I was thinking over the topic and try to put it on its feet.
The paradigm is that \( \lim_{x\to\infty}\text{slog}_a(\text{sexp_b}(x)) - x \) shall exist for each \( a,b>\eta \).
So if this limit exists, we let \( x=x+n \), \( a=e \)
\( c_{a,b}=\lim_{n\to\infty}\text{slog}_a(\text{sexp}_b(x+n)) - (x+n) \)
\( \lim_{n\to\infty}\log_b^{\circ n}(\text{sexp}_a(x+n+c_{a,b}))=\text{sexp}_b(x) \) (Jay's change of base formula)
\( \lim_{n\to\infty}\log_b^{\circ n}(\text{sexp}_a(x+n))=\text{sexp}_b(x-c_{a,b}) \)
\( \lim_{n\to\infty}\log_b^{\circ n}({\exp_a}^{\circ n}(y)) =\text{sexp}_b(\text{slog}(y)-c_b) \)
\( \kappa_{b,a}(y) = \lim_{n\to\infty} {\log_b}^{\circ n}({\exp_a}^{\circ n}(y)) \), \( \kappa_{a,b}={\kappa_{b,a}}^{-1} \)
Change of base is just the application of a function (I think one could show that \( \kappa \) is analytic):
\( \text{slog}_b(\kappa_{b,a}(x)) = \text{slog}_a(x)-c_{a,b} \)
\( \kappa_{b,a}(\text{sexp}_a(x))=\text{sexp}_b(x-c_{a,b}) \)
While \( \kappa_{a,b} \) does not depend on any \( \text{sexp} \) or \( \text{slog} \), \( c_{a,b} \) does. we can define it by setting \( x=c_{a,b} \) in the second equation:
\( \kappa_{b,a}(\text{sexp}_a(c_{a,b}))=1 \)
\( c_{a,b}=\text{slog}_a(\kappa_{a,b}(1)) \).
So we need to show in your case that the limit:
\( \lim_{b\to\eta^+}\text{slog}_b(\kappa_{b,a}(x)) \) exists for \( x \) in some initial range, where \( \text{slog}_b \) is your linear approximation.
To show this we could use the Cauchy criterion. It should work in the form \( \lim_{b\to\eta} f_b \) exists, if for each \( \epsilon>0 \) there exists a \( \delta>0 \) and a \( b_0>\eta \)such that for all \( \eta < b,b' <b_0 \) and \( |b-b'|<\delta \): \( |f_b\circ f_{b'}^{-1}-\text{id}|<\eps \).
Where we put \( f_b=\text{slog}_b\circ \kappa_{b,a} \).
Suprisingly but happily the compositional difference is independent on \( a \):
\( f_b\circ f_{b'}^{-1} = \text{slog}_b\circ \kappa_{b,a}\circ \kappa_{a,b'}\circ \text{sexp}_b = \text{slog}_b\circ \kappa_{b,b'}\circ \text{sexp}_b \).
Well, I dont pretend that it helps you, but it helped me at least somewhat in understanding
The paradigm is that \( \lim_{x\to\infty}\text{slog}_a(\text{sexp_b}(x)) - x \) shall exist for each \( a,b>\eta \).
So if this limit exists, we let \( x=x+n \), \( a=e \)
\( c_{a,b}=\lim_{n\to\infty}\text{slog}_a(\text{sexp}_b(x+n)) - (x+n) \)
\( \lim_{n\to\infty}\log_b^{\circ n}(\text{sexp}_a(x+n+c_{a,b}))=\text{sexp}_b(x) \) (Jay's change of base formula)
\( \lim_{n\to\infty}\log_b^{\circ n}(\text{sexp}_a(x+n))=\text{sexp}_b(x-c_{a,b}) \)
\( \lim_{n\to\infty}\log_b^{\circ n}({\exp_a}^{\circ n}(y)) =\text{sexp}_b(\text{slog}(y)-c_b) \)
\( \kappa_{b,a}(y) = \lim_{n\to\infty} {\log_b}^{\circ n}({\exp_a}^{\circ n}(y)) \), \( \kappa_{a,b}={\kappa_{b,a}}^{-1} \)
Change of base is just the application of a function (I think one could show that \( \kappa \) is analytic):
\( \text{slog}_b(\kappa_{b,a}(x)) = \text{slog}_a(x)-c_{a,b} \)
\( \kappa_{b,a}(\text{sexp}_a(x))=\text{sexp}_b(x-c_{a,b}) \)
While \( \kappa_{a,b} \) does not depend on any \( \text{sexp} \) or \( \text{slog} \), \( c_{a,b} \) does. we can define it by setting \( x=c_{a,b} \) in the second equation:
\( \kappa_{b,a}(\text{sexp}_a(c_{a,b}))=1 \)
\( c_{a,b}=\text{slog}_a(\kappa_{a,b}(1)) \).
So we need to show in your case that the limit:
\( \lim_{b\to\eta^+}\text{slog}_b(\kappa_{b,a}(x)) \) exists for \( x \) in some initial range, where \( \text{slog}_b \) is your linear approximation.
To show this we could use the Cauchy criterion. It should work in the form \( \lim_{b\to\eta} f_b \) exists, if for each \( \epsilon>0 \) there exists a \( \delta>0 \) and a \( b_0>\eta \)such that for all \( \eta < b,b' <b_0 \) and \( |b-b'|<\delta \): \( |f_b\circ f_{b'}^{-1}-\text{id}|<\eps \).
Where we put \( f_b=\text{slog}_b\circ \kappa_{b,a} \).
Suprisingly but happily the compositional difference is independent on \( a \):
\( f_b\circ f_{b'}^{-1} = \text{slog}_b\circ \kappa_{b,a}\circ \kappa_{a,b'}\circ \text{sexp}_b = \text{slog}_b\circ \kappa_{b,b'}\circ \text{sexp}_b \).
Well, I dont pretend that it helps you, but it helped me at least somewhat in understanding
