11/11/2009, 11:58 PM
could it be true that by andrew robbins method :
f(x) = 2^f(x-1) f(0) = 1 ( coo base 2 tetration thus )
g(x) = df^2/dx^2 f(x)
g(0) = 0 ???
regards
tommy1729
f(x) = 2^f(x-1) f(0) = 1 ( coo base 2 tetration thus )
g(x) = df^2/dx^2 f(x)
g(0) = 0 ???
regards
tommy1729

