Let $f(x,y)$ be a symmetric real function and a formal group law

$$G(x + y) = f(G(x),G(y)). \tag{1}$$

Consider the equation

$$ h(2x) = f(h(x),h(x)) = A(h(x)). \tag{2}$$

This equation has many solutions.

Compute a solution to that equation with the fixed point at $0$ and its Koenigs function, and call the solution $k(x)$.

So

$$k(2x) = A(k(x)). \tag{3}$$

Then it seems it is always true that

Conjecture $T$:

$$ G(x) = k(x), \tag{4}$$ $$ G(x+y) = k(x+y) = f(k(x),k(y)). \tag{5}$$

Of course we can not use The Koenigs function if its conditions are not met, and there is no way around it. In other words The fixed point of $A(x)$ (at $0$) should not be parabolic and must be strictly positive.

I searched the Internet for Koenigs function and formal group law but did not find them combined.

Is conjecture $T$ true? How is it proved?

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After hesitation I considered posting A generalisation that might be helpful.

Generalization

Let $ g(2x) = A(g(x)) [1]$

and

$ g(x+1) = B(g(x)) [2] $.

( [3] note : $[1]+[2]$ define $g$ uniquely and equivalently they define $f(x,y)$ uniquely )

Let $ K_1(x) $ be the solution to $ K_1(2x) = A(K_1(x)) $ obtained by using the associated koenigs function.

Likewise Let $ K_2(x) $ be the solution to $ K_2(x+1) = B(K_2(x)) $ obtained by using the associated koenigs function.

Conjectures :

A) $K_1(x) $ satisfies $[2]$.

B) $K_2(x) $ satisfies $[1]$.

C) $K_1 = K_2$

D) $ K_1 = g $

E) $ K_2 = g $

Ofcourse these conjectures relate to each other and proving a few is equivalent to proving them all because of note [3].

Ps: it might be intresting to consider ( or generalize the conjecture ) replacing [2] with $g(x + q) = B_q(g(x)) $ for Some ( or any ) values of $q$ say eg 2 or e.

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