Finding f(0) When f(f(x))=x^2-x+1
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- Опубликовано: 12 июн 2022
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Is the problem from Baltic Way?
No idea
@@SyberMath I found, Baltic Way 2011, A-5
@@alfreds1347 Wow! Nice! Thanks for the find 🤗🥰
If f(f(x)) = x+1/(x-1) then Find f(0)
Only need to notice that f(x)^2-f(x)+1=f(f(f(x)))=f(x^2-x+1). Settinh x=1 gives f(1)=1, and setting x=0 gives f(0)=1 or 0. Then do the final check.
Wow!!! 🤩
That's what I did too
Also how I found it. Seems to be less convoluted altogether
I haven’t rigorously checked this, but it seems like this might just be really close to what’s shown in the video, but without the relabeling
You got that beautiful buddy.. Seems we all are attracted to SyberMath's graphics so much that we binge some old videos too😁
Function cannot have multiple results
But a question can have multiple functions as the result
That was my first thought, too. But @SyberMath has a point about always checking your results against the intitial question.
E.g: if the question is to find the side length of a square with an area of 4, you can write an equation of x^2=4 which has multiple solution but only one of them is a valid answer to the question.
But who said it's a function, not a relation?
Great video and a really clear explanation. I love how you almost walked through the thought process for this, explaining each stage step-by-step with no huge leaps in logic. I guess I'll put it like this, I did not have to pause once in this video because of an unclear explanation. Keep up the great work!
Thank you for the kind words! 🥰
Important to note with functional equations is that multiple values for an input can also just imply multiple functions complying with the equation
That said, you would still check both cases if you reach this point
I try so hard to create math teaching videos for the student in Cambodia, Although there are no more subscriptions or views, I do it from my heart to be an introductory video when they need it at one time. Thank SyberMath for a goodvideo.
I think we can all appreciate your dedication. Surely you've helped at least one person, and it probably made a difference for him!
Bro , your video quality is excellent , but the problem is we don't understand the language you are communicating.
@@safwanuddin8888 thank brother, it is a khmer language, from cambodia, Angkor wat temple.
@@mismis3153 thank😍
Bro it looks like you put a lot of effort doing your videos, I would watch them if I knew the language but they do have a high quality from what I saw
The solution can be simpler:
let x=0,
f(f(0))=0*0-0+1=1,
let f(0)=a, therefore f(a)=1
also, f(f(a))=a*a-a+1 = f(1) =1
then a(a-1)=0, aka a=0 or a=1
if f(0)=0, f(f(0))=0-0+1=1=f(0), contradict!
if f(0)=1, f(f(1)) = 1-1+1=1=f(0), check!
therefore, f(0)=1
You are right
It’s not simpler, it’s the same thing. you’re still missing the proof f(1)=1.
However it is the correct way to start by x=0 and then you have to find out you have to calculate for x=1.
The way it is presented in the video (starting by calculating f(1), f(0) is asked) is based on hindsight.
The solution as presented in the video (starting by calculating f(1) when f(0) is asked) is based on hindsight.
The correct way is to start by x=0 and then you find out you have to calculate f(1), which is solved by setting x=1 and solving f°f°f(1)=f(1)=f^2(1)-f(1)+1
Very cool! As for the question at 7:47, “Does that automatically imply f(0)=1?”, the answer should be “Yes, it does”. Unless I misunderstood the problem, its very statement ensures the existence of function f.
Sometimes problems have no solution. Even if the problem presupposes the existence of f(0) it could be that this supposition is wrong and f(0) actually does not exist. A good example of this is x^2=(-1) solution does not exist for real numbers and we have to construct a whole new number system with imaginary numbers to force a solution to exist.
@@kazedcat Yeah. But in this case Syber didn't prove that f(x) is well-defined, only that the argument that shows that f(0)=0 doesn't work does not also show that f(0)=1 doesn't work. It could still be the case that no function satisfies the equation, so the extra work has not achieved anything.
A more rigorous way to finish the argument would be to give any example function that satisfies the functional equation, if we don't think the question allows us to assume one exists.
Seeing this remembered me my course on functional programming and the concept of streams
I boycott all products advertised by RUclips. Have a nice day.
Wow sir. Your videos teach me a new thing every time!
The thing that I like most about you is that you explain the techniques neatly which is really helpful for students.
You preparing for jee dude??
@@Vishnu-dk2rc yes bro. Qualified mains now let's see what happens in advanced. You?
@@arpangoswami5760 i got 15000 in adv last year....and now in a tier 3 college.....
I thought I could get in without proper chem 😂😂😂...
@@Vishnu-dk2rc why Tier 3.....? I mean 15,000 rank is not bad. Literally a shit load of students appear for the exam. So securing 15000 rank is quite good
@@arpangoswami5760 I'm general merit dude 😂😂. I got 95% in mains but no seat in ece or eee which I wanted. So I joined a state govt college through kcet...
What is an example function that satisfies this constraint?
This is incredible. Could we come up with a general method of applying any function 0.5 times? The "square root" of any function?
Easy peasy!
Let some function say g(x) = 2x+1,
⇒ g(g) = 2g+ 1
now let f(x) be the composition, f(x) = g○g(x) = g(g(x))
here f(x) = 2(2x +1) + 1 = 4x+ 3.
so we may state that "2x+ 1" is the _functional_ square root of "4x +3"
(note it is not the arithmetical √(4x +3) however, be careful)
@@tomctutor what you demonstrated is to find a function composed with itself. Tim is asking for a generalized method to find the g(x) from g(g(x)).
For example, if g(g(x)) = x^3 - 2x^2 + 1, what’s g(x)?
And to Tim, it is generally not possible to determine all possible f(x) based on just the expression of f(f(x)). Let me give a short example and then talk about the general reason this is difficult.
Say, f(f(x)) = x, the identity function, and we are looking for f(x). Now both f(x) = x and f(x) = -x will satisfy this requirement. But I can also define a piece-wise function f(x) = “if |x| < 3, then -x; if |x| >= 3, then x”, and upon inspection, this also will make f(f(x)) = x true. Obviously the choice of 3 is arbitrary, so you can produce as many such f as you want, and they can even be broken down into many pieces.
Now you may ask, then how come the problem in this video makes sense? In fact, this is exploring what’s known as a fixed point in a continuous function. If we call f compose f a new name, h; then we have h(1) = 1 for the f in this video. This is a fixed point of h. Now f(f(f(x))) can be interpreted as both f(h(x)) and h(f(x)). This problem is crafted so that f(h(1)) = f(1) = h(f(1)) can be solved, and it actually yields another fixed point for f. And that fixed point happens to be 1 again. (Try replacing the prompt’s right-hand side for a different polynomial, you will see that it falls apart here) The fact that we have a fixed point for f is the only way we are able to determine something non-trivial about the nature of f.
Personally, I find the solution of this problem intentionally hiding away a lot of intricate (but interesting nonetheless!) details that makes this problem work.
@@gal6145 I know just clarifying his question. you could try g(x)= P(x) a polynomial, in my case assume g(x)= (ax+b) works, but its a bit like integration, just a matter of trial and error.
@@gal6145 Is there a name for the solvability criterion or some keyword that can be looked into?
I have stopped studying math for a long time, but it was really fun to watch this.
Arkadaşlar bu ortalama bir yks öğrencisinin çözmesi gereken soru
Hi,
my approach to find the solution is as follow:
F(x) = x^2 -x +1
F´(x) = f(x) = 2x-1
F´( x) =0 Tangente and per definition @ x of 0 , therefor :
f(x of 0 ) is set to Zero, meaning ------> 2x -1 = 0 ----> x of Zero = 1/ 2 -------> which refers to the Point P ( 1/2 ; 0) or point of Tangency to the curve F(x).
Now :
By substitution x=0 and evaluating f(0) = 2( 0) -1------> f(0 ) = - 1
Conclusion :
By plotting the graf of F( x) and considering P (1/2 ; 0) as condition for Tangent Line and ONLY in Domain [0
Not really useful here because your F'(x) = f(x) = 2x- 1 leads to
f(f(x)) = 2(2x -1) -1 = 4x - 1 which is not that originally given.
Or are you referring to a general method using calculus with some other example?
stupid solution
@@tomctutor hello there. i´m sorry for not seeing your solution until now( today )🙏.
so, Jes, in this sence < ... it is a general form/ equation in order to have any answer with F(x). As you put very nicely, my calculation shows , if & only if there is a solution. As you see it does NOT work with initial condition given at the beginning either.
if it helps just imagin : you have a Circle & F(x) toches it down at one small rigion, bc...----> we ´ re talking about < very smal section of F(x) > and Not the whole F(x). ✌
these damn composite functions, always had tricks up their sleeves that had nothing to do with anything
This problem, like most other functional equations you posted on your channel suffers from one major weakness. You do not specify over what domain the functional equation holds. In your reasoning, while assuming the existenance of a and b, you assume that they might have any value and then limit the possible values at the end. However in the end, your reasoning only holds if there is really a function f, and you would need to know the domain of it. Of course, if you limit the domain to the numebrs 0 and 1, then f exists and f(x)=1 is a function that satisfies the equation for 0 and 1. It would however be much more interessting to find out if there is a function f that is defined either for all real numbers, or at least for a non trivial subset that stisfies the functional equation. And then of course, the problem could also be extended to complex numbers.
Good point. Check this out:
yaroslavvb.com/papers/rice-when.pdf
@@SyberMath Interresting! So in essnce, it has been proved that f does not exist. So we calculted f(0) for a non existing function f
@@MarcelCox1 No. Read the paper carefully. It has only been proved that f does not exist if you take the domain to be an algebraically closed field. The paper above actually proves, near the end, that there are indeed solutions to the equation over certain domains, particularly the real numbers. For the real numbers, given that (f°f)(x) = a·x^2 + (b + 1)·x + c, roots exist if and only if b^2 - 4·a·c > 1 is false. Here, a = 1, b = -2, and c = 1. So b^2 - 4·a·c = 4 - 4 = 0 > 1, which is indeed false. Therefore, solutions do exist over the real numbers, but not over the complex numbers.
I agree with your criticisms of the video, though, and I do fundamentally disagree with the mathematical community of YT as a whole on establishing this type of video as a trend without providing proper mathematical rigor for it. There is nothing that you or I can do about it, though.
@@angelmendez-rivera351 You are right. I have to admit I did not fully the study, and as a well versed, but non professional in maths, the article touches things that might be a bit challenging for me.
Anyway, to bounce back on your last comment, I follow the channel of Michaell Penn which overall posts slightly more difficult problems, but for functional equations, he gives more context and overall he has more rigor.
In general though we kind of typically assume that f(x) is a function on R and f(z) is a function on C. I think it's fine for this type of content.
If the problem were like: prove that any function that satisfies this equation satisfies also that f(0) = 1 than your solution is ok. But the question is to find f(0) that means that if there is no such functions than the answer is: the set of possible values for f(0) is empty. So you should find an example of such f.
very nice work sir!
This is a perfect explanation. Syber Math, you are the best.
Wow, thanks! 🥰
How did you avoid ending up with garbage as you often do when you keep inserting an equation into itself?
Obviously no mathematician here, but this was amazing, thanks!
My solution:
f(f(x))=x²-x+1
f(f(0))=1 ==> f(0)=f⁻¹(1)
f(f(1))=1 ==> f(1)=f⁻¹(1)
The intersection of a function with its inverse is always on the y=x line, except for when the function is itself its own inverse. We know that f(x) is not its own inverse, because then f(f(x)) would be equal to x, but that is not the case.
Therefore, f(1)=f⁻¹(1)=1.
But f(1)=f⁻¹(1)=f(0).
Thus, f(0)=1.
We don't know that f has an inverse
@@TheEternalVortex42 Even worse, we know that f is not invertible because f(0)=f(1)
Hi Syber,
I really like this problem too since it demonstrates a very nice technique in solving fe. Actually my second fe video was discussing this problem😁
Nice video. I enjoyed this question in Japan. Math can spread it overseas.
I did this question with a different approach:-
I first replaced all f(x) by x and all x by f(x), in the given equation, so it became,
f(x) = (f(x))² - f(x) +1
We then get,
(f(x))² + 1 - 2f(x) = 0
Therefore,
(f(x) - 1)² = 0
Square rooting both sides we finally get,
f(x) = 1
Therefore, for all values of x belonging to it's domain, f(x) = 1
Hence, f(0) = 1
Wow. This is so meta, it’s like trying to see into the fourth dimension.
Great comment! 🤩
Great explanation as always
Thank you!!! 🥰
Nice! If only one could solve and find what f is.... (without making adsumptions about f.).
I had a shorter solution, not sure if anything is wrong with this?
I called the unknown f(0)=y
f(y) = f(f(0)) = 1 (by putting x=0 in the original equation)
f(f(y)) = y²-y+1 = f(1) (follows from previous)
f^3(y) = f(f(1)) = 1 (by putting x=1 in the original equation)
f^4(y) = f(1) = y²-y+1
So we have
f(y) = 1
f^2(y) = y²-y+1
f^3(y) = 1
f^4(y) = y²-y+1
and so on. It's clear that operation f will always place 1 -> y²-y+1 and y²-y+1 -> 1. Therefore y = y²-y+1 (because f(y)=1), which gives y=1.
I used a slightly different method to solve this.
I started by setting x=1. That means f(f(1))=1^2-1+1=1.
I then set a=f(1), and x=a. Then, f(f(a))=a^2-a+1. But we know that f(a)=f(f(1))=1. So, f(f(a))=f(1)=a. Therefore, a=a^2-a+1, which means a^2-2a+1=0. You can factor this to get that (a-1)^2=0. This means that a=1, which implies that f(1)=1.
Next, let's set x=0. That means f(f(0))=0^2-0+1=1. Now, set b=f(0), and set x=b. So, f(f(b))=b^2-b+1. However, we know that f(b)=f(f(0))=1, which means f(f(b))=f(1)=1. So, 1=b^2-b+1, which means b^2-b=0. By factoring, b(b-1)=0. That means that either b=0, or b=1. We cannot have both, since f(x) is a function. So, which is it?
We know that f(f(1))=1=f(f(0)). This implies that f(1)=f(0). Since we know that f(1)=1, that implies that f(0)=1.
"We know that f(f(1))=1=f(f(0)). This implies that f(1)=f(0)."
@@jmiki89 Huh. I guess I'm a dumbass then.
@@nachocheeseonpizzawithextr6899 I wouldn't say that, everybody makes mistakes, but you need to be mindful with your hidden assumtions. I sincerelly hope you've learned something from this mistake which will help you avoid it next time. Because it a great thing that you tried to come up with your own solution, keep it up!
Your proof is almost correct, except for the mistake at the end, which you already acknowledged. However, you said f(0) cannot be both 0 or 1. But it can be, and all it really means is that there are multiple functions f such that (f°f)(x) = x^2 - x + 1 everywhere.
It is not entirely true that function always have one value, example being lambert w function :)
How about if use the inverse identity..?
I think in this way:
Define g(x) = x² - x so:
f(f(x)) = g(x) + 1 => f(f(0)) = g(0) + 1 => f(f(0)) = 1.
But, by definition
f(f(0)) = [f(0)]² - f(0) + 1 = 1 => [f(0)]² - f(0) = 0, and, therefore, f(0) = 0 or f(0) = 1.
Suppose f(0) = 0 so
0 = f(0) = f(f(0)) = 0² - 0 + 1 = 1. Contradiction!!!
Therefore f(0) = 1.
You assumed, in different points, that the equation has a function that solves it, and it is unique. As many pointed out, the initial functional equation can have multiple solutions, some with F(0)=1 and some with F(0)=0. OK, but you did prove that F(0) cannot be zero. However, does that imply that it is equal to one? well, you did not prove that the initial equation has a solution at all.
In short, what you proved is that, if there exist a function that solves the equation, then for that solution F(0)=1
Question from a Math-Dodo: Does looking at "nested" functions, like f(f(x)), have any practical application? To me, it appears to rather be a theoretical discussion.
In deep learning, the final class label is a nested function of sigmoids, tanhs and all other sorts of functions.
Also, in functional analysis, there are many theorems on the composition of T and S (2 linear functionals- also a type of function) which bring conclusions on S or T.
years ago this problem made for "iran university entrance exam", so many 18 years old answer it in iran, but last year in 2 international compatition noone answer it. lol 😂
Nice videos ! I just realized that we were actually trying to find f(0) not f(x), is it possible to find f(x) from just f(f(x)) = x²-x+1 ? Also, can u plz explain the principes of solving some equations like what do we need to know before getting into the resolution, anyways appreciate it, have a good day !
f(x) is not polynomial and neither is possible to find its form.
Depending on what the domain is, there may or may not be a function satisfying the equation. However, even if it does exist, it cannot be expressed using any familiar functions.
@@angelmendez-rivera351 f(f(x)) and f(x) and x are 3 variables, and we are lookin for f(x) (in terms of x) so we need (n ( which is the number of variables) -1) equations which should be 2 equations but we have one equation, and what if u suppose that f(x) = x .. Also, if f(f(x)) is not a constant so f(x) also is not a constant (yes ?) So f(x) qhould be in the form of (d°(f(f(x))) - 1) = 2-1 = 1 so it is in the form of ax + b ? Is that helpful to just find f(x) ?
@@hacker64xfn99 What makes you think that d°(f°f) = 2? Also, supposing f(x) = x obviously does nothing, since then it follows that f(f(x)) = x.
Thinking of this in terms of variables is the wrong way to approach it, since this is a functional equation, where f°f, f, and id are not all independent of each other. The only variable here is f, and f°f is equal to the fixed object id·id - id + 1, which is the function analogue of x^2 - x + 1. id is also a fixed object: the identity function. So the approach that we must take to solve the equation must acknowledge this much.
As for your question of f(x) = A·x + B, we can easily prove this cannot be the case, because then f(f(x)) = A^2·x + A·B + B, and as you can see, this can never be a quadratic for all x, unless the domain is restricted to be the set of all x in a ring such that x^2 - x + 1 = A^2·x + A·B + B, but this has several problems. Meanwhile, f cannot be a quadratic function, because if it were, then f°f would be a quartic. So, f is in general not a polynomial or power-based function. This is the issue at hand.
@@angelmendez-rivera351 d°(f(f(x)) = 2 is the degree of that function ,that is what I meant, and supposing f(x) = x will lead to : f(x) = f(x)²-f(x)+1 therefore we have an equation containing f(x)
And if we suppose that f(x)= ax+b : x = (f(x) - b)/a so f(f(x)) = ((f(x)-b)/a)² - ((f(x)-b)/a) + 1 we can see from this last equation that f(x) on both sides if we replace f(x) by x (we did not suppose that f(x) = x, but we can still do that for functions) we get f(x) = ((x-b)/a)² - ((x-b)/a) + 1 and as a result we find that :
ax + b = ((x-b)/a)² - ((x-b)/a) + 1
I don't know how u find ffx in ur comment and not sure why f(x) is linear at first (ax+b) and then turns to a quadratic (((x-b)/a)² - ((x-b)/a) + 1) ?
thanks for solution
Just a question about f(x) given f(f(x)). For examples:
if f(f(x))=x we find f(x)=x and f(x)=1/x also.
If f(f(x))=x⁴ we can find f(x)=x² and f(x)=1/x².
Which function f(x) can lead to f(f(x))=x² ?
That depends on the domain of the function. For example, you said that the reciprocal function solves the equation (f°f)(x) = x, but this is not true if the domain of f°f is the set of integers. The definition of a function requires that the domain of a function be specified. Otherwise, whatever you are dealing with is not actually a function, it is some ill-defined mathematical idea. As such, with functional equations, the domain must be specified. This is one criticism I have always had of mathematics on YT: very often, videos about solving equations are made, but they are made without the proper mathematical rigor to validate any interesting results you can find.
Now, let me get more specific. Suppose F is an algebraically closed field. Then if f°f : F -> F, it can be proven that (f°f)(x) = x^2 has no solutions. On the other hand, for arbitrary rings, you have more freedom. For example, if F = R, the field of real numbers, then a solution does exist: f : R -> R given by f(x) = |x|^sqrt(2), since (f°f)(x) = ||x|^sqrt(2)|^sqrt(2) = (||x||^sqrt(2))^sqrt(2) = |x|^(sqrt(2)·sqrt(2)) = |x|^2 = x^2.
@@angelmendez-rivera351 thanks Angel for your explanations.
@@pageegap If R is the ring R = {-1, 0, 1} where 1 + 1 = -1, then (f°f)(-1) = (-1)^2 = 1, so f(1) = f[(f°f)(-1)] = (f°f)[f(-1)] = f(-1)^2, so if f(1) = 0, then f(-1) = 0. Therefore, one possibility is that f(-1) = f(1) = 0, f(0) = 1. Another possibility is that f(1) = 1, so f(-1) = -1 or f(-1) = 1. If f(-1) = -1, then (f°f)(-1) = f(-1) = -1, which contradicts (f°f)(-1) = 1, so f(-1) = 1. Therefore, the other possibility is that f(-1) = f(1) = 1, f(0) = 0. These are the two functions f : R -> R such that (f°f)(x) = x^2 everywhere. These are given by f(x) = 1 - x^2 and f(x) = x^2, respectively.
If R = {0, 1, α, α + 1}, where 1 + 1 = 0 and α^2 = α + 1, then (f°f)(α) = α^2 = α + 1, so f[(f°f)(α)] = f(α + 1) = (f°f)[f(α)] = f(α)^2, and (f°f)(α + 1) = (α + 1)^2 = α^2 + α + α + 1 = α^2 + 1 = α + 1 + 1 = α, so f(α) = f[(f°f)(α + 1)] = (f°f)[f(α + 1)] = f(α + 1)^2 = f(α)^4, hence f(α) = 0, or f(α)^3 = 1. If f(α) = 0, then f(α + 1) = 0, but this contradicts f(0) = 0 or f(0) = 1, since (f°f)(α) = f(0) = α + 1. Similarly, f(α) = α contradicts (f°f)(α) = α + 1, and f(α) = α + 1 implies f(α + 1) = α + 1, which contradicts (f°f)(α) = α + 1, so f does not exist.
If, instead, R = {-1, 0, 1, 1 + 1}, where (1 + 1) + (1 + 1) = 0, then (f°f)(1 + 1) = (1 + 1)^2 = (1 + 1) + (1 + 1) = 0, so f(0) = f[(f°f)(1 + 1)] = (f°f)[f(1 + 1)] = f(1 + 1)^2, so if f(0) = 0, then f(1 + 1) = 0, which does help satisfy the functional equation. Therefore, f defined by f(-1) = f(1) = 1, f(0) = f(1 + 1) = 0 is a solution. If f(0) = 1, then f(1 + 1)^2 = 1 implies f(1 + 1) = -1 or f(1 + 1) = 1, so the other solutions are f(1 + 1) = -1, f(-1) = f(1) = 0, f(0) = 1, or f(1 + 1) = f(0) = 1, f(-1) = f(1) = 0.
Hopefully this gives you another idea of how it works.
x^(sqrt(2)) , x to the power of sqrt(2) or inverse of it as you did with square number powers..
So x=a=b=1. Why not introduce even more variables? a0=b0 for example
I had gotten the right answer accidentally with wrong logic by taking the square root of "x^2 - x +1" for x=0, but obviously that was not right. If you have a simple function like f(x] = x^2, then f(f(x) = x^4, and you could work backwards to f(x) by taking square root, but not for a polynomial with mutiple terms. For this video, if f(x) had been "x^2 - x + 1", it would have been easy to work forward to f(f(x)) = ( (x^2 - x + 1)^2) - (x^2 -x + 1) + 1. but trying to go backward from f(f(x)) to f(x) seems much more difficult, and wondering if there is any generalized approach. Seems a bit like an egg, much easier to break than to unbreak.
Why can't a function have multiple outputs? If f(x) = sqrt(x) then the answer has a =/-. For example f(25) could be 5 or -5.
Sqrt(x) is defined as the positive part of the inverse function of x^2 which is a multivalueted function; therefore Sqrt is a single-valueted function, and the inverse for x^2 is +/-Sqrt(x).
@@Quved thanks for explaining my example! Why does every function absolutely have to have a single output though?
@@the_otter5936 You're welcome. Multivalueted functions are functions, therefore a function can actually have multiple outputs, using single-valueted functions makes the understanding of functions easier, just as you were told that you can do a - b if b is greater than a, later they told you negative number exists.
Is that function f can exists ¿? ...so that f²(x) is polynomial of degree 2 ??...
What is the algebric expression of such a fonction ? 🤔
I've just started up my own math and science RUclips channel. How do you progress as well as you have?
Can you give me a hint how to solve this one if a=5?
Given:
f(1/2f(f(a))))=c
I just put the zeros in to the function….
rất bổ ích khi kết luận f(0)=1.
Is it even possible to get a quadratic from the composition of a function with itself?
Good question. I'd love to see a closed form for such a function. Such a function is certainly not a polynomial. The degree of a polynomial composed with itself is necessarily a perfect square.
Since both f(f(0)) and f(f1)) both equal 1, you can conclude that f(0) equals f(1) right unless f(×) is some periodic function like sine or cosine right or like the square root function but besides something like that..since you are applying applying the same function twice..why not proceed this way..then you see a pattern f-1) equals f(2) and f(-2) equals f(3) and so on.
f(a) = f(b) does not imply a = b, even when f is not periodic.
@@howareyou4400 I don't see why not? The inputs would need to be the same to get the same output if you think about it..unless it's some periodic or funky function like one of the trig functions or something like that or a power function like x^2 or the like otherwise..yea most of the time you would get the same output hence the same input...
@@leif1075 No, f(x) = f(y) can be true for some x and y. It does not require periodic.
Think about it, quadratic function has that.
@@howareyou4400 you're literally agreeing with what I'm saying bkw..like I said it is not all functuons..I said some ..actually inthink it's a good majority if not most..and indid mention the quadratic function as a counterexample well or the square function if that's what you mean how x squared can produce the same output with two different inputs like 4 is either 2 or negstive 2 squared..yea so that type of function would be the exception..but those are rare compared to many other functions..
@@leif1075 There is no easy way to quantify "majority". However, by intuition I think most people would agree that reverse-able functions are a very special case. Functions in general are not reverse-able.
Hi, at 6:05, can you just divide both sides by b? Just asking cuz I suk at math
nvm 0/0 is undefined im trash
We can make this where b≠0, then we get b-1=0 b=1, then we think what would be if b=0 0×(-1)=0 this also satisfies that equation, so we conclude it too. Solutions are: b=0 and b=1
I think you made it more complicated then it needs to be.
That’s me 😜
x=1,5 per delle possibilità matematiche , perché è una funzione e non solo un'equazione.
Can an explicit form for f(x) be found?
I don't think so
Hello, you proved that *if* such a function exists, then f(0) = 1. But does really such a function exist?
Good question. Check this out:
yaroslavvb.com/papers/rice-when.pdf
@@SyberMath That paper is for complex domain. But for positive reals, x to the power of root two, is a function which iterates to x squared. So it still may be possible on a large domain.
We can't express the function by any polynomial. Just check coefficients and you will get contradiction.
@@kezza7773 I think you need to read the paper carefully. The paper clarifies exactly the type of domain for which the results proven hold, and for which they do not.
This is a largely an issue of domain. It depends on what the domain of f°f is taken to be. Technically, the domain can be an arbitrary subset of an arbitrary ring. This complicates the issue significantly.
очень хороший способ
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how can solve this f(f(x)) = x^2 +x+3
Is it possible to find f and just put x=0?
This is only tangently related to the problem, but does there even exist a function f:R->R satisfying f(f(x))=x^2-x+1?
math.stackexchange.com/questions/1158619/do-there-exist-functions-f-such-that-ffx-x2-x1-for-every-x
@@SyberMath Thank you very much. Sorry for the late response, for some reason youtube didn't notify me of your comment.
Мне батя такую задал и сказал: «Запомни, сын, любое уравнение второй степени можно решить, даже если оно функциональное». Я сразу понял о чем он. И сделал замену: f(x)=x; f(x)=(f(x))^2-f(x)+1 (f(x))^2-2*(f(x))+1=0 (f(x)-1)^2=0 f(x)=1 => f(0)=1, т.к. f(x)=k*x+b, где k=0, т.е. функция линейна, но не зависит от переменной “x”. Батя мне сказал: «Я не знаю прав ты или нет, вышли это в интернет».
And what's f(x)?
You've shown well that if a solution exists then it is equal to one. But what if there is no function f s.t. f(f(x)) = x^2 - x + 1? Without finding an example of of such an f, or otherwise proving it exists, you can never be sure that there is any meaning in assigning f(0) a value
Good point. There was an article that claimed there were no solutions for f:
yaroslavvb.com/papers/rice-when.pdf
Take a look at this, though:
math.stackexchange.com/questions/1158619/do-there-exist-functions-f-such-that-ffx-x2-x1-for-every-x
Also check (page 5):
www.math.olympiaadid.ut.ee/eng/archive/bw/bw11sol.pdf
Is there a way to find f(x)?
Isn't f(f(1)) = f(1)^2 - f(1) + 1? How do you get to f(f(1)) = 1 when we don't know what f(1) is?
just replace x with 1
so cool!
I liked it.
I solved this question 😏. Preparing for IIT exam.
2min question no need for 8mins
Wow! You're good!
Nice solution, but I really do prefer the equivalent solution of instead substituting f(x) as the variable. I find teh extra variables slightly confusing. Here's the solution written out this way. We start with subbing in f(x) to the functional equation to get
f(f(f(x))) = f(x^2 - x + 1) = f(x)^2 - f(x) + 1
Now we can inspect just plugging in x = 0 to see what we have:
f(1) = f(0)^2 - f(0) + 1
Which then motivates asking what happens when x = 1 is to yield ((1)^2 - (1) + 1 = 0)
f(1) = f(1)^2 - f(1) + 1
=> f(1)^2 - 2f(1) + 1 = 0
=> (f(1) - 1)^2 = 0
=> f(1) = 1
Then you may sub back in to see
1 = f(0)^2 - f(0) + 1
=> f(0)^2 - f(0) = 0
=> f(0)(f(0) - 1) = 0
=> f(0) = 0 or f(0) = 1
Then it is the same elimination of the case f(0) = 0 by checking f(f(x)) = x^2 - x + 1 for x = 0.
Why b^2=b >>> b= square root of b ....does not correct ... can you proove it.?
what's the problem? both 0 and 1 are their own square roots
thank you for all of you'r good work. you are the best. ... nice vid_
Thank you! 🥰
@@SyberMath most welcome
You haven't shown that a function can exist satisfying your requirements. Is there any function f such that f(f(x)) = x^2 - x + 1 for all x? We don't know. You've just shown that, if such a function exists, then f(0) = 1.
this might help:
math.stackexchange.com/questions/1158619/do-there-exist-functions-f-such-that-ffx-x2-x1-for-every-x
Man wtf are you doing to my brain i guessed F(0)=1 Literally halfway through the video..
My head hurts😵💫
Sorry
really complicated way to get the answer. There is a much shorter way
a=x=1=0. This is ridicuuuuuulous
Imagine you are in the Mathe Abitur 2022
why cant you just put x=0?
And what is f(x)?
How do you prove that the function f surely exists ?
If the equation has no solution, then the argument is useless.
For example, we don't talk about the solution of the equation x + 1 = x .
I think that f dose not exist (at least if we talk about continues function),but still i think it is nice problem,and the logic he used to solve it,is nice to watch.
@@yoav613 -- * does * continuous
Good point. Check this out:
yaroslavvb.com/papers/rice-when.pdf
@@yoav613 bruh u present everywhere 😂😂😂
It is easy to prove that f(0) is invariant (fixed point) of operation composition on space of polynomials. So f ∘ f ∘ ... ∘ f(0) = f(0). But in this exercise f(x) is NOT a polynomial and is neither analytic function. So it either doesn't exist or it can represent for example a linear operator with functions in its coeffiecients.
Your reasoning is fallacious and highly flawed, since f(0) is not required to be a fixed point.
Thanks! So, what actual is f(x)?
Nobody knows 😄
@@SyberMath I hate to be "that guy", but one could argue, if such function even exists at all :) Then this video is a moot point, heh.
I got it right!!
Yay!
there is only one answer on the function......just cancel out b=1...
ok i am stupid
Hello sir great video but i have a suggestion for the solution
can we put f(x)=y and then we can find inverse function so directly we get value of y which is automatically equal to f(x) and then put x=0 in the inverse function and get your answer..
Hi! I don't think that'll work! Any details?
f ( f j aweSUM ) ) = aweSUM
1?
what is f(x) then?
Whether it exists or not, and what it is, depends on the domain. In the simplest possible case, where the domain is the ring {0, 1} with 1 + 1 = 0, f is comletely defined by f(0) = f(1) = 1. So f is the unit constant function. However, if R = {-1, 0, 1}, where 1 + 1 = -1, then there are no functions f : R -> R such that (f°f)(x) = x^2 - x + 1 is satisfied everywhere. If R = {-1, 0, 1, 1 + 1}, where 1 + (1 + 1) = -1, then f[f(-1)] = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = -1, and so f(-1) = f[(f°f)(-1)] = [f°(f°f)](-1) = [(f°f)°f](-1) = (f°f)[f(-1)] = f(-1)^2 - f(-1) + 1, which means f(-1) = 1. Notwithstanding, -1 = f[f(-1)] = f(1) = 1 is a contradiction, so there is no f : R -> R in this case such that (f°f)(x) = x^2 - x + 1 everywhere.
On the other hand, if R is simply the ring of integers, then there are no issues, and various solutions f do exist to the equation.
Question, is it possible to find f(x)?
I don't think so
f(x) = 1/(x - 2), f' * (3) =؟
F(0)=2
Thanks. Anyone know f(x) = ?
1
f(0)= 1
Здравствуйте, как найти f(x)=?
Невозможно, я думаю
Türk müsün? Dersler çok başarılı
Saolasin
@@SyberMath keep it up.
-3/2 ?
F(x)=0 Quadratic formula
answer is (1+-3i)/2
So put it x that f(f(x))=x^2-x+1
F(0)=-3/2
من فضلك ، أريد صيغة حل معادلة من الدرجة الثالثة
Very nice answer, but unfortunately not a complete answer. You showed that "0" is not possible, which is great, since one method of driving inconsistency is sufficient. You showed "1" is possible, but you checked one method of inconsistency, but that may not be the only inconsistency.
1. It is possible that there is no such function.
2. So your answer is, if there is such a function, the only answer can be 1. In that case, you did not have to check to one method of inconsistency to rule out 1.
3. The only way to completely answer this question is to show an example function, which satisfies the given condition.
*The only way to completely answer this question is to give an example function which satisfies the equation.*
No. Providing an example function is not needed. What is needed is only a proof of the existence of a function satisfying the equation, not a construction of it.
Regardless, whether such a function exists or not, as well as what type of function is, depends on the domain chosen.
@@angelmendez-rivera351 i agree. what i am trying to say that not able to rule out "1" does not mean you can't rule out "1" by another method. so as a promise problem, once you have ruled out other answers you did not even have to rule out "1" (that is my second point). the third point, if you can prove existence non-constructively that is fine, but that is rare for such problems. if that is the case, this will be publishable result. your domain must include "0" , "f(0)", "f(f(0))", "f(f(f(0)))", and so on.
here is an example function which satisfied the problem.
domain = {0,1}
function is: f(0) = f(1) = 1.
@@kamaljain5228 *The third point, if you can prove existence non-constructively, that is fine, but that is rare for such problems.*
No, it is not. Almost all functional equations cannot be solved constructively. For every constructively solvable equation, there are more than uncountably infinite amount of non-constructively solvable ones.
*If that is the case, this will be a publishable result.*
No, because there is very little remarkable about it. But, there actually are papers written about this. See the paper that SyberMath has been sharing in the replies to the top comments.
@@angelmendez-rivera351 there are many more unsolvable problems than solvable one, many more unprovable mathematical statement than what we can prove, and many more tasks without algorithms than the ones which have algorithms, and these are all theoretical statements. the problems humans are interested are those which have semantics, and they are so far only finite in numbers, just because humans have typed finite number of letters. on one thing you are wrong theoretically is "uncountably" infinite. as long as the problem description is finite, there are only countably many possibilities to describe these problems. yeah unless you call a single problem as uncountably infinite, e.g., for every real y, show that there is an x such that x+x=y. i call this as a single problem, but you may count it as uncountably infinite problem, one for every real y.
the remarkable part is that such an unremarkable problem being done through a non-constructive existence rather than showing a constructive function.
i will look at the paper. thanks.