08/22/2007, 11:28 PM
Well, I've prepared some graphs in the hope that I can address the question about my super-logarithm extension. The question was "are you sure?" about my super-logarithm definition not working for bases between \( 1<b<\eta \). My only answer is to look at the graphs.
These are some graphs of the first three coefficients of the super-logarithm, i.e. the Abel function of \( b^x \). In each graph there are multiple curves, each curve corresponds to a specific approximation; the approximations shown are roughly \( n=\{2, 3, 4, ..., 10\} \) and the independant axis represents the base \( 1 < b < 4 \). What follows are graphs of \( A_k \) as functions of b for \( k=\{1, 2, 3\} \) for the functional equation \( A(b^x) = A(x) +1 \) with \( A(0) = -1 \).
As you can see, when the base is between \( 1<b<\eta \), although the matrix equation has a solution, the solutions (coefficients) do not seem to converge. This is what leads me to beleive that my super-logarithm definition will only work for \( b>\eta \).
Andrew Robbins
These are some graphs of the first three coefficients of the super-logarithm, i.e. the Abel function of \( b^x \). In each graph there are multiple curves, each curve corresponds to a specific approximation; the approximations shown are roughly \( n=\{2, 3, 4, ..., 10\} \) and the independant axis represents the base \( 1 < b < 4 \). What follows are graphs of \( A_k \) as functions of b for \( k=\{1, 2, 3\} \) for the functional equation \( A(b^x) = A(x) +1 \) with \( A(0) = -1 \).
As you can see, when the base is between \( 1<b<\eta \), although the matrix equation has a solution, the solutions (coefficients) do not seem to converge. This is what leads me to beleive that my super-logarithm definition will only work for \( b>\eta \).
Andrew Robbins