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 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 \(f(x)^4=f(f(x))\);
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)