@@MichaelRothwell1 Actually, that is not quite true. 0 = 1 is true in the trivial ring. Hence, if R is the trivial ring, then there does exist exactly one f : R -> R such that f(x) - f(1 - x) = x: namely, the only function R -> R that exists at all, in this case.
I like that you point this out because after discovering the problem with 0, 1, and 1/2 my first question was whether there was a solution if we excluded those points. Obviously this proves there isn't.
f(x) - f(1-x) = x Sub 1-x in: f(1-x) - f(1 - (1-x)) = 1-x f(1-x) - f(x) = 1-x f(x) - f(1-x) = x-1 x = x-1 This is nonsensical, so no solution exists. Note that the thing you did only shows that, if f(x) exists, it is not defined at x=0, 1/2, and 1
I mean, half way into the video he said we have f: R -> R The domain also includes 0, 1 and ½. But since it isn't defined at those points, there doesnt exist such a function (given the domain)
@@methatis3013 This is fair. It's still more thorough to show that there aren't any solutions of the form A/(x-1) + B/x + c/(2x-1) or the like, even though they aren't f: R -> R.
@@chaosredefined3834 Right. I think the point of this functional equation is that it does not matter what the domain is, it cannot have a solution at all, unless the domain is the trivial ring, where 0 = 1 actually is true.
The solution to this equation is of the form: (1/2)*(x-(ln(x - (1/2))/ln(-1)) + T(x)) Where T(x) is an arbitrary symetric function over x=1/2 axis, so we have a logarithm of a negative number which is a bit bad... because one has that ln(-1) is πi but it is also 3πi, 5πi, ... This problem would vanish though if one wanted to know (-1)^f(x)
But asuming f(x) is complex valued and defining ln(z) for just one possible branch this solution would be a valid solution to this equations given that logarithmic function properties are not changed
No. The functional equation has no solutions. Here is a simple proof: f(x) - f(1 - x) = x everywhere implies f(1 - x) - f(x) = 1 - x. f(x) - f(1 - x) = x & f(1 - x) - f(x) = 1 - x imply f(x) - f(1 - x) + f(1 - x) - f(x) = x + 1 - x = 0 = 1. 0 = 1 is FALSE. Therefore, f(x) - f(1 - x) = x everywhere is FALSE. Therefore, there are no functions that satisfy it.
@@angelmendez-rivera351 Then what do we do with ln(-1), should we just ignore it? I know that the proof is correct and that there is no function that solves this equation but then, why does this expression works?
The method in the video for analyzing the functional equation is unnecessarily complicated, and also not valid. The method only establishes that 0, 1/2, 1 are not in the domain of f. The method does not establish that there is no f : R -> R such that f(x) - f(1 - x) = x everywhere. f(x) - f(1 - x) = x everywhere implies f(1 - x) - f(x) = 1 - x everywhere, via x |-> 1 - x. The conjunction of the above implies f(x) - f(1 - x) + f(1 - x) - f(x) = 0 = x + 1 - x = 1 everywhere. Therefore, f(x) - f(1 - x) = x everywhere implies 0 = 1. This is true if and only if the domain of f is the trivial ring. Otherwise, the functional equation is false for every function f on the chosen ring.
Hello Sybermath you are explain well but I want you speak slowly It is easy to understand well cause I'm not strong at english but I try to watch everyday about your video .Thanks.
Well f(x) = x + f(1 - x) apparently converges because |x| > |1-x| and sgn(x)≠sgn(1-x), therefore it's a proper recursive solution. But that breaks down at x = 1-x => x= ½.
The question seems to be: If f(x) - f(1-x) = x for all x, find the function f. f(x) = x + f(1-x) = x + f(y) (where y = 1 - x) = x + y + f(1-y) = 1 + f(x). So 0 = 1, impossible. ...
I wondered if there could be a function that is defined on complex numbers that satisfies this, but isn't defined on any real number... but unfortunately no. Note that f(z) - f(1-z) = z [By definition] Define a new complex number y = 1-z Then f(y) - f(1-y) = y [By definition] f(1-z) - f(1-(1-z)) = (1-z) [Substitute y -> 1-z] f(1-z) - f(z) = 1-z [1-(1-z) = z] f(z) - f(1-z) = z-1 [Multiply both sides by -1] This contradicts the first statement that was true by definition, so thus no such complex function f can be defined anywhere.
Since you're interested in functional equations, here's a problem I was defeated by a while back, in case you come up with something: f: R -> R such that f(f(x)) = f(x+1) There are lots of solutions, the simplest being x+1. It's easy to show that f injective => f(x) = x+1 Now my problem is, to prove or disprove the conjecture: f surjective => f(x) = x+1.
WTF? My solve: -(f(x) - f(1 - x)) = -(f(x + 1) - f(-x); f(x) - f(1 - x) = f(x + 1) - f(1 - x - 1); if we said that x = x + 1, f(x) - f(1 - x) equals f(x - 2) + f(1 - x + 2); ect for infinity. so f(x - y) - f(1 - x + y) = x; if y = x, f(0) - f(1) = x. But how.... x can be bigger or smaller, but f(0) and f(1) cannot, cause it's constant...
Oh my god! I watched the video. I thought that I did mistake, but this is was a no result's task! And SyberMath getting it in other ways... Interesting
It's simple to show that f(x) is not actually a function of x: f(x) = x + f(1-x) f(1-x) = 1-x + f(1-1+x) Substitute f(1-x) in the first equation: f(x) = 1 + f(x) => f(x) = ∞ or, if you prefer, f(x) = x/0
For f(x)=ax one finds x=a/(2a-1). Thus, for a given a, there is just one point x, where this works. For a polynomial f, there will be multiple such points, as many as one wants.
This was a really cool demonstration, and since I don’t want to give any spoilers, I’ll just say that I was surprised by the result in the first half of the video!
Well, if you think about it, the first equation tells you that f(0)=f(1), which yields 1=f(1)-f(1)=0 in the second equation - contradiction. So your initial guess was right.
try substitution u := 1-x so x == 1-u f(1-u) - f(u) = 1-u; times.[-1]:: f(u) - f(1-u) = u-1 & f(u) - f(1-u) = u [from definition] this concludes u = u-1 a non reconcilable in most fields {presuming the REAL's here}
Plug in x=0 and you find f(0)-f(1) = 0 So f(0) = f(1) If you plug in x=1, you find f(1)-f(0) = 1 But f(1)-f(0) = 0, so this is a contradiction. So no solution exists.
nonsense. too much wishful thinking here. There is no solution f(x) that satisfies the given equation. Just replace x by 1-x in the original equation and add up the equations.
Another quick way to show a contradiction is to substitute 1-x for x and add the equations, you get 0=1 as well
Agreed, that was my solution. This shows that you can't even fix the problem by restricting the domain - unless you restrict to the empty set!
@@MichaelRothwell1 Actually, that is not quite true. 0 = 1 is true in the trivial ring. Hence, if R is the trivial ring, then there does exist exactly one f : R -> R such that f(x) - f(1 - x) = x: namely, the only function R -> R that exists at all, in this case.
@@angelmendez-rivera351 True, if you don't limit yourself to the real numbers.
@@angelmendez-rivera351 Luckily,this channel doesn’t cover rings, trivial or otherwise. It’d make me feel extra stupid.
I like that you point this out because after discovering the problem with 0, 1, and 1/2 my first question was whether there was a solution if we excluded those points. Obviously this proves there isn't.
f(x) - f(1-x) = x
Sub 1-x in:
f(1-x) - f(1 - (1-x)) = 1-x
f(1-x) - f(x) = 1-x
f(x) - f(1-x) = x-1
x = x-1
This is nonsensical, so no solution exists.
Note that the thing you did only shows that, if f(x) exists, it is not defined at x=0, 1/2, and 1
I mean, half way into the video he said we have
f: R -> R
The domain also includes 0, 1 and ½. But since it isn't defined at those points, there doesnt exist such a function (given the domain)
Oh, good!
@@methatis3013 This is fair. It's still more thorough to show that there aren't any solutions of the form A/(x-1) + B/x + c/(2x-1) or the like, even though they aren't f: R -> R.
@@chaosredefined3834 Right. I think the point of this functional equation is that it does not matter what the domain is, it cannot have a solution at all, unless the domain is the trivial ring, where 0 = 1 actually is true.
If you just pmug I vsluesof x equals 1 and zero you get the contradiction 1 equals zero..didnt anyone else do tbis..
(1) : f(x) - f(1-x) = x
Substitute x: 1-x
(2) : f(1-x) - f(x) = 1-x
Equation (1) + (2) : 0 = 1
Contradiction.
Explain about some paradoxes in mathematics....
The solution to this equation is of the form:
(1/2)*(x-(ln(x - (1/2))/ln(-1)) + T(x))
Where T(x) is an arbitrary symetric function over x=1/2 axis, so we have a logarithm of a negative number which is a bit bad... because one has that ln(-1) is πi but it is also 3πi, 5πi, ... This problem would vanish though if one wanted to know (-1)^f(x)
But asuming f(x) is complex valued and defining ln(z) for just one possible branch this solution would be a valid solution to this equations given that logarithmic function properties are not changed
I think in this case the paradox is caused by the different branches of ln(z) and that is why one reaches 0=1
No. The functional equation has no solutions. Here is a simple proof: f(x) - f(1 - x) = x everywhere implies f(1 - x) - f(x) = 1 - x. f(x) - f(1 - x) = x & f(1 - x) - f(x) = 1 - x imply f(x) - f(1 - x) + f(1 - x) - f(x) = x + 1 - x = 0 = 1. 0 = 1 is FALSE. Therefore, f(x) - f(1 - x) = x everywhere is FALSE. Therefore, there are no functions that satisfy it.
@@angelmendez-rivera351 Then what do we do with ln(-1), should we just ignore it? I know that the proof is correct and that there is no function that solves this equation but then, why does this expression works?
@@ismaelperbech ln(-1) is ill-defined, technically. If f(x) = y and f(x) = z, but y = z is false, then f(x) simply does not exist.
The method in the video for analyzing the functional equation is unnecessarily complicated, and also not valid. The method only establishes that 0, 1/2, 1 are not in the domain of f. The method does not establish that there is no f : R -> R such that f(x) - f(1 - x) = x everywhere.
f(x) - f(1 - x) = x everywhere implies f(1 - x) - f(x) = 1 - x everywhere, via x |-> 1 - x. The conjunction of the above implies f(x) - f(1 - x) + f(1 - x) - f(x) = 0 = x + 1 - x = 1 everywhere. Therefore, f(x) - f(1 - x) = x everywhere implies 0 = 1. This is true if and only if the domain of f is the trivial ring. Otherwise, the functional equation is false for every function f on the chosen ring.
I went straight in with x=0 and x=1 which is the kind of thing I usually do with functional equations. Got lucky.
Thank you. I am sure it will be instructive to show that such a simply defined problem can have no solution.
I really appreciate the explanation of how you come up with homemade functional equations, SyberMath!
Glad to hear that!
The Jamie Oliver of mathematics
Hello Sybermath you are explain well but I want you speak slowly It is easy to understand well cause I'm not strong at english but I try to watch everyday about your video .Thanks.
Sorry about that! I need to remind myself to slow down
I keep getting 0=1 no metter what i do. I've been doing this for 20 minutes and finally give up-It must be impossible.
This functional equation has no solution.
If you take x=0 then f(0)-f(1)=0
If you take x=1 then f(1)-f(0)=1
This is contradictory.
No solution, syber i like it when you fool us!🤣💯
Hehe! 😜
Well f(x) = x + f(1 - x) apparently converges because |x| > |1-x| and sgn(x)≠sgn(1-x), therefore it's a proper recursive solution. But that breaks down at x = 1-x => x= ½.
The equation has no solutions, though.
a special case solution exists for integer x > 0
f(x) = the summation from i=0 to x of the absolute value of x
example: x=4 -> (1+2+3+4) - (1+2+3) = 4
f(0.4) - f(0.6) = 0.4
so we should have
f(0.6) - f(0.4) be the negative of that but instead the formula says it equals 0.6. So no function can do this
🤣🤣 you don't talk too much, rather you love teaching.
Thanks for everything ❤️
You are so welcome! Thank you! 🥰
Just by looking at the inconsistency in the equation, I could see that it would be difficult to solve, and I know nothing about it, and remain so
0:56 If f(x) = ax + b, then f(1 - x) = f(x) - x = ax + b - x = (a - 1)x + b.
f(x) - f(1-x) = x
x = 0 => f(0) - f(1) = 0
x = 1 => f(1) - f(0) = 1
conflict, hence no exists
Nice video. Keep it up! I am learning new approaches! One day I ll be able to solve these myself!
@@English_shahriar1 Ok i ll check it out!!
The question seems to be: If f(x) - f(1-x) = x for all x, find the function f.
f(x) = x + f(1-x) = x + f(y) (where y = 1 - x) = x + y + f(1-y) = 1 + f(x). So 0 = 1, impossible.
...
I wondered if there could be a function that is defined on complex numbers that satisfies this, but isn't defined on any real number... but unfortunately no.
Note that f(z) - f(1-z) = z [By definition]
Define a new complex number y = 1-z
Then f(y) - f(1-y) = y [By definition]
f(1-z) - f(1-(1-z)) = (1-z) [Substitute y -> 1-z]
f(1-z) - f(z) = 1-z [1-(1-z) = z]
f(z) - f(1-z) = z-1 [Multiply both sides by -1]
This contradicts the first statement that was true by definition, so thus no such complex function f can be defined anywhere.
Congratulations for hitting 100k subs. I wish you more and more subs.
Thank you so much 😀
Likewise! 🥰🤗
Since you're interested in functional equations, here's a problem I was defeated by a while back, in case you come up with something:
f: R -> R such that f(f(x)) = f(x+1)
There are lots of solutions, the simplest being x+1. It's easy to show that
f injective => f(x) = x+1
Now my problem is, to prove or disprove the conjecture:
f surjective => f(x) = x+1.
Here is another solution to your problem: f(x)=42.
Apply this at x=0 and x=1 leads to an immediate contradiction.
Sometimes i feel like hes speed running a game
WTF? My solve: -(f(x) - f(1 - x)) = -(f(x + 1) - f(-x); f(x) - f(1 - x) = f(x + 1) - f(1 - x - 1); if we said that x = x + 1, f(x) - f(1 - x) equals f(x - 2) + f(1 - x + 2); ect for infinity. so f(x - y) - f(1 - x + y) = x; if y = x, f(0) - f(1) = x. But how.... x can be bigger or smaller, but f(0) and f(1) cannot, cause it's constant...
Oh my god! I watched the video. I thought that I did mistake, but this is was a no result's task! And SyberMath getting it in other ways... Interesting
😁
offhand with x = 1/2 you get f (1/2) - f (1/2) = 1/2, or 0 = 1/2. seems unlikely.
It's simple to show that f(x) is not actually a function of x:
f(x) = x + f(1-x)
f(1-x) = 1-x + f(1-1+x)
Substitute f(1-x) in the first equation:
f(x) = 1 + f(x) => f(x) = ∞
or, if you prefer, f(x) = x/0
I can't find my x now.. he just left me in pieces.
😁
For f(x)=ax one finds x=a/(2a-1). Thus, for a given a, there is just one point x, where this works. For a polynomial f, there will be multiple such points, as many as one wants.
In one line: f(x) = x + f(1-x) = x + (1-x) + f(1 - (1 - x)) = 1 + f(x), i.e. no solutions.#
There are infinitely many solutions. In the linear case f(x)=a x will give x = a/(2 a-1) where a neq 1and so on for f(x)=a x^2+b same solution etc
If you consider x=1/2 you get f(1/2) -- f(1/2)= 1/2, i.e., 0 = 1/2
greetings Syber!, it is incredible that you do this every day!..
Thank you, Juan! 🥰
This was a really cool demonstration, and since I don’t want to give any spoilers, I’ll just say that I was surprised by the result in the first half of the video!
Thank you!
It's right the on the left of the equation!
if x=1/2 you get : f(1/2) - f(1/2) = 1/2. Therefore no such function can exist.
My first instinct from the thumbnail was just to substitute 1-x in place of x and I already saw weird things happening.
for x=1/2 we have 1/2=0 not possible so if one solution exists, it's a very special function. Finished.
Great! The first part was a blackhole today 😂😂.. but the second was golden (ratio). Learnt something today 😉
Glad you enjoyed it!
but shouldn't f(y) actually be the inverse of f(x)?
Thanks so much sir!!! Now i know how to solve this:)
Glad to hear that! 🤩
@@SyberMath 👍👍😁😁
0=1❓
Well suppose f(x) = a x + b
Plug it in the équation
ax + b - a + ax - b = x
2ax - a = x
a = x / ( 2x - 1 ) b whatever
--> f(x) = ( x² / ( 2x-1 )) + b.
f(x) - f(1-x) = x
f(1-x) - f(x) = 1-x
adding one gets 0 = 1
so no feasible solution does exist
Great video, thanks for the explanation at the end.
Glad it was helpful!
x=0:
f(0) - f(1) = 0
x=1:
f(1) - f(0) = 1
I can't think of any values for f(0) and f(1) that would make this possible.
Well, if you think about it, the first equation tells you that f(0)=f(1), which yields 1=f(1)-f(1)=0 in the second equation - contradiction. So your initial guess was right.
Question to solve : If f(f(x))=2^x, What function is f(x)? What is f(4)?
First thing i did i tried zero and one and it gives me 0=1 then i stopped thinking of a solution and complete the video and tada the same result 🤣
Plugging 0 and 1 in for x, i found f(0)=f(0)+1 a contradiction.
Simple f(x) = x + f(1 - x(
Just set x = 1/2, and you get f(1/2) - f(1/2) = 1/2
0 = 1/2 ??
@@Achmd that’s the point. There is no such function f(x).
@@NightSkyJeff are you meaning f(1/2) is NOT equal f(1/2) ?
@@Achmd No, I am saying the same thing the video is saying. Watch 4:30 onward.
Maby it has just infinities at those points?
Just put 1/2
Can f be a complex function?
the proof only proves that the function is not defined in 0, 1/2, 1
I don't think this is a right question. when x = 1/2 => f(x) - f(1-x) = f(1/2) - f(1/2) = 0, but per the question, the result shall be 1/2.
Or, you can take x = 0 => f(0) - f(1) = 0, when x = 1 => f(1) - f(0) = 1 => f(0) - f(1) = -1, which contradict with the first equation.
What if f() was NOT a function?!🤔
set x=.5:
f(.5) - f(1 - .5) = f(.5) - f(.5) = 0 but this is not = .5
so there is no such function f()
Can’t you just substitute x=1/2 and get 0=1/2
You certainly can!
A tricky functional equation elegantly explained, I really enjoyed it.
Glad you liked it!
try substitution u := 1-x so x == 1-u f(1-u) - f(u) = 1-u; times.[-1]:: f(u) - f(1-u) = u-1 & f(u) - f(1-u) = u [from definition] this concludes u = u-1 a non reconcilable in most fields {presuming the REAL's here}
To be exact, the equation is impossible in every nontrivial ring. You do not have to restrict yourself to a field.
Я что-то не понял. Такой функции не существует?
f(x)=0
Plug in x=0 and you find
f(0)-f(1) = 0
So f(0) = f(1)
If you plug in x=1, you find
f(1)-f(0) = 1
But f(1)-f(0) = 0, so this is a contradiction.
So no solution exists.
That only proves there is no solution for x = 0 or 1
Thats just an error problem. Absurd.
He is a bad teacher. Just because he knows it he doesn't give time to the students to follow
👍
nonsense. too much wishful thinking here. There is no solution f(x) that satisfies the given equation. Just replace x by 1-x in the original equation and add up the equations.
Lol "nonsense" isn't quite the way I'd put it; the problem is worth working through because it's important to know when a solution doesn't exist 😁
You got it! Thanks 💖
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