$f(a(x))=f(x)$ - useful formula

I read "Functional Equations and also How to Solve Them" by Small and also the list below comment turns up without much validation on p. 13:

If $a(x)$ is an involution, after that $f(a(x))=f(x)$ has as remedies $f(x) = T\,[x,a(x)]$, where $T$ is an approximate symmetrical function of $u$ and also $v$.

I was asking yourself why this held true (it benefits instances I've attempted, yet I am not exactly sure $(1)$ just how to confirm this and also $(2)$ if there's anything noticeable looking at me in the face below).

0
2019-05-05 23:21:53
Source Share
Answers: 1

Any function f (x) that is a remedy to your useful formula f (a (x)) = f (x) have to please the building that it is unmodified when you connect in a (x) as opposed to x. On top of that, f (x) plainly has to be a function f (x) = T [x, a (x) ] relying on the x and also a (x) ; the inquiry is to see why T has to be symmetrical.

Currently, T [x, a (x) ] = f (x) = f (a (x)) = T [a (x), a (a (x)) ] = T [a (x), x ] given that a (x) is an involution. Specifically, this suggests that T has to be symmetrical in its 2 variables.

0
2019-05-08 14:08:47
Source