10/20/2009, 08:42 PM
Just had another thought on this matter.
In order for this interpolation to work, you should also check to see if it applies for ranks above 3, for example (to use a combination Knuth-Bromer-Mueller arrow system), we have the first 5 lower hyperoperations are: \( \{a + b, ab, a{\uparrow}b, a{\downarrow}{\uparrow}b, a{\downarrow}{\downarrow}{\uparrow}b\} \), and so we should check that:
\( \mathcal{S}[{\downarrow}{\uparrow}b](x) = \log_a(\log_a(x)) \), and
\( \mathcal{S}[{\downarrow}{\downarrow}{\uparrow}b](x) = \log_a(\log_a(\log_a(x))) \)
and I have a feeling that this is not the case.
In order for this interpolation to work, you should also check to see if it applies for ranks above 3, for example (to use a combination Knuth-Bromer-Mueller arrow system), we have the first 5 lower hyperoperations are: \( \{a + b, ab, a{\uparrow}b, a{\downarrow}{\uparrow}b, a{\downarrow}{\downarrow}{\uparrow}b\} \), and so we should check that:
\( \mathcal{S}[{\downarrow}{\uparrow}b](x) = \log_a(\log_a(x)) \), and
\( \mathcal{S}[{\downarrow}{\downarrow}{\uparrow}b](x) = \log_a(\log_a(\log_a(x))) \)
and I have a feeling that this is not the case.
