what f satisfy the following equation: f(f(x)) = 2^x?
g(2)=2^2=4 so 2 is not a fixed point. I don't think that 2^x has a fixed point in the reals.Let g(x) = 2^x
x = 2 is a fixed point of g(x), so it makes sense to define
oops! yeah, the second to the last line is incorrect. as you said, it should be A(16)=A(2^4)=A(4)+1=s+5. then the last line should be, as you said, A(65536)=A(2^16)=A(16)+1=s+7.A(2^(-oo))=A(-oo)+1 yields A(0)=s+1.
A(2^0)=A(0)+1 yields A(1)=s+2.
A(2^1)=A(1)+1 yields A(2)=s+3.
A(2^2)=A(2)+1 yields A(4)=s+4.
A(2^3)=A(3)+1 yields A(8)=s+5.
A(2^4)=A(4)+1 yields A(16)=s+6.
a similar thing was done with abel's equation. now i get how definining f or the solution to abel's/schroeder's equation on some interval will then lock in place all other values but i don't think this can be done in an arbitrary way. how is it ensured, for example, that the arbitrary assignment of values to f over the range [-oo,b] will lead to satisfaction of f(f(x))=2^x?Assign any values you like to f(x) over the range [-oo, b] satisfying f(-oo) = b, f(b) = 0, and f continuous on [-oo, b].