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 -

sorry, took a long time to answer.
The coefficients occur as sums; your formula contains the x-parameter, maybe if you insert x=-1 we get identity.

[update] I've the same coefficients as yours, just evaluated at x=-1;see next post [/update]

Let me explain in my matrix/Pari-GP-notation how I got the coefficients:

Assume the notation of variables as usual: the variables b,t,u encoding the base-parameters, where for our example t=2, u=log(t) , b=t^(1/t) = sqrt(2).

Also let exp_b°h(x) the h'th iteration of exp-function to base b, and dxp_t°h(x) the h'th iteration of the decremented exp-function to base t where we use the identity (if we have a real fixpoint)

\( \hspace{24} \exp_b^{\circ h}(x) = ( dxp_t^{\circ h} (\frac{x}{t}-1) +1 )*t = t + t* dxp_t^{\circ h} (\frac{x}{t}-1) \)

which will be numerically with x=1, t=2

\( \hspace{24} \exp_{\sqrt2}^{\circ h}(1) = 2 + 2* dxp_2^{\circ h} (-\frac12) \)

The function dxp to base t (=2) gets its coefficients by the triangular Bell-matrix Ut, where we implement the h'th fractional power using diagonalization, denoting the eigenmatrices as W and WInv (=W^-1)

Code:

´  Ut^h = W * dV(u^h) * WInv

and the function is in general finally computed by

Code:

´  V(y)~ = V(x/2-1)~ * Ut^h

Here   y = V(y)[1]  , meaning y is the second element of V(y)
and is also y = dxp°h(-1/2) by the construction of the formula.
Then

   exp_b°h (1) = 2 + 2*y    = 2+ 2*dxp_t°h(-1/2)

Keeping the iteration-parameter variable we have, using the eigenmatrices

Code:

´   V(y)~ = V(x/2-1)~ * W * dV(u^h) * WInv

Since we assume x being constant x=1, we can precompute the rowvector S~

Code:

´     S~ = V(-1/2)~ *W

which implements also a Schröder-function (need not be the principal one) for dxp_t°h(x) at the constant x=-1/2 . This is also the schröder-function for exp_b°h(x) at x=1, as fas as we look only on the coefficients of the 2'nd column of W.

So we can write

Code:

´     V(y)~ = S~ * dV(u^h) * WInv

WInv provides the inverse of the schröder-function, and if we extract only the scalar result we can write ( with the notation W[,1], meaning the second column of a marix W )

Code:

´    y  = S~ * dV(u^h) * WInv [,1]

Here we can interchange the the order of multiplication of the last two factors and precompute the constant coefficients of a powerseries in the resultvector M of

Code:

´   M~ = S~ * diag(WInv[,1])

and get

\( \hspace{24} y = M\sim * V(u^h) = \sum_{k=0}^{\infty} m_k * (u^h)^k \)

and according to the above

Code:

´  exp_b°h (1) = 2 + 2*y

we have the source of my coefficients in the vector M:

Code:

´   exp_b°h (1) = 2 + 2*y = 2 + 2* M~ * V(u^h)

So the coefficients, which I provided are just the precomputed coefficients in M, which represent the evaluation of the coefficients of the Schröder-functions for dxp, including a fixpointshift of the x-parameter and of the function-value.

--------------------------------

In short, omitting matrices:

Let C(x) denote a schröder-function for dxp_t(x), c_k the k'th coefficient of its powerseries,
D(x) the inverse and d_k the k'th coefficient of its powerseries,
for brevity C and D the functions at x=-1/2
and v = u^h, the h'th power of u.

Then we can write
\(
\hspace{48} C = \sum_{j=0}^{\infty} c_j*(-1/2)^j \\
\hspace{48} b\^\^^h = \exp__{\sqrt2}^{\circ h}(1) =2 + \sum_{k=1}^{\infty} 2* C^k * d_k * v^k \\
\)
and the k'th coefficient in my first mail is just 2* C^k * d_k in the formula above.


Gottfried