In this video, I showed how to solve a functional equattions. I did not show at the end that it is the only solution but this is the general way to approach a composition of functions.
Getting f(x) = x + 1/2 is easy, the hard part is proving that is the only solution. I figured out f(x) = x + 1/2 in a few seconds in my head without even watching the video. I just need to build up my skills on proving a discovered solution is unique.
@@PrimeNewtons Okay. I think this works, but I'm not sure. Let e be a real number between 0 and 1. You already showed that f(x + 1) = f(x) + 1. This is true for any value of x, including n + e, where n is a positive integer. So, f(n + e + 1) = f(n + e) + 1. Then f(n + e) + 1 = f(n-1 + e) + 2 = f(n-2 + e) + 3 = ... = f(n-n + e) + n. So, f(n+e) + 1 = f(e) + n. So, f(n+e) = f(e) + n-1. But then, f(n+e) = n + e + f(e) - 1 - e. But f(e) - 1 - e is constant. So, f(n+e) = n+e + c. So, for any e, it is true for all numbers of the form n+e, where n is a natural number. I think it's fairly trivial to extend that to n+e where n is an integer, we just need to go in the opposite direction.
@@chaosredefined3834That doesn't prove what we need, and it can't as f(x + 1) = f(x) + 1 alone does not imply f(x) is linear. If you take any function [0,1) -> R, you can extend it to a function R->R that satisfies f(x + 1) = f(x) + 1 by saying that for values greater than 1 (or smaller than 0), you add an integer N to bring it to [0,1) and then use the original function + N. The equation f(x + 1) = f(x) + 1 alone is weaker than f(f(x))=x+1
A counterexample from R to R is a piecewise function where {x} is the fractional part of x (eg: {2.66} = .66). Let f(x) = x + 3/2 if {x} = 0 f(x) = x -1/2 if {x} = .5 f(x) = x + 1/2 otherwise Then, f(1) = 2.5, f(2.5) = 2, and you can check for any number you choose, f(f(x)) = x+1.
This is not the only solution for f:R->R however it is the only continuous solution. As continuity is not stated, take h:[0,1)->[0,1) be any involution without a fixed point, and g:[0,1)->Z s.t. g(h(e))= 1-g(e). Then for any x in R let x = z + e where z is in Z and e is in [0,1) (i.e. integer and decimal parts) then define f(x) = z + g(e) + h(e) and this will also be a solution. Conceptually, as f(x+1) = f(x) + 1 leaves the only freedom of such a function to be on the a unit interval which can always be translated to [0,1). Hence pair off elements in [0,1) with the involution h (can't have fixed points otherwise it is not a pair) and make sure the pair {e,h(e)} give a images with difference of one without effecting the decimal part (i.e. stay in the integers) which is the the job of the function g.
Let's f(x)=ax+b We get x+1 = a^2*x + ab + b By identification a^2 = 1 => a= 1 or -1 And b(a+1)= 1 which let us give up the -1 option and we have b=1/2 Hence f(x)= x+1/2
This is true, but it's assuming f(x) is a linear function. It makes sense for this equation, but it's possible for some other functional equations to include non-linear functions as answers.
@@МартинАндреев-ы4л It could probably be proved that any function which satisfies the functional equation must be linear, then there wouldn't be a problem.
Starting from fof(x) = x + 1 Let t = f^-1(x) [inverse] so x = f(t) substitute for x in original equation fof(f(t)) = f(t) + 1 this can be rewritten as f(fof(t)) = f(t) + 1 from original equation, fof(t) = t + 1 so f(t+1) = f(t) + 1 which gives f(x+1) = f(x) + 1
@@UltraAryan10 hello. I have a question about. Please don't understeand it like a critic. Im learning about it.😊 I have troubles when you said "from original equation: fof(t)= t +1" how can be the same that fof(x)=x+1 if you said t=f^-1(x). In fact im having troubles to understand the same step than the fellow of the original question. Thanks for all.
As it was observed earlier, the general solution to the equation f(x+1) = f(x) + 1 is f(x) = x + g(x) with a periodic function g of a period 1, that is, g(x) = g({x}). Then ff(x) = f(x + g({x})) = x + g({x}) + g( x + g({x}) ) = x + g({x}) + g( {x} + g({x}) ). So we conclude that the only condition on g is g(θ) + g( θ + g(θ) ) = 1 for 0≤ θ < 1. This does not allow for a quick and simple general formula. However, it allows for construction of example. One was given earlier, the other is here: g(θ) = 1.5, 0≤θ
F is a polynomial, say of degree n (positive n) So f will be of degree n^2. But rhs is of degree 1. So n^2=1 => n=1 So assume f(x)= ax+b So f(f(x)) = a^2*x+ab+b= x+1 => a^2=1 Let a=1 => b+b=1 => b=1/2 => f(x)=× + 1/2 Let a=-1 => -b+b=1 => rejected So one solution
No, actually not! They let x = t - 1, so they only need to change all x by "t-1". You are getting mixed with setting the function g(x) = f(x+1) f(x) = g(x-1), which would result in changing all functions "f(z)" by "g(z-1)" whatever each z is. In our case, we would have f(1) = f(x + 1) - x g(0) = g(x) - x. Don't hestiate to take some time to really grasp the difference between changing the function, and changing the variable. This is not easy, but it is definitely something great when you understand :)
It's not the only solution, but all the other ones are piecewise. It's not hard to show that f(t+1) = f(t) + 1, you did that. So the only place we really need to define our function is when t is between 0 and 1. Lets say f(0) = C. Then f(C) = 1 by f(f(0)) = 1. We can prove that f is an injective function, i.e. if f(t) = f(u) then t = u. f(f(t)) = t + 1 and f(f(u)) = u + 1. because f(t) = f(u), f(f(t)) = f(f(u)), and t + 1 = u + 1. t = u. we can also show that f is surjective, i. e. for all values K there exists J such that f(J) = K. f(f(K-1)) = K. Let J = f(K-1), and the proof is done. So we know our function is both injective and surjective. It is therefore bijective, so now we can start playing with inverses. Define some function h from the domain [0, C) to the codomain [C, 1) that is bijective. Could be whatever, really, could be h(x) = x + 1/2 on [0, 1/2), could be h(x) = sqrt(x - x^2) + 1/2 on [0, 1/2), could be h(x) = 1/3 + 2x on [0, 1/3), whatever. As long as h(0) = C and h(C) = 1. Now we want some function g(x) on the rest of the interval [C, 1) such that g(h(x)) = x + 1. g is gonna be bijective so we can use inverses to our advantage. h(x-1) = g¯¹(x) if two functions are the same and they're both bijective, then their inverses are the same. g(x) = h¯¹(x)+1 Note that the range of h¯¹(x) is the domain of h(x): [0, C). The range of g(x) is therefore [1, C + 1). We now have a base pattern f(x) = h(x) when x is in [0, C) and g(x) when x is in [C, 1). For all other x, f(x+Z) = f(x) + Z for any integer Z. Now we just need to check that this actually works. Case 1: x is in [0, C) f(x) = h(x), is in [C, 1) f(f(x)) = g(h(x)) = h¯¹(h(x)) + 1 = x + 1 Case 2: x is in [C, 1) f(x) = g(x), is in [1, C + 1) f(x) - 1 = g(x) - 1, is in [0, C) f(f(x)) = f(f(x) - 1) + 1 = h(g(x) - 1) + 1 = h(h¯¹(x) + 1 - 1) + 1 = h(h¯¹(x)) + 1 = x + 1 THOSE are all the possible functions f such that f(f(x)) = x + 1.
I like this approach, I always just guess by letting f(x) equal mx + b , then solve the system of equation to find m and b. If that doesn't work ill try letting f(x) equal exponential function or quadratic etc. The benefit of the method I use is that its 'easier' but doesn't guarantee a solution
If we imagine f(x) = ax + b & find fof(x) then we get a²x+ab+b . Comparing this with x+1 we get the values of a & b easily Murthy retired maths teacher
f(f(x)) = x + 1 by definition. Let u = x So, f(f(u)) = u + 1. Now if you let u = f(x), you get f(f(f(x))) = f(x) + 1 ... Eq1 But, as f(f(x)) = x + 1, then f°(f(f(x))) = f(f(f(x))) = f(x + 1) ... Eq2, then as Eq1 = Eq2, so, f(x + 1) = f(x) + 1.
@@davidbrisbane7206That doesn't really help me. Your last two lines you say: .. f(f(f(x))) = f(x+1) // just as he said in the video. So, f(x+1) = f(x) +1 // Same 'miracle occurs' with no supporting reason Let u = x But then you say u = f(x) // which is it?? u=x or u=f(x)??? I don't see how this helps?
Isn’t true that if x is different from y, we have m=(f(x)-f(y))/(x-y)=((f(x)+1)-(f(y)+1))/(x-y)=1? That means f(x) must be a linear function; and so f(x)=x+1/2 must be true.
This seems to be the only solution, because f(x) = -x+c leads to f(f(x)) = x identically regardless of the value of c. The t-substitution becomes unnecessary if you sum only up to f(x-1) = f(x) -1.
f(f(x)) = x +1 means that f(x) must be its own inverse function with a constant transfer. In the case of zero transfer f(x) = -x would work because f(f(x)) would flip back to x, but nonzero transfer doesn't work because flipping nullified the transfer to zero. Non-linear functions are not their own inverse functions and that's why only linear solutions are possible here. If we had f(f(x)) = x^4, then f(x) = x^2 would be a solution. @@sobolzeev
@@pojuantsalo3475 First, the equation does not suppose that f is invertible. It is you who has assumed it is, to apply inv f to both sides of the equation. Second, as you observed, even if f is invertible, it is not its own inverse but inv f(y) = f(y-1). Finally, can you prove that it is only a linear function which is its own inverse? Say, in case when the function is not differentiable?
If we assume f is a polynomial of degree n, then f.f must be of degree n^2. The degree of f.f is 1, so n^2 = 1, therefore n=1. So we know f is linear and can easily show that f(x) = x + 1/2. I don't know how to prove there are no non-polynomial solutions though.
An equation f(x+1) = f(x) + 1 has infinitely many solutions. Indeed, subtract (x+1) from both sides. Then you have f(x+1) - (x+1) = f(x) - x. That is, g(x) = f(x) - x is periodic with period 1. If you assume that g is constant, you obtain g=½. What if it is not?
There’s a Quora post showing a complicated continuous function which satisfies this functional equation. We need a strong condition if we hope to prove uniqueness. Perhaps differentiability?
@@RU-vid_username_not_found the one titled “If f(f(x)) = x+1, how do I know that f(x) must be x+0.5? I can guess it, but is there any method to work it out?” The first answer uses absolute values and the fractional part operator. It doesn’t quite show the derivation but it still works I believe.
@@GreenMeansGOF I found it. At first I was checking the answers to a similar question and for some reason Google didn't give me the other question until I ssearched for it directly. Now its my turn to give you something interesting. check the comment written by md2perpe under the following video. He presented a general method for solving the equation f(f(x)) = g(x) for arbitrary g.
I didn't send the video link because my reply kept being hidden. The video is on Sybermath channel and it's about the functional equation f(f(x)) = exp(-x)
Can't asume that f(x) = ax+b Because we are getting linear function when composing f(f(x))= a(ax+b)+b x+1 =a^2 x + ab+ b a=+1 or -1 But a!=-1 because when we sub b dispear but we have 1 in the right side a=1 2b =1 b=1/2 f(x)= x+ 1/2
Dang, I'm really stumped by functional equations. Looks like I have to go back and find some earlier and simpler examples to be able to understand this stuff.
I spent a bit of time trying to show the uniqueness of the solution to this equation, but it appears to be quite hard. This discussion is quite long, but there is a TL;DR at the end if you don't want to read it all :) Note that, for some reasons, my comment did not appear to other users. Let me thus split its paragraphs into answers to this comment.
For this discussion, we will use n = floor(x) to simplify our notation. We moreover need to define the fractional part function {x} = x - floor(x) = x - n. It outputs the fractional part of a positive number: {1.3} = 0.3 or {324.1415} = 0.1415. For negative numbers, we can use the property {-x} = 1 - {x} for any x, that directly comes from the definition; yielding for instance {-0.6} = 0.4. This is important for this function to have a period of 1: {x + k} = {x} for any integer k, no matter if x is positive or negative.
So, using the same method as in the video, we get f(x) = f(n + {x}) = f(n-1 + {x}) + 1 = ... = f({x}) + n = f({x}) - {x} + n + {x} = f({x}) - {x} + x = g(x) + x (*), for g(x) = f({x}) - {x}. This g(x) is the equivalent of the constant found in the video: if you suppose it is constant, you will get the same result. We can find a constraint on this function. We indeed already found at (*) that f(x) = g(x) + x, so composing it with f(x), we get f(f(x)) = g(f(x)) + f(x) x + 1 = g(g(x) + x) + g(x) + x g(g(x) + x) + g(x) = 1. We have to solve this equation for a function that has a period 1, g(x + 1) = g(x) for any x (this directly comes from g(x) = f({x}) - {x}, since {x} is 1-periodic); so I considered using Fourier series, but this did not seem to bring me anywhere.
Instead, let's now make some linearity assumption on g(x): g(x) = a{x} + b. The constraint g(g(x) + x) + g(x) = 1 yields a{a{x} + b + x} + a{x} + b = 1. Let us look more closely at {a{x} + b + x} = {a{x} + floor(b) + {b} + n + {x}} = {(a+1){x} + {b} + floor(b) + n} = {(a + 1){x} + {b}} (since {x + k} = {x} for any integer k, and floor(b) + n is an integer). We also take the assumption that {(a + 1){x} + {b}} = (a + 1){x} + {b} (i.e. that 0
At that point, I stopped working a bit on this problem, and started doing something else. When I came back on it, I got the following idea. We consider the more general problem to find functions f such that f(f(x)) = x + A where A = p/q for integers p, q. It is possible to find that g(x) = f(x) - x must solve the constraint g(g(x) + x) + g(x) = A. We guess that g(x) is of the form g(x) = -2/o {o*x} + b/o. The idea here is that we take a = -2 as found previously, but we add some ability to change the frequency of g using o. Looking at the left hand side of the constraint, we see that g(g(x) + x) + g(x) = -2/o {og(x) + ox} + b/o - 2/o {ox} + b/o = -2/o {-2{ox} + b + ox} - 2/o {ox} + 2b/o. We look only at the term {-2{ox} + b + ox} = {-2(ox - floor(ox)) + b + ox} = {-2ox + 2floor(ox) + b + ox} = {-ox + 2floor(ox) + b}. Since 2floor(ox) + b is an integer, this is equal to {-ox}, which is equal to 1 - {ox} as explained in the very beginning of this discussion. This yields that our constraint is equal to g(g(x) + x) + g(x) = -2/o (1 - {ox}) - 2/o {ox} + 2b/o = -2/o + 2/o {ox} - 2/o {ox} + 2b/o = 2(b - 1)/o. We want this to be equal to A, yielding that A = 2(b-1)/o oA/2 = b-1 b = oA/2 + 1 = (op)/(2q) + 1. We want b to be an integer, so we need 2q to divide op. Supposing p and q are coprime (i.e. that A = p/q is expressed in its reduced form), this yields two possibilities: if p is a multiple of 2, then we can take any o = q*k; if p is not a multiple of 2, we can take any o = 2*q*k. In both cases k can be any integer, except for 0 since we divide by o in the definition of g. In fact, we can find that the solutions with negative k are equivalent to solution with a positive one, so we only need to non-zero positive integers k. In our case, we have A = 1, and thus p = q = 1. This yields that our solution is given by o = 2*k and b = oA/2 + 1 = 2k/2 + 1 = k + 1. This yields g(x) = -1/k {2kx} + (k+1)/(2k) and therefore f(x) = g(x) + x = x - 1/k {2kx} + (k+1)/k is a family of solutions to f(f(x)) = 1. The solution presented in the video is therefore not unique! :) I invite you to try those functions on a graphic calculator, with different values of A, o and k (even non-integers one), it is quite impressive that f(f(x)) makes a straight line with very specific values for those variables, and otherwise an incredible mess.
There are still other questions left (as always with science): - Does there exist any other solution than the solutions I found, together with the solution f(x) = x + A/2, to the equation f(f(x)) + A when A is rational? - Is the solution f(x) = x + A/2 unique for the equation f(f(x)) = x + A where A is irrational? - Is the solution f(x) = x + A/2 unique for the equation f(f(x)) = x + A if we also require that f is continuous, or even differentiable? My conjecture is that there exist many more solutions, to both when A is rational and when A is irrational. I have to admit I don't have any idea for a continuous or differentiable f.
If we accept f²:=f°f , for some f lets say it is in the set C(M), and the set M is in IR . Then we may think of equations of the form: f² = g, for some g in C(IR°) given. This equation may pr may not have a solution, or not but it could have many. For example the equation: f²=I has the identity function as a solution but also f=-I is another solution and also f: x---> 1/x if it is defined on IR*.
It is possible to derive a closed form for f^(n)(x), where f(x) = ax+b and f^(n)(x) means f composed n times on x. This can be done working with only non-negative n. And then be amazed when feeding in n = -1 and the inverse function of f pops out. This then leads to the supposition that n=fractional or even n=complex applications of f are also possible, whatever that might mean.
Your solution seems way too complicated. From the equation at 3:32 f(x+1) = f(x) + 1, it is clear that a solution is linear ie, f(x) = x + C. Therefore, f(x + 1) = (x + 1) + C. So f(x + 1) = (x + 1) + C = (x + C) + 1 = f(x) + 1, satisfying the original equation f(x+1) = f(x) + 1
This approach can be easily extended by principles of mathematical induction to prove f(x)=x+1/2 for rational x. However, it seems that continuity may be required to prove its uniqueness for all real numbers, particularly irrational numbers. I'm not sure whether pathological discontinuous f(x) exist which satisfy f(f(x))=x+1. (Consider pathological solution of Cauchy's equation, assuming the axiom of choice)
@@RU-vid_username_not_found The counterexample posted by @zachgurwitz7078 shows it's not the case: Let f(x) = x + 3/2 if {x} = 0 f(x) = x -1/2 if {x} = .5 f(x) = x + 1/2 otherwise
@zachgurwitz7078 posted a counterexample of your claim for rational x: Let f(x) = x + 3/2 if {x} = 0 f(x) = x -1/2 if {x} = .5 f(x) = x + 1/2 otherwise
I assume f(x)=ax+b then f(f(x))=af(x)+b= a(ax+b)+b= a^2x+ab +b=x+1 then a^2=1 and ab+b=1 then a=-1 is not acceptable while a=1 implies b=1/2 then f(x) = x + 1/2
@@sobolzeevcan you give me a counterexample? that is, can you give me a nonlinear or other type of function (of higher degree, irrational, fractional, transcendent ...) that applied twice gives x+1? it certainly can't be of the type ax^2+bx+c ... try ... have fun!
@@annacerbara4257 I saw a counterexample on Quora. But this is not the point. The point is it is you who should prove your solution is unique, to have the problem solved. As a hint I can give you the general solution to the functional equation f(x+1) = f(x) + 1. It is f(x) = x + g(x), where g is a periodic function with period 1. For instance, f(x) = x + cos(2πx). Enjoy!
