Does anyone know of a function , such that ?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Functional Square Root

Does anyone know of a function , such that ?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
06/08/2022, 12:14 PM
Yes x^4 obviously.
Regards Tommy1729
06/08/2022, 02:54 PM
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.
MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\) (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 piecewise analytic.), and square roots each term of the Taylor series about any point. If , then would have to equal . But is not the answer.
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
06/08/2022, 10:31 PM
(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 piecewise 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
06/08/2022, 11:45 PM
(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 piecewise 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
On the topic of function composition. Is the successor function the only function such that ?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ (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.
Thank you for helping me.
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ 
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