A nice series for b^^h , base sqrt(2), by diagonalization

[Gottfried]

Here for base b = sqrt(2) :
\(
\begin{tabular}{llll}
\exp_{\sqrt{2}}^h(1)
& = 2 \\
& - 0.6321 u_h \\
& - 0.2253 u_h^2 \\
& - 0.08541 u_h^3+ \cdots
\end{tabular}
\)
(I rewrote this in TeX)

I used regular iteration, and I did not get these expansions at all. I got
\(
\begin{tabular}{llll}
\exp_{\sqrt{2}}^h(x+2)
& = 2 \\
& + u_h x \\
& + (0.5647 u_h - 0.5647 u_h^2) x^2 \\
& + (0.2296 u_h - 0.6378 u_h^2 + 0.3382 u_h^3) x^3 + \cdots
\end{tabular}
\)
so what this series represents would be regular iteration, evaluated at \( x=(-1) \), and expanded about \( u_h \). Is that right? How did you get this?

Andrew Robbins

Hi Andrew,
2'nd note. I just looked at my coefficients. If I do not apply the summation with constant x. I get the same numbers as you gave (just looked at 4 decimals and handful of coefficients). I think we do the same computation except I changed order of summation.

Gottfried