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+--- Thread: composition lemma (/showthread.php?tid=734)
composition lemma - tommy1729 - 04/21/2012
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
RE: composition lemma - tommy1729 - 04/29/2012
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