A Functional Equation from Putnam and Beyond

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[@ishaanlakhera4077]

It was so easy, tho so difficult to think about that idea

[@SyberMath]

I agree

[@user-un6ib1ce6s]

It is so easy for our Chinese students

[@jmk6696]

It may be easy like an egg of Columbus.

[@user-bp9qp2xh9m]

I'm Japanese. this video is very helpful because I can improve my English listening skills as well as learn math.

[@SyberMath]

That's great!

[@fredthelegend7673]

I really liked this question. It was the perfect sort of difficulty for me, as I was able to solve it, but at the same time, it was still difficult enough that I had to think about it for a while in order to work out how to answer it, as I am still very new to functional equations. I referred back to other functional equation videos in order to work this one out. Great video anyway!

[@SyberMath]

Thank you! I'm glad to hear that! 😊

[@fredthelegend7673]

@@SyberMath No worries at all! It was awesome 👏, I really love all these sorts of problems you always present to us, and for me at least, most of them are challenging but still doable, but the odd one is impossible for me. Also, when I do manage to solve one, you always seem to have this neat, clever trick that is like 10x quicker than the way I did it. And I love seeing those neat tricks, it’s so amazing seeing how they work, and now I’m finally beginning to use them myself which is even better. So thank you very much for helping me with that Anyway, congratulations on recently hitting 20,000 subscribers, and I, along with plenty of others, am really, really enjoying these videos, so thank you very much for continuously making them and continually stretching my mathematical abilities every day. Sorry for the long essay. Two more quick things: 1. I really admire the fact that you reply to basically everyone’s comments in the comment section, it is very nice to see that you have taken the time to do that, and it is very nice to get a response from you, it’s very nice to see 2. Do you have any book recommendations for questions like the algebraic equation solving/number theory type problems, particularly like the ones that you have done on your channel? I would like to practice them some more Sorry for the really long comment Thanks again and congrats!

[@txikitofandango]

Same here, I really had to fight (and cheat a little with graphing) but it was worth it!

[@fredthelegend7673]

@@txikitofandango Yeah exactly! It’s so satisfying once you get the answer!

[@manojsurya1005]

Adding the 2 equations was a brilliant idea ,I did not think that it wud simplify to x, great work

[@SyberMath]

Glad you liked it

[@TDRinfinity]

A less clever and more systematic way to think about it would be to include the original equation in the system of equations. Then there are 3 equations and 3 unknowns, and you can use well known methods like gaussian elimination to solve for f(x)

[@p4vector]

Here's a slight reformulation of the solution. If you set g(x) = (x-3) / (x +1), then the original equation can be rewritten as f(g(x)) + f(g(g(x)) = x By substituting x = g(x) and x = g(g(x)), and using the curious fact that g(g(g(x))) = x we also have f(g(g(x)) + f(x) = g(x) f(x) + f(g(x)) = g(g(x)) So we have three linear equations with three unknowns: f(x), f(g(x)), and f(g(g(x)). For example by adding the second and the third, and subtracting the first we get: 2f(x) = g(x) + g(g(x)) - x EDIT: I just realized that several other posters proposed the same idea before. Well, I'm keeping the post in case someone finds this mini-writeup useful.

[@benkahtan6802]

This is a really elegant approach. Thanks for sharing. When I did the problem, I saw that we had f(g(x)) + f(g⁻¹(x)), and that g(g(x)) = g⁻¹(x), and that g⁻¹(g⁻¹(x)) = g(x), but didn't think to put it all in terms of g(x), g(g(x)), and g(g(g(x))). I think your approach is much tidier.

[@diogenissiganos5036]

Woah, that's a really interesting one!

[@satyapalsingh4429]

Wow ! My heart is filled with joy .Marvellous . Your method of solving is unique .God bless you !

[@snejpu2508]

Wow. That's the first functional equation I have ever solved and it's correct. What can I say, I did exactly what you do in the video. I even used the same letters for a substitution. : )

[@SyberMath]

Excellent!

[@adandap]

I used s and t, so I must be on a different wavelength. :)

[@leif1075]

@@SyberMath Why not just replace the second expression with something in terms of y so you dont have to introduce a third variable..more streamlined and elegantno? That's how I did it.

[@gianantoniosongia8722]

you' re a genius

[@user-zi2ld3dq4b]

This video made my day. Every good math video makes my day. Thank you for bringing this nice problem. Good luck.

[@SyberMath]

Happy to help! 😊

[@txikitofandango]

Holy crap I found the answer by trial and error. I reasoned through and evaluated f at 0 and plus or minus 1/3, 2, 3, and 5. Can prove that f is odd. Plotted the points. Noted that there's probably vertical asymptotes at plus and minus 1, so put an (x-1)^2 in the denominator. Fiddled around with cubic equations in the numerator. Presto. Okay, now to figure out how to actually do it.

[@SyberMath]

🤩

[@paulortega5317]

Nice problem. Same kind of idea. Let A=(x-3)/(x+1) and B=(x+3)/(1-x). If you repeatably substitute x in A with A you cycle through A ---> B, B ---> x, x ---> A, A---> B,... So, doing the same repeated substitution for this problem you get the 3 equations f(A)+f(B) = X, f(B)+f(X)=A, and f(X)+f(A)=B which quickly reduces to f(x)=(A+B-x)/2=( 8x/(1-x^2) - x)/2

[@jasonleelawlight]

I spent a good amount of time figuring out how to plug in the right values to get enough equations to cancel out the noisy parts. It’s essentially the same as what you did.

[@timeonly1401]

^^^ Great use of word 'noisy'! 👍

[@Z_o_r_r_o1267]

I verified that the function you found does indeed work, but I like to try to be rigorous in my approach to these problems. I understand the concept of a dummy variable. However I am not convinced that you can always do this in general. x, y and z are all distinct values. In this specific case, one of the inner x expressions was the inverse of the other one. I am not sure this technique would work in a more general case. To understand what I am getting at, make a table of values for x, y and z. When x=0, y = -3 and z = 3. So what you then have is that f(-3) + f(3) = 0. Now pick x=3. When x=3, y=0, and z = -3 So what you get in that case is f(0) + f(-3) = 3. I am not convinced that you can just casually replace y and z with the same dummy variable t and still have the the table of values work out the way I described. I think it works only in very specific cases, like this one, where one inner expression was the inverse of the other inner expression.

[@harshchoudhary279]

replacing is ok coz after those are functional eqn and it doesn't matters if all variables are x or x1 or y

[@ED-iq3mv]

Finally i found technique for equations like this…..thanks i have been looking for this for along time…..this is really really useful for someone like me

[@SyberMath]

Glad it was helpful!

[@ED-iq3mv]

Hahaha lol....i'm trying to study English now thanks for your help...

[@yoav613]

So nice problem! I solved it by my self and i am happy i got the same answer:)

[@SyberMath]

Great!

[@esteger1]

I've only started watching recently, and this is my favorite so far. Before viewing, I was able to solve the problem after much trial and error, with a similar but less efficient method. In other words, it took more algebra. Anyway, I've learned a lot about solving functional equations. Thanks!

[@juniorcandelachillcce1255]

Bonito ejercicio, me hizo recordar a mi primer año en la universidad. Saludos desde Perú.

[@SyberMath]

Greetings from the United States! 💖

[@aashsyed1277]

i hope you get 1 million subscribers at the end of this year :) you are awesome :) take care :)

[@SyberMath]

Thank you so much 😀💖

[@MathZoneKH]

I think so! I’m waiting for to see there!! Your content is the best!

[@SyberMath]

@@MathZoneKH Thanks! I appreciate it! 💖

[@aashsyed1277]

@@SyberMath you are probably the most underrated YT Channel.

[@ronbannon]

Love the problem and plan to share it with my students. You may note that the two arguments are inverses, so there's another way to find f(x)! I'll post a video if anyone is interested.

[@ronbannon]

Just posted a video illustrating the use of inverses. kzbin.info/www/bejne/pKOcaJ6ka7qcmtksi=A8sPlJv6CU9rVZ8Y

[@mahajankeshav14]

This channel is really very interesting and informative.

[@SyberMath]

Glad you think so! 💖

[@babitamishra524]

You really make a good recap and recreational stuff ,really enjoy solving and learning.

[@SyberMath]

Thanks! I'm glad to hear that! 🥰

[@c8h182]

Nice solution.Thank you @syberMath.

[@SyberMath]

You're welcome!

[@resilientcerebrum]

Sybermath can you suggest me how can I start learning functional equations, techniques and approaches from the very basic?

[@SyberMath]

Good question! There are some books but they do not start at the basic level unfortunately. I'm planning to come up with a document that explains the basics but who knows when I can write it up

[@3r3nite98]

I solved f(2x)=f(x)^2 the Hard way given f(x)≠0 and f'(0)=alpha and I can thank 2^n for that, without 2^n I wouldnt have solved it,2^n is my New Best friend also the equations that I Just solved has e^(xalpha) as solution,also It solves f(x+c)=f(x)f(c) since if c=x,then the solution is the same.

[@resilientcerebrum]

@@SyberMath : (

[@kaskilelr3]

@@resilientcerebrum some tips to get you going: something like f(x+3/1-x) has an expression inside the function argument and takes the general form of f(g(x)) So if you have an equation like f(g(y)) = h(y), you can solve it by defining x = g(y) F(x) = h(g^-1(x))

[@jasonleelawlight]

In general I always get started with this kind of problems by trying plugging in some special values and see if I can find any patterns or get some insights during the course of the calculation. For this problem I tried 0, 3, -3 and found it's always about f(3) f(-3) and f(0) and so I can solve all 3 of them. Then I tried 5, -5 and found f(1/3), f(-1/3) f(2) and f(-2) also showed up, so I added in 1/3, -1/3, 2 and -2, then I found it's all about f(5) f(-5) f(2) f(-2) f(1/3) and f(-1/3) and all these 6 can be solved. This gave me some insights as my gut feeling was that this pattern probably can be generalized for any given values, i.e. if I start with 1 or 2 values and keep adding the new ones showing up in the f(), then with a few iterations I should be able to get back and form a closed domain. So I did the same again but with the letters this time, and guess what, this ended up with essentially the same methodology presented in the video.

[@christopherrice4360]

SyberMath you never fail to impress me and astound me with the math problem videos you come up with. Keep up the awesome content👏👏👏👏!!!!

[@SyberMath]

Thank you!!! 💖

[@miloradtomic]

One of wonderful Functional Equation. Congratulations on being methodical. Thanks a lot.

[@SyberMath]

You are most welcome

[@aashsyed1277]

YOU ARE SO AWESOME you are the best pro i have seen!

[@SyberMath]

Thanks 😅

[@federicopagano6590]

Does it help you in anything to notice the input of the 2 functions are the inverse of each other? In order to generalize the result? I can't find yet the connection if they are inverse functions like this case

[@Hanible]

that's the first thing I noticed, but it's more than that: f(y(x))+f(z(x)) = x (y and z are functions here) since z(x) = y-1(x) we have: f(y(x))+f(y-1(x)) = x but notice how y(y(x)) = y-1(x) and similarly y-1(y-1(x)) = y(x) This is the crucial property y has, so by doing; f(y(y-1(x))+f(y-1(y-1(x))) = y-1(x) you get: f(x)+f(y(x)) = y-1(x) and by doing: f(y(y(x))) + f(y-1(y(x))) = y(x) you get: f(y-1(x)) + f(x) = y(x) by adding the 2 together you get: 2f(x) + x = y-1(x) + y(x) So the real challenge in this problem was finding this niche function y props to the author of the book once you figure out this trick you can instantly write 2f(x) + x = y-1(x) + y(x) and solve!

[@advaykumar9726]

Cool problem and solution

[@user-wp1hh1vj1b]

Wow idea is interesting.

[@leecherlarry]

my compi can't do it. no surprise!

[@SyberMath]

Yay! Finally!!! 😜😁

[@leecherlarry]

@@SyberMath lol!! haha

[@TheSimCaptain]

This was ok until the 7.00 minute mark, where it turned into a mathematical "Shell Game". When Y was replaced by X, that X was different to the X we started with. Same thing happened later when Z was replaced by another X. So we have three different values of X which are represented by the same X in the following equations. How does that make sense?

[@0404pipe]

Exactly! I don't get why those replacements are allowed.

[@SyberMath]

They are not the same x. They don't have to be. You can choose x to be whatever you want for that particular occasion

[@TheSimCaptain]

@@SyberMath Why did you use X and not another letter such as W?

[@elisabetk2595]

Keep in mind that the x here isn't the unknown to be solved for, it's just the placeholder used to describe a rule. The unknown the problem is asking us to solve for in this case is a particular rule, a function, namely, what is the rule for f? What rule describes it? Commonly we write f(x) when describing the rule but we could use f(z) or f(w) or f(whatever). It's how we manipulate thing in the parenthesis that counts.

[@michalchik]

At 7:50 by using X to replace both Y and Z isn't he supposedly assuming that y equals Z? What justification do we have to do that? They were solved against different X expressions.

[@SyberMath]

You can make any replacement you want. No justification needed

[@atlasufo7367]

thanks a lot for this beautifull work , i enjoy it .maths make me feel good best wishes from Algeria

[@SyberMath]

Hi! Thank you for the kind words! 💖

[@stellacollector]

Beautiful!

[@medmoh2390]

On doit vérifier que la fonction f satisfait à l'équation de départ puisque dans les étapes de la résolution on a procédé par implication et non par équivalence. Merci

[@drmathochist06]

SO much algebra could be dispensed with by using matrix multiplication and inversion on the linear fractional transformations.

[@teddyong4829]

yep, mobius transformations

[@justinnitoi3227]

This was a very easy question from putnam.

[@giorgiogasbarrini5438]

Formally there is an omissis. After the expression of X as a function of Y, you need to set out the condition that Y is different from 1. Same applies to Z different from -1.

[@aymanalgeria7302]

This channal is growing fast .. I hope the best for you ❤

[@SyberMath]

Thank you! 💖

[@timeonly1401]

Beautiful & clever! Love it. 😍

[@SyberMath]

Thank you! 😊

[@user-lh1yx6sb9x]

I solved problem in the same way, assuming g(x) = (x-3)/(x+1). Then g^{-1}(x)=(x+3)/(1-x), g(g(x))=g^{-1}(x).

[@SyberMath]

Nice! I like that!

[@unacademians6249]

@@SyberMath it's just the same thing just with a better tip of understanding..

[@nicogehren6566]

great solution sir thanks

[@SyberMath]

Most welcome

[@tonyhaddad1394]

Awesome !!! Good job

[@SyberMath]

Thank you so much 😀

[@tonyhaddad1394]

@@SyberMath 😍😍

[@sgdufbaoaah8692]

good

[@michaelbaum6796]

Very tricky, brilliant solution 👍

[@hsjkdsgd]

Nice one

[@Physicsnerd1]

Excellent!

[@SyberMath]

Many thanks!

[@sidimohamedbenelmalih7133]

This is really nice

[@SyberMath]

Thank you! 💖

[@MathZoneKH]

That’s great! It’s what I want functional equation!❤️❤️❤️pretty cool 😎 solution

[@SyberMath]

Glad you like it! 💖

[@laurentreouven]

It's f(g(x))+f(g-1(x))=x with g(x)=(x-3)/(1+x) maybe more simple like that ?

[@SyberMath]

I thought about it

[@tahasami5606]

Thank for sybermath

[@SyberMath]

You’re welcome, Taha!

[@imonkalyanbarua]

Amazing work sir! 😇🙏

[@SyberMath]

Thank you!

[@imonkalyanbarua]

@@SyberMath most welcome 😇🙏

[@vuyyurisatyasrinivasarao3140]

Super

[@statusking-514]

I like functional equations👍

[@SyberMath]

😊

[@buntheitkayowe]

I enjoyed

[@SyberMath]

Glad to hear that

[@kylekatarn1986]

The only part I do not understand is how you can return from the variables y and z to x just like that...

[@SyberMath]

You can use any variable you want

[@kylekatarn1986]

@@SyberMath and we find the solution in that variable?

[@tahasami597]

Thank you very good

[@SyberMath]

thanks

[@user-eh2ec3rn6w]

Great Solutions

[@SyberMath]

Thanks

[@sidimohamedbenelmalih7133]

I love it

[@Germankacyhay]

Як завжди вподобайка.

[@amh3139]

Wonderful I love this question

[@math_person]

I did it, phew!! Although I'll admit that wasn't sure I was going anywhere with it initially.

[@SyberMath]

Nice!

[@hsnrachid8299]

This mathematics is for the middle school student in Moroccan before you get to the secondary school so it's so easy even for me unfortunately I was very very bad at math

[@SPVLaboratories]

me and my homies HATE functional equations

[@SyberMath]

Sorry to hear that! :(

[@akechaijantharopasakorn2897]

Calculus is so beautiful.

[@Jon-xu7jl]

11:10 when you ask your girlfriend how many exes she's had...

[@nonoobott8602]

Clean solution

[@mvrpatnaik9085]

The way the rigorous problem is solved is quite effective

[@SyberMath]

Thank you!

[@pradyumnakumarnayak9384]

Namaste.

[@SyberMath]

Namaste

[@riyadkhalid3220]

Good method!

[@SyberMath]

Glad you think so!

[@jameshe3710]

It is a nice lecture. But is there a general solution if 3 and 1 are replaced by arbitrary a and b? Or it works just accidentally for this choice of numbers?

[@techysubham1939]

Wow

[@carlosrivas2012]

Excelente. Gracias.

[@SyberMath]

Thank you! 💕

[@carlosrivas2012]

@@SyberMath Todos los que miran tus videos, aprenden un montón. Yo estoy en ese grupo.

[@juanmolinas]

Beautifull solution!

[@SyberMath]

Thank you! 🤩

[@65ankitgujar40]

Can someone tell me why we can replace z with x after getting the second expression

[@platformofscience9790]

Thank you Please keep it up.

[@SyberMath]

Thank you, I will

[@alexandermorozov2248]

Занимательное решение! Как так удачно получилось, что из суммы двух уравнений с «y» и «z» вышла часть исходного уравнения, дающая в сумме икс? *** An interesting solution! How did it happen so successfully that a part of the original equation came out of the sum of two equations with "y" and "z", giving the sum of x?

[@aashsyed1277]

please do videos with matrices!

[@JACK-ho6cj]

VERY BASIC PROBLEM AND YOU TAKING SO MUCH TIME ON THIS

[@SyberMath]

Sorry

[@yemsan6843]

Good video!

[@SyberMath]

Glad you enjoyed it

[@route66math77]

Nice problem, thanks for sharing!

[@SyberMath]

No problem 👍 Thanks for watching! 😊

[@Darkev77]

I don’t see how at one step you used: “z = (x+3)/(1-x)” then when you arrived at a simplified version you were able to plug in “x” for “z” implying “x = z”. I think perhaps a 4th variable would make more sense?

[@SyberMath]

No you can use any variable you want

[@nitayweksler3051]

Am i suppose to solve it by using logic or is it supposed to be taught at some level of math?

[@mattdd9027]

Really pretty!

[@SyberMath]

Thank you! 😊

[@atanuroy3548]

Good math tricks.

[@bacanigoodvibes7795]

You're very good! Thanks for sharing :)

[@SyberMath]

Thank you too!

[@user-uo3ow1us7f]

Fuck, I spent an hour solving this. Just forgot to write some necessary step and covered the entire A4 page starting over and over. So stupid. I lost my mind in these scribbles but it came back on the next blank page. In an hour. Damn it, I am a furious anger.

[@mathisnotforthefaintofheart]

This is an interesting approach, but I don't think this method works in general, in other words, the Putnam problem is "rigged". I mean I tried my own problem: Find f(x) if f(4x+3)+f(2x-7)=x+3. Going through exactly the same process I arrive (after addition) 2f(x)+f(0.5x-8.5)+f(2x+17)=0.75x+15.75 which leads me nowhere

[@juuso4939]

It works here because the two terms are inverse function of each other, so in general: f(g(x)) + f(g-1(x)) =x -> f(x) = g(x) + g-1(x) - x It would be interesting to learn how to solve equations like your example.

[@arghamallick2126]

Found you today 😄... Looks very interesting !!

[@SyberMath]

Welcome aboard! 😁

[@3r3nite98]

Oh summation im late,but still ur awesome and entertaining us. Also here are some variants of me. Ok Now Im Ready Ok Now Your Ready Ok Now Hes Ready Ok Now -Shes- Ready Ok now were ready Ok now Theyre ready Ok Now Its Ready

[@SyberMath]

😂

[@nurettinsarul]

We can use that way if and only if y and z are the inverses of eachother.

[@farhadbahari720]

If f(g)+f(g Invers)=K then f(x)=[g+g Invers+K]/2

[@jlbarba2910]

Please could you elaborate on this. Thanks.

[@otisccx]

The substitution you do at around 7:00 seems dubious. y and z have already defined in terms of x, and x equals neither y nor z. How can you make this direct substitution?

[@abusoumaya8469]

no. you can replace variable with other as you wish you need just to respect the domaine. in other term, x and z and y and whatever are unkown variables so they can take all possible values

[@Lionroarr]

Good problem.

[@sarojtarei8446]

nice one

[@SyberMath]

Thanks

[@cuongtu6088]

good verry good

[@murusiva1549]

In what level does this mathematics come?it is amazing sim. iam maths graduate.I did not know theses types of sums.I am not teaching maths.

[@murusiva1549]

Which country r u from?your voice is nice .I am from srilanka.What is your exact. Profession?

[@SyberMath]

These are competition level problems ranging from middle school to college

[@user-lg6fv8mp7r]

老了，听着都头疼

[@MS-cj8uw]

Beautiful..

[@SyberMath]

Thank you! 😊

[@kt8362]

Love it