A special equation : f^[t](x) = f(t x)/t

[tommy1729]

For instance solve for

\( f(f(x)) = ( f(2x)/2 + f(4x)/4 )/2. \)

ok lets try.

f(0) = 0
f(1) = a
f(f(1)) = f(a) = f(2)/4 + f(4)/8.
f(f(2x)) = f(4x)/4 + f(8x)/8.

f(2) = b, f(4) = c,
f(f(2)) = f(b) = f(4)/4 + f(/8.
f(a) = b/4 + c/8.
f(b) = c/4 + f(/8.

f(a)/f(b) = (2b + c)/(2c + f()

hmmm
seems like a pattern but idk.

lets use differentiation

f(f(2x)) = f(4x)/4 + f(8x)/8.

f ' (f(2x) ) f ' (2x) 2 = f ' (4x) + f ' (8x).

2 f ' (f(x)) f ' (x) = f ' (2x) + f ' (4x)

hmm

taylor at 0

\( f(f(x)) = ( f(2x)/2 + f(4x)/4 )/2. \)

This gives us a system of equations.
Not sure were it leads us but ...

the first term :

f(x) = a x + ...

f(f(x)) = a^2 x + ...

f(f(x)) = a^2 x + ... = 2 x /4 + ... + 4 x / 8 + ... = 1/2 x + 1/2 x + ... = x + ...

so a = 1.

But we already knew this : f ' (0) = 1 as condition.

But anyways this is solvable.

I think I solved it before actually hmm

well im just thinking out loud.

regards

tommy1729