07/01/2022, 12:10 AM
Idk Tommy. In my attempt I tried to set it up as \(Dg=C_fg\) i.e. \[g'(x)=g(f(x))\] where we solve in the \(g\)
I did that only because pre-composition by \(f\) is linear. Only for that reason. In fact \(g'(x)-g(f(x))=0\) can be expressed in linear operators as \((D-C_f)[g(x)]=0\)...
Anyways I remember I read something about \(g^{-1}(g'(x))\)... somewhere... maybe something about the legendre transform? Too bad I'm ignorant about that.
Just a note: define \(\mathcal F[g]=g^{-1}\circ Dg\) then what we are loking for are fixed points \[\mathcal F[f]=f\] So maybe we could unlock some iterative method: consider
\[f_{n+1}(x)=f_n^{-1}(f_n'(x))\]
and take the limit \(\lim f_n\)
I did that only because pre-composition by \(f\) is linear. Only for that reason. In fact \(g'(x)-g(f(x))=0\) can be expressed in linear operators as \((D-C_f)[g(x)]=0\)...
Anyways I remember I read something about \(g^{-1}(g'(x))\)... somewhere... maybe something about the legendre transform? Too bad I'm ignorant about that.
Just a note: define \(\mathcal F[g]=g^{-1}\circ Dg\) then what we are loking for are fixed points \[\mathcal F[f]=f\] So maybe we could unlock some iterative method: consider
\[f_{n+1}(x)=f_n^{-1}(f_n'(x))\]
and take the limit \(\lim f_n\)
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)