A functional equation from the Philippines.

We look at a nice functional equation from the 2011 Philippine Mathematics Olympiad.
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[@Opalescencz]

Its all fun until there's no function to find

[@mepoor761]

couldn't agree more

[@a.osethkin55]

exactly

[@gigelfranaru]

It's all FUNctions until there's NO FUNctions

[@mepoor761]

@@gigelfranaru literature xxxxx math

[@leif1075]

WAIT there is an answer that works of f(x) equals -×^2 + 1..bit you read thst not as squaring x and then multiplying by negstive 1 but rather multiply the input of the function by -x and then adding 1 .if you interpret it that way you get x^3 - x + 1 -x^3 + x which equals 1 and I don't see why this isn't valid..

[@bahrss]

Do not be afraid. That's the best hint in my life

[@konstantinoslazos9902]

Philippine Math Olympiad: Can you find all the functions? Me: I can't Michael Penn: True and that's a good place to stop

[@sadagoapan]

Problem: Find all functions Me: I can't Michael Penn: Yeah, you can't!

[@orenfivel6247]

Michael:... good place to stop

[@JasonOvalles]

I couldn't believe it! The first of these functional equations that I try and I swore I messed up. Spent an entire day working on it (off and on) to see what I did wrong. Then I come back and realize there was no solution. Well, at least I'm glad I got it correct!

[@kirill_glide]

Problem: prove it!

[@nasekiller]

not being able to find solutions is different from proving there are no solutions.

[@adityansingla5656]

6:09 we can directly use the first equation to show that it implies f(0)=1 as contradiction

[@pablomartinsantamaria8689]

But that's not a contradiction, is it?

[@adityansingla5656]

@@pablomartinsantamaria8689 we get f(0)=1 in case f(0)=0 which is a contradiction Here is the working if you want Case 1: f(0)=0 Putting it in f(f(0))=1 (first eq) We get f(0)=1 which contradicts that f(0)=0

[@pablomartinsantamaria8689]

Oh idk why but I though you were talking about case 2

[@adityansingla5656]

@@pablomartinsantamaria8689 Oh lol, happens sometimes :)

[@adude6568]

The worst part is that, assuming I'd be able to come up with this solution, I'd be insecure about there being no answers and I'd spend almost all of the exam time double checking the solution

[@tamarpeer261]

Same, just did

[@lychenus]

this is actually pretty fundamental. the solution can be a null set. not easy for junior high, but thats an important education

[@lexhariepisco2119]

Filipino subscriber here, ever since 10k subscribers thank you so much prof penn, im still a highschool student, and I really learned a lot ever sinced this 'michael-penn' community emerged

[@moonlightcocktail]

I am happy you finally covered my country, for one. I think...

[@KarlHsia]

but that does not exist.

[@leadnitrate2194]

@@catmeaow2484 how is it a wrong question? If there are no functions, that's the answer

[@leadnitrate2194]

@@catmeaow2484 but the proof was so interesting. It's not like they just made a question because it looked good, and put it in the paper.

[@targetiitbcse1761]

Penn is mightier than the sword!

[@prathmeshraut1616]

😂

[@ohgosh5892]

Clearly you have never met Mike Sword, known as Mike 'PennCrusher' Sword to his mates.

[@xxxx015]

That second hint is all what a man kind needs to succeed . btw i learn a lot of things here as a highschool student Some of them seem complicated for me such as switching the integrant and the summation or but mostly the content comes out self contained .lots of love sir🥰

[@andreben6224]

I love how you can prove there is no function, just by looking for the values at 0 and 1. There is no order or deep structure of R involved either and I believe this can be generalized to any function from any field (or ring) to any other field (or ring). Great find and great video ^_^

[@jimallysonnevado3973]

No you can't do it in any ring there are rings in which ab=0 does not imply a=0 or b=0. I think you need atleast a division ring.

[@toricon8070]

@@jimallysonnevado3973 I think you just need a domain (non-zero ring with no non-zero zero divisors). Division is unnecessary for this proof.

[@adityaekbote8498]

@@jimallysonnevado3973 Z/2Z?

[@UltraTrash2]

Dual numbers:

[@jonathanbonicel1654]

Love these kids noises in the background arguing whether the 127th decimal of pi is an even or an odd number. So cute.

[@littlefermat]

I agree with you sir! FE is my favourite topic in math Olympiad. That's why I started my KZbin channel with a functional equations tutorial 😁

[@philippezevenberg1332]

FE? wutisit?

[@chamsderreche5750]

@@philippezevenberg1332 abbreviation for Functional Equations, you might also see Alg(algebra), Geo(Geometry), NT(number theory) and combi(combinatorics)

[@Soul-cu8zn]

Hey i love your videos

[@SlidellRobotics]

6:06: Use the first identity: f(0)=0 implies f(0)=1. Toss it!

[@mistycremo9301]

For some reason, I proved that f(x) had to be injective and used that. That only made the lack of a solution a bit more disappointing, as that was the first time I'd ever proven a function had to be injective

[@Pacuvio25]

Nice problem! I used a different pattern: I showed that f(x) can't be 0, then that f is injective, and eventually proved that those two facts lead to a contradiction. Thanks!

[@drsonaligupta75]

I understood the first 2 points, but how did that lead to a contradiction?

[@aadfg0]

@@drsonaligupta75 f(f(f(x))) = f(1-xf(x)) = 1-f(x)f(f(x)) = 1-f(x) + xf(x)^2 by applying f in 2 different ways. Easy to prove f(f(0)) = 1, now put x = f(0) to get f(1-f(0)) = 1-f(f(0)) + f(0)f(f(0))^2 = f(0), so 1-f(0) = 0 -> f(0) = 1 -> f(1) = f(f(0)) = 1 = f(0), contradicting injectivity.

[@Pacuvio25]

@@drsonaligupta75 I chose a long route: I calculated f^n(0) until I got a contradiction (I found it for n=4)

[@shloksharma328]

They had us in the first half , not gonna lie 😂😂

[@mathkul3t]

Nice one sir🙂 ur fans from the philippines

[@thanosxypolytos4093]

I think the hardest thing for me is trying to understand the question, but in the end, it all made sense.

[@rainerzufall42]

The cases are a bit complicated! On the board before at 5:27, you can easily see, that f(0) != 0, because with f(0) = 0 and f(f(0)) = 1, you get f(0) = 1. That means, that f(1) = 0 and f(0) = 1 - f(1) = 1. But f(f(0)) = f(1) = 0, contradicting f(f(0)) = 1.

[@MathElite]

Awesome video, I learn a lot from these

[@adrien7933]

I think that exercice is very good for people who want to learn how to solve functional equations!

[@profesorjan7614]

Great, after me spending 30 minutes scribbling all kinds of nonsense on my whiteboard it turns out the answer to the question is that there is no answer. Great video as always by the way

[@backyard282]

"Well we did find all of the functions, it's just that there are no functions." This is such a typical mathematician kind of thinking

[@funkygawy]

i can't decide if this statement implies 0! = 1 or 0/0 = 1.

[@cross4326]

@@funkygawy 0/0 isn't 1.

[@funkygawy]

We found x functions, there are a total of y such functions. So both x and y = 0 (there are no such functions, we found no functions). If that proportion is x/y, and that is equal to "all", i.e., 1, then it follows that x/y=1 qed ;) perhaps I should express this as a limit

[@michalbreznicky7460]

​@@funkygawy All doesn't necessarily mean 100%. I guess what Michael means is this: (for all f) (if f satisfies the functional equation, then we found it).

[@funkygawy]

@@michalbreznicky7460 This was mostly tongue in cheek of course.

[@filippocona6232]

Fun thing is that if in that exam you left the solution blank you should have been awarded at least some points for finding the right solution without proving it thoroughly

[@nasekiller]

not really, since you would have to actually write out "there are no solutions"

[@babakbeheshti1]

@@nasekiller actually the problem doesn’t want to determine whether there’s solutions or not. It simply says find all solutions. 😂

[@ALC100percent]

"Don't be afraidfor possibly failing". Except your lovly professor surprises you with such a problem at his math tests...

[@brinzanalexandru2150]

This problem could be done in fewer steps:1) suppose we have some real number t such that f(t)=0 now plug x=t in yhe orihinal equation and we easily get f(0)=1 now replace in the original eqn. x with f(x) so after some simplification we get f(1-xf(x))=1-f(x)+xf(x)² now plug here x=0 and get f(1)=0(using f(0)=1) but plugging x=0 in the original implies f(f(0))=1 which is equivalent to f(1)=1-contradiction.

[@mathfincoding]

Oop, Philippine flag was posted the wrong way. If the red stripe is on the left side when the flag is displayed vertically, it means the country's at war.

[@jaykenarn6223]

war on drugs

[@mrpenguin815]

The country is at war with functional equations which have a valid function.

[@taufiqutomo]

Wait, what if I was viewing it from the other side?

[@mohammadshaanmdsalamsaikh6010]

I love how he replaced a mistake on board with *ine* :)

[@goodplacetostart9099]

Magandang Lugar Upang Magsimula 0:01

[@martothelfriky]

Interestingly, the solution doesn't involve the real numbers at all, we just operated on the assumption that f is a function from a commutative domain onto itself. Is there a non-trivial commutative ring with unity, R, where this function exists?

[@dirk_math6794]

When you put f(0)=A, you'll see A²=A. So just choose a ring with an idempotent element, e.g. 3 in Z/6Z (f(0)=f(2)=f(4)=3, f(1)=4, f(3)=1, f(5)=2).

[@nasekiller]

my solutions was a little similar (concerning the values i plugged in for x) but still different. first, lets note, that if f(x)=f(y) ≠ 0, then the functional equation gives us f(f(x)) +x f(x) = 1 = f(f(y))+yf(y), however, since f(x)=f(y) ≠ 0, this implies x=y (so the function is basically "injective", when the function value is not 0) now, lets denote a := f(0) (I) setting x=0, the equation yields 1 = f(f(0))+0f(0) = f(a) (II) this also implies a≠0, since otherwise we would have 0 = f(0)=f(a)=1, which is a contradiction. setting x=a, the equation yields 1= f(f(a))+af(a) = f(1)+a => f(1) = 1-a (III) setting x=1, we get 1 = f(f(1))+f(1) = f(1-a)+1-a => f(1-a) = a (IV) now, since a is not 0, we can combine (I) and (IV) with our "injectivity" to conclude 0=1-a, which implies a=1. however, plugging this into equation (II), we get f(1)=1, and in equation (III), we get f(1)=1-a=0, which is a contradiction. so there are no functions, which fulfill this equationl.

[@NonTwinBrothers]

Turns out we had all the functions all along

[@jatloe]

For the f(0)=0 case, that would make 1=f(f(0))=f(0)=0, which would be a contradiction.

[@gabrielbarrantes6946]

Couldn't we have guessed that by the structure of the problem? Usually this type of things are posted to be solved in Z and for R is quite weird.

[@brabhamfreaman166]

5:49 Pretty sure you meant to say “We figured out f(0)f(1) = 0 so at least one of f(0) or f(1) is zero”, rather than vice versa. But the meaning was clear and you’d stated it on the previous board; I guess the only difference is the possibility (up to that point) that 0 = f(0) = f(1).

[@captainsnake8515]

Beautiful solution!

[@BigDBrian]

for case one, since f(0)=0, f(f(0))=f(0)=0 which contradicts the first statement.

[@richardlongman5602]

The hardest part would be writing the answer symbolically. I would be inclined to set up a set stating the problem and equal to the empty set.

[@richardfarrer5616]

case 1 (6:20 or so) is quicker by noting that f(0) = 0 -> f(f(0)) = 0 which is a contradiction.

[@hamzasaleh8570]

We need to create a new subset and call it the nice set [NOICE]

[@evanmurphy4850]

Wondering if this is valid reasoning: Given the conditions, there must be some a in R with f(a)=0 Set x=a Then f(f(a))+af(a)=f(0)=1 Set x=0 Then f(f(0))=1 So f(1)=1 Set x=1 f(f(1))+f(1)=f(1)+f(1)=2f(1)=1 So f(1)=1/2 (contradiction)

[@ricardocavalcanti3343]

It's a proof that there is no a in R such that f(a)=0.

[@floriannom6691]

Great video. Funny problem. Thank you.

[@pratikmaity4315]

Here is a nice functional equation. Michael please try this and make a video. The problem goes like this: Show that there exists no function f such that f(f(x))=(x^2)-2. And enjoying your videos a lot!!

[@FadkinsDiet]

Almost all of these functional equations define a function that's linear (or constant). I assumed it was linear, and very quickly arrived at a contradiction. Then I assumed it was a finite polynomial and that also leads to the same contradiction. Realized that was the wrong path to go down because it doesn't prove anything, but it strongly suggests there's no solution.

[@juuso4939]

I prefer problems when there is indeed a function that might be something more complicated. For this problem the function was at least something else than linear or quadratic: kzbin.info/www/bejne/pX-3ZZytipuDm9ksi=zK8yuDmxrHzI8Ed9

[@edwardlulofs444]

Very nice. Good work.

[@soranuareane]

Exercise in pattern matching. This sounds like a PROLOG candidate. Thoughts?

[@MadaxeMunkeee]

That's kinda cool. I kinda expected there would be something of that form but apparently not.

[@cactusmanfr6900]

If f(x) is going from R to R, there should be a value y so that f(y)=0 . If we evaluate f at x=y, we have : f(f(y))+y*f(y)=1 => f(0)+y*0=1 => f(0)=1 Then we have f(f(0))=1 => f(1)=1 (from evaluating f at x=0). But we also have (x=1) : f(f(1))+f(1)=1 => f(1)+1=1 => 1+1=1 , which is a contradiction. Is there any error ?

[@Junieper]

You can’t assume there is a y such that f(y)=0. F(x)=1/2 is a function from the reals to the reals

[@HagenvonEitzen]

7:35 Finding all such functions is easy, but finding any such function is impossible :)

[@emilkaczmarek4641]

I think I came up with a nice problem, based on this one. Find all functions satisfying the equation f(f(x)) + xf(x) = 2f(x), domain not specified. Any chance you could solve it?

[@bot24032]

6:12 Set a = f(0) = 0 (did this for better understanding) 1st equation f(f(0))=1 f(a)=1 f(0)=1 0=1 Wrong

[@ramgobindram7402]

How about replacing x with f inverse x .getting f(x)=1-x^2 which is not invertible in real domain so it contradicts so no soln.

[@rubetz528]

I bet it says 2011 Philippenn Mathematics Olympiad in the corner xD

[@1ab1]

write f(f(x))+xf(x)=1 as f(f(x))=1-xf(x). so (plug x to be f(x)): f(f(f(x)))=1-x(f(f(x))) therefore: f(1-xf(x))=1-x(1-xf(x)) now plug x=0: f(1)=1 and we get: 1=f(1)=f(f(1))=1-f(1)=0 and that's a contradiction. so there is no function f:R->R satisfying the conditions.

[@digxx]

f(f(f(x)))=1-x(f(f(x))) is wrong.

[@shankhadeepghosh8574]

Just awesome solution it was

[@raygean1639]

After watching this for a week till today I tried to solve it myself and I found that I got that f(x)=2^1/2 but that doesn’t seem right

[@orenfivel6247]

the question is: can we find such a function in different domains eg complex numbers, square matrices or any Algebraic structure etc?

[@evannibbe9375]

Complex number functions might as well be considered as a function with two inputs. A function that takes a vector is a function with an arbitrary number of inputs.

[@johannesh7610]

The same reasoning works in any unitary ring without zero divisors instead of the reals. But probably not with matrices (zero divisors)

[@Andreyy98]

I passed through the exact steps. I had even written them in points. I checked 3 times for mistakes as I didn't expect to be no answer. In the end I gave up and watched the video to see there are no solutions which kind a left me with mixed feelings :D

[@goodplacetostop2973]

7:41 Magandang lugar upang huminto No homework today but if you want a particular topic for them in the next days, tell me

[@pbj4184]

Could you tell what's the text etched on the stone in your profile picture? It's impossible to read on a mobile

[@goodplacetostop2973]

@@pbj4184 This is the Rifle Mountain Park.

[@pbj4184]

@@goodplacetostop2973 Photos of it look beautiful. Would you mind if I asked why this photo for your profile picture?

[@goodplacetostop2973]

@@pbj4184 Since you like to dig up and find hidden things, it’s gonna be your homework to find why. It’s not that hard 😛

[@pbj4184]

@@goodplacetostop2973 Is it because it has many climbing routes? I tried to post the link to their website but it wouldn't post the comment

[@JohnSmith-nx7zj]

So if you leave your answer sheet blank for this question do you get full marks?

[@yashvardhan6521]

I used a different approach sir but arrived ultimately at the same conclusion as u. Firstly I did the same steps and got f(f(0)) =1 and f(0) +f(1) =1 Then I got the equation from the first that f(f(f(x))) =1-f(x) +x[f(x) ]² Then we get f(0) =1-f(1) =1-1.f(1) =f(f(1)) Then applying f on both sides we get f(f(f(1))) =1 and using an above equation we get f(1) as 1 or 0 in both cases which f(0) is 0 or 1 giving a contradiction and hence no solutions. Though we can arrive at the contradiction fairly easily, there is yet another method I found. First examine case when f(0) =0 Then if we let c be a root other than 0 , then after using c in the given equation we get a contradiction. Thus zero is the only root and we get f(x) =x^n but using this in the above equation we see that x^(n²) +x^(n+1) is not constant for all x belonging to the reals and thus a contradiction. And the second Case is proved ordinarily.

[@AlexAnastaso]

Can we make equations like that by violating bolzano ?The equations needs to violates bolzano for every set of points (x,y) or (x,f(x)) ofc we need to force the function to be finite so we can have Bolzano in a weak sense (u can connect the points somehow)

[@aditaggarwal3486]

Nice and easy problem to start my day. Took me a good 5 minutes.

[@tushroy81]

Love you bro

[@enisheadpay]

All of this is based on the very first assumption that when x=0, x*f(x)=0. But what if f(0)->infinity? Then x*f(x) is not necessarily zero. I wonder if a function of the type ln(x) would give some kind of results that work.

[@Junieper]

F is a function from the reals to the reals, and infinity is not a real number.

[@Asterism_Desmos]

I’ll leave finding the functions as an exercise to the reader.

[@SurfinScientist]

Didn't Michael only prove that functions satisfying the functional equation are not defined for x=0 and x=1? That would not necessarily mean that there are no functions satisfying the functional equation.

[@tisyarawat3638]

very elegant solution :)

[@EebstertheGreat]

This video only asks me to "find" all solutions, not to prove that there are no other solutions. So I guess I succeeded before I even clicked the video.

[@criskity]

There seems to be some commotion in the background!

[@dirk_math6794]

If you want a solution: When you put f(0)=A, you'll see A²=A. So just choose a ring with an idempotent element, e.g. 3 in Z/6Z (f(0)=f(2)=f(4)=3, f(1)=4, f(3)=1, f(5)=2).

[@bjornfeuerbacher5514]

Shouldn't Michael's argument work also in a ring with an idempotent element? I.e. in that case, there also shouldn't exist any solutions.

[@aakashtiwari3223]

Certainly thought that at the starting of the video that there'd exist atleast one function satisfying the given relation considering the language of the question ("find all..."). He he, anyway the number of all the functions are zero.

[@williamrutherford553]

That first case is more complicated than it needs to be. Just look at the first equation, f(f(0)) = 1. If f(0)=0, we can plug that in to get f(0)=1. Immediate contradiction!

[@tokajileo5928]

should not you consider that f(x) may be undefined at x, like 1/x at x=0? ?

[@rafael7696]

Very good explication

[@conrad5342]

Well, this solution path implies there is no pole at x=0. Is there a function g(x) when you substitute f(x)=g(x)/x ? ... OK, now I see why this has not been discussed previously. The function f(x) needs to be defined for all real numbers.

[@JI-cp4pk]

Alright Filipino Represent

[@chispun2]

How is it called this arcane art within mathematics?

[@icfj77]

If we take f(f(x))+xf(x)=a, we have that a=0 if we want this equation to be solvable. There is at least one solution for a=0: f(x)=0.

[@riccardofroz]

I spend one hour thinking I was doing something wrong. And when I watched the video I realised that the contraddictions I was having were because there are no solutions!!! f(f(x))+xf(x)=1 f(a)=1 x=a f(1)+a=1 f(1)=1-a x=1 f(1-a)+1-a=1 f(1-a)=a x=1-a f(a)+(1-a)a=1 (1-a)a=0 a=0 f(1)=1-a f(1)=1 f(1-a)=a f(1)=0 a=1 f(1)=1 f(1)+a=1 f(1)=0 contraddiction in both cases

[@udic01]

6:06 f(f(0))=f(0)=1. It's shorter (by one line ;-) )

[@juuso4939]

These kind of problems are too often "no solution" or trivial solution like a constant. It would more fun to actually get something more complex out.

[@jucom756]

I reached the conclusion by fist saying take f(x)=0 and by that equation getting f(1)=1 and f(0)=1, then just replacing with either and getting f(1)+1*1=1+1=2=1

[@digxx]

How do you know, that x€R s.t. f(x)=0 exists?

[@prithujsarkar2010]

Great! plz make more videos on functional equations :D

[@ludviglagerstedt5726]

Does this mean that if you didn't solve the question you technically got it right cause you found all the functions?

[@musaburakerdihan5148]

Micheal gives hints, Eminem : Not afraid !

[@jackhandma1011]

There are no functions. - Master Oogway

[@davedave3520]

There are no Oogways - Master function

[@josash]

That was really heplful.

[@shaiavraham2910]

In terms of functions, we have no functions

[@ARKGAMING]

That really is a good place to stop

[@0dWHOHWb0]

So the answer is just { } ?

[@nodamnnickname]

yes

[@TJStellmach]

Pretty sure this one requires only that f be a function on *any* field.

[@amit2.o761]

If a/(b-8)=18 then what will be a/b ? And also if à/(ß+8)=18 than what will be à/ß ? Can u help me

[@flowerwithamachinegun2692]

You can't really say. That's because 18 can be written as a fraction in infinitely many ways, so you have infinitely many pairs (a,b) which satisfy the equation, but their ratio, a/b, differs. In fact, and if I'm not mistaken, a/b can take almost any real value, except 0 (in which case you'd get a=0, which leads to 18=0 absurd) and 18. This can be easily seen if you take a=b*k, where k is some real constant, and you plug this back in the original equation.

[@amit2.o761]

@@flowerwithamachinegun2692 thanks

[@digxx]

Once you have f(f(1))=f(0) you can also prove the following: f is injective: Suppose x,y€R s.t. f(x)=f(y), then 1-yf(y)=f(f(y))=f(f(x))=1-xf(x) i.e. (x-y)f(x)=0. So either f(x)=f(y)=0 for those inputs, or x=y (i.e. f is injective). Now suppose there existed some x€R s.t. f(x)=0. Using the equation, we find f(0)=1=f(f(0)). Using the injectivity gives f(0)=0, a contradiction. Thus no x€R s.t. f(x)=0 exists. Now using the above result f(0)=f(f(1)) and the injectivity gives f(1)=0, again a contradiction by the just proven fact. Thus, a function that outputs a value for every real input does not exist.

[@nanachichi1044]

I thought it's gonna be about the geometry of the flag

[@QuaDue]

very nice one even with the small complication in case 1. Forests hiding trees I guess

[@gigagrzybiarz]

Very interesting problem

[@klementhajrullaj1222]

Who's screaming there?! 😲😲😲

[@mstarsup]

case 1 not valid from your first equation alone: f(0)=0 ==> f(f(0))=1 not true.