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 composition lemma tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 04/21/2012, 05:04 PM 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 tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 04/29/2012, 08:32 PM 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 « Next Oldest | Next Newest »

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