02/22/2008, 07:17 PM
quickfur Wrote:Hmm, interesting. I was going to ask what happens if we consider f(x)=c#x instead, but it doesn't change the conclusion because we assume the same operator #. So we established that a higher operation cannot distribute over exponentiation without becoming trivial.
I wonder what is the root cause of this... maybe it has something to do with the non-associativity of exponentiation?
Who knows, but the proof can be similarly done for other distributivity type functional equations, for example
\( f(xy)=f(x)f(y) \) with the only continuous solutions \( f(x)=x^c \) or \( f(x)=0 \)
\( f(x+y)=f(x)+f(y) \) with the only continuous solutions \( f(x)=cx \).
