arriving at a such a powerful condition. (though here the 'constant term' sum is presented very explicitly)
Hey, I'm training for mathematical olympiads of my country, National Math Olympiad is totally easy, so I actually train for the internationals, so I just wanted to ask you if you know more youtube channels that solve this kind of problems. I hope I get to the IMO next year.
And yes, sorry if I make some mistakes in my english, I speak spanish (it would be very helpful is you correct me if I said something wrong)
@@trefoil2938 Yes.
3b1b: Uploads a vid on IMO
Next day
MYD:
Coincidence? I think not 🤔
Yeah, it may be coincidence as there are comments by patreons a week ago
Don't. Next thing you know, MYD will be posting a video where we have to solve for the probability of a colab.
@@ConnorSmith-lh7uw 😂😂😂
It's because IMO just happened. Many other channels uploaded all the solutions much before 3blue1brown.
why did i read 3b1b as 2b2t
One further way to know f(x) has linear shape, is to differentiate the whole equation once by a and once by b. The right term is the same on both equations, so you can equate the left terms and get f'(2a) = f'(b). This can only be true if f'(x)=const, so f(x) must be linear.
I did the same , putting the const back into the eqn gave me zero , still cudnt get the other linear solution tho.
It's not given that the function is differentiable
since the function maps Z to Z it is more appropriate to use the discrete derivative which gives the same conclusion
*sees equation*
"Yea, can I forfeit life real quick?"
*sees answer*
"Yea, can I forfeit life real quick?"
Ashes cringe
@@Andrew-ri5vs thanks denton
This was a very challenging and interesting problem!
Very nice. This is the type of problem that Osman Nal often shows on his channel. There's always some kind of obscure manipulation that you need to be able to "see".
Me after graduating from highschool. "Takes math book" " set them on fire"
when you start with as and bs, and somehow your solution is in x's and n's
You can keep the a's and b's, if you want, they're just arbitrary representations of numbers :/
You have to find the funtions ( f(x) ) for which this equation is true. f(x) is the general representation of an unknown/unspecified function that is used to state this equation. a and b are just random variables, like x, that replace the x in this unspecified function. For example f(x) can be f(x) = x + 1. If you place in x=2 you write f(2) = 2 + 1 = 3. Like that f(2a) would he f(2a) = 2a + 1. However, you have to find the functions for which this left side of the equation f(2a) + 2f(b) is equal to f(f(a+b)), which is only true in certain forms of function f(x)
I read that as Ass and BS! I was thinking you’re right, then finished the rest of the sentence, and realized you were talking about letters.
@@ivovandeboom6767 still lost on the function of a function, never mind, no need to reply, anyway what happend to the idea that X=1 as fitting in the first function found ?
@@highpath4776 Not sure what you mean, the first function found is f(x)=0. There is no x in the function, so f(1)=0.
As my physics professor tells us, "You're doing voodoo math."
Now, that's a good way to summarize it in four words.
Before you solved the problem, I got f (x) = 2x, it wasn't too far
Yeah, not hard to guess that f(x) = 2x was *an* f that worked, but I had no idea how to find any others or how to prove those were the only ones.
Bro I get 2x+y,for all y belongs to Z,
BUT not identify Zero function...
Try this problem
ruclips.net/video/igdy05LZj90/видео.html
@@pylavenkatesh8739 but for all points, I think you should also prove that there is no other solutions.
Me: Not able to sleep.
RUclips: would you like to solve an international math problem?
I managed a positive score on a corona test recently..my best,so far!
So I mostly managed to solve it, but only by making some assumptions and I couldn’t determine if my two solutions were the only solutions, showing it had to be an arithmetic progression was amazing.
I had to assume analytic at x=0. What did you do?
Solving this problem by progression method was very good. I realised the function to be linear as in LHS f(x) is present while in RHS f(f(x)) so composition of only a linear function will result in a function having same degree or a constant function. Anyways, your method was more thought provoking !
Wouldn't that only work if it is a polynomial specified??
If the function was for example 1/x the composition would give x
4:32 Here's my intuition: (f(2)-f(0))/2 is the slope of the secant line from x=0 to 2, while f(x+1)-f(x) is that from x to x+1, since both are the same for all x in Z, that means the function has a const. slope. So we can set f '(x) = m, and integrate to get our solution: f(x) = mx+c.
I was learning a lot of competition maths this summer (almost everything I know). And I can tell you that I'm able to easy understand this solution. I'm sure that before personal training I couldn't be able able to understand it. So, keep study, work hard - push your limits!
More for-them-its-easy IMO question please
I can't believe I solved it on my own. I'm preparing for JEE Advanced and every now and then I try solving such questions, not only for experience but also for my own love for Maths. I was so happy to solve it under like 3-4 minutes and I believe that Maths is all about rational approach rather than cramming all those bookish data. Self-approaches unlock incredible brain potential which is hidden in all of us. Maths is truly a gem of an art ❤
Good ha bro
I will keep training to become way smarter than you one fateful day, to make you shut up for good.
This is way tougher than jee advanced dude u didn't solve it liar
All the best for your preparation
@@floppathebased1492it was in imo
Just awesome Presh Talwalkar, do bringing such videos among us, it took my 2days to solve this problem, untill I didn't reach to apex , but i have solve 80% , I have cracked RMO, now I am pursuing B. Sc maths honours,, 🤙🤙👍👍❤❤
I don’t even know why I watch this stuff, I barely understand most of it
You didn't understand how to speak when you were born, but through exposure you learned. Keep watching.
@@demonzabrak this inspired me, thanks :))
I am struggling with the assumption at 5:38 that 2m=m^2 and 3n = mn+n. I can't figure out why that has to me true. It seems like it's one possible case but how do we know there aren't other possibilities?
He's equating the co-efficients on both sides, it's a fairly helpful and common trick. If you were to expand you'd notice there is no other way get those mixture of terms on both sides due to the + sign in the middle. Therefore it's completely reasonable to equate them
"I hope this video gave you some sense of how to solve--"
Nope.
Total noob!!!
@@wakingfromslumber9555 Your the one who also cant do it? So what are you doing? You cant do it yourself? And dont start saying „SaYs ThE RoBlOx PrOfIlE“
Edit: You probably will you son of a OOF
@@karl-heinz5924 OOF size, Large.
@@karl-heinz5924 Roast Acceleration: Y E S
Easy function
at the beginning if you let Fx be mx+n and do the rest which is putting the values in the equation you get 2m=msquared so either m is zero or 2 so you don't have to do all the changing values stuff. if you think its going to take long no it took like 3 minutes. sorry for my bed england its not my first language
i honestly dont understand how this is a ''Insanely Hard Problem'' im not kidding i could solve this 8th grade im not flexing btw this is seriously not that hard
Agreed this is consired easy-medium in my country
But how do you know that Fx is of the form mx+n?
@@hayimshamir8426 no reason not to because generally it’s not a parabola type function or any higher
It will probably be the same answer if you let Fx be axsquared + bx+c
PRANJAL SRIVASTAVA FROM INDIA WAS THAT 13 YEAR OLD STUDENT WHO GOT GOLD MEDAL IN IMO 2019 UK!!!
An australian mathematician not only got an gold medal he was the highest scoring individual at the age of 12-13.
Still lives with his mom I'll bet.
He is 15 actually
He is 15 this year(2019) and got gold(was the only one in his team). Last year he was 14 and got silver.
@shubham sharma India's rank is 18 in the world and won 1Gold and 4 silvers in IMO'19 and ALL of them are 100% Indian citizens
IMO that question is hard
Pun intended ofc
stop
@@sanchitverma2892 oh hey how nice of you stopping by!
@@kcChicken123 *slow clap...*
Nice
You lost me after “Hey”.
Incredibly, MYD formulates the answer in under 5 minutes using only a pencil, a piece of paper, and the answer.
Mind your decisions, Would be nice if you explain the question before explaining the answer
Lmao
Here is my attempt at an explanation:
1st) What is a function (from the integers to the integers)?
-> You can think of functions as "mapping" or "connecting with a directed line" every integer with some other (no necessarily different) integer.
-- Examples are f(n) = n for all integers n, or f(2*n) = n and f(2*n+1) = 0 for all integers n.
----- In the first example you would "map" 0 to 0, 4 to 4, 103 to 103, -20 to -20 and so on
----- In the second example you would "map" 0 to 0, 3 to 0, 8 to 4, -100 to -50, -2021 to 0, and so on.
Functions can describe any sort of mapping. The most important bit here is that while *every number maps to ONLY ONE number*, it is also true that *two different numbers ARE ALLOWED to map to the same number* (as is the case in the second example)
2nd) How do you even begin to solve a functional equation like the one in the problem?
-> Here one way to think about this is that you are "given a property that holds for all integers" and your goal is to "narrow down what the function can be"
Q: How do you "narrow down what the function can be"?
A: You do this by plugging in specific values in the function, noticing what properties must hold true once you plug stuff in, and trying to come up with conclusions from there.
So you can think of it as:
1) you give me a rule the function must follow for all integers
2) I plug in specific values and INFER what also must be true for such function
3) I keep plugging stuff in and continue doing inference until I've forced the function to be described in a certain way
4) I prove that this function which satisfies all the equations I got from my inferences and plugged in numbers, satisfies the functional equation in general
I'll try to make a video about this problem in the next month and hopefully explain the logic a bit better there
1:15 is the full question. Better explain it? Just read the question
Its easy because its normal and participants have seen many many similar problems
Difficult ones are those that need creativity and real skill to solve.
I haven't seen such a simple IMO problem for a few years...
IMO 2018 P1
@@mironemiron2454 this is easier I think.
@@mironemiron2454 IMO 2018 P1 isn't that easy at all, I'm quite good at geometry I think but I still spent 30 minutes on that problem. Maybe you meant IMO 2017 P1?
@@desarguesbaptiste55772017 P1 was easy too, but i'm good at geo so 2018 P1 was very simple for me. Actually, it's just angle chasing
@@mironemiron2454 P6 IMO 2011 is also just angle chasing but it's quite hard I think. I have not seen a simple solution P1 2018, but I guess it was considered easy because it has a lot of solution.
I solved by
Using f(2) + 2f(x) = f(0) +2(x+1) -->(explained in vid)
And then put the values of x as 1,2,3
And found that f(0),f(1),f(2)... Were in ap
So fx = nx+k
And solved to find fx = 2x + k
This went from hard to easy once I realized it was for integers rather than for reals.
I found the two functions... but I would have not been able to proof that these are the only ones...
Same here. I considered a very general mapping of Z in itself as a linear form f(n)=mn+k and found easely the solutions. But unfortunately then I thought that a integer power could also teasform an integer in another integer and that the most general form should be f(n) = n^r+mn+k (r, m and k all integers) and I lost myself as the calculations become difficult. The focal point was to realize that the function has to be an arithmetic progression (linear). I failed the IMO test :)
The video shows a direct proof that f(x) = mx+n and concludes the values for m and n.
@@Sletty73 Although this is an extremely difficult starting point, a small correction: 'r' will have to be a POSITIVE integer and the expression will also have to include r-1, r-2, r-3, .... , 2. When r is negative the result may not be an integer.
I worked backwards assuming it has to be linear. Because I thought if it isn't linear, then f(a+b)has to have a term of ab. But on the left side, 2a and b are not terms of ab, therefore it had to be linear. Once that is confirmed, the rest is easy.
@@NVDAbets I think that doesn't work, because a function doesn't have to be a polynomial. Exemple: f(x)=1 if x is odd and f(x)=0 if x is even.
Finally,hard problem
do more IMO problems pls
Hey , actually from the first step when we substitute a = 0 , we can differentiate it be assuming b as x . And we'll see the function f '(f(x))=2 has a constant slope , so therefore it can only be a linear solution and the rest follows . Just wanted to show it can be done using calculus .
It's on integrers so you can't differentiate
And also f(0)= is not equal to 0
I didn’t even have to do any work, I just instantly thought “f(x)=0”. XD
The question asks for ALL solutions, not just the trivial though.
Same
I would be amazed if a question was given for something like this and the answers were not low single digits, zero or algebra. I don't think I have ever seen an given question like this where the answer would be a multiple digit significant figure with decimils.
After watching the video, I was like "damn! That was easy!"
LOL
I am pretty sure you will not say the same thing about this question
ruclips.net/video/h4WFGo9OP8w/видео.html
I solved it almost immediately by guessing that the function was linear, look:
if f is linear then
2f(a+b)=f(f(a+b)) the only way this is true is if f(x) is equal to 2x
Takes about 30 seconds to come up with this assuming f is a linear operator.
Same here
I agree. It's almost like looking at 2 + 2 = 4. Do you actually have to prove it somehow if you see solution (almost) at once.?
Woah! You are genius! Why you wasted this talent for damn like this problem? I really sorry
But that's not the answer. You've only found one function out of an infinite set of functions...
I got surprisingly (and frustratingly) close. I knew it had to be linear by a factor of 2, 2x, but I forgot the "+n" and f(x)=0.
This is why you always plug in the definitions of equations first. Awesome video!