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Hi

while trying to make a numerical method i stumbled upon a perhaps classical and/or old question :

" composition lemma "

f: R-> R
g: R->R

Let f [ g^(-1)[x] ] + g^(-1) [ f[x] ] = 2 g[x]
(?) ==> (?) g^(2) [x] = f [x]

regards

tommy1729
Let f [ g^(-1)[x] ] + g^(-1) [ f[x] ] = 2 g[x]

replace x with g(x)

hence we get for x e image g(x)

f(x) + g^(-1)[ f[g(x)] ] = 2 g[g(x)]

if f and g are real analytic then for all real x

f(x) + g^(-1)[ f[g(x)] ] = 2 g[g(x)]

Let g(g(x)) = f(x) + r(x)/2

hence

f(x) + g^(-1)[ f[g(x)] ] = 2 f(x) + r(x)

g^(-1)[ f(g(x)) ] = f(x) + r(x)

thus r(x) = 0 for all x iff f(g(x)) = g(f(x))

which is the condition we needed.

QED

tommy1729