A Curious Functional Equation | Math Olympiads
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- Опубліковано 15 вер 2024
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f(f(x))=e^{-x}
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You concluded that f is neither increasing nor decreasing... but that doesn't mean that it doesn't exist.
If you assume f is continuous, then sure, an injective continuous function on R is either strictly increasing or strictly decreasing. But that wasn't stated in the problem.
You could still have some crazy function that's injective, and neither increasing nor decreasing.
Not in the real world. This would imply a constant function, which clearly isn‘t a solution. As Simon Clarkstone there might be a solution in the complex world.
@@thomaslangbein297 It would not. Consider for instance the function f(x) = 1/x for x nonzero and f(0) = 0. This function is clearly injective, but is not monotone.
@@sjoerdo6988 does your function applied twice lead to e^-x? Besides, it‘s not continuous, so, in each interval monotonous.
@@thomaslangbein297 look at the top comment. A construction of a real-valued, injective, non-constant function satisfying the functional equation is given.
@@nagamanikomarla5376 and you should look whom I responded to LOL
Let g(x) = exp(-x).
The real numbers can be partitioned into an uncountable number of sets of the form { x, g(x), g(g(x)), g(g(g(x))), ... }. One of these sets is singleton containing the solution to g(x) = x. Let c be this value and set f(c) = c.
Since we have an infinite number of such sets, we can pair them (this will require the axiom of choice, I think). For every such couple (A, B), with A = { a, g(a), g(g(a)), g(g(g(a))), ... } and B = { b, g(b), g(g(b)), g(g(g(b))), ... }, set
f(a) = b, f(b) = g(a), f(g(a)) = g(b), f(g(b)) = g(g(a)), f(g(g(a))) = g(g(b)), ...
This defines a function f : ℝ → ℝ such that f(f(x)) = g(x) for every x ∈ ℝ.
You have explicitly highlighted the problem that I spotted: it's only true that injective => monotone for continuous functions.
Just as an aside: If sets of this form are additionally required to be maximal, they always start with a negative number x (easy to see); these x (together with c) can be taken as representatives for the partition and any pairing gives rise to a particular f. These are the only solutions to the functional equation. If the problem was indeed an olympiad problem, your solution is the intended one (I guess).
You didn't finish your thought. You saidbthst defines a function but then what? Where do you go from here..since thst function does not satisfy this relation so there is no solution to this equation..
@@Nepenth I'm not sure what you mean by that. Pairings of what x and c. But those don't satisfy this equation..there is no solution since if f(×)= x or even x plus c or something you will never wring an e out of that . Because then you would have e^e^xc or e twice. Ergo there are no solutions to this..at least no real functions don't ask Me about complex function solutions or whatever..
@@leif1075 What makes you think that this doesn't satisfy the functional equation in the video?
Ooh, trick question!
It is not necessary for f(x) to be uniformly increasing or decreasing on the real numbers. As long as f(x) is continuous and non-constant there will be some interval
on the real numbers where f(x) is monotonic, and on this interval the logic in the video applies. You can't compose a monotonic function with itself on some real interval
and produce a decreasing function. It's sort of analogous to the real numbers: a negative times a negative is positive. and a positive times a positive is positive.
There is still an unexplored posibility if we can redefine the domain of f as a single value. It would have to satisfy x=e^-x , which has the solution W(1) or about 0.567.
This is incoherent
@@angelmendez-rivera351Petru was saying that there is a solution at a single value of x. The function f(f(x)) in the video works at the point Petru noted.
Good point!! Just because the function isn’t valid everywhere doesn’t mean that it isn’t valid anywhere.
This proof assumes that f is continuous, but we only know that f \circ f is continuous. There's a missing step there
A very interesting problem ! Thank you 😊
I was expecting some kind of complex solution. Your proof as written uses ">" and "
If you look at the complex version of this problem and compare the absolute values (the modulus) of both functions, it may work! I didn't do the calculation, but I suppose the logic could be similar because e^x is analytic
finding f, when f•f=g is still pretty unknown territory in general. There might indeed exist a complex solution, but who knows?😅
Plus, the proof that f(x)=f(y) => x=y doesn't work if complex values are allowed. Instead, x=y-2πni for integers n.
je ne comprends pas la langue anglaise mais votre explication est tellement logique que je l'approve . BRAVO :)
I have never learnt French, but I understood that sentence without using a translation tool.
@@md2perpe Comment avez-vous déterminé que c'est une langue française si vous ne l'avez jamais apprise ? )) MDR
I can finally understand french not from a textbook
@alexeyalex8482 You don't need to learn a given language (it means knowing the vocabulary and grammar) to recognize it. It is enough to know its characteristic features.
@@piotrmiska3361 фраза "I have never learnt French, but I understood" это оксюморон.
If we have increasing right side f(f(x))=e^x instead, then f(x) can be monotonically increasing or decreasing. Now, what is the solution of this functional eq? Is it analytically solvable ?
there is no elementary function that solves this. The solution would have to grow faster than any polynomial but slower than any exponential functions. And it is shown that such a function cannot be written using our normal tools...
But there are solutions nonetheless. It is related to finding a continous version of f^n(x) where
f^0(x)=x
f^1(x)=f(x)
f^2(x)=f(f(x))
etc..
for real or complex values of n. in this case n=1/2. Fascinating topic
Don't you also need to prove or assume that f is continuous before you can demonstrate that it must be monotone?
Yes. The entire video of his rests on the incorrect assumption that f must be monotonic/antitonic solely because it is injective.
3:04 - 3:07 False. This is incorrect. A trivial example is the function f : R -> R, such that for all x in R, f(x) = -x. f is injective, and even continuous, but it is antitonic, not monotonic. Also, there are examples of functions which are injective, and are neither monotonic nor antitonic. For example, consider a Hamel function f : R -> R such that for all x, y in R, f(x + y) = f(x)•f(y). Such a Hamel function is injective, but not monotonic, and not antitonic. Injectivity does not imply monotonicity or antitonicity.
3:53 - 4:05 This is not mathematically correct terminology. Given two posets (X, = X' is called monotonic if and only if x =< y implies f(x) =
Do you suggest any books to learn these functional topics in particular or are they just included as part of analysis?
Great proof! And just simple and straightforward. People cannot except the idea that the half-exponential doesn’t exist. It‘s the same or similar with the idea that 1 to the x equals 2 has no solution.
Check out the comment made by md2perpe where he gives the general method of solving the equation f(f(x)) = g(x)
Also, half exponential does indeed exist. It is just impossible to express it using elementary functions. You can look that up on internet.
Yeah, your function is not differentiable at any point. You cannot give the output of any value. It‘s a function in the same sense as x = ”Hamburg“ g(x) = ”Thursday“ x = ”green“ g(x) = ”black“ x = 3 g(x) = ”Betelgeuze“. This is a totally valid function!!
@@thomaslangbein297 "Yeah, your function is not differentiable at any point"
But it is still a function.
Also, we can give the output of some particular values. For example, if we choose f(-1) to be -2, we can know the values of g(-1), g(-2l , g(g(-1) , g(g(-2) and so on .. (I suppose you have read the comment
However, it's impossible to determine the output of every input, much like how one can't determine all the digits of an uncomputable number.
@@UA-cam_username_not_foundIt‘s a chimera. You know there must be a function, but you can never find it! It‘s like the mystic ElDorado. You must make assumptions to get a little bit further. Let me put it like this: You don’t know the output of g(Hamburg), but if you assume f(Hamburg) is 3 then you know that g(Hamburg) is Thursd… (you don’t know whether it‘s Thursdok or Thursdff or Thursday, and you had to make the assumption in the first place. Like I don’t know the googleth cipher of pi but under the condition (purely hypothetical) the google-1th cipher is 3 it will be 7. Bravo!
@@thomaslangbein297 I don't think things are how you describe them.
It's more like "there is a function but we can't totally describe it"
The fact that we cannot know all the digits of an uncomputable number doesn't mean they don't exist. The digits in the decimal base are still some numbers between 0 and 9.
the assumption I made about f(-1) is crucial because there isn't just a unique solution to the equation, it's like imposing an initial condition on a differential equation in order to select one of its infinite solutions.
As for Pi, it is a computable number, there is a way to determine any digit we like given its position . However the fact that we don't know the digits a priory doesn't mean that they don't exist.
Hmmm, interesting. I was looking to see how it was going to work out because ln doesn’t work and e^x has no more cousin.
Thanks! This is great!
You're welcome!
Just 24 seconds after starting it
Supponendo una f(x)=Ae^(-x)...risulta f(x)=-W(-e^(-2x))
こういう異質感ある問題好き！
私はそれを聞いてうれしい！
Can you solve f(f(x))=exp(x)?
just answered that to another comment:
there is no elementary function that solves this. The solution would have to grow faster than any polynomial but slower than any exponential functions. And it is shown that such a function cannot be written using our normal tools...
But there are solutions nonetheless. It is related to finding a continous version of f^n(x) where
f^0(x)=x
f^1(x)=f(x)
f^2(x)=f(f(x))
etc..
for real or complex values of n. in this case n=1/2. Fascinating topic
What is the point of replacing x with y like at 1:52 I don't see what insight or progress you make from that?
We are not replacing x with y. That's the result that follows from f(x)=f(y)
@@SyberMath but what is f(y) then if x equals y.that just follows from what I am saying you said.isn't that the same thing or sorry maybe I misunderstood..the notation can make it more confusing
3:17 you still have x^2= y^2 => x=y. "=>" needs to be crossed out.
If f is injective that dosn't mean that f is monotone.
Prove it
@@ht2897 f(x)=x on [0,1] and f(x) = 4-x on ]1,2] is injective but not monotone. We need the continuity of
f for the monotonous
@@ht2897 f(x) = x if x is rational and f(x)=-x if x is irrational. Injective (bijective in fact), not continuous on any interval, and not monotonous on any interval.
As a bonus, f(f(x))=x. I know, not a solution to this problem, but proof that a crazy function might still behave nicely when composed with itself.
Unexpected!
Brilliant
Lovely!
@ 2:52 does this mean f is injective
No second method?
What if there are complex valued solutions
That becomes significantly harder
It's suppose that what's you are presented to be a mathematical solution !!!!!!!!
Nice
رائع
لكن هل هناك تكافؤ يسمح لنا بحسم الأمر!
güzel soru hocam
It's a too complicated way. I've another one which is much easy.
f(f(x)) = exp(-x);
[x=0]: f(f(0)) = 1;
[x=1]: f(f(1)) = e⁻¹;
[x=f(0)]: f( f(f(0)) ) = f( exp(-f(0) );
⇒ f( 1 ) = f(exp(-f(0));
[x=f(1)]: f( f(f(1)) ) = f( exp(-f(1) );
⇒ f( e⁻¹ ) = f( exp(-f(1) ) [ same as: f( exp( -1 ) ) = f( exp( -f(1) ) ) ]
⇒ -1 = -f(1) ⇒ f(1) = 1;
BUT: f(f(1)) = e⁻¹ ⇒ f(1) = e⁻¹ while f(1) = 1. So, f(x) does not exist.
Mistake in the 4th line, it should be f( f(f(0)) ) = exp(-f(0) ); .
By the way, md2perpe has used a similar reasoning to find the method to solve the equation. Check out his comment!
This dude is taking too long.
Here's my solution
Let f(x)=y
Then x=inv(f(y))=g(y)
==> f(y)=e^(-g(y))
Differentiating wrt y
Therefore, f'(y)= -e^(-g(y)) .g'(y)
Multiplying on both side by g'(y)
f'(y).g'(y)= -e^(-g(y)).(g'(y))²
But f'(y).g'(y)=1
(Because f'(x).(inv f'(x))=1)
Hence -e^(-g(y)).(g'(y))²=1
Which implies (g'(y))²= -e^(g(y))
But in real world square of anything cannot be negative
Hence g'(y) does not exist
Which ultimately implies f(x) does not exist
(Please comment if I made some mistake)
What you show then is that there is no differentiable solution. There might still exist non-differentiable solutions.
Not necessarily f'(y)*g'(y) is 1, but f'(x)*g'(y)=1. This is because of the chain rule (g(f(x))'=g'(f(x))*g'(x).
f'(y)*g'(y) = 1 (with g = f^-1) is not always true. Take f(x) = x^2 and g(x) = x ^ (1/2) for example. It doesn't work.