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Functional Square Root - Catullus - 06/08/2022
Does anyone know of a function , such that ? RE: Functional Square Root - tommy1729 - 06/08/2022
Yes x^4 obviously. Regards Tommy1729 RE: Functional Square Root - MphLee - 06/08/2022
The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\) If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example. - if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
- if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
RE: Functional Square Root - Catullus - 06/08/2022
(06/08/2022, 02:54 PM)MphLee Wrote: The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If , then would have to equal . But is not the answer. RE: Functional Square Root - tommy1729 - 06/08/2022
(06/08/2022, 09:47 PM)Catullus Wrote:(06/08/2022, 02:54 PM)MphLee Wrote: The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and (x) is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If f(x) = x^0+x+x^/2, (x) would have to equal sqrt(x^2)+sqrt(x)+sqrt(x^2/). But f(x)=x^0+x+x^2/2 is not the answer. Well x^4 is analytic everywhere. And cosh(x) also not a solution. Let g(x) = c * x^a. f(x) = g(g(x)) = c * ( c^a x^(a^2) ) = c^(a+1) x^(a^2) f(sqrt(x)) = c^(a+1) x^(a^2 /2) = g(x) so c^(a+1) = c , a^2 = 2 * a. So a = 0 or a = 2. So we get g(x) = C for any C. or g(x) = c x^2 for c^3 = c. c^3 = c implies c = -1 or 0 or 1. regards tommy1729 RE: Functional Square Root - tommy1729 - 06/08/2022
(06/08/2022, 10:31 PM)tommy1729 Wrote:(06/08/2022, 09:47 PM)Catullus Wrote:(06/08/2022, 02:54 PM)MphLee Wrote: The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and (x) is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If f(x) = x^0+x+x^/2, (x) would have to equal sqrt(x^2)+sqrt(x)+sqrt(x^2/). But f(x)=x^0+x+x^2/2 is not the answer. also Let g(x) = c(x) x^a f(x) = g(g(x)) = c(g(x)) * g(x)^a = c(g(x)) * c(x)^a * x^(a^2) f(sqrt(x)) = c(g(sqrt(x)) * c(sqrt(x))^a * x^(a^2/2) f(sqrt(x))/g(x) = 1 = c(g(sqrt(x)) * c(sqrt(x))^a / c(x) * x^(a^2 / 2 - a) 1 = c(g(x))* c(x)^a / c(x^2) * x^(a^2 - 2a) WLOG we can take a = 0 ( and consider c a function of x and a !! ) so we get : c(x^2) = c(g(x)) c(x^2) = c(c(x)) d (x^2)^2 = d (d (x^2))^2 = d^3 x^4 = d x^4. just as before. So it seems we have uniqueness ; there are no other solutions. regards tommy1729 RE: Functional Square Root - Catullus - 06/10/2022
On the topic of function composition. Is the successor function the only function such that ? RE: Functional Square Root - JmsNxn - 06/11/2022
(06/10/2022, 11:35 PM)Catullus Wrote: On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1? This depends on how you want to phrase this question. I believe you could probably find solutions to this equation in tropical algebra, and crazy areas like that. In Complex Function theory, the answer is yes, but with a caveat. If you assume that \(x \in \widehat{\mathbb{C}}\), then \(f = \infty\) is also a solution. But, there's much much much more from this equation. $$ f(f(x)) = f(x) + 1\\ $$ Since \(\infty\) is a fixed point (assuming \(f(\infty) = \infty\)), then... let's do leibniz variable change: \(F(x) = 1/f(1/x)\). This function satisfies \(F(0) = 0\), assuming that \(f\) is meromorphic at infinity. Now take \(F'(0) = C\). We are asking that: $$ F(F(x)) = F(x) + 1\\ $$ In a neighborhood of \(x \approx 0\). This is kind of like a golden ratio/fibonacci equation, which I've never seen before. I think It's unique though. $$ \begin{align*} F'(0)\cdot F'(0) = F'(0) = 1\\ F'(F(x))\cdot F'(x) = F'(x)\\ \end{align*} $$ Therefore \(F' =1\). Therefore \(F(x) = x+1\). Therefore there are two solutions in complex analysis. There's the constant solution \(\infty\), and the successor function \(x \mapsto x+1\). So you are right, but there's an exception in complex analysis. The "successorship" hyperoperator could be an essential singularity... it looks like how \(e^{1/z}\) looks like \(0\) and \(\infty\) near \(z\approx 0\). We can have a hyperoperator successorship that looks like \(\infty\) and \(x \mapsto x +1\). So, SINGULAR solutions to the above question can exist. RE: Functional Square Root - Catullus - 06/11/2022
Thank you for helping me. RE: Functional Square Root - Catullus - 06/11/2022
What about that ? Is the only answer to it? |