12/21/2016, 01:27 PM

Some credit Goes to Gottfried.

Assume a strictly increasing analytic function f(x) with fixpoints A and B such that

B > A > 0 and

F(x) > 0 for x > 0

And also the Period of the first fixpoint = the Period of the second fixpoint.

( notice this is possible because Period Q = Period - Q !!! Example f ' (A) = 1/ f ' (B) )

Now we construct a function that can be used as a kind of slog type , including if necc BRANCHES for the FUNCTIONAL inverse to make iterations of f.

Call the slog type function T

Then we have for a suitable C whenever f is not growing too fast :

( C depends on the growth of f and the derivate at the fixpoints )

T(x) = ... Rf^[-2] c^2 + Rf^[-1] C + R ( x ) + Rf(x)/C + R f^[2](x)/c^2 + ...

Where Rf^[y] means R ( f^[y] (x) ) and R(x) = (x-a)(x-b).

Now we split Up t(x) into t1(x) + t2(x) in the obvious way ( separate positive and negative iterates )

Such that we arrive at

T(x) = t1(x) + t2(x)

T(f(x)) = t1(x) C + t2(x) / C.

So finally ( again for inverse use branches )

We get the slog type equation

T( f(x)^[y] ) = t1(x) C^y + t2(x) C^{-y}

As desired.

Since the periods agree and we can use analytic continuation we have a solution that is analytic in the interval [A, +oo] , assuming that T is analytic in a nonzero radius.

So we have a solution agreeing on 2 fixpoints !!

Regards

Tommy1729

Assume a strictly increasing analytic function f(x) with fixpoints A and B such that

B > A > 0 and

F(x) > 0 for x > 0

And also the Period of the first fixpoint = the Period of the second fixpoint.

( notice this is possible because Period Q = Period - Q !!! Example f ' (A) = 1/ f ' (B) )

Now we construct a function that can be used as a kind of slog type , including if necc BRANCHES for the FUNCTIONAL inverse to make iterations of f.

Call the slog type function T

Then we have for a suitable C whenever f is not growing too fast :

( C depends on the growth of f and the derivate at the fixpoints )

T(x) = ... Rf^[-2] c^2 + Rf^[-1] C + R ( x ) + Rf(x)/C + R f^[2](x)/c^2 + ...

Where Rf^[y] means R ( f^[y] (x) ) and R(x) = (x-a)(x-b).

Now we split Up t(x) into t1(x) + t2(x) in the obvious way ( separate positive and negative iterates )

Such that we arrive at

T(x) = t1(x) + t2(x)

T(f(x)) = t1(x) C + t2(x) / C.

So finally ( again for inverse use branches )

We get the slog type equation

T( f(x)^[y] ) = t1(x) C^y + t2(x) C^{-y}

As desired.

Since the periods agree and we can use analytic continuation we have a solution that is analytic in the interval [A, +oo] , assuming that T is analytic in a nonzero radius.

So we have a solution agreeing on 2 fixpoints !!

Regards

Tommy1729