Again we begin with some standard notation:
f(x) = cx, c > 0, x in R.
f(n)(x) = {f(x), iff n = 1, f(f(n-1)(x)) iff n > 1}.
Lemma:
If k>=2, then for all n >=1, all the real fixed points of the function f(n)(x), are also real fixed points of the functions f(kn)(x).
Proof:
Trivial, by induction.
Corollary 1:
The real fixed points of the function f(x) are also real fixed points of all the functions f(n)(x), and the real fixed points of the function f(2)(x) are also real fixed points of all the functions f(2k)(x).
Corollary 2:
If t odd, m even and m > t, then all the real fixed points of the functions f(t)(x), are also real fixed points of the functions f(m)(x).
Proof:
Follows from the two cases n=1, n=2 from the previous articles: Solving the First Auxiliary Real Exponential Equation, Solving the Second Auxiliary Real Exponential Equation, where it is clear that the real fixed points of the function f(x) are real fixed points of the function f(2)(x), along with the Lemma, above and induction.
Theorem:
If n=2k or n=2k+1, then the functions f(n)(x), have exactly the same real fixed points (in other words: If n=2k or n=2k+1, then the n-th Auxiliary Exponential Equations f(n)(x) - x = 0, have exactly the same real roots).
Proof:
By induction on odd and even n. Steps n = 1 and n = 2 have been addressed in the articles: Solving the First Auxiliary Real Exponential Equation, Solving the Second Auxiliary Real Exponential Equation. For the inductive step: 1) For even n=2k. Assume the theorem true for k=m, and show that it is true for k=m+1. That is: We need then to show that the functions f(2m)(x) and f(2(m+1))(x) have exactly the same real fixed points. ->: Follows from the Corollary 1. <-: Assume not. Let x0 be a fixed point of f(2(m+1))(x) that is not a fixed point of f(2m)(x). Then: f(2(m+1))(x0) = x0 and f(2m)(x0) <> x0. => f(2m)(f(2(m+1))(x0)) = f(2m)(x0), => f(2t)(x0) = f(2m)(x0). By Corollary 1, t=2m+1 > m+1, so x0 is a fixed point of the function f(2t)(x), so => x0 = f(2m)(x0), and this is a contradiction. 2) For odd n=2k+1. Assume the theorem true for k=m, and show that it is true for k=m+1. That is: We need then to show that the functions f(2m+1)(x) and f(2(m+1)+1)(x) have exactly the same real fixed points. ->: Follows from the Corollary 1. <-: Assume not. Let x0 be a fixed point of f(2(m+1)+1)(x) that is not a fixed point of f(2m+1)(x). Then: >f(2(m+1)+1)(x0) = x0 and f(2m+1)(x0) <> x0. => f(2m+1)(f(2*(m+1)+1)(x0)) = f(2m+1)(x0), => f(2t)(x0) = f(2m+1)(x0). By Corollary 2, 2(m+1)+1 is odd, so x0 is a fixed point of the function f(2(m+2))(x), and t =2*2*(m+1) > 2*(m+2), so x0 is a fixed point of the function f(2t)(x), so => x0 = f(2m)(x0), a contradiction, and the theorem follows.
Note that the above proof cannot be used for the complex case, because there exp and W have multiple branches, so it doesn't follow in that case that fixed points of successor iterates are also fixed points of predecessor iterates. The lemma, however, still holds there as well.
Le's now see some Maple code that will verify the above:
> f:=x->c^x;
> i:=(n,x)->(f@@n)(x)-x;
On the interval (e(1/e),+oo):
> c:=exp(exp(-1))+0.01;
> plot([seq(i(2*n,x),n=1..4)],x=1..3); >
plot([seq(i(2*n+1,x),n=1..4)],x=1..3);
For c = e(1/e):
> c:=exp(exp(-1));
> plot([seq(i(2*n,x),n=1..4)],x=2..3.2);>
plot([seq(i(2*n,x),n=1..4)],x=2..3.2);
On the interval (1, e(1/e)):
> c:=exp(exp(-1))-0.01;
> plot([seq(i(2*n,x),n=1..4)],x=2..3.4);
>plot([seq(i(2*n+1,x),n=1..4)],x=2..3.4);
On the interval (e-e,1):
> c:=exp(-exp(1))+0.01;
> plot([seq(i(2*n,x),n=1..4)],x=0..1);
>plot([seq(i(2*n+1,x),n=1..4)],x=0..1);
For c=e-e:
> c:=exp(-exp(1));
> plot([seq(i(2*n,x),n=1..4)],x=0..1);
>plot([seq(i(2*n+1,x),n=1..4)],x=0..1);
On the interval (0, e-e):
> c:=exp(-exp(1))-0.01;
> plot([seq(i(2*n,x),n=1..4)],x=0..1);
>plot([seq(i(2*n+1,x),n=1..4)],x=0..1);
The real bonus here is that our Maple code that solves the First and Second Real Auxiliary Exponential Equations, solves all the Real Auxiliary Exponential Equations.
The final question is what are the real fixed points of the function F(x) = limn->+oof(n)(x)? But this has been answered with the first article on Infinite Exponentials and we have come full circle.