Functional Square Root Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   06/08/2022, 06:06 AM (This post was last modified: 07/01/2022, 12:07 AM by Catullus.) Does anyone know of a function $f(x)$, such that $\sqrt{f}(x)=f(\sqrt{x})$? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ tommy1729 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 06/08/2022, 12:14 PM Yes x^4 obviously. Regards Tommy1729 MphLee Long Time Fellow Posts: 368 Threads: 28 Joined: May 2013 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. 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 $$f(x)^4=f(f(x))$$; MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 06/08/2022, 09:47 PM (This post was last modified: 07/01/2022, 12:11 AM by Catullus.) (06/08/2022, 02:54 PM)MphLee Wrote: 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 $$f(x)^4=f(f(x))$$; Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and $f(x)$ is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If $f(x)=x^0+x+x^2/2$, then $\sqrt{f}(x)$ would have to equal $\sqrt{x^0}+\sqrt{x}+\sqrt{x^2/2}$. But $f(x)=x^0+x+x^2/2$ is not the answer. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ tommy1729 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 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 $$s=\sqrt{}$$ 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 $$f(x)^4=f(f(x))$$; Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and $\sqrt{f}$(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, $\sqrt{f}$(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 tommy1729 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 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 $$s=\sqrt{}$$ 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 $$f(x)^4=f(f(x))$$; Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and $\sqrt{f}$(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, $\sqrt{f}$(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 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 Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   06/10/2022, 11:35 PM (This post was last modified: 07/01/2022, 12:13 AM by Catullus.) On the topic of function composition. Is the successor function the only function $f(x)$ such that $f(f(x))=f(x)+1$? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow Posts: 1,064 Threads: 121 Joined: Dec 2010 06/11/2022, 12:14 AM (This post was last modified: 06/11/2022, 12:18 AM by JmsNxn.) (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. Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   06/11/2022, 12:36 AM (This post was last modified: 07/11/2022, 11:11 PM by Catullus.) Thank you for helping me. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   06/11/2022, 03:10 AM (This post was last modified: 07/01/2022, 12:15 AM by Catullus.) What about that $\sqrt{f}(x)=f(\sqrt{x})$? Is $f(x)=x^4$ the only answer to it? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ « Next Oldest | Next Newest »

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